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lora16 [44]
4 years ago
15

Dr. John Paul Stapp was U.S. Air Force officer who studied the effects of extreme deceleration on the human body. On December 10

, 1954, Stapp rode a rocket sled, accelerating from rest to a top speed of 282m/s (1015km/h) in 5.00s, and was brought jarringly back to rest in only 1.40s! Calculate the magnitude of his (a) acceleration as he sped up and (b) acceleration as he slowed down. Express each in multiples of g (9.80 m/s2) by taking its ratio to the acceleration of gravity.
Physics
1 answer:
Nuetrik [128]4 years ago
3 0

Answer:

acceleraions 5.76g and 20.55g

Explanation:

This constant acceleration exercise can be solved using the kinematic equations in one dimension

    Vf = Vo + a t

As part of the rest Vo = 0

    a = Vf / t

    a = 282/5

    a = 56.4 m / s2

In relation to the acceleration of gravity

    a ’= a / g = 56.4 / 9.8

    a ’= 5.76g

To calculate the acceleration to stop we use the same formula

     a2 = 282 / 1.40

     a2 = 201.4 m / s2

 This acceleration of gravity acceleration function is

     a2 ’= 201.4 / 9.8

     a2 ’= 20.55g

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forsale [732]

Answer:

(1) V = 0.2 J (2) 0.05J

Explanation:

Solution

Given that:

K = 160 N/m

x = 0.05 m

Now,

(1) we solve for the  initial potential energy stored

Thus,

V = 1/2 kx² = 0.5 * 160 * (0.05)²

Therefore V = 0.2 J

(2)Now, we solve for how much of the internal energy is produced as the toy springs up to its maximum height.

By using the energy conversion, we have the following

ΔV = mgh

=(0.1/9.8) * 9.8 * 1.5 = 0.15J

The internal energy = 0.2 -0.15

=0.05J

8 0
3 years ago
Current that moves in one direction from negative to positive. May be created by a battery. Is generally NOT found in U.S. elect
Alla [95]

That's "<em>DC</em>" . . . Direct Current .

8 0
3 years ago
Read 2 more answers
A particle traveling in a circular path of radius 300 m has an instantaneous velocity of 30 m/s and its velocity is increasing a
hodyreva [135]

Answer:

5 m/s2

Explanation:

The total acceleration of the circular motion is made of 2 components: centripetal acceleration and linear acceleration of 4 m/s2. They are perpendicular to each other.

The centripetal acceleration is the ratio of instant velocity squared and the radius of the circle

a_c = \frac{v^2}{r} = \frac{30^2}{300} = \frac{900}{300} = 3 m/s^2

So the magnitude of the total acceleration is

a = \sqrt{a_c^2 + a_l^2} = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5 m/s^2

4 0
4 years ago
What is the mass of a 3,500-N rock?
Viktor [21]
350kg because to get Newton’s it’s mass x Gravity, earths gravity is x10 so 3500 divided by 10 is 350
6 0
3 years ago
The radius of the earth is 6378 km. What is the diameter of the earth in meters?
Nesterboy [21]
To determine the diameter of the earth in metres first multiply the original value by 2.

6378 X 2 = 12 756 km.

Then convert km - m

1 km = 1000 m
12 756 km = ? m

12 756 • 1000 = 12 756 000 = 12 756 000 m or 1.2756 X 10 ^ 7 m

The final solution for the diameter is 1.2756 X 10 ^ 7 m.
7 0
4 years ago
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