A baseball flying through the air has what kind of energy?

A parachutist falls 50.0 m without friction. When the parachute opens, he slows down at a rate of 67 m/s*2. If he reaches the gr

KIM [24]

Answer:

3.49 seconds

3.75 seconds

-43200 ft/s²

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

$s=ut+\frac{1}{2}at^2\\\Rightarrow 50=0t+\frac{1}{2}\times 9.81\times t^2\\\Rightarrow t=\sqrt{\frac{50\times 2}{9.81}}\\\Rightarrow t=3.19\ s$

Time the parachutist falls without friction is 3.19 seconds

$v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times 9.81\times 50+0^2}\\\Rightarrow v=31.32\ m/s$

Speed of the parachutist when he opens the parachute 31.32 m/s. Now, this will be considered as the initial velocity

$v=u+at\\\Rightarrow 11=31.32+9.81t\\\Rightarrow t=\frac{11-31.32}{-67}=0.3\ s$

So, time the parachutist stayed in the air was 3.19+0.3 = 3.49 seconds

$s=ut+\frac{1}{2}at^2\\\Rightarrow \frac{s}{2}=0t+\frac{1}{2}\times a\times t^2\\\Rightarrow \frac{s}{2}=\frac{1}{2}at^2$

$s=ut+\frac{1}{2}at^2\\\Rightarrow \frac{s}{2}=u1.1+\frac{1}{2}\times a\times 1.1^2$

Now the initial velocity of the last half height will be the final velocity of the first half height.

$v=u+at\\\Rightarrow v=at$

Since the height are equal

$\frac{1}{2}at^2=u1.1+\frac{1}{2}\times a\times 1.1^2\\\Rightarrow \frac{1}{2}at^2=at1.1+\frac{1}{2}\times a\times 1.1^2\\\Rightarrow 0.5t^2-1.1t-0.605=0\\\Rightarrow 500t^2-1100t-605=0$

$t=\frac{11\left(1+\sqrt{2}\right)}{10},\:t=\frac{11\left(1-\sqrt{2}\right)}{10}\\\Rightarrow t=2.65, -0.45$

Time taken to fall the first half is 2.65 seconds

Total time taken to fall is 2.65+1.1 = 3.75 seconds.

When an object is thrown with a velocity upwards then the velocity of the object at the point to where it was thrown becomes equal to the initial velocity.

$v^2-u^2=2as\\\Rightarrow a=\frac{v^2-u^2}{2s}\\\Rightarrow a=\frac{0^2-240^2}{2\times \frac{8}{12}}\\\Rightarrow a=-43200\ ft/s^2$

Magnitude of acceleration is -43200 ft/s²

5 0

3 years ago

Based on the data, which prediction should he expect to occur? A2 repels B1. C2 attracts B2. B1 repels C1. A1 attracts C2.

VladimirAG [237]

I believe the answer is a1

6 0

2 years ago

Someone please help

saul85 [17]

Based on the attached image:

The name of the longitude line that passes through point A is the International Date Line
The longitude 180° is experiencing solar noon because the rays of the sun are parallel to it.
The longitude for 6 pm is 90° W, 12 midnight is 0°, and 6 am is 90° E
Longitude 120° is B
Solar time at Point B is 4 pm
The location will correspond to any point on the same latitude as A

<h3>What are lines of longitude?</h3>

Lines of longitude are imaginary lines which run along the earth from the North pole. to the South pole.

Longitude lines divide the earth into semi-circles.

Longitude lines are known as meridians and each meridian measures one arc degree of longitude.

Considering the attached image:

The name of the longitude line that passes through point A is the International Date Line
The longitude 180° is experiencing solar noon because the rays of the sun are parallel to it.
The longitude for 6 pm is 90° W, 12 midnight is 0°, and 6 am is 90° E
Longitude 120° is B
Solar time at Point B is 4 pm
the location will correspond to any point on the same latitude as A

In conclusion, longitude lines are imaginary lines and run from North to South on the earth.

Learn more about lines of longitude at: brainly.com/question/1939015

#SPJ1

8 0

1 year ago

An engine is used to lift a beam weighing 9,800 N up to 145 m. How much work must the engine do to lift this beam? How much work

Arturiano [62]

Explanation:

Given that,

Weight of the engine used to lift a beam, W = 9800 N

Distance, d = 145 m

Work done by the engine to lift the beam is given by :

W = F d

$W=9800\ N\times 145\ m\\\\W=1421000\ J\\\\W=1421\ kJ$

Let W' is the work must be done to lift it 290 m. It is given by :

$W'=9800\ N\times 290\ m\\\\W'=2842000\ J\\\\W'=2842\ kJ$

Hence, this is the required solution.

5 0

3 years ago

A 10 kg mass rests on a table. What acceleration will be generated when a force of 5 N is applied? a 0.5 m/s2 b 3.5 m/s2 c 5 m/s

rosijanka [135]

Answer:

a= 0.5m/s^2

Explanation:

Force applied on an object is known as

F=m.a (Newton's second law states it)

a=F/m

a=5/10=0.5m/s^2

3 0

3 years ago