If light moves at a speed of 299,792,458 m/s, how long will it take light to move a distance of 1,000 km

In a jump spike, a volleyball player slams the ball from overhead and toward the opposite floor. Controlling the angle of the sp

MAVERICK [17]

Answer:

The ball would have landed 3.31m farther if the downward angle were 6.0° instead.

Explanation:

In order to solve this problem we must first start by doing a drawing that will represent the situation. (See picture attached).

We can see in the picture that the least the angle the farther the ball will go. So we need to find the A and B position to determine how farther the second shot would go. Let's start with point A.

So, first we need to determine the components of the velocity of the ball, like this:

$V_{Ax}=V_{A}cos\theta$

$V_{Ax}=(21m/s)cos(-14^{o})$

$V_{Ax}=20.38 m/s$

$V_{Ay}=V_{A}sin\theta$

$V_{Ay}=(21m/s)sin(-14^{o})$

$V_{Ay}=-5.08 m/s$

we pick the positive one, so it takes 0.317s for the ball to hit on point A.

so now we can find the distance from the net to point A with this time. We can find it like this:

$x_{A}=V_{Ax}t$

$x_{A}=(20.38m/s)(0.317s)$

$x_{A}=6.46m$

Once we found the distance between the net and point A, we can similarly find the distance between the net and point B:

$V_{Bx}=20.88 m/s$

$V_{By}=-2.195 m/s$

$y_{Bf}=y_{B0}+V_{0}t-\frac{1}{2}at^{2}$

$0=2.1m+(-2.195m/s)t-\frac{1}{2}(-9.8m/s^{2})t^{2}$

$-4.9t^{2}-2.195t+2.1=0$

$t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$

$t=\frac{-(-2.195)\pm\sqrt{(-2.195)^{2}-4(-4.9)(2.1)}}{2(-4.9)}$

t= -0.9159s or t=0.468s

we pick the positive one, so it takes 0.468s for the ball to hit on point B.

so now we can find the distance from the net to point B with this time. We can find it like this:

$x_{B}=V_{Bx}t$

$x_{B}=(20.88m/s)(0.468s)$

$x_{B}=9.77m$

So once we got the two distances we can now find the difference between them:

$x_{B}-x_{A}=9.77m-6.46m=3.31m$

so the ball would have landed 3.31m farther if the downward angle were 6.0° instead.

7 0

3 years ago

3. The direction of the centripetal force acting on a ball on a string being spun in a circle around a person is

ruslelena [56]

D. center seeking. Good Luck

4 0

3 years ago

Read 2 more answers

I would like help with this physics problem

Darina [25.2K]

(a) This is a freefall problem in disguise - when the ball returns to its original position, it will be going at the same speed but in the opposite direction. So the ball's final velocity is the negative of its initial velocity.

Recall that

$v_f=v_i+at$

We have $v_f=-v_i$ , so that

$-2v_i=at\implies-2\left(8\,\dfrac{\mathrm m}{\mathrm s}\right)=\left(-2\,\dfrac{\mathrm m}{\mathrm s^2}\right)t\implies t=8\,\mathrm s$

(b) The speed of the ball at the start and at the end of the roll are the same 8 m/s, so the average speed is also 8 m/s.

(c) The ball's average velocity is 0. Average velocity is given by $\dfrac{v_i+v_f}2$ , and we know that $v_f=-v_i$ .

(d) The position of the ball $x_f$ at time $t$ is given by

$x_f=x_i+v_it+\dfrac12at^2$

Take the starting position to be the origin, $x_i=0$ . Then after 6 seconds,

$x_f=\left(8\,\dfrac{\mathrm m}{\mathrm s}\right)(6\,\mathrm s)+\dfrac12\left(-2\,\dfrac{\mathrm m}{\mathrm s^2}\right)(6\,\mathrm s)^2=42\,\mathrm m$

so the ball is 42 m away from where it started.

We're not asked to say in which direction it's moving at this point, but just out of curiosity we can determine that too:

$x_f-x_i=\dfrac{v_i+v_f}2t\implies42\,\mathrm m=\dfrac{8\,\frac{\mathrm m}{\mathrm s}+v_f}2(6\,\mathrm s)\implies v_f=6\,\dfrac{\mathrm m}{\mathrm s}$

Since the velocity is positive, the ball is still moving up the incline.

8 0

3 years ago

As __________ increases, the humidity of that area will __________. temperature; also increase dew; also increase rainfall; alwa

Nady [450]

Always increase temperature

4 0

3 years ago

Name four types of friction

Shtirlitz [24]

Answer:

Static Friction-Friction that acts when the two objects are not moving

Sliding Friction-Two solid surfaces slide across each other

Rolling Friction-An object rolls across each other

Fluid Friction-Friction that occurs as an object moves through a fluid

Explanation:

6 0

3 years ago