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Neko [114]
3 years ago
8

At an air show, a stunt pilot performs a vertical loop-the-loop in a circle of radius 3.63 x 103 m. During this performance the

pilot whose weight is 676 N, maintains a constant speed of 2.25 x 102 m/s. At what speed, in m/s, will the pilot experience weightlessness
Physics
1 answer:
san4es73 [151]3 years ago
5 0

Answer:

189 m/s

Explanation:

The pilot will experience weightlessness when the centrifugal force, F equals his weight, W.

So, F = W

mv²/r = mg

v² = gr

v = √gr where  v = velocity, g = acceleration due to gravity = 9.8 m/s² and r = radius of loop = 3.63 × 10³ m

So, v = √gr

v = √(9.8 m/s² × 3.63 × 10³ m)

v = √(35.574 × 10³ m²/s²)

v = √(3.5574 × 10⁴ m²/s²)

v = 1.89 × 10² m/s

v = 189 m/s

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Science help please!
OLga [1]

Answer:

104 N

Explanation:

m = 1300 kg

a = 0.08m/s^2

F = 1300*0.08

F = 104 N

Newtons is the unit of force.

3 0
3 years ago
A 61.0-kg person jumps from rest off a 10.0-m-high tower straight down into the water. Neglect air resistance. She comes to rest
9966 [12]

Answer:

Explanation:

In this case, law of conservation of energy will be implemented. It states that "the energy of the system remains conserved until or unless some external force act on it. Energy of the system may went through the conversion process like kinetic energy into potential and potential into kinetic energy.But their total always remain the same in conserved systems."

Given data:

Height of tower = 10.0 m

Depth of the pool = 3.00 cm

Mass of person = 61.0 kg

Solution:

Initial energy = Final energy

U_{i} =  (K.E) + U_{f}

As the person was at height initially so it has the potential energy only.

mg(h_{1} +h_{2}) = K.E + mgh_{2}

K.E = mgh_{1}

K.E = (61.0)(9.8)(10)\\K.E = 5978 J

Lets find out the magnitude of the force that the water is exerting on the diver.

W =ΔK.E

F.h_{2} = 5978\\

F = \frac{5978}{3}

F = 1992.67 N

7 0
3 years ago
A 75Kg man jumps from a window 1.0m above the sidewalk. If the man falls with his knees and ankles locked, the only cushion for
liq [111]

Answer:

150153.06122 N

Explanation:

m = Mass of person = 75 kg

h = Height of fall = 1 m

g = Acceleration due to gravity = 9.81 m/s²

F = Force

s = Displacement = 0.49 cm

Potential energy is given by

P=mgh\\\Rightarrow P=75\times 9.81\times 1\\\Rightarrow P=735.75\ J

Work is given by

W=Fs\\\Rightarrow F=\frac{W}{s}\\\Rightarrow F=\frac{735.75}{0.0049}\\\Rightarrow F=150153.06122\ N

The average force exerted is 150153.06122 N

8 0
3 years ago
Which of these is a benefit of nuclear energy?
Mademuasel [1]
The answer i think would be D 

4 0
3 years ago
Read 2 more answers
A car and a train move together along straight, parallel paths with the same constant cruising speed v(initial). At t=0 the car
kogti [31]

Answer:

t_1 = \frac{v_i}{a_i}

t_2 = \frac{v_i}{a_i}

Δd = v_it_1 = v_i^2/a_i

Explanation:

As v(t) = v_i + at, when the car is making full stop, v(t_1) = 0 . a = -a_i . Therefore,

0 = v_i - a_it_1\\v_i = a_it_1\\t_1 = \frac{v_i}{a_i}

Apply the same formula above, with v(t_2) = v_i and a = a_i, and the car is starting from 0 speed,  we have

v_i = 0 + a_it_2\\t_2 = \frac{v_i}{a_i}

As s(t) = vt + \frac{at^2}{2}. After t = t_1 + t_2, the car would have traveled a distance of

s(t) = s(t_1) + s(t_2)\\s(t_1) = (v_it_1 - \frac{a_it_1^2}{2})\\ s(t_2) = \frac{a_it_2^2}{2}

Hence s(t) = (v_it_1 - \frac{a_it_1^2}{2}) + \frac{a_it_2^2}{2}

As t_1 = t_2 we can simplify s(t) = v_it_1

After t time, the train would have traveled a distance of s(t) = v_i(t_1 + t_2) = 2v_it_1

Therefore, Δd would be 2v_it_1 - v_it_1 = v_it_1 = v_i^2/a_i

8 0
3 years ago
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