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Ber [7]
3 years ago
15

The acceleration of a cart rolling down a ramp depends on __________.

Physics
1 answer:
zmey [24]3 years ago
8 0

The angle that the cart rolls with the horizontal. The closer the ramp gets to 90 degrees the faster the cart will accelerate.

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Two astronauts (each with mass 100 kg) are drifting together through space. They are connected to each other by a rope 5 m in le
Nana76 [90]

Answer:

1000 kgm²/s, 400 J

1000 kgm²/s, 1000 J

600 J

Explanation:

m = Mass of astronauts = 100 kg

d = Diameter

r = Radius = \frac{d}{2}

v = Velocity of astronauts = 2 m/s

Angular momentum of the system is given by

L=mvr+mvr\\\Rightarrow L=2mvr\\\Rightarrow L=2\times 100\times 2\times 2.5\\\Rightarrow L=1000\ kgm^2/s

The angular momentum of the system is 1000 kgm²/s

Rotational energy is given by

K=I\omega^2\\\Rightarrow K=\frac{1}{2}(mr^2)\left(\frac{v}{r}\right)^2\\\Rightarrow K=mv^2\\\Rightarrow K=100\times 2^2\\\Rightarrow K=400\ J

The rotational energy of the system is 400 J

There no external toque present so the initial and final angular momentum will be equal to the initial angular momentum 1000 kgm²/s

L_i=L_f\\\Rightarrow 2mv_ir_i=2mv_fr_f\\\Rightarrow v_f=\frac{v_ir_i}{r_f}\\\Rightarrow v_f=\frac{2\times 2.5}{0.5}\\\Rightarrow v_f=10\ m/s

Energy

E_2=mv_f^2\\\Rightarrow E_2=100\times 10\\\Rightarrow E_2=1000\ J

The new energy will be 1000 J

Work done will be the change in the kinetic energy

W=E_2-E\\\Rightarrow W=1000-400\\\Rightarrow W=600\ J

The work done is 600 J

5 0
3 years ago
A closed, uninsulated system fitted with movable piston, so no matter is exchanged with the surroundings, was assembled. Introdu
xeze [42]

Answer: You do not specify what is being asked for. ∆E? ∆H?

∆E = (430 - 238) J = 192 J

∆H = 430 J

Explanation:

If asked for the value of ∆H the answer is simply the change in heat, and in the question, it states introduction of 430 J of heat is causing the system to expand.

Therefore ∆H = 430 J

If asked for ∆E, we know that ∆E = ±q (heat) + work (-P∆V) = ±q + w

The question states that 238 J of work are done AND the system expanded

(work is negative because expansion means work is done BY the system, releasing energy/heat... Conversely, if the system were compressed, work is done ON the system, absorbing heat/energy)

Therefore, ∆E = (430 - 238) J = 192 J

8 0
2 years ago
Convert the speed of light, 3.0 x 108 m/s, to km/s.
zhuklara [117]

Answer:

Answer:

I don't know

Explanation:

I don't know

Explanation:

Answer:

I don't know

Explanation:

I don't know

3 0
3 years ago
A merry-go-round of radius 2 m is rotating at one revolution every 5 s. A
galben [10]

Answer:

a) The angular speed of the child is approximately 1.257 rad/s

b) The angular speed of the teenager is approximately 1.257 rad/s

c) The tangential speed of the child is approximately 1.257 m/s

d) For the child, r = 2 m

The tangential speed of the teenager is approximately 2.513 m/s

Explanation:

The revolutions per minute, r.p.m. of the merry-go-round = 1 revolution/(5 s)

The radius of the merry-go-round = 2 m

The location of the child = 1 m from the axis

The location of the teenager = 2 m from the axis

1 revolution = 2·π radians

Therefore, we have;

The angular speed, ω = (Angle turned)/(Time elapsed) = (2·π radians)/(5 s)

∴ The angular speed of the merry-go-round, ω = 2·π/5 radians/second

a) The angular speed of the child = The angular speed of the merry-go-round = 2·π/5 radians/second ≈ 1.257 rad/s

b) The angular speed of the teenager = The angular speed of the merry-go-round = 2·π/5 radians/second ≈ 1.257 rad/s

c) The tangential speed, v = r × The angular speed, ω

Where;

r = The radius of rotation of the object

For the child, r = 1 m

The tangential speed of the child = 1 m × 2·π/5 radians/second = 2·π/5 m/s ≈ 1.257 m/s

d) For the child, r = 2 m

The tangential speed of the teenager = 2 m × 2·π/5 radians/second = 4·π/5 m/s ≈ 2.513 m/s

8 0
2 years ago
If a builder of mass 75kg climbs a vertical ladder of 25m how much energy has she gained ?
erica [24]
So,

GPE (graviational potential energy) = mass x g x height

GPE is depends on where zero height is defined.  In this situation, we define h = 0 as the initial height.

GPE = 75 \ kg*9.8 \ \frac{m}{s^2}*25 \ m

GPE = 18,375 \frac{kg*m^2}{s^2}

GPE = 18,375 \ joules(J) \ or \ 18.375 \ kilojoules(kJ)

The builder has gained 18.375 kJ of PE.
4 0
3 years ago
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