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notsponge [240]

2 years ago

11

You toss a tennis ball straight upward. At the moment it leaves your hand it is at a height of 1.5 m above the ground, and it is

moving at a speed of 7.0 m/s.
(a) How much time does it take for the tennis ball to reach its maximum height? (s)

(b) What is the maximum height above the ground that the tennis ball reaches? (m)

(c) When the tennis ball is at a height of 2.4 m above the ground, what is its speed? (m/s)

1 answer:

ollegr [7]2 years ago

4 0

My answer was incorrect, please disregard.

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The Eiffel tower in Paris is 300 meters tall on a cold day (T = -24 degrees Celsius), what is its height on a hot day when the t

Varvara68 [4.7K]

Answer:

Length of Eiffel tower, when the temperature is 35 degrees = 300.21 m

Explanation:

Thermal expansion is given by the expression

$\Delta L=L\alpha \Delta T \\$

Here length of Eiffel tower, L = 300 m

Coefficient of thermal expansion, α = 0.000012 per degree Celsius

Change in temperature, = 35 - (-24) = 59degrees Celsius

Substituting

$\Delta L=L\alpha \Delta T= 300\times 0.000012\times 59=0.2124m \\$

Length of Eiffel tower, when the temperature is 35 degrees = 300 + 0.2124 = 300.21 m

3 0

3 years ago

shows a conical pendulum, in which the bob (the small object at the lower end of the cord) moves in a horizontal circle at const

Contact [7]

Answer:

a) $T=0.40 N$

b) $T=1.9 s$

Explanation:

Let's find the radius of the circumference first. We know that bob follows a circular path of circumference 0.94 m, it means that the perimeter is 0.94 m.

The perimeter of a circunference is:

$P=2\pi r=0.94$

$r=\frac{0.94}{2\pi}=0.15 m$

Now, we need to find the angle of the pendulum from vertical.

$tan(\alpha)=\frac{r}{L}=\frac{0.15}{0.90}=0.17$

$\alpha=9.44 ^{\circ}$

Let's apply Newton's second law to find the tension.

$\sum F=ma_{c}=m\omega^{2}r$

We use centripetal acceleration here, because we have a circular motion.

The vertical equation of motion will be:

$Tcos(\alpha)=mg$ (1)

The horizontal equation of motion will be:

$Tsin(\alpha)=m\omega^{2}r$ (2)

a) We can find T usinf the equation (1):

$T=\frac {mg}{cos(\alpha)}=\frac{0.04*9.81}{cos(9.44)}=0.40 N$

We can find the angular velocity (ω) from the equation (2):

$\omega=\sqrt{\frac{Tsin(\alpha)}{mr}}=3.31 rad/s$

b) We know that the period is T=2π/ω, therefore:

$T=\frac{2\pi}{\omega}=\frac{2\pi}{3.31}=1.9 s$

I hope it helps you!

8 0

3 years ago

A man pulls a sled at a constant velocity across a horizontal snow surface. if a force of 80 n is being applied to the sled rope

balandron [24]

A constant velocity implies the two forces must be equal and opposite.
Friction acts horizontal to the ground, therefore we must find the force applied to the sled rope that acts horizontal to the ground.
Do this by resolving:
Force = 80cos53
The force opposing this is equal, and so also = 80cos53 = 48 N (2 sig. fig.)

3 0

4 years ago

Read 2 more answers

Please help, i’ll give brainliest!!

mrs_skeptik [129]

Answer:

duty h gucuuvu h just hc i oicuxp o cut o icucj x uc jo 8cuc8c

5 0

3 years ago

Suppose a baseball pitcher throws the ball to his catcher.

amm1812

a) Same

b) Same

c) Same

d) Throw the ball takes longer

e) F is larger when the ball is catched

Explanation:

a)

The change in speed of an object is given by:

$\Delta v = |v-u|$

where

u is the initial velocity of the object

v is the final velocity of the object

The change in speed is basically the magnitude of the change in velocity (because velocity is a vector, while speed is a scalar, so it has no direction).

In this problem:

- In situation 1 (pitcher throwing the ball), the initial velocity is

u = 0 (because the ball starts from rest)

while the final velocity is v, so the change in speed is

$\Delta v=|v-0|=|v|$

- In situation 2 (catcher receiving the ball), the initial velocity is now

u = v

while the final velocity is now zero (ball coming to rest), so the change in speed is

$\Delta v =|0-v|=|-v|$

Which means that the two situations have same change in speed.

b)

The change in momentum of an object is given by

$\Delta p = m \Delta v$

where

m is the mass of the object

$\Delta v$ is the change in velocity

If we want to compare only the magnitude of the change in momentum of the object, then it is given by

$|\Delta p|=m|\Delta v|$

- In situation 1 (pitcher throwing the ball), the change in momentum is

$\Delta p = m|\Delta v|=m|v|=mv$

- In situation 2 (catcher receiving the ball), the change in momentum is

$\Delta p = m\Delta v = m|-v|=mv$

So, the magnitude of the change in momentum is the same (but the direction is opposite)

c)

The impulse exerted on an object is equal to the change in momentum of the object:

$I=\Delta p$

where

I is the impulse

$\Delta p$ is the change in momentum

As we saw in part b), the change in momentum of the ball in the two situations is the same, therefore the impulse exerted on the ball will also be the same, in magnitude.

However, the direction will be opposite, as the change in momentum has opposite direction in the two situations.

d)

To compare the time of impact in the two situations, we have to look closer into them.

- When the ball is thrown, the hand "moves together" with the ball, from back to ahead in order to give it the necessary push. We can verify therefore that the time is longer in this case.

- When the ball is cacthed, the hand remains more or less "at rest", it doesn't move much, so the collision lasts much less than the previous situation.

Therefore, we can say that the time of impact is longer when the ball is thrown, compared to when it is catched.

e)

The impulse exerted on an object can also be rewritten as the product between the force applied on the object and the time of impact:

$I=F\Delta t$

where

I is the impulse

F is the force applied

$\Delta t$ is the time of impact

This can be rewritten as

$F=\frac{I}{\Delta t}$

In this problem, in the two situations,

- I (the impulse) is the same in both situations

- $\Delta t$ when the ball is thrown is larger than when it is catched

Therefore, since F is inversely proportional to $\Delta t$ , this means that the force is larger when the ball is catched.

6 0

4 years ago