Can someone please help me ASAP???!!!

While skydiving, your parachute opens and you slow from 50.0 m/s to 8.0 m/s in 0.75 s . Determine the distance you fall while th

Anestetic [448]

Answer:

21.75 m

Explanation:

t = Time taken for the car to slow down = 0.75 s

u = Initial velocity = 50 m/s

v = Final velocity = 8 m/s

s = Displacement

a = Acceleration

Equation of motion

$v=u+at\\\Rightarrow a=\frac{v-u}{t}\\\Rightarrow a=\frac{8-50}{0.75}\\\Rightarrow a=-56\ m/s^2$

Acceleration is -56 m/s²

$v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{8^2-50^2}{2\times -56}\\\Rightarrow s=21.75\ m$

The distance covered in the 0.75 seconds is 21.75 m

8 0

3 years ago

A 17.6-kg block rests on a horizontal table and is attached to one end of a massless, horizontal spring. By pulling horizontally

leonid [27]

In order get the block up to a speed of 3.58 m/s in 1.77 s, it must undergo an acceleration a of

a = (3.58 m/s) / (1.77 s) ≈ 2.02 m/s²

When the spring is getting pulled, Newton's second law tells us

• the net vertical force is

∑ F = n - mg = 0

where ∑ F is the net force, n is the magnitude of the normal force, and mg is the weight of the block - it follows that n = mg ; and

• the net horizontal force is

∑ F = F - f = ma

where F is the applied force, f is kinetic friction, m is the block's mass, and a is the acceleration found earlier. F stretches the spring by x = 0.250 m, so we have

F - f = kx - µn = kx - µmg = ma

where k is the spring constant and µ is the coefficient of kinetic friction.

When the block is being pulled at a constant speed, Newton's second law says

• the net vertical force is still

∑ F = n - mg = 0

so that n = mg again; and

• the net horizontal force is

∑ F = F - f = 0

This time, F stretches the spring by y = 0.0544 m, so we have

F - f = ky - µmg = 0

Solve the equations in boldface for k and µ :

kx - µmg = ma

ky - µmg = 0

==> k (x - y) = ma

==> k = ma / (x - y)

==> k = (17.6 kg) (2.02 m/s²) / (0.250 m - 0.0544 m) ≈ 182 N/m

Then

ky - µmg = 0

==> µ = ky / (mg)

==> µ = (182 N/m) (0.0544 m) / ((17.6 kg) g) ≈ 0.0574

3 0

3 years ago

A runner with a mass of 80kg accelerates from 0 to 9 m/s in 3 s find the net force

frosja888 [35]

So, first you find your acceleration which is 3m/s^2, using the acceleration formula.
Now set up your equation, F=ma, so put in the stuff, F=80kg·3m/s^2. Then solve your equation by multiplying, and you get F=240N, since newtons are your measurement.
Hope this helps

4 0

3 years ago

A Tennis ball falls from a height 40m above the ground the ball rebounds

worty [1.4K]

If the ball is dropped with no initial velocity, then its velocity v at time t before it hits the ground is

v = -g t

where g = 9.80 m/s² is the magnitude of acceleration due to gravity.

Its height y is

y = 40 m - 1/2 g t²

The ball is dropped from a 40 m height, so that it takes

0 = 40 m - 1/2 g t²

==> t = √(80/g) s ≈ 2.86 s

for it to reach the ground, after which time it attains a velocity of

v = -g (√(80/g) s)

==> v = -√(80g) m/s ≈ -28.0 m/s

During the next bounce, the ball's speed is halved, so its height is given by

y = (14 m/s) t - 1/2 g t²

Solve y = 0 for t to see how long it's airborne during this bounce:

0 = (14 m/s) t - 1/2 g t²

0 = t (14 m/s - 1/2 g t)

==> t = 28/g s ≈ 2.86 s

So the ball completes 2 bounces within approximately 5.72 s, which means that after 5 s the ball has a height of

y = (14 m/s) (5 s - 2.86 s) - 1/2 g (5 s - 2.86 s)²

==> (i) y ≈ 7.5 m

(ii) The ball will technically keep bouncing forever, since the speed of the ball is only getting halved each time it bounces. But y will converge to 0 as t gets arbitrarily larger. We can't realistically answer this question without being given some threshold for deciding when the ball is perfectly still.

During the first bounce, the ball starts with velocity 14 m/s, so the second bounce begins with 7 m/s, and the third with 3.5 m/s. The ball's height during this bounce is

y = (3.5 m/s) t - 1/2 g t²

Solve y = 0 for t :

0 = (3.5 m/s) t - 1/2 g t²

0 = t (3.5 m/s - 1/2 g t)

==> (iii) t = 7/g m/s ≈ 0.714 s

As we showed earlier, the ball is in the air for 2.86 s before hitting the ground for the first time, then in the air for another 2.86 s (total 5.72 s) before bouncing a second time. At the point, the ball starts with an initial velocity of 7 m/s, so its velocity at time t after 5.72 s (but before reaching the ground again) would be

v = 7 m/s - g t

At 6 s, the ball has velocity

(iv) v = 7 m/s - g (6 s - 5.72 s) ≈ 4.26 m/s

4 0

3 years ago

The pH of acid rain would be _____. greater than 7 less than 7 exactly 7 exactly 14

Nookie1986 [14]

Less than 7, Ph of 7 is neutral Acid Rain would be between 4 and 0.

7 0

2 years ago

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