All categories

3 years ago

Why is the answer C?

Physics

1 answer:

4vir4ik [10]3 years ago

7 0

Explanation:

We want to find the statement that is proven by the fact that the balls reach the same height.

A isn't supported by the evidence. Balls can reach the same height without having the same initial speed.

B isn't supported by the evidence. Balls can reach the same height without having the same launch angle.

C is supported. Projectiles spend the same amount of time going up as they do coming down, so if two projectiles reach the same height, then they must spend the same amount of time in the air.

D isn't supported by the evidence. Balls thrown at the same speed and complementary angles have the same range but different heights.

E isn't supported by the evidence. The mass of the ball doesn't affect the height.

You might be interested in

A 39-foot ladder is leaning against a vertical wall. If the bottom of the ladder is being pulled away from the wall at the rate

Viefleur [7K]

Answer:

The rate of change of the area when the bottom of the ladder (denoted by $b$ ) is at 36 ft. from the wall is the following:

$\frac{dA}{dt}|_{b=36}=-571.2\, ft^2/s$

Explanation:

The Area of the triangle is given by $A=h\times b$ where $h=\sqrt{l^2-b^2}$ (by using the Pythagoras' Theorem) and $b$ is the length of the base of the triangle or the distance between the bottom of the ladder and the wall.

The area is then

$A=\sqrt{l^2-b^2}b$

The rate of change of the area is given by its time derivative

$\frac{dA}{dt}=\frac{d}{dt}\left(\sqrt{l^2-b^2}\cdot b\right)$

$\implies \frac{dA}{dt}=\frac{d}{dt}\left(\sqrt{l^2-b^2}\right)\cdot b+\frac{db}{dt}\cdot\sqrt{l^2-b^2}$

$\implies\frac{dA}{dt}=\frac{1}{2\sqrt{l^2-b^2}}\frac{d}{dt}(l^2-b^2)\cdot b+\sqrt{l^2-b^2}}\cdot \frac{db}{dt}$ Product rule

$\implies\frac{dA}{dt}=-\frac{1}{2\sqrt{l^2-b^2}}\cdot 2\cdot b^2\cdot \frac{db}{dt}+\sqrt{l^2-b^2}}\cdot \frac{db}{dt}$ Chain rule

$\implies\frac{dA}{dt}=-\frac{1}{\sqrt{l^2-b^2}}\cdot b^2\cdot \frac{db}{dt}+\sqrt{l^2-b^2}}\cdot \frac{db}{dt}$

$\implies\frac{dA}{dt}=\frac{db}{dt}\left(-\frac{1}{\sqrt{l^2-b^2}}\cdot b^2+\sqrt{l^2-b^2}}\right)$

In here we can identify $b=36\, ft$ , $l=39$ and $\frac{db}{dt}=8\,ft/s$ .

The result is then

$\frac{dA}{dt}=8\left(-\frac{1}{\sqrt{39^2-36^2}}\cdot 36^2+\sqrt{39^2-36^2}}\right)=-571.2\, ft^2/s$

3 0

3 years ago

What is the purpose of a rearview mirror in a car?

Shalnov [3]

The third choice.

The driver wants to see the object that is behind him. The light reflects off the mirror into the eyes of the driver portraying the object behind him

5 0

3 years ago

Read 3 more answers

If a pitcher throws a baseball 10 m in 30 seconds - what would be the average speed?

allsm [11]

Answer:

0.333 m/s

Explanation:

avg speed= (total distance)/(total time)

10m/30s

0.333 m/s

7 0

3 years ago

Read 2 more answers

The si unit for acceleration is blank

True [87]

Answer:

meters per second squared

Explanation:

Because acceleration is equal to velocity over time where velocity is meters per second and time is measured in seconds making acceleration ms-2

4 0

3 years ago

Which diagram represents an open circuit?

Kruka [31]

Answer : D) Circuit A

This circuit is the only circuit where it is not complete, having and open spot towards the bottom of it, making it and open circuit.

6 0

3 years ago

Read 2 more answers