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Svetach [21]
3 years ago
14

Two identical balls are at rest and side by side at the top of a hill. You let one ball, A, start rolling down the hill. A littl

e later you start the second ball, B, down the hill by giving it a shove. The second ball rolls down the hill along a line parallel to the path of the first ball and passes it. At the instant ball B passes ball A:
a. it has the same position and the same velocity as A.b. it has the same position and the same acceleration as A.c. it has the same velocity and the same acceleration as A.d. it has the same displacement and the same velocity as A.e. it has the same position, displacement and velocity as A.
Physics
1 answer:
ICE Princess25 [194]3 years ago
3 0

Answer:

Option b. it has the same position and the same acceleration as A

Explanation:

Let's analyze every statement:

a. it has the same position and the same velocity as A

In the instant where B passes A, they Do have the same position. Velocity however, cannot be the same because if they were, ball B would never pass ball A. So, this is false.

b. it has the same position and the same acceleration as A

As we said in the previous option, the position is the same. The acceleration is gravity for both balls, so this is true.

c. it has the same velocity and the same acceleration as A

Acceleration is the same but velocities are not, so this is false.

d. it has the same displacement and the same velocity as A

The distance they have traveled is the same, so the displacement is the same, but the velocity is not, so this is false.

e. it has the same position, displacement and velocity as A

The position and displacement is the same but not velocity, so this is false.

Only option b is true.

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A boat traveling across a river has a resultant velocity of 10 km/h and travels 34 degrees with respect to the shore. A) What is
vodka [1.7K]

Answer:

a) 1.55 m/s

b) 2.3 m/s

Explanation:

We know that the boat travels across the river, if we define the river as the x-axis, then the velocity of the boat is only on the y-axis.

Then we can write the velocity of the boat in still water as:

S = (0, B)

Now, when the boat is on the river, the velocity of the boat will be equal to the velocity of the boat in still water plus the velocity of the river.

The velocity of the river is:

v = (R, 0).

Then the velocity of the boat in that river is:

V' = (0, B) + (R, 0) = (R, B)

Now, we know that the velocity of the boat is 10km/h, and it travels at an angle of 34° with respect to the shore.

We can use the Pythagoreans theorem to write the components of this velocity as:

x-axis component = 10km/h*cos(34°) = 8.29 km/h

y-axis component = 10km/h*sin(34°) =  5.59 km/h

Then the velocity of the boat can be written in components as:

velocity = ( 8.29 km/h,  5.59 km/h)

And we knew that the velocity of the boat was written as  (R, B)

Then we must have:

R = 8.29 km/h

B = 5.59 km/h

a) The speed of the boat in m/s:

We know that the speed of the boat is 5.59 km/h.

First, we know that:

1km = 1000m, then:

5.59 km/h = 5.59*(1000m)h = 5,590 m/h

And we know that:

1h = 3600s

Then we can write:

5,590 m/h = 5,590 m/(3600s) = 1.55 m/s

b) The speed of the river in m/s:

We know that the speed of the river is 8.29 km/h

Using the same reasoning as above, we can do the change of units as follows:

8.29 km/h = 8.29 (1000m)/(3600s) = 2.3 m/s

6 0
3 years ago
A power boat pulls a water skier 2.6 km maintaining a
Luba_88 [7]

Answer:

W = 650 [kJ]

Explanation:

The definition of work is denoted by the product of force by the distance traveled by the body, this distance traveled corresponds to the direction of the force.

In this case we have:

d = distance = 2.6[km] = 2600 [m]

F = force = 250 [N]

W = F*d = 250 * 2600 = 650000 [J] or 650 [kJ]

7 0
3 years ago
Two firefighters are trying to break through a door. One firefighter is heavy, and the other is light. If they run at the same s
frez [133]

Answer:

The Heavier Firefighter

Explanation:

Generally, more massive objects will have more intertia than less massive objects.  As such it takes more force to halt a more massive object if its moving at the same speed as a smaller object. This can also be thought of in the context of Newton's second law. The more force needed to accelerate an object means the more force the object will have.

6 0
2 years ago
A ball is thrown horizontally from the top of a 55 m building and lands 150 m from the base of the building. Ignore air resistan
PtichkaEL [24]

Answer:

a) t =3.349 s

b) V_x,i = 44.8 m/s

c) V_y,f = 32.85 m/s

d)  V = 55.55 m/s

Explanation:

Given:

- Total throw in x direction x(f) = 150 m

- Total distance traveled down y(f) = 55 m

Find:

a) How long is the rock in the air in seconds.  

b) What must have been the initial horizontal component of the velocity, in meters per second?

c) What is the vertical component of the velocity just before the rock hits the ground, in meters per second?

d) What is the magnitude of the velocity of the rock just before it hits the ground, in meters per second?

Solution:

- Use the second equation of motion in y direction:

                                 y(f) = y(0) + V_y,i*t + 0.5*g*t^2

- V_y,i = 0 (horizontal throw)

                                 55 = 0 + 0 + 0.5*(9.81)*t^2

                                 t = sqrt ( 55 * 2 / 9.81 )

                                 t =3.349 s

- Use the second equation of motion in x direction:

                                 x(f) = x(0) + V_x,i*t

                                 150 = 0 + V_x,i*3.349

                                  V_x,i = 150 / 3.349 = 44.8 m/s

- Use the first equation of motion in y direction:

                                 V_y,f = V_y,i + g*t

                                 V_y,f = 0 + 9.81*3.349

                                 V_y,f = 32.85 m/s

- The magnitude of velocity of ball when it hits the ground is:

                                 V^2 = V_y,f^2 + V_x,i^2

                                 V = sqrt (32.85^2 + 44.8^2)

                                 V = 55.55 m/s

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