All categories

3 years ago

Calculate the maximum height a ball reaches when thrown straight up at a velocity of 10m/s

Physics

1 answer:

murzikaleks [220]3 years ago

7 0

Answer:

5.1m

Explanation:

Use the fourth key equation of accelerated motion

vf^2=vi^2+2aΔd

we know vf is = 0 because that is when the ball reaches it's peak (not going up anymore)

0=vi^2+2aΔd

and we can rearrange for Δd, which is what you need

-vi^2=2aΔd

Δd = -vi^2/2a

Now we can plug in the values

Δd = -vi^2/2a

Δd = -(10m/s)^2/2(-9.8)

Δd = -100/-18.6

Δd = 5.10m

You might be interested in

a Porsche 911 accelerates from rest to 27 metre per second due north in 5.8 seconds the mass of the car is 1400 kg what is the m

user100 [1]

Answer:

The average force has a magnitude 6524 N due north.

Explanation:

The average net force F = ma where m = mass of car = 1400 kg and a = acceleration.

a = (v - u)/t where u = initial velocity of car = 0 m/s (since it starts from rest)

v = final velocity of car = 27 m/s due north and t = time of motion = 5.8 s

a = (27 m/s - 0 m/s)/5.8 s = 27 m/s ÷ 5.8 s = 4.66 m/s

Since the direction of the velocity change is the direction of the acceleration, the acceleration is 4.66 m/s due north.

The average force, F = ma = 1400 kg × 4.66 m/s = 6524 N

Since the acceleration is due north, the average force takes the direction of the acceleration.

So the direction of the average force is due north

The average force has a magnitude 6524 N due north.

7 0

3 years ago

The topsoil layer has the greatest concentration of organic matter. The subsoil generally lacks organic matter, but it does rece

zmey [24]

Answer:

C. the C horizon likely has a rockier texture than the topsoil and subsoil.

Explanation:

because i did it on study island

3 0

3 years ago

Select the correct answer. A ball is thrown straight down from the top of a building at a velocity of 16 ft/s. The building is 4

BartSMP [9]

Answer:

The ball takes 5s to reach the ground

Explanation:

in order to solve this problem we use the kinematics equation with gravity as acceleration:

$h=v_0t+1/2*gt^2$

we replace the values

$0=1/2*32t^2+16t-480$

We solve this quadratic equation:

t=5s

t=-6s (this solution has not physical sense)

8 0

3 years ago

Tidal Forces near a Black Hole. An astronaut inside a spacecraft, which protects her from harmful radiation, is orbiting a black

arlik [135]

Answer:

$833.4801043*10^6N$ on ear that is closer to Black hole.

$13.83803929*10^6N$ On ear that is farther from Black hole.

Explanation:

This problem can be solved as two masses that are at two different location from a bigger mass whose gravity affects both.

tension is an equal and opposite force that is exerted in response to applied force.

so on ear that is closer to black hole would have tension that is equal in magnitude and opposite in direction to gravitational force that ear experience due to the black hole at that location.

this true for ear that is further away from black hole as well.

(1) Force on ear that is closer to black hole.

$F =\frac{m_{1}*m_{2} }{r^2} G$

$m_{1}= 5*1.989*10^30kg$ is mass of Black hole.

$m_{2}= 0.030kg$ is mass of ear that is close to black hole.

$G = 6.679*10^-11m^3*kg^-1*s^-2$

r = $120km-\frac{6}{100000} km=119.99994km=119999.94m$

Note, we have subtracted because ear is closer to black hole.

plugging all this in formula gives.

$F = 833.4801043*10^6N$

That is tension of ear.

(2) Force on ear that is further from black hole.

$F =\frac{m_{1}*m_{2} }{r^2} G$

$m_{1}= 5*1.989*10^30kg$ is mass of Black hole.

$m_{2}= 0.030kg$ is mass of ear that is close to black hole.

$G = 6.679*10^-11m^3*kg^-1*s^-2$

this time r is further away from black hole so it would be.

$r = 120km+\frac{6}{100000} km = 120000.06m$

Plugging this all in we get

$F = 13.83803929N$

and that is tension on ear that is further from black hole.

Notice the tension difference, and order of magnitude of tension,it is enormous .

this astronaut is lethally close to black hole.

5 0

3 years ago

Assume your car reaches a speed of 21.7 m/s at a steady rate for 5.05 s after the light turns green. (a) What distance have you

boyakko [2]

Answer:

The distance and average speed are 54.79 m and 10.85 m.

Explanation:

Given that,

Speed = 21.7 m/s

Time = 5.05 s

(a). We need to calculate the distance

Firstly we will find the acceleration

Using equation of motion

$v = u+at$

$a = \dfrac{v-u}{t}$

Where, v = final velocity

u = initial velocity

t = time

Put the value in the equation

$a = \dfrac{21.7-0}{5.05}$

$a = 4.297 m/s^2$

Now, using equation of motion again

For distance,

$s = ut+\dfrac{1}{2}at^2$

$s = 0+\dfrac{1}{2}\times4.3\times(5.05)^2$

$s=54.79\ m$

The distance is 54.79 m.

(b). We need to calculate the average speed during this time

$v_{avg}=\dfrac{D}{T}$

Where, D = total distance

T = time

Put the value into the formula

$v_{avg}=\dfrac{54.79}{5.05}$

$v_{avg}=10.85\ m/s$

Hence, The distance and average speed are 54.79 m and 10.85 m.

3 0

3 years ago