Answer:
Its called PSY
Explanation: I so do not know why they named it this way but, hope i helped.
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]
Learn more about the acceleration of object here:
https://brainly.com/question/460763
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
If you want to learn more about forces, you can read:
https://brainly.com/question/13370981
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
Two identical objects are pressed against two different springs so that each spring stores 55.0J of potential energy. The objects are then released from rest. One spring is quite stiff (hard to compress), while the other one is quite flexible (easy to compress).Which of the following statements is or are true? (More than one statement may be true.)A. Both objects will have the same maximum speed after being released.B. The object pressed against the stiff spring will gain more kinetic energy than the other object.C. Both springs are initially compressed by the same amount.D. The stiff spring has a larger spring constant than the flexible spring.E. The flexible spring must have been compressed more than the stiff spring.
Answer:
A , D , E
Explanation:
Solution:-
- Consider the two identical objects with mass ( m ).
- The stiffness of the springs are ( k1 and k2 ).
- Both the spring store 55.0 J of potential energy.
- We will apply the principle of energy conservation on both the systems. In both cases the spring stores 55.0 Joules of energy. Once released, the objects gain kinetic energy with a consequent loss of potential energy in either spring.
- The maximum speed ( v ) is attained when all the potential energy is converted to kinetic energy.
- Apply Energy conservation for spring with stiffness ( k1 ).
ΔU = ΔEk
55.0 = 0.5*m*v^2
v = √ ( 110 / m )
- Apply Energy conservation for spring with stiffness ( k2 ).
ΔU = ΔEk
55.0 = 0.5*m*v^2
v = √ ( 110 / m )
Answer: Both objects will have the same maximum speed ( A )
- We are told that one spring is more stiff as compared to the other one. The measure of stiffness is proportionally quantified by the spring constant. To mathematically express we can write it as:
k1 > k2
Where,
k1: The stiff spring
k2: The flexible spring
Answer: The stiff spring has a larger spring constant than the flexible spring. ( D )
- We will assume that the spring with constant ( k1 ) undergoes a displacement ( x1 ) and the spring with constant ( k2 ) undergoes a displacement ( x2 ). The potential energy stored in both spring is the same. Hence,
U1 = U2
0.5*( k1 ) * ( x1 )^2 = 0.5*( k2 ) * ( x2 )^2
[ k1 / k2 ] = [ x2 / x1 ]^2
Since,
k1 > k2 , then [ k1 / k2 ] > 1
Then,
[ x2 / x1 ]^2 > 1
[ x2 / x1 ] > 1
x2 > x1
Answer: The flexible spring ( x2 ) was compressed more than the stiff spring ( x1 ). ( E )
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]
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
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
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:
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.
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.
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 .
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
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
The Nardo ring is a circular test track for cars. It has a circumference of 12.5km. Cars travel around the track at a constant speed of 100km/h. A car starts at the easternmost point of the ring and drives for 15 minutes at this speed.
1. What distance, in km, does the car travel?
2. What is the magnitude of the car's displacement, in km, from its initial position?
3. What is the speed of the car in m/s?
Answer:
1. 25 Km
2. zero
3. 27.7 m/s
Explanation:
Data provided in the question:
Circumference of the track = 12.5 km
Speed of the car = 100 Km/h
Time for which car travels = 15 minutes = [tex]\frac {15}{60}[/tex] hr
Now,
1. Distance traveled = Speed × Time
= 100 × [tex]\frac{15}{60}[/tex]
= 25 Km
2. The distance traveled is 2 times the circumference of the track (i.e 2 × 12.5 = 25 Km)
Which means that the car is again at the initial position
Therefore, The displacement is zero.
3. Speed of car in Km/hr = 100 Km/h
now,
1 Km = 1000 m
1 hr = 3600 seconds
therefore,
100 Km/h = [tex]100\times\frac{1000}{3600}[/tex] m/s
= 27.7 m/s
Hence, the speed of car in m/s = 27.7
Help with this answer please
Answer:
Everytime you do an experiment you need something that is regular. For example if you try and measure how much germs spread in bread. you need 1 bread thats clean and 3 different breads for different molds. So thats called a CONTROL
AAAAAAAAAAAA is the answer
Calculate the potential difference across a 25-Ohm. resistor if a 0.3-A current is flowing through it.
V
Answer:7.5V
Explanation:
Ohm's law, V=IR
so, V=0.3×25
V=7.5V
Answer:
7.5 V
Explanation:
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.
Learn more about Velocity here:
https://brainly.com/question/18084516
#SPJ2
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.
For more details regarding soundproofing, visit:
https://brainly.com/question/8980142
#SPJ2
Answer: C.D
Explanation:...
Two workers are sliding 330 kg crate across the floor. One worker pushes forward on the crate with a force of 430 N while the other pulls in the same direction with a force of 330 N using a rope connected to the crate. Both forces are horizontal, and the crate slides with a constant speed. What is the crate's coefficient of kinetic friction on the floor?
Answer:
Coefficient of kinetic friction = 0.235
Explanation:
Given:
Mass of crate = 330 kg
1st force = 430 N
2nd force = 330 N
Find:
Coefficient of kinetic friction.
Computation:
We know that, velocity is constant.
So, acceleration (a) = 0
So, net force (f) = 430 N + 330 N
Net force (f) = 760 N
F = μmg
μ = f / mg [∵ g = 9.8]
μ = 760 / [330 × 9.8]
μ = 760 / [3,234]
μ = 0.235
Coefficient of kinetic friction = 0.235
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]
A ball with a mass of 275 g is dropped from rest, hits the floor and rebounds upward. If the ball hits the floor with a speed of 3.30 m/s and rebounds with a speed of 1.60 m/s, determine the following. (a) magnitude of the change in the ball's momentum in kg · m/s (Let up be in the positive direction.)
Answer:
[tex]\Delta p=1.3475\ kg-m/s[/tex]
Explanation:
The computation of magnitude of the change in the ball's momentum in kg · m/s is shown below:-
We represent
The ball mass = m = 275 g = 0.275 kg
Thus it goes to the floor and resurfaces upward.
The ball hits the ground at 3.30 m/s speed that is
u = -3.30 m/s which represents the Negative since the ball hits the ground)
It rebounds at a speed of 1.60 m / s i.e. v = 1.60 m/s (positive as the ball rebounds upstream)
[tex]\Delta p=p_f-p_i[/tex]
[tex]\Delta p=m(v-u)[/tex]
[tex]\Delta p=0.275\ kg(1.60\ m/s-(-3.30\ m/s))[/tex]
[tex]\Delta p=1.3475\ kg-m/s[/tex]
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
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 is the goal of the Standing Waves lab? Group of answer choices To determine how frequency changes with mode number. To determine the velocity of a wave traveling on string. To determine wavelength of a wave on a string. To be the very best like no one ever was.
Answer:
To determine wavelength of a wave on a string.
Explanation:
The Standing Waves lab study the parameters that affect standing waves in various strings. The effects of string tension and density on wavelength and frequency will be studied.
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.
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.
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
A carousel has a diameter of 6.0-m and completes one rotation every 1.7s. Find the centripetal acceleration of the traveler in m / s2.
Answer:
The centripetal acceleration of the traveler is [tex]40.98\ m/s^2[/tex]
Explanation:
It is given that, A carousel has a diameter of 6.0-m and completes one rotation every 1.7 s.
We need to find the centripetal acceleration of the traveler. It is given by the formula as follows :
[tex]a=\dfrac{v^2}{r}[/tex]
r is radius of carousel
[tex]v=\dfrac{2\pi r}{T}[/tex]
So,
[tex]a=\dfrac{4\pi ^2r}{T^2}[/tex]
Plugging all the values we get :
[tex]a=\dfrac{4\pi ^2\times 3}{(1.7)^2}\\\\a=40.98\ m/s^2[/tex]
So, the centripetal acceleration of the traveler is [tex]40.98\ m/s^2[/tex].
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
a body with v=20m/s changes its speed to 28m/s in 2sec. its acceleration will be
Answer:
Explanation:
Givens
vi = 20 m/s
vf = 28 m/s
t = 2 seconds
Formula
a = (vf - vi) / t
Solution
a = (28 - 20)/2
a = 8/2
a = 4 m/s^2