What conversion takes place in a motor?

Which type of heat transfer takes place in a vacuum? (3 points)

tiny-mole [99]

Radiation is the only transfer in a vacuum

3 0

3 years ago

What was the 2nd deepest ship wreck that James Cameron visited (from WWII)?

True [87]

Answer: :) i think its the t

8 0

2 years ago

Two balls of mass m1 and m2, with velocities v1 and v2 collide head on. Is there any way for both balls to have zero velocity af

nordsb [41]

Answer:

Explanation:

As the final Kinetic energy is zero or less than initial kinetic energy, the collision must be inelastic.

In Inelastic collision both the bodies must stick together as final velocity is zero for both the bodies.

To conserve the momentum, momentum associated before the collision of first must be equal and opposite to the momentum associated with the second ball.

i.e.

$m_1v_1=m_2v_2$

8 0

3 years ago

A tourist stands at the top of the Grand Canyon, holding a rock, overlooking the valley below. Find the final velocity and displ

Anestetic [448]

Answer:

a. -39.2 m/s; -78.4 m

b. -31.2 m/s; -46.4 m

c. -47.2 m/s; -110.4 m

Explanation:

We are given/can infer these variables:

t = 4.0 s
a = -9.8 m/s²
v_0 = 0 m/s

We want to find the displacement and the final velocity of the rock.

Δx = ?
v = ?

We can use this equation to find the final velocity:

v = v_0 + at

Plug in the known variables into this equation.

v = 0 + (-9.8)(4.0)
v = -9.8 * 4.0
v = -39.2 m/s

The final velocity of the rock is -39.2 m/s.

Now we can use this equation to find the displacement of the rock:

Δx = v_0 t + 1/2at²

Plug in the known variables into this equation.

Δx = 0 * 4.0 + 1/2(-9.8)(4.0)²
Δx = 1/2(-9.8)(4.0)²
Δx = -4.9 * 16
Δx = -78.4 m

The displacement of the rock is -78.4 m.

We are given/can infer these variables:

v_0 = 8.0 m/s
a = -9.8 m/s²
t = 4.0 s

We can use this equation to find the final velocity:

v = v_0 + at

Plug in the known variables into this equation.

v = 8.0 + (-9.8)(4.0)
v = 8.0 + -39.2
v = -31.2 m/s

The final velocity of the rock is -31.2 m/s.

We can use this equation to find the displacement:

Δx = v_0 t + 1/2at²

Plug in known variables:

Δx = 8.0(4.0) + 1/2(-9.8)(4.0)²
Δx = 32 - 4.9(16)
Δx = -46.4 m

The displacement of the rock is -46.4 m.

We are given/can infer these variables:

v_0 = -8.0 m/s
a = -9.8 m/s²
t = 4.0 s

We can use this equation to find the final velocity:

v = v_0 + at

Plug in the known variables into this equation.

v = -8.0 + (-9.8)(4.0)
v = -8.0 - 39.2
v = -47.2 m/s

The final velocity of the rock is -47.2 m/s.

We can use this equation to find the displacement:

Δx = v_0 t + 1/2at²

Plug in known variables:

Δx = -8.0(4.0) + 1/2(-9.8)(4.0)²
Δx = -32 - 4.9(16)
Δx = -110.4 m

The displacement of the rock is -110.4 m.

8 0

3 years ago

The horizontal bar rises at a constant rate of three hundred mm/s causing peg P to ride in the quarter circular slot. When coord

SVETLANKA909090 [29]

Answer:

Explanation:

Given

Horizontal bar rises with 300 mm/s

Let us take the horizontal component of P be

$P_x=rcos\theta$

$P_y=rsin\theta$

where $\theta$ is angle made by horizontal bar with x axis

Velocity at y=150 mm

$150=300sin\theta$

thus $\theta =30^{\circ}$

position of $P_x=rcos\theta =300\cdot cos30=300\times \frac{\sqrt{3}}{2}$

$P_x=259.80 mm$

$P=259.80\hat{i}+150\hat{j}$

Velocity at this instant

$u_x=-rsin\theta =300\times sin30=-150 mm/s$

$u_y=rcos\theta =300\times cos30=259.80 mm/s$

4 0

3 years ago