Answer:
1. The Universe is Expanding
2. It’s temperature is it’s uniform
3. Cosmic background radiation
4. I will give a hint for this one, since I don’t know, the hint is the universe is cooling.
5. It provides evidence of universe expansion.
6. Sorry I don’t know the rest
Explanation:
The universe is the collection of every item in space and time as well as the contents of those items
The correct options are as follows;
1. The universe is expanding
2. Its temperature is uniform
3. Composition of matter in the universe, cosmic background radiation
4. The universe is cooling which, according to the Big Bang Theory, is expected to happen as the cosmos disperses
5. It provides evidence of universe expansion
6. Nuclear fission involves one large atom splitting into two smaller atoms
Nuclear fission takes place in the nucleus of an atom
7. A distant object travels rapidly towards an observer
8. The fuel in nuclear fusion is often uranium
Nuclear fusion is used to generate electricity at nuclear power plants
9. The universe began as a very high density singularity
Dark matter makes up majority of the universe
10. Acceleration expansion — Dark energy
Structure forming in the early universe — Dark matter
11. It causes the expansion of the universe to accelerate
The reasons for selecting the above options are as follows;
1. The universe is expanding
The redshift of light from galaxies indicates that that are moving away
2. Its temperature is uniform
The uniform temperature of the microwave background suggest a common source
3. Composition of matter in the universe, cosmic background radiation
The matter present in the universe are characteristically similar in their origins
The cosmic background provides evidence of the existence of a singularity
4. The universe is cooling which, according to the Big Bang Theory, is expected to happen as the cosmos disperses
Based on the Big Bang Theory, the temperature of the universe is reducing as the universe expands, compared to the initial temperature
5. It provides evidence of universe expansion
The background radiation coming from a single source as the rest of the universe is expected to spread throughout the universe
6. Nuclear fission involves one large atom splitting into two smaller atoms
Nuclear fission takes place in the nucleus of an atom
Nuclear fusion involves the joining of small atoms to form a larger atom
7. A distant object travels rapidly towards an observer
The redshift is the opposite, indicating that the object is moving further away
8. The fuel in nuclear fusion is often uranium
Nuclear fusion is used to generate electricity at nuclear power plants
Nuclear fusion usually consists of joining small atoms together. It has not been used for commercial energy production
9. The universe began as a very high density singularity
According to the Big Bang Theory, the universe started from the dense, high temperature singularity
Dark matter makes up majority of the universe
10. Acceleration expansion — Dark energy
Dark energy causes expansion
Structure forming in the early universe — Dark matter
Dark matter is instrumental to the formation of structures in the universe
11. It causes the expansion of the universe to accelerate
Dark energy is seen as the cause of the accelerating expansion of the universe
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physics I need help :(
Some cats can be trained to jump from one location to another and perform other tricks. Kit the cat is going to jump through a hoop. He begins on a wicker cabinet at a height of 1.765 m above the floor and jumps through the center of a vertical hoop, reaching a peak height 3.130 m above the floor. (Assume the center of the hoop is at the peak height of the jump. Assume that +x axis is in the direction of the hoop from the cabinet and +y axis is up. Assume g = 9.81 m/s2.)
(a) With what initial velocity did Kit leave the cabinet if the hoop is at a horizontal distance of 1.560 m from the cabinet?
v_0 = m/s
(b) If Kit lands on a bed at a horizontal distance of 3.582 m from the cabinet, how high above the ground is the bed?
m
Answer:
a. the initial velocity of the cat is 5.95 m/s at 60.2° from the horizontal
b. 0.847 m
Explanation:
a. Using v² = u² + 2as, we find the initial vertical velocity of the cat. Now at the peak height, v = final velocity = 0, u = initial velocity and a = -g = 9.8 m/s², s vertical distance travelled by the cat from its position on the cabinet = Δy = 3.130 m - 1.765 m = 1.365 m.
Substituting these variables into the equation, we have
0² = u² + 2(-9.8m/s²) × 1.365 m
-u² = -26.754 m²/s²
u = √26.754 m²/s²
u = 5.17 m/s
To find its initial horizontal velocity, u₁ we first find the time t it takes to reach the peak height from
v = u + at. where the variables mean the same as above.
substituting the values, we have
0 = 5.17 m/s +(-9.8m/s²)t
-5.17 m/s = -9.8m/s²t
t = -5.17 m/s ÷ (-9.8m/s²)
= 0.53 s
Now, the horizontal distance d = u₁t = 1.560 m
u₁ = d/t = 1.560 m/0.53 s = 2.96 m/s
So, the initial velocity of the cat is V = √(u² + u₁²)
= √((5.17 m/s)² + (2.96 m/s)²)
= √(26.729(m/s)² + 8.762(m/s)²)
= √(35.491 (m/s)²)
= 5.95 m/s
its direction θ = tan⁻¹(5.17 m/s ÷ 2.96 m/s) = 60.2°
So, the initial velocity of the cat is 5.95 m/s at 60.2° from the horizontal
(b)
First, we find the time t' it takes the cat to land on the bed from d' = u₁t'
where d' = horizontal distance of cabinet from bed = 3.582 m
u₁ = horizontal velocity = 2.96 m/s
t' = d'/u₁
= 3.582 m/2.96 m/s
= 1.21 s
The vertical between the bed and cabinet which is the vertical distance moved by the cat is gotten from Δy = ut' +1/2at'²
substituting u = initial vertical velocity = 5.17 m/s, t' = 1.21 s and a = -g = -9.8 m/s² into Δy, we have
Δy = ut' +1/2at'² = 5.17 m/s × 1.21 s +1/2(- 9.8 m/s²) × (1.21 s)² = 6.256 - 7.174 = -0.918 m
Δy = y₂ - y₁
Since our initial position is the position of the cabinet above the ground = y₁ = 1.765 m
y₂ = position of bed above ground.
Δy = y₂ - y₁ = -0.918 m
y₂ - 1.765 m = -0.918 m
y₂ = 1.765 m - 0.918 m
= 0.847 m
Sr-90 has a half-life of T1/2 = 2.85 a (years). How much Sr-90 will remain in a 5.00 g sample after 5.00 a? Show all of your work. (2 marks)
Answer:
1.48 g
Explanation:
A = A₀ (½)^(t / T)
where A is the final amount,
A₀ is the initial amount,
t is time,
and T is the half life.
A = (5.00 g) (½)^(5.00 a / 2.85 a)
A = 1.48 g
A constant force applied to object A causes it to accelerate at 5 m/s2. The same force applied to object B causes an acceleration of 3 m/s2. Applied to object C, it causes an acceleration of 7 m/s2.
A. Which object has the largest mass?B. Which object has the smallest mass?C. What is the ratio of mass A to mass B?
Answer:
(A) object B has the largest mass because it has the least acceleration
(B) object C has the smallest mass because it has the largest acceleration
(C) mass A : mass B = 3 : 5
Explanation:
Given;
acceleration of object A = 5 m/s²
acceleration of object B = 3 m/s²
acceleration of object C = 7 m/s²
A constant force, F
According to Newton's second law of motion;
F = ma
m = F / a
Mass of object A:
m = F / 5
Mass of object B:
m = F / 3
Mass of object C:
m = F / 7
(A). Which object has the largest mass:
object B has the largest mass because it has the least acceleration
(B). Which object has the smallest mass:
object C has the smallest mass because it has the largest acceleration
(C). What is the ratio of mass A to mass B;
mass A = F / 5
mass B = F / 3
[tex]mass \ A : \ mass \ B = \frac{F}{5} : \frac{F}{3} \\\\\frac{mass \ A}{mass \ B} = \frac{F}{5} * \frac{3}{F}= \frac{3}{5} \\\\mass \ A : \ mass \ B = 3: 5[/tex]
A. The Object B has largest mass.
B. The Object A has smallest mass.
C. The ratio of mass A to mass B is, [tex]\frac{3}{5}[/tex]
Newton second law of motion:The second law states that the acceleration of an object is dependent upon two variables - the net force acting upon the object and the mass of the object.
[tex]F=ma\\\\m=\frac{F}{a}[/tex]
For constant force, mass is inversely proportional to acceleration of object.Given that, acceleration of object A is [tex]5m/s^{2}[/tex] and object B is [tex]3m/s^{2}[/tex]Thus, Object B has largest mass.Object A has smallest mass.the ratio of mass A to mass B is,[tex]\frac{m_{A}}{m_{B}} =\frac{a_{B}}{a_{A}} =\frac{3}{5}[/tex]
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Inside a stereo speaker, you will find two permanent magnets: one on the cone and one near the cone. True of false?
Answer:
false
Explanation:
A rod of mass M = 154 g and length L = 35 cm can rotate about a hinge at its left end and is initially at rest. A putty ball of mass m = 11 g, moving with speed V = 9 m/s, strikes the rod at angle θ = 29° a distance D = L/3 from the end and sticks to the rod after the collision.Calculate the rotational kinetic energy, in joules, of the system after the collision.
Answer:
Explanation:
moment of inertia of the rod = 1/3 mL² , m is mass and L is length of rod.
1/3 x .154 x .35²
= .00629
moment of inertia of putty about the axis of rotation
= m d² , m is mass of putty and d is distance fro axis
= .011 x( .35 / 3 )²
= .00015
Total moment of inertia I = .00644 kgm²
angular momentum of putty about the axis of rotation
= mvRsinθ
m is mass , v is velocity , R is distance where it strikes the rod and θ is angle with the rod at which putty strikes
= .011 x 9 x .35 / 3 x sin 29
= .0056
Applying conservation of angular momentum
angular momentum of putty = angular momentum of system after of collision
.0056 = .00644 ω where ω is angular velocity of the rod after collision
ω = .87 rad /s .
Rotational energy
= 1/2 I ω²
I is total moment of inertia
= .5 x .00644 x .87²
= 2.44 x 10⁻³ J .
Which of the following is often found in individuals who are active and eating a healthy diet?
Answer:
Increased blood circulation to the body.
Explanation:
plato/edmentum
As your bus rounds a flat curve at constant speed, a package with mass 0.900 kg , suspended from the luggage compartment of the bus by a string 50.0 cm long, is found to hang at rest relative to the bus, with the string making an angle of 30.0 â with the vertical. In this position, the package is 55.0 m from the center of curvature of the curve.
Required:
a. What is the radial acceleration of the bus?
b. What is the radius of the curve?
Answer:
a.[tex]5.66ms^{-2}[/tex]
b.55 m
Explanation:
We are given that
Mass ,m=0.9 kg
Length of string,l=50 cm=[tex]\frac{50}{100}=0.50 m[/tex]
1 m=100 cm
[tex]\theta=30^{\circ}[/tex]
R=55 m
a.Centripetal acceleration
[tex]a_c=gtan\theta[/tex]
[tex]a_c=9.8tan30^{\circ}[/tex]
[tex]a_c=5.66 m/s^2[/tex]
Hence, the radial acceleration of the bus=[tex]5.66m/s^2[/tex]
b. Radius of curve,R=55 m
Question 7 of 10
The coefficient of kinetic friction between a couch and the floor is 0.4. If the
couch has a mass of 35 kg and you push it with a force of 200 N. what is the
net force on the couch as it slides?
O A. 337 N
B. 143 N
O C. 343 N
O D. 63 N
Answer:
D
Explanation:
Now the net force is the applied force minus the frictional force; this is expressed mathematically as:
Fnet= Fappplied - Ffrictional
Now the frictional force is given as ;
Coefficient of friction × normal reaction
Normal reaction is the weight of the human acting in opposite direction.
Normal reaction of the human is ;
35 × 9.8 = 343N { note that weight = m× g and g= 9.8m/S2, a known standard }
Hence the Frictional force =343×0.4 =137.20N
Hence Fnet = 200-137.20 = 62.8N
Fnet = 63N to the nearest whole
The net force on the couch as it slides is 63N.
What is frictional force?
When an object is moving on a rough surface, it experiences opposition. This opposing force is called the friction force.
The friction force is given by
f = coefficient of friction x Normal force
Given, the coefficient of kinetic friction between a couch and the floor is 0.4. If the couch has a mass of 35 kg and you push it with a force of 200 N.
Normal reaction is the weight of the human acting in opposite direction.
Normal reaction N =35 × 9.8 = 343N
Frictional force f =0.4 x 343
f =137.20N
The net force will be
Fnet= Fappplied - Ffrictional
Fnet = 200-137.20 = 62.8N
Fnet = 63N
Thus, the net force on the couch as it slides is 63N.
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The current through an inductor of inductance L is given by I(t) = Imax sin(ωt).
(a) Derive an expression for the induced emf in the inductor as a function of time.
(b) At t = 0, is the current through the inductor increasing or decreasing?
(c) At t = 0, is the induced emf opposing or aiding the flow of the charge carriers? (Remember that the direction of a positive induced emf is the same as the current direction and the direction of a negative induced emf is opposite the current direction.)
(d) How are the answers to parts b and c consistent with the behavior of inductors discussed in the text?
Answer:
(a) [tex]emf_L=-LI_{max}\omega cos(\omega t)[/tex]
(b) neither increasing or decreasing
(c) opposite to the flow of charge carriers
Explanation:
The current through an inductor of inductance L is given by:
[tex]I(t)=I_{max}sin(\omega t)[/tex] (1)
(a) The induced emf is given by the following formula
[tex]emf_L=-L\frac{dI}{dt}[/tex] (2)
You derivative the expression (1) in the expression (2):
[tex]emf_L=-L\frac{d}{dt}(I_{max}sin(\omega t))\\\\emf_L=-LI_{max}\omega cos(\omega t)[/tex]
(b) At t=0 the current is zero
(c) At t = 0 the emf is:
[tex]emf_L=-\omega LI_{max}[/tex]
w, L and Imax have positive values, then the emf is negative. Hence, the induced emf is opposite to the flow of the charge carriers.
(d) read the text carefully
At t zero, the current through the inductor neither increasing nor decreasing because current is zero.
The current through an inductor of inductance L can be calculated by
[tex]\bold {I_t = I_m_a_x sin (\omega t)}[/tex].........1
(a) The induced emf can be calculated by
[tex]\bold {emf_L = - L \dfrac {dI}{dt}}[/tex]............2
Derivative the equation (1) in the equation (2)
[tex]\bold {emf _L= -L \dfrac {d (I _m_a_x sin (\omega t)} {dt}}\\\\\bold {emf _L= -L (I _m_a_x \omega cos( \omega t) }[/tex]
(b) At t=0 the current is zero,
(c) At t = 0 the emf is:
[tex]\bold {emf_L = -\omega LI _m_a_x}[/tex]
Therefore, at t zero, the current through the inductor neither increasing nor decreasing because current is zero.
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ii.
The drift velocity
(b) A 1800w toaster, a 1.3KW electric frying pan, and a 100w lamp are plugged to the same
20A, 120V circuit.
i.
What current is drawn by each device and what is the resistance of each device?
State whether this combination will blow the fuse or not.
Answer:
toaster -- 15 A, 8 Ωfry pan -- 10.83 A, 11.08 Ωlamp -- 0.83 A, 144 Ωfuse will blowExplanation:
P = VI
I = P/V = P/120
R = V/I = V/(P/V) = V^2/P = 14400/P
Toaster: I = 1800/120 = 15 . . . amps
R = 14400/1800 = 8 . . . ohms
Fry pan: I = 1300/120 = 10.833 . . . amps
R = 14400/1300 = 11.08 . . . ohms
Lamp: I = 100/120 = 0.833 . . . amps
R = 14400/100 = 144 . . . ohms
The total current exceeds 20 A, so will blow the fuse.
What's a line of best fit? Will give BRAINLIEST
A line of best fit expresses the relationship between the points.
Explanation:
It does not go through all the points but goes through most of them and it is like a hardrawn curve
A fuel pump sends gasoline from a car's fuel tank to the engine at a rate of 5.37x10-2 kg/s. The density of the gasoline is 739 kg/m3, and the radius of the fuel line is 3.37x10-3 m. What is the speed at which gasoline moves through the fuel line
Answer:
Speed v = 2.04 m/s
the speed at which gasoline moves through the fuel line is 2.04 m/s
Explanation:
Given;
Mass transfer rate m = 5.37x10^-2 kg/s.
Density d = 739 kg/m3
radius of pipe r = 3.37x10^-3 m
We know that;
Density = mass/volume
Volume = mass/density
Volumetric flow rate V = mass transfer rate/density
V = m/d
V = 5.37x10^-2 kg/s ÷ 739 kg/m3
V = 0.00007266576454 m^3/s
V = 7.267 × 10^-5 m^3/s
V = cross sectional area × speed
V = Av
Area A = πr^2
V = πr^2 × v
v = V/πr^2
Substituting the given values;
v = 7.267 × 10^-5 m^3/s/(π×(3.37x10^-3 m)^2))
v = 0.203678639672 × 10 m/s
v = 2.04 m/s
the speed at which gasoline moves through the fuel line is 2.04 m/s
A horizontal force of 150 N is used to push a 40.0-kg packing crate a distance of 6.00 m on a rough horizontal surface. If the crate moves at constant speed, find (a) the work done by the 150-N force and (b) the coefficient of kinetic friction between the crate and surface.
Answer:
a. 900 J
b. 0.383
Explanation:
According to the question, the given data is as follows
Horizontal force = 150 N
Packing crate = 40.0 kg
Distance = 6.00 m
Based on the above information
a. The work done by the 150-N force is
[tex]W = F x = \mu N x = \mu\ m\ g\ x[/tex]
[tex]W = 150 \times 6[/tex]
= 900 J
b. Now the coefficient of kinetic friction between the crate and surface is
[tex]\mu = \frac {F}{m\timesg}[/tex]
[tex]= \frac{150}{40\times 9.8}[/tex]
= .383
We simply applied the above formulas so that each one part could calculate
We want to find the work and kinetic friction for the given situation. The solutions are:
a) W = 900 N*mb) μ = 0.38Here we have a horizontal force of 150N pushing a 40.0 kg packing crate a distance of 6.00m at a constant speed.
a) First we want to find the work, it is given by the force applied times the distance moved, so the work is just:
W = 150N*6.00m = 900 N*m
b) Now we want to find the coefficient of kinetic friction, it must be such that the kinetic friction force is equal to the pushing force, in this way there is no net force, and then there is no acceleration.
Remember that the friction force is:
F = m*g*μ
Where:
m = mass of the box = 40 kgg = gravitational acceleration = 9.8m/s^2μ = coefficient of kinetic friction.Then we must solve:
150N = 40kg*(9.8 m/s^2)*μ = 392N*μ
150N/392N = 0.38
So the coefficient of kinetic friction between the crate and the surface is 0.38
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How many significant figures does 0.09164500561 have?
Answer:
10 Sig Figs
Explanation:
Just start counting at the first non zero after the decimal so in this case the nine, and count all of the numbers including zeros after that.
A factory worker pushes a 30.0 kg crate a distance of 3.7 m along a level floor at constant velocity by pushing downward at an angle of 30∘ below the horizontal. The coefficient of kinetic friction between the crate and floor is 0.25.
Required:
a. What magnitude of force must the worker apply?
b. How much work is done on the crate by this force?
c. How much work is done on the crate by friction?
d. How much work is done on the crate by the normal force? By gravity?
e. What is the total work done on the crate?
Answer:
a) [tex]F = 210.803\,N[/tex], b) [tex]W_{F} = 779.971\,J[/tex], c) [tex]W_{f} = 235.683\,J[/tex], d) [tex]W_{N} = 0\,J[/tex]; [tex]W_{g} = 544.289\,J[/tex], e) [tex]W_{net} = 0\,J[/tex]
Explanation:
a) The net force exerted on the crate is:
[tex]\Sigma F = F - m\cdot g \cdot \sin \theta - \mu_{k}\cdot m\cdot g \cdot \cos \theta = 0[/tex]
The magnitud of the force that the work must apply to the crate is:
[tex]F = m\cdot g \cdot \sin \theta + \mu_{k}\cdot m\cdot g \cdot \cos \theta[/tex]
[tex]F = (30\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot \sin 30^{\circ} + 0.25 \cdot (30\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot \cos 30^{\circ}[/tex]
[tex]F = 210.803\,N[/tex]
b) The work done on the crate due to the external force is:
[tex]W_{F} = (210.803\,N)\cdot (3.7\,m)[/tex]
[tex]W_{F} = 779.971\,J[/tex]
c) The work done on the crate due to the external force is:
[tex]W_{f} = (63.698\,N)\cdot (3.7\,m)[/tex]
[tex]W_{f} = 235.683\,J[/tex]
d) The work done on the crate due the normal force is zero, since such force is perpendicular to the motion direction.
[tex]W_{N} = 0\,J[/tex]
And, the work done by gravity is:
[tex]W_{g} = (147.105\,N)\cdot (3.7\,m)[/tex]
[tex]W_{g} = 544.289\,J[/tex]
e) Lastly, the total work done is:
[tex]W_{net} = W_{F} - W_{f} - W_{g} - W_{N}[/tex]
[tex]W_{net} = 779.971\,J - 235.683\,J - 0\,N - 544.289\,J[/tex]
[tex]W_{net} = 0\,J[/tex]
A 888 kg car is driven clockwise around a flat circular track of radius 59 m. The speed of the car is a constant 7 m/s. Which factor, when doubled, would produce the greatest change in the centripetal force acting on the car? A. Radius of the track B. Weight of the car C. Mass of the car D. Velocity of the car
Answer:
D. Velocity of the car
Explanation:
The centripetal force acting on the car is given by the following formula:
[tex]F_c=ma_c=m\frac{v^2}{r}[/tex] (1)
m: mass of the car = 888 kg
v: tangential speed of the car = 7 m/s
r: radius of the flat circular track = 59 m
By the form of the equation (1) you can notice that the greatest change in the centripetal force is obtained when the velocity v is twice. In fact, you have:
[tex]F_c=m\frac{(2v)^2}{r}=4m\frac{v^2}{r}=4F_c[/tex]
Then, the greatest values of the centripetal force is:
[tex]F_c=4(888kg)\frac{(7m/s)^2}{59m}=2949.96N[/tex]
The greatest change in Fc is obtained by changing the value of the speed
answer
D. Velocity of the car
Why do bears activity increase as certain points during the day
Because they are well rested and have to work to get food in their system.
How much work is done by 0.30 m of gas if its pressure increases by 8.0 x105 Pa and the volume remains constant
Salerno
Answer:
0
Explanation:
if the volume remains constans, the works is 0 because the equation
W = P . ∆V
P = pressure
∆V = change in volume
g it as been suggested that solar powered space ships could get a boost from a laser either on earth or in orbit around earth. the laser would have to be very powerful to give any measurable benefit to the ship. if the laser produces a 0.18-m diameter beam of 490-nm light, what is the minimum angular spread of the beam?
Answer:
The minimum angular spread of the laser beam is 3.32 × [tex]10^{-6}[/tex] radians (or 3.32 μrad).
Explanation:
The minimum angular spread of a wave is the ratio of its narrowest diameter to its wavelength.
From Rayleigh's formula,
Angular spread = 1.22 (wavelength ÷ diameter)
= 1.22 (λ ÷ D)
Given that:
diameter, D = 0.18 m and wavelength, λ = 490 nm, then;
Angular spread of the laser beam = 1.22 (λ ÷ D)
= 1.22[tex](\frac{490*10^{-9} }{0.18})[/tex]
= 1.22× 2.7222 × [tex]10^{-6}[/tex]
= 3.3211 × [tex]10^{-6}[/tex] rad
The minimum angular spread of the laser beam is 3.32 × [tex]10^{-6}[/tex] radians.
Calculate the maximum deceleration (in m/s2) of a car that is heading down a 14° slope (one that makes an angle of 14° with the horizontal) under the following road conditions. You may assume that the weight of the car is evenly distributed on all four tires and that the static coefficient of friction is involved—that is, the tires are not allowed to slip during the deceleration.
The question is incomplete. Here is the complete question.
Calculate the maximum deceleration of a car that is heading down a 14° slope (one that makes an anlge of 14° with the horizontal) under the following road conditions. You may assum that the weight of the car is evenlydistributed on all four tires and that the sttic coefficient of friction is involved - that is, the tires are not allowed to slip during the deceleration. (Ignore rolling) Calculate for a car: (a) On a dry concrete. (b) On a wet concrete. (c) On ice, assuming that μs = 0.100, the same as for shoes on ice.
Answer: (a) a = - 11.05 m/s²; (b) a = - 10.64 m/s²; (c) a = - 9.84m/s²
Explanation: The image in the attachment describe the forces acting on the car. Observing that, we know that:
[tex]F_{net}[/tex] = - [tex]W_x[/tex] - [tex]f_s[/tex]
The [tex]W_x[/tex] is a x-component of force due to gravity (W) and, in this case, is given by: [tex]W_x[/tex] = W.sin(14)
W is described as: W = m.g
Force due to friction ([tex]f_s[/tex]) is given by: [tex]f_s[/tex] = μs.N
N is the normal force and, in the system, is equivalent of [tex]W_y[/tex], so:
[tex]W_y[/tex] = m.g.cos(14)
Therefore, the formula will be:
[tex]F_{net}[/tex] = - [tex]W_x[/tex] - [tex]f_s[/tex]
m.a = - (m.g.sin14) - (μs.mg.cos14)
a = - g (sin14 + μscos 14)
a) For dry concrete, μs = 1:
a = - g (sin14 + μscos 14)
a = - 9.8 (sin14 + 1.cos14)
a = - 11.05 m/s²
b) For wet concrete, μs = 0.7:
a = - g (sin14 + μscos 14)
a = - 9.8 (sin 14 + 0.7.cos14)
a = - 10.64 m/s²
c) For ice, μs = 0.1:
a = - g (sin14 + μscos 14)
a = - 9.8 (sin14 + 0.1cos14)
a = - 9.84 m/s²
can a body be in equilibrium if only one external force act on its ? explain
Answer:
Explanation:
If there is only 1 force, the body can never be in equilibrium, providing that the force is not zero (and that would hardly be a force. Zero is possible in math and it means something. It is debatable in physics).
You cannot think of a condition where something is stationary on planet earth and there are not 2 forces or more forces involved.
Think of something like a block of wood sitting on a table. It is not moving, we'll say. Gravity is holding it down, but what is pushing up on it?
The table is. There are 2 forces and they are equal in magnitude, but opposite in direction. That matters.
The starships of the Solar Federation are marked with the symbol of the Federation, a circle, whereas starships of the Denebian Empire are marked with the Empire's symbol, an ellipse whose major axis is n times its minor axis (a=nb in the figure ).
How fast, relative to an observer, does an Empire ship have to travel for its markings to be confused with those of a Federation ship? Use c for the speed of light in a vacuum.
Express your answer in terms of n and c.
Complete question
The complete question is shown on the first uploaded image
Answer:
The velocity is [tex]v = c* \sqrt{1 - \frac{1}{n^2} }[/tex]
Explanation:
From the question we are told that
a = nb
The length of the minor axis of the symbol of the Federation, a circle, seen by the observer at velocity v must be equal to the minor axis(b) of the Empire's symbol, (an ellipse)
Now this length seen by the observer can be mathematically represented as
[tex]h = t \sqrt{1 - \frac{v^2}{c^2} }[/tex]
Here t is the actual length of the major axis of of the Empire's symbol, (an ellipse)
So t = a = nb
and b is the length of the minor axis of the symbol of the Federation, (a circle) when seen by an observer at velocity v which from the question must be the length of the minor axis of the of the Empire's symbol, (an ellipse)
i.e h = b
So
[tex]b = nb [\sqrt{1 - \frac{v^2}{c^2} } ][/tex]
[tex][\frac{1}{n} ]^2 = 1 - \frac{v^2}{c^2}[/tex]
[tex]v^2 =c^2 [1- \frac{1}{n^2} ][/tex]
[tex]v^2 =c^2 [\frac{n^2 -1}{n^2} ][/tex]
[tex]v = c* \sqrt{1 - \frac{1}{n^2} }[/tex]
The density of atmosphere (measured in kilograms/meter3) on a certain planet is found to decrease as altitude increases (as measured from the planet's surface). What type of relationship exists between the altitude and the atmospheric density, and what would the atmospheric density be at an altitude of 1,291 kilometers?
A.
inverse plot, 0.45 kilograms/meter3
B.
inverse plot, 0.51 kilograms/meter3
C.
quadratic plot, 1.05 kilograms/meter3
D.
inverse plot, 1.23 kilograms/meter3
E.
inverse plot, 0.95 kilograms/meter3
' A ' looks like the best choice.
Answer:
B. inverse plot, 0.51 kilograms/meter3
Explanation:
A metal ring 4.60 cm in diameter is placed between the north and south poles of large magnets with the plane of its area perpendicular to the magnetic field. These magnets produce an initial uniform field of 1.12 T between them but are gradually pulled apart, causing this field to remain uniform but decrease steadily at 0.280 T/s.
A. What is the magnitude of the electric field induced in the ring?
B. In which direction (clockwise or counterclockwise) does the current flow as viewed by someone on the south pole of the magnet?1. Counterclockwise2. Clockwise
Answer:
A. Ein = 8.05*10^-4 V/m
B. Clockwise sense
Explanation:
A. the magnitude of the electric field induced in the ring is obtaind by using the following formula:
[tex]\int E_{in} \cdot ds=-\frac{d\Phi_B}{dt}[/tex] (1)
Ein: induced electric field
ds: differential of a path of the ring
ФB: magnetic flux in the ring
The Ein vector is parallel to ds in the complete ring. Furthermore, the area of the ring is constant, hence, you have in the equation (1):
[tex]\int E_{in}ds=E_{in}(2\pi r)=-A\frac{dB}{dt}\\\\E_{in}=-\frac{A}{2\pi r}\frac{dB}{dt}[/tex] (2)
dB/dt = -0.280T/s (it is decreasing)
A: area of the ring = π(r/2)^2= (π/4) r^2
r: radius of the ring = 4.60/2 = 2.30 cm
Then, you replace the values of all variables in the equation (2):
[tex]E_{in}=-\frac{(\pi/4)r^2}{2\pi r}\frac{dB}{dt}=\frac{r}{8}\frac{dB}{dt}\\\\E_{in}=-\frac{0.0230m}{8}(-0.280T)=8.05*10^{-4}\frac{V}{m}[/tex]
hence, the induced electric field is 8.05*10^-4 V/m
B. The induced current in the ring produced a magnetic field that is opposite to the magnetic field of the magnet. The, in this case you have that the induced current is in a clockwise sense.
One car travels 40. meters due east in 5.0 seconds, and a second car travels 64 meters due west in 8.0 seconds. During their periods of travel, the cars definitely had the same
Answer:
They had the same speed.
Explanation:
It won't be velocity, because velocity is a vector quantity. Speed is scalar.
Velocity is the rate of change of displacement. During their periods of travel, the cars definitely had the same velocity.
What is Velocity?Velocity is the directional speed of a moving object as an indicator of its rate of change in location as perceived from a certain frame of reference and measured by a specific time standard.
Given that the first car travels 40 meters due east in 5 seconds. Therefore, we can write,
Distance = 40 meters
Time = 5 seconds
Velocity = Distance / Time = 40 meter/ 5 sec = 40 m/sec
Also, given that the second car travels 64 meters due west in 8 seconds. Therefore, we can write,
Distance = 64 meters
Time = 8 seconds
Velocity = Distance / Time = 64 meter/ 8 sec = 8 m/sec
Hence, During their periods of travel, the cars definitely had the same velocity.
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A car of mass 410 kg travels around a flat, circular race track of radius 83.4 m. The coefficient of static friction between the wheels and the track is 0.286. The acceleration of gravity is 9.8 m/s 2 . What is the maximum speed v that the car can go without flying off the track
Answer:
The maximum speed v that the car can go without flying off the track = 15.29 m/s
Explanation:
let us first lay out the information clearly:
mass of car (m) = 410 kg
radius of race track (r) = 83.4 m
coefficient of friction (μ) = 0.286
acceleration due to gravity (g) = 9.8 m/s²
maximum speed = v m/s
For a body in a constant circular motion, the centripetal for (F) acting on the body is given by:
F = mass × ω
where:
F = maximum centripetal force = mass × μ × g
ω = angular acceleration = [tex]\frac{(velocity)^2}{radius}[/tex]
∴ F = mass × ω
m × μ × g = m × [tex]\frac{v^{2} }{r}[/tex]
410 × 0.286 × 9.8 = 410 × [tex]\frac{v^{2} }{83.4}[/tex]
since 410 is on both sides, they will cancel out:
0.286 × 9.8 = [tex]\frac{v^{2} }{83.4}[/tex]
2.8028 = [tex]\frac{v^{2} }{83.4}[/tex]
now, we cross-multiply the equation
2.8028 × 83.4 = [tex]v^{2}[/tex]
[tex]v^{2}[/tex] = 233.754
∴ v = √(233.754)
v = 15.29 m/s
Therefore, the maximum speed v that the car can go without flying off the track = 15.29 m/s
HELPP MEE
Which image illustrates the desired interaction of a sound wave with
soundproofing material in a recording studio?
Soundproofing material is required for blocking sound during some works like recording voice in the studio. Image D represents the interaction of a sound wave with soundproofing material in a recording studio.
What is the basis of soundproofing?Soundproofing is done by absorbing the sound. A very much used material for this is a dense foam.
Foam and like materials absorbs sound and it travels directly into the soft surface resulting in soundproofing.
Thus, the correct option is C, as the D image is showing the absorption.
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Answer: C.D
Explanation:...
Espresso is a coffee beverage made by forcing steam through finely ground coffee beans. Modern espresso makers generate steam at very high pressures and temperatures, but in this problem we'll consider a low-tech espresso machine that only generates steam at 100?C and atomospheric pressure--not much good for making your favorite coffee beverage.The amount of heat Q needed to turn a mass m of room temperature ( T1) water into steam at 100?C ( T2) can be found using the specific heat c of water and the heat of vaporization Hv of water at 1 atmosphere of pressure.Suppose that a commercial espresso machine in a coffee shop turns 1.50 kg of water at 22.0?C into steam at 100?C. If c=4187J/(kg??C) and Hv=2,258kJ/kg, how much heat Q is absorbed by the water from the heating resistor inside the machine?Assume that this is a closed and isolated system.Express your answer in joules to three significant figures.Q = _________________ J
Answer:
Q = 3877 KJ
Explanation:
Since, the system is closed and isolated. Therefore, the law of conservation of energy can be written as:
Heat Absorbed By Water (Q) = Heat required to raise the temperature of water (Q₁) + Heat required to convert water to steam (Q₂)
Q = Q₁ + Q₂ ----- equation (1)
Now, for Q₁:
Q₁ = m C ΔT
where,
m = Mass of Water = 1.5 kg
C = Specific Heat of Water = 4187 J/kg.°C
ΔT = Change in Temperature of Water = T₂ - T₁ = 100°C - 22°C = 78°C
Therefore,
Q₁ = (1.5 kg)(4187 J/kg.°C)(78°C)
Q₁ = 490 x 10³ J =490 KJ
Now, for Q₂:
Q₂ = m H
where,
m = Mass of Water = 1.5 kg
H = Heat of Vaporization of Water = 2258 KJ/kg
Therefore,
Q₂ = (1.5 kg)(2258 KJ/kg)
Q₂ = 3387 KJ
Substituting the values in equation (1), we get:
Q = Q₁ + Q₂
Q = 490 KJ + 3387 KJ
Q = 3877 KJ
A block is supported on a compressed spring, which projects the block straight up in the air at velocity VVoj The spring and ledge it sits on then retract. You can win a prize by hitting the block with a ball. When should you throw the ball and in what direction to be sure the ball hits the block?
A. At the instant when the block is at the highest point, directed at the spring.
B. At the instant when the block is at the highest point, directed at the block.
C. At the instant when the block leaves the spring, directed at the spring.
D. At the instant when the block leaves the spring, directed at the block.
E. When the block is back at the spring's original position, directed at that position.
Answer:
the correct answer is B
Explanation:
We analyze this exercise a little, the block goes into the air and is under the acceleration of gravity. The ball is fired by the hand and is describing a parabolic movement, subjected to the acceleration of gravity.
For the ball to hit the block we must have the distance the ball goes up equal to the distance the block moves, therefore we must shoot the ball at the block at its highest point.
Let's write the kinematic equation for the two bodies
The block. At the highest point of the path
y = - ½ g t2
The ball, in its vertical movement
y = vo t - ½ g t2
therefore the correct answer is B