Answer:
1.25kg
Explanation:
Simply multiply volume and density together
Having aced your Physics 2111 class, you get a sweet summer-job working in the International Space Station. Your room-mate, Cosmonaut Valdimir tosses a banana at you at a speed of 16 m/s. At exactly the same instant, you fling a scoop of ice cream at Valdimir along exactly the same path. The collision between banana and ice cream produces a banana split 8.2 m from your location 1.4 s after the banana and ice cream were launched.
1. How fast did you toss the ice cream?
2. How far were you from Valdimir when you tossed the ice cream?
Answer:
a
The speed is [tex]s = 5.857 m/s[/tex]
b
The distance is [tex]D = 22.4 \ m[/tex]
Explanation:
From the question we are told that
The speed of the banana is [tex]v = 16 \ m/s[/tex]
The distance from my location is [tex]d = 8.2 \ m[/tex]
The time taken is [tex]t = 1.4 \ s[/tex]
The speed of the ice cream is
[tex]s = \frac{d}{t}[/tex]
substituting values
[tex]s = \frac{8.4}{1.4}[/tex]
[tex]s = 5.857 m/s[/tex]
The distance of separation between i and Valdimir is the same as the distance covered by the banana
So
[tex]D = v * t[/tex]
substituting values
[tex]D = 16 * 1.4[/tex]
[tex]D = 22.4 \ m[/tex]
28 points!! please help
1. Calculate the centripetal force exerted on a 900kg900kg car that rounds a 600m600m radius curve on horizontal ground at 25.0m/s25.0m/s. 2. Static friction prevents the car from slipping. Find the magnitude of the frictional force between the tires and the road that allows the car to round the curve without sliding off in a straight line.
Explanation:
It is given that,
Mass of a car is 900 kg
Radius of curve is 600 m
Speed of the car in the curve is 25 m/s
We need to find the centripetal force exerted on a car. The formula used to find the centripetal force is given by :
[tex]F=\dfrac{mv^2}{r}\\\\F=\dfrac{900\times (25)^2}{600}\\\\F=937.5\ N[/tex]
So, the centripetal force exerted on a car is 937.5 N.
Static friction prevents the car from slipping. It means that the magnitude of centripetal force is balanced by the frictional force. So, the frictional force of 937.5 N is acting on the car.
The Gulf Stream off the east coast of the United States can flow at a rapid 3.8 m/s to the north. A ship in this current has a cruising speed of 8.0 m/s . The captain would like to reach land at a point due west from the current position.
At this heading, what is the ship's speed with respect to land?
Answer:
61.6° west of South
Explanation:
The ship goes to the south at an equal rate just like water flows to the north. Thus, the velocities would balance making the ship move towards the west.
Since we're dealing with water, the ship goes 3.8 m / s to the South, but a lot still remains to the west. Finding this would require us drawing a triangle. 3.8 m/s point down side and the hypotenuse is 8
cos(θ) = [adjacent/hypotenuse]
Cos θ = 3.8/8
Cos θ = 0.475
θ = cos^-1 (0.475)
θ = 61.6°
Therefore the angle is 61.6° west of South.
Which of the following best describes the current age of the Sun?
A.) It is near the end of its lifespan.
B.) It is about halfway through its lifespan.
C.) It is early in its lifespan.
D.) We do not have a good understanding of the Sun's age.
Answer: Its b, The only problem with this is is there supposed to be a picture?
Explanation: NASA has used there fancy gadgets to figure this out but if there was a picture, this answer could be different.
19. After a snowstorm, you put on your frictionless skis and tie a rope to the back of your friend’s truck. Your total mass is 70 kg and the truck exerts a constant force of 20 N. How fast will you be going after 15 seconds, in m/s and MPH?
Explanation:
It is given that,
Total mass is 70 kg
The truck exerts a constant force of 20 N.
Then the net force is given by :
F = ma
a is acceleration of rider
[tex]a=\dfrac{F}{m}\\\\a=\dfrac{20}{70}\\\\a=\dfrac{2}{7}\ m/s^2[/tex]
Initial velocity of rider is 0. So, using equation of kinematics to find the final velocity as :
[tex]v=u+at\\\\v=at\\\\v=\dfrac{2}{7}\times 15\\\\v=4.28\ m/s[/tex]
Since, 1 m/s = 2.23 mph
4.28 m/s = 9.57 mph
So, the speed of the rider is 4.28 m/s or 9.57 mph.
A car moving in a straight line starts at X=0 at t=0. It passesthe point x=25.0 m with a speed of 11.0 m/s at t=3.0 s. It passes the point x=385 with a speed of 45.0 m/s at t=20.0 s. Find the average velocity and the average acceleration between t=3.0 s and 20.0 s.
Answer:
Average velocity v = 21.18 m/s
Average acceleration a = 2 m/s^2
Explanation:
Average speed equals the total distance travelled divided by the total time taken.
Average speed v = ∆x/∆t = (x2-x1)/(t2-t1)
Average acceleration equals the change in velocity divided by change in time.
Average acceleration a = ∆v/∆t = (v2-v1)/(t2-t1)
Where;
v1 and v2 are velocities at time t1 and t2 respectively.
And x1 and x2 are positions at time t1 and t2 respectively.
Given;
t1 = 3.0s
t2 = 20.0s
v1 = 11 m/s
v2 = 45 m/s
x1 = 25 m
x2 = 385 m
Substituting the values;
Average speed v = ∆x/∆t = (x2-x1)/(t2-t1)
v = (385-25)/(20-3)
v = 21.18 m/s
Average acceleration a = ∆v/∆t = (v2-v1)/(t2-t1)
a = (45-11)/(20-3)
a = 2 m/s^2
A surveyor measures the distance across a river that flows straight north by the following method. Starting directly across from a tree on the opposite bank, the surveyor walks distance, D = 130 m along the river to establish a baseline. She then sights across to the tree and reads that the angle from the baseline to the tree is an angle θ = 25°. How wide is the river?
Answer:
The width of the river is [tex]z = 60.62 \ m[/tex]
Explanation:
From the question we are told that
The distance of the base line is D = 130 m
The angle is [tex]\theta = 25^o[/tex]
A diagram illustration the question is shown on the first uploaded image
Applying Trigonometric Rules for Right-angled Triangle,
[tex]tan 25 = \frac{z}{130}[/tex]
Now making z the subject
[tex]z = 130 * tan (25)[/tex]
[tex]z = 60.62 \ m[/tex]
An engine draws energy from a hot reservoir with a temperature of 1250 K and exhausts energy into a cold reservoir with a temperature of 322 K. Over the course of one hour, the engine absorbs 1.37 x 105 J from the hot reservoir and exhausts 7.4 x 104 J into the cold reservoir.
1) What is the power output of this engine?
2) What is the maximum (Carnot) efficiency of a heat engine running between these two reservoirs?
3) What is the actual efficiency of this engine?
Answer:
The power output of this engine is [tex]P = 17.5 W[/tex]
The the maximum (Carnot) efficiency is [tex]\eta_c = 0.7424[/tex]
The actual efficiency of this engine is [tex]\eta _a = 0.46[/tex]
Explanation:
From the question we are told that
The temperature of the hot reservoir is [tex]T_h = 1250 \ K[/tex]
The temperature of the cold reservoir is [tex]T_c = 322 \ K[/tex]
The energy absorbed from the hot reservoir is [tex]E_h = 1.37 *10^{5} \ J[/tex]
The energy exhausts into cold reservoir is [tex]E_c = 7.4 *10^{4} J[/tex]
The power output is mathematically represented as
[tex]P = \frac{W}{t}[/tex]
Where t is the time taken which we will assume to be 1 hour = 3600 s
W is the workdone which is mathematically represented as
[tex]W = E_h -E_c[/tex]
substituting values
[tex]W = 63000 J[/tex]
So
[tex]P = \frac{63000}{3600}[/tex]
[tex]P = 17.5 W[/tex]
The Carnot efficiency is mathematically represented as
[tex]\eta_c = 1 - \frac{T_c}{T_h}[/tex]
[tex]\eta_c = 1 - \frac{322}{1250}[/tex]
[tex]\eta_c = 0.7424[/tex]
The actual efficiency is mathematically represented as
[tex]\eta _a = \frac{W}{E_h}[/tex]
substituting values
[tex]\eta _a = \frac{63000}{1.37*10^{5}}[/tex]
[tex]\eta _a = 0.46[/tex]
PLEASE HELP !
Complete the following sentence. Choose the right answer from the given ones. The internal energy of the body can be changed A / B / C. A. only when the body is warmed or cooled B. when work is done on the body or heat flow C. only when the body does work
B
HOPE IT HELPS LET ME KNOW IF U NEED EXPLANATION
If a light is moved twice (2x) as far from a surface, the area the light covers is ___ as big.
- 2x
- 1/4
- 1/2
- 4x
Answer:
The correct option is;
- 4x
Explanation:
From the inverse square law, as the distance from the source of a physical quantity increases, the intensity of the source is spread over an area proportional to the square of the distance of the object from the source
The inverse square law can be presented as follows;
[tex]I = \dfrac{S}{4\times \pi \times r^2 }[/tex]
As the distance, r, increases, the surface it covers also increases by the power of 2
Therefore, where the distance increases from r to 2·r, we have;
When, I, remain constant
[tex]I = \dfrac{4\times S}{4\times \pi \times (2\cdot r)^2 } = I = \dfrac{4\times S}{4\times 4\times \pi \times r^2 } = \dfrac{S}{4\times \pi \times r^2 }[/tex]
The surface increases to 4·S by the inverse square law
Therefore, the correct option is 4 × x.
A 72.0 kg swimmer jumps into the old swimming hole from a tree limb that is 3.90 m above the water.
A. Use energy conservation to find his speed just as he hits the water if he just holds his nose and drops in.
b) Use energy conservation to find his speed just he hits the water if he bravely jumps straight up (but just beyond the board!) at 2.90 m/s .
c) Use energy conservation to find his speed just he hits the water if he manages to jump downward at 2.90 m/s .
Answer:
Explanation:
The Law of Energy Conservation states that K1 + U1 = K2 + U2
m= 72.0 kg
h= 3.90 m
a)
K1 + U1 = K2 + U2
0 + mgh = 1/2mvf^2 + 0
mass cancels out so gh=1/2vf^2
(9.8 m/s^2)(3.9 m)=(.5)(vf^2)
vf= 8.74 m/s
b)
1/2mv^2 + mgh = 1/2mv^2 + 0
mass cancels again
(.5)(2.9^2 m/s) + (9.8 m/s^2)(3.9 m) = (.5)(vf^2)
vf= 9.21 m/s
c)
This would be the same as the past problem as the velocity gets squared so direction along the axis doesn't matter. Thus, vf= 9.21 m/s
calculate the volume of marble if its diameter is 10mm
Answer:
The volume of the marble is [tex]523.33\ mm^2[/tex].
Explanation:
Marble is spherical in shape. The diameter of marble is 10 mm. It radius will be 5 mm.
The volume of spherical shaped object is given by :
[tex]V=\dfrac{4}{3}\pi r^3[/tex]
Plugging all the values, we get :
[tex]V=\dfrac{4}{3}\times 3.14\times (5)^3\\\\V=523.33\ mm^2[/tex]
So, the volume of the marble is [tex]523.33\ mm^2[/tex].
A uniform disk with a 25 cm radius swings without friction about a nail through the rim. If it is released from rest from a position with the center level with the nail, then what is its angular velocity as it swings through the point where the center is below the na
Answer:
Explanation:
During the swing , the center of mass will go down due to which disc will lose potential energy which will be converted into rotational kinetic energy
mgh = 1/2 I ω² where m is mass of the disc , h is height by which c.m goes down which will be equal to radius of disc , I is moment of inertia of disc about the nail at rim , ω is angular velocity .
mgr = 1/2 x ( 1/2 m r²+ mr²) x ω²
gr = 1/2 x 1/2 r² x ω² + 1/2r² x ω²
g = 1 / 4 x ω² r + 1 / 2 x ω² r
g = 3 x ω² r/ 4
ω² = 4g /3 r
= 4 x 9.8 / 3 x .25
= 52.26
ω = 7.23 rad / s .
Mr. Patel is photocopying lab sheets for his first period class. A particle of toner carrying a charge of 4.0 * 10^9 C in the copying machine experiences an electric field of 1.2 * 10^6 N/C as it’s pulled toward the paper. What is the electric force acting on the toner particle?
Answer:
4.8 × 10^15 N
Explanation:
Electric Field is defined as Force per unit Charge.
This is expressed mathematically as;
E= F/Q
Where E- Electric Field
F- Force
Q- charge
From the expression above by change of subject of formula for F, we have;
F=E×Q
= 1.2 * 10^6 ×4.0 * 10^9
= 4.8 × 10^15 N
I really need help with this question someone plz help !
Answer:weight
Explanation:weight
Chapter 24, Problem 20 GO A politician holds a press conference that is televised live. The sound picked up by the microphone of a TV news network is broadcast via electromagnetic waves and heard by a television viewer. This viewer is seated 2.9 m from his television set. A reporter at the press conference is located 4.1 m from the politician, and the sound of the words travels directly from the celebrity's mouth, through the air, and into the reporter's ears. The reporter hears the words exactly at the same instant that the television viewer hears them. Using a value of 343 m/s for the speed of sound, determine the maximum distance between the television set and the politician. Ignore the small distance between the politician and the microphone. In addition, assume that the only delay between what the microphone picks up and the sound being emitted by the television set is that due to the travel time of the electromagnetic waves used by the network.
Answer:
Therefore, the distance between politician and TV set is 2536kmExplanation:
Assuming that the TV signal is sent in a straight line from the camera to the TV receiver, which is very far from the truth.
The reporter hears the sound is
4.1 / 343 = 0.01195 s later
The viewer hears the sound from the TV is
2.9 / 343 = 0.00845s
the difference is 0.00845 sec
the question is how far the TV signal can travel in that time.
the distance between politician and TV set is
= 0.00845 * 3*10^8 m
= 2536 km
d = 2536km
Therefore, the distance between politician and TV set is 2536kmA solid sphere has a temperature of 556 K. The sphere is melted down and recast into a cube that has the same emissivity and emits the same radiant power as the sphere. What is the cube's temperature in kelvins
Answer:
Cube temperature = 526.83 K
Explanation:
Volume of the cube and sphere will be the same.
Now, volume of cube = a³
And ,volume of sphere = (4/3)πr³
Thus;
a³ = (4/3)πr³
a³ = 4.1187r³
Taking cube root of both sides gives;
a = 1.6119r
Formula for surface area of sphere is;
As = 4πr²
Also,formula for surface area of cube is; Ac = 6a²
Thus, since a = 1.6119r,
Then, Ac = 6(1.6119r)²
Ac = 15.5893r²
The formula for radiant power is;
Q' = eσT⁴A
Where;
e is emissivity
σ is Stefan boltzman constant = 5.67 x 10^(-8) W/m²k
T is temperate in kelvin
A is Area
So, for the cube;
(Qc)' = eσ(Tc)⁴(Ac)
For the sphere;
(Qs)' = eσ(Ts)⁴(As)
We are told (Qc)' = (Qs)'
Thus;
eσ(Tc)⁴(Ac) = eσ(Ts)⁴(As)
eσ will cancel out to give;
(Tc)⁴(Ac) = (Ts)⁴(As)
Since we want to find the cube's temperature Tc,
(Tc)⁴ = [(Ts)⁴(As)]/Ac
Plugging in relevant figures, we have;
(Tc)⁴ = [556⁴ × 4πr²]/15.5893r²
r² will cancel out to give;
(Tc)⁴ = [556⁴ × 4π]/15.5893
Tc = ∜([556⁴ × 4π]/15.5893)
Tc = 526.83 K
1. How is it possible to use pools to model apparent weightlessness, similar to what astronauts
experience on the Moon or on the space station? Explain
Answer:
by using it's buoyant or floating effect by Archimedes.
the buoyant force act on the astronauts body and make he/ she feels like in low gravity.
the buoyant force equation is
F = Density of liquid x earth gravitational field x volume of astronauts body and suit.
the Weight of astronauts in the pools will be less than in the land or air.
Weight in water = weight in air/land - buoyant force
so the astronauts will feel like in the outer space with low gravity.
A 0.150 kg lump of clay is dropped from a height of 1.45 m onto the floor. It sticks to the floor and does not bounce.
What is the magnitude of the impulse imparted to the clay by the floor during the impact? Assume that the acceleration due to gravity is =9.81 m/s2.
Answer:
J = 0.800 kg m/s
Fmax = 291 N
Explanation:
During the fall, energy is conserved.
PE = KE
mgh = ½ mv²
v = √(2gh)
v = √(2 × 9.81 m/s² × 1.45 m)
v = 5.33 m/s
Alternatively, you can use kinematics to find the velocity.
Impulse = change in momentum
J = Δp
J = mΔv
J = (0.150 kg) (5.33 m/s)
J = 0.800 kg m/s
Impulse = area under F vs t graph
J = ∫ F dt
J = ½ Fmax Δt
(0.800 kg m/s) = ½ Fmax (0.0055 ms)
Fmax = 291 N
Two parallel, vertical, plane mirrors, 38.8 cm apart, face each other. A light source at point P is 30.1 cm from the mirror on the left and 8.7 cm from the mirror on the right.
(a) How many images of point P are formed by the mirrors?
(b) Find the distance from the mirror on the right to the two nearest images behind the mirror.
first nearest image=
second nearest image=
(c) Find the number of reflections of light rays for each of these images.
first nearest image=
second nearest image=
Answer:
Explanation shown below.
Explanation:
1.The number of images formed by 2 parallel mirrors is an infinite number of images.
2. The characteristics of a plane mirror is such that the object distance equals the image distance.
Hence the object distance is 8.7cm from the right; the image formed would be 8.7cm behind the mirror.
Now a second image is going to be formed by the left mirror which is going to have an image distance of 30.1cm behind the mirror.
Now this image would be reflected on the right side to form a new image which is going to be seen as 38.8 +30.1 = 68.9cm behind the right Mirror .
Hence the shortest distances are 8.7cm and 68.9cm
3. The number of reflections is infinite for both cases.
Water is traveling through a horizontal pipe with a speed of 1.7 m/s and at a pressure of 205 kPa. This pipe is reduced to a new pipe which has a diameter half that of the first section of pipe. Determine the speed and pressure of the water in the new, reduced in size pipe.
Answer:
The velocity is [tex]v_2 = 6.8 \ m/s[/tex]
The pressure is [tex]P_2 = 204978 Pa[/tex]
Explanation:
From the question we are told that
The speed at which water is travelling through is [tex]v = 1.7 \ m/s[/tex]
The pressure is [tex]P_1 = 205 k Pa = 205 *10^{3} \ Pa[/tex]
The diameter of the new pipe is [tex]d = \frac{D}{2}[/tex]
Where D is the diameter of first pipe
According to the principal of continuity we have that
[tex]A_1 v_1 = A_2 v_2[/tex]
Now [tex]A_1[/tex] is the area of the first pipe which is mathematically represented as
[tex]A_1 = \pi \frac{D^2}{4}[/tex]
and [tex]A_2[/tex] is the area of the second pipe which is mathematically represented as
[tex]A_2 = \pi \frac{d^2}{4}[/tex]
Recall [tex]d = \frac{D}{2}[/tex]
[tex]A_2 = \pi \frac{[ D^2]}{4 *4}[/tex]
[tex]A_2 = \frac{A_1}{4}[/tex]
So [tex]A_1 v_1 = \frac{A_1}{4} v_2[/tex]
substituting value
[tex]1.7 = \frac{1}{4} * v_2[/tex]
[tex]v_2 = 4 * 1.7[/tex]
[tex]v_2 = 6.8 \ m/s[/tex]
According to Bernoulli's equation we have that
[tex]P_1 + \rho \frac{v_1 ^2}{2} = P_2 + \rho \frac{v_2 ^2}{2}[/tex]
substituting values
[tex]205 *10^{3 }+ \frac{1.7 ^2}{2} = P_2 + \frac{6.8 ^2}{2}[/tex]
[tex]P_2 = 204978 Pa[/tex]
To get up on the roof, a person (mass 69.0 kg) places a 6.40 m aluminum ladder (mass 11.0 kg) against the house on a concrete pad with the base of the ladder 2.00 m from the house. The ladder rests against a plastic rain gutter, which we can assume to be frictionless. The center of mass of the ladder is 2 m from the bottom. The person is standing 3 m from the bottom. What are the magnitudes (in N) of the forces on the ladder at the top and bottom
Answer:
N = 243.596 N ≈ 243.6 N
Explanation:
mass of person = 69 kg ( M )
mass of aluminium ladder = 11 kg ( m )
length of ladder = 6.4 m ( l )
base of ladder = 2 m from the house (d )
center of mass of ladder = 2 m from the bottom of ladder
person on ladder standing = 3 m from bottom of ladder
Calculate the magnitudes of the forces at the top and bottom of the ladder
The net torque on the ladder = o ( since it is at equilibrium )
assuming: the weight of the person( mg) acting at a distance x along the ladder. the weight of the ladder ( mg) acting halfway along the ladder and the reaction N acting on top of the ladder
X = l/2
x = 6.4 / 2 = 3.2
find angle formed by the ladder
cos ∅ = d/l
∅ = [tex]cos^{-1][/tex] 2/6.4 = [tex]cos^{-1}[/tex]0.3125 ≈ 71.79⁰
remember the net torque around is = zero
to calculate the magnitude of forces on the ladder we apply the following formula
[tex]N = \frac{mg(dcosteta)+ Mgxcosteta}{lsinteta}[/tex]
m = 11 kg, M = 69 kg, l = 6.4 , x = 3, teta( ∅ )= 71.79⁰, g = 9.8
back to equation N = [tex]\frac{11*9.8(2*cos71.79)+ 69*9.8*3* cos71.79}{6.4sin71.79}[/tex]
N = (67.375 + 633.938) / 2.879
N = 243.596 N ≈ 243.6 N
In a circuit, a 100.-ohm resistor and a 200.-ohm resistor are connected in parallel to a 10.0-volt battery.
Calculate the equivalent resistance of the circuit. [Show all work, including the equation and substitution with units.]
Answer:
Explanation:
The equivalent resistance of resistor connected parallel in the circuit is [tex]66.66 ohm[/tex]
What is equivalent resistance?The equivalent resistance is the total resistance measured in a parallel or series circuit. If several resistors are connected together and connected to a battery, the current supplied by the battery depends on the equivalent resistance of the circuit.
What is equivalent resistance in series?Resistors are in series whenever the current flows through the resistors sequentially. It is given by
[tex]R_{eq} = R_{1} + R_{2} + ....[/tex]
What is equivalent resistance in parallel?Resistors are in parallel when one end of all the resistors are connected by a continuous wire and the other end of all the resistors are also connected to one another through a continuous wire.
The equivalent resistance is the total resistance measured in a parallel. It is given by
[tex]\frac{1}{R_{eq} } = \frac{1}{R_{1} } + \frac{1}{R_{2} }+ ....[/tex]
Given:
Resistor, [tex]R_{1} = 100 ohm[/tex]
Resistor, [tex]R_{2} = 200 ohm[/tex]
Voltage, [tex]V = 10 Volt[/tex]
Since, resistors are connected in parallel, the equivalent resistor is given by,
[tex]\frac{1}{R_{eq} } = \frac{1}{R_{1} } + \frac{1}{R_{2} }[/tex]
[tex]\frac{1}{R_{eq} } = \frac{1}{100 } + \frac{1}{200 }[/tex]
[tex]R_{eq} = \frac{100*200}{100+200} \\R_{eq} = 66.66 ohm[/tex]
Hence, the equivalent resistor is [tex]66.66 ohm[/tex].
To learn more about equivalent resistor here
https://brainly.com/question/113987
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A rocket rises vertically, from rest, with an acceleration of 5.0 m/s2 until it runs out of fuel at an altitude of 960 m . After this point, its acceleration is that of gravity, downward.
(A) What is the velocity of the rocket when it runs out of fuel?
(B) How long does it take to reach this point?
(C) What maximum altitude does the rocket reach?
(D) How much time (total) does it take to reach maximum altitude?
(E) With what velocity does it strike the Earth? () How long (total) is it in the air?
a) 70.427m/s
b) 22 m
c) 1027.8m
d) 29.179 s
e) 142m/s
f ) 43.654s
Answer:
a) 98 m/s
b) 19.6 s
c) 1449.8 m
d) 29.6 s
e) 168.6 m/s
f) 46.8 s
Explanation:
Given that
Acceleration of the rocket, a = 5 m/s²
Altitude of the rocket, s = 960 m
a)
Using the equation of motion
v² = u² + 2as, considering that the initial velocity, u is 0. Then
v² = 2as
v = √2as
v = √(2 * 5 * 960)
v = √9600
v = 98 m/s
b)
Using the equation of motion
S = ut + ½at², considering that initial velocity, u = 0. So that
S = ½at²
t² = 2s/a
t² = (2 * 960) / 5
t² = 1920 / 5
t² = 384
t = √384 = 19.6 s
c)
Using the equation of motion
v² = u² + 2as, where u = 98 m/s, a = -9.8 m/s², so that
0 = 98² + 2(-9.8) * s
9600 = 19.6s
s = 9600/19.6
s = 489.8 m
The maximum altitude now is
960 m + 489.8 m = 1449.8 m
d)
Using the equation of motion
v = u + at, where initial velocity, u = 98 m, a = -9.8 m/s. So that
0 = 98 +(-9.8 * t)
98 = 9.8t
t = 98/9.8
t = 10 s
Total time then is, 10 + 19.6 = 29.6 s
e) using the equation of motion
v² = u² + 2as, where initial velocity, u = o, acceleration a = 9.8 m/s, and s = 1449.8 m. So that,
v² = 0 + 2 * 9.8 * 1449.8
v² = 28416.08
v = √28416.08
v = 168.6 m/s
f) using the equation of motion
S = ut + ½at², where s = 1449.8 m and a = 9.8 m/s
1449.8 = 0 + ½ * 9.8 * t²
2899.6 = 9.8t²
t² = 2899.6/9.8
t² = 295.88
t = √295.88
t = 17.2 s
total time in air then is, 17.2 + 29.6 = 46.8 s
A 2-kilogram toy car is traveling forward at 1 meter per second when it is hit in the rear by a 3-kilogram toy truck that was traveling at 3 meters per second just before impact. If the two toys stick together, their speed immediately after the collision is
Answer:
v = 1.4 m/s
Explanation:
This problem is about an inelastic collision. The total momentum before the collision is equal to total momentum after (because of the conservation of momentum law):
[tex]m_1v_1-m_2v_2=(m_1+m_2)v[/tex] (1)
m1: mass of the toy car = 2 kg
m2: mass of the toy truck = 3 kg
v1: speed of the toy car = 1 m/s
v2: speed of the truck car = 3 m/s
v: speed of both car and truck after the collision = ?
In the equation (1) the negative sign of m2v2 is because of the opposite direction of the toy truck respect to the toy car.
You solve the equation (1) for v, and you replace the values of all variables involved:
[tex]v=\frac{m_1v_1-m_2v_2}{m_1+m_2}\\\\v=\frac{(2kg)(1m/s)-(3kg)(3m/s)}{2kg+3kg}=-1.4\frac{m}{s}[/tex]
this velocity is negative, then, the direction of motion of both car and truck is in the direction of the truck
Hence, the speed of both car and truck toys is 1.4 m/s
During last year’s diving competition, the divers always pull their limbs in and curl up their bodies when they do flips. Just before entering the water, they fully extend their limbs to enter straight down as shown. Explain the effect of both actions on their angular velocities and kinetic energy (support your answer with working). Also explain the effect on their angular momentum.
Answer:
the angular speed of the person increases, being able to make more turns and faster.
K₂ = K₁ I₁ / I₂
Explanation:
When the divers are turning the system is isolated, so all the forces are internal and therefore also the torque, therefore the angular momentum is conserved
initial, joint when starting to turn
L₀ = I₁ w₁
final. When you shrink your arms and legs
Lf = I₂ w₂
L₀ = Lf
I₁ w₁ = I₂ w₂
when you shrink your arms and legs the distance to the turning point decreases and since the moment of inertia depends on the distance squared, the moment of inertia also decreases
I₂ <I₁
w₂ = I₁ / I₂ w₁
therefore the angular speed of the person increases, being able to make more turns and faster.
When it goes into the water it straightens the arm and leg, so the moment of inertia increases
I₁> I₂
w₁ = I₂ / I₁ w₂
therefore we see that the angular velocity decreases, therefore the person trains the water like a stone and can go deeper faster.
In both cases the kinetic energy is
K = ½ I w²
the initial kinetic energy is
K₁ = ½ I₁ w₁²
the final kinetic energy is
K₂ = ½ I₂ w₂²
we substitute
K₂ = ½ I₂ (I₁ / I₂ w1² 2
K₂ = ½ I₁² / I₂ w₁² = (½ I₁ w₁²) I₁ / I₂
K₂ = K₁ I₁ / I₂
therefore we see that the kinetic energy increases by factor I₁/I₂
Someone plzzz helpppppp with this last question
Answer:
I dont know someone deleted answers. But they were wrong. INERTIA IS CORRECT I DID THIS IN MY SCHOOL
C IS CORRECT
A rigid tank contains 2 kg of an ideal gas at 4 atm and 40 C. Now a valve is opened, and half of the mass of the gas is allowed to escape. if the final pressure in the tank is 2.2 atm. The final temperature in the tank is: Hint: make sure you convert the units of temperature and pressure to the proper units
Answer:
Final Temperature = 71 °C
Explanation:
In this case, the ideal gas equation is written as;
PV = mRT
Where;
P is pressure
V is volume
m is mass
R is gas constant
T is temperature
We will take the volume to be constant.
So, in the initial state, we have;
P1•V = m1•R•T1 - - - eq(1)
In the final state, we have;
P2•V = m2•R•T2 - - - - eq(2)
Combining eq (1) and eq(2),we have;
P1•m2•R•T2 = P2•m1•R•T1
Dividing both sides by R gives;
P1•m2•T2 = P2•m1•T1
Making T2 the subject gives;
T2 = (P2•m1•T1)/(P1•m2)
Now, we are given;
m1 = 2kg
m2 = ½*2 = 1kg
P1 = 4 atm
P2 = 2.2 atm
T1 = 40°C = 273 + 40 K = 313K
Plugging in this values into the T2 equation, we have;
T2 = (2.2 × 2 × 313)/(4 × 1)
T2 = 344 K
Converting to °C, we have;
T2 = 344 - 273 = 71 °C
1. (a) The battery on your car has a rating stated in ampere-minutes which permits you to
estimate the length of time a fully charged battery could deliver any particular current
before discharge. Approximately how much energy is stored by a 50 ampere-minute 12
volt battery?
Answer:
Energy Stored = 36000 J = 36 KJ
Explanation:
The power of a battery is given by the formula:
P = IV
where,
P = Power delivered by the battery
I = Current Supplied to the battery
V = Potential Difference between terminals of battery = 12 volt
Now, we multiply both sides by the time period (t):
Pt = VIt
where,
Pt = (Power)(Time) = Energy Stored = E = ?
It = Battery Current Rating = 50 A.min
Converting this to A.sec;
It = Battery Current Rating = (50 A.min)(60 sec/min) = 3000 A.sec
Therefore,
E = (12 volt)(3000 A.sec)
E = 36000 J = 36 KJ