Help with question b

A ball rolls down a ramp. A small amount of friction heats up the ball. What energy transformations are taking place?

Natasha2012 [34]

Answer:kinetic energy converted to heat energy

Explanation:

As the ball rolls down kinetic energy is converted to heat energy

6 0

3 years ago

Read 2 more answers

Two spectators at a soccer game in Montjuic Stadium see, and a moment later hear, the ball being kicked on the playing field. Th

Akimi4 [234]

Answer:

a) The distance of spectator A to the player is 79.2 m

b) The distance of spectator B to the player is 43.9 m

c) The distance between the two spectators is 90.6 m

Explanation:

a) Knowing the time it takes the sound to reach both spectators, we can calculate their position relative to the player, using this equation:

x = v * t

where:

x = position of the spectators

v = speed of sound

t = time

Then, the position for spectator A relative to the player is:

x = 343 m/s * 0.231 s = 79.2 m

b)For spectator B:

x = 343 m/s * 0.128 s

x = 43.9 m

The distance of spectator A and B to the player is 79.2 m and 43.9 m respectively.

c) To calculate the distance between the spectators, please see the attached figure. Notice that the distance between the spectators is the hypotenuse of the triangle formed by the sightline of both. We already know the longitude of the two sides. Then, using Pythagoras theorem:

(Distance AB)² = A² + B²

(Distance AB)² = (79.2 m)² + (43.9 m)²

Distance AB = 90. 6 m

6 0

3 years ago

An uncharged metal sphere, A, is on an insulating base. A second sphere, B, of the same size, shape, and material carrying charg

solong [7]

Answer:

0

Explanation:

If we bring the charged sphere B close to, but not touching it , to the uncharged sphere A, as charges can move freely on the conductor, a charge -Q will be built on the outer surface of the sphere A, facing to sphere B.
As the sphere A must remain neutral, a charge Q will be built on the surface, on the side farther to the sphere B, as the following condition must be met:

Q +(-Q) =0.

If we now remove sphere B, and place it far away, there will be a charge redistribution within sphere A, making to disappear the separation between Q and -Q.
The total charge on sphere A must be 0, as there is no charge transfer from sphere B to sphere A.

3 0

3 years ago

A migrating robin flies due north with a speed of 12 m/s relative to the air. The air moves due east with a speed of 6.7 m/s rel

mafiozo [28]

Here it is given that speed of migrating Robin is 12 m/s relative to air

so we can say that

$\vec v_{ra} = 12 m/s$ North

so it will be

Let North direction is along Y axis and East direction is along X axis

$\vec v_{ra} = 12\hat j$

also it is given that speed of air is 6.7 m/s relative to ground

$\vec v_a = 6.7 \hat i$

now as we know by the concept of relative motion

$\vec v_{ab} = \vec v_a - \vec v_b$

$\vec v_{ra} = \vec v_r - \vec v_a$

now by rearranging the terms

$\vec v_r = \vec v_{ra} + \vec v_a$

$\vec v_r = 12 \hat j + 6.7 \hat i$

now we need to find the speed of Robin which means we need to find the magnitude of its velocity which we found above

So here we will say

$v_r = \sqrt{12^2 + 6.7^2}$

$v_r = 13.7 m/s$

so the net speed of Robin with respect to ground will be 13.7 m/s

7 0

3 years ago

A train has an acceleration of magnitude 0.90 m/s2 while stopping. A pendulum with a 0.55-kg bob is attached to a ceiling of one

anygoal [31]

The angle of the pendulum with the vertical is $5.2^{\circ}$

Explanation:

As the train decelerates, the bob of the pendulum will feel a force given by

$F=ma$

where

m = 0.55 kg is the mass of the bob

$a=0.9 m/s^2$ is the magnitude of the acceleration

In the horizontal direction.

The pendulum will be inclined at an angle $\theta$ from the vertical, so it will be in equilibrium, and therefore the horizontal component of the tension in the string must be equal to the net force F of the previous equation:

$T sin \theta = ma$ (1)

where T is the tension in the string.

We also know that the bob is in equilibrium along the vertical direction: so the vertical component of the tension must be equal to the weight of the bob,

$T cos \theta = mg$ (2)