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TiliK225 [7]
4 years ago
9

A 72 kg skydiver is descending on a parachute. His speed is still increasing at 1.2 m/s2. What are the magnitude and direction o

f the force of the parachute harness on the diver? What are the magnitude and direction of the net force on the diver?
Physics
1 answer:
denis23 [38]4 years ago
5 0

Answer:

Parachute exerts a force of 619.2 N upward

The net force is 86.4 N acting downward

Explanation:

As the gravitational acceleration g = 9.8 m/s2, the parachute help reduces the net acceleration to 1.2m/s. So it must exerted an upward acceleration on the skydiver of

9.8 - 1.2 = 8.6 m/s2

Since the skydiver mass is 72 kg, we can use Newton's 2nd law to calculate the force that causes this acceleration of 8.6

F = ma = 8.6*72 = 619.2 N acting upward

The net force is also the product of net acceleration and mass

= 1.2 * 72 = 86.4 N acting downward

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3 years ago
Samara stands on the ground. Gravity is applying a force to pull
sineoko [7]

Answer:

action reaction

Explanation:

newtons third law

3 0
3 years ago
A record turntable is rotating at 33 rev/min. A watermelon seed is on the turntable 2.3 cm from the axis of rotation. (a) Calcul
VLD [36.1K]

Answer:

a) a_{r} = 0.275\,\frac{m}{s^{2}}, b) \mu_{s} = 0.028, c) \mu_{s} = 0.036

Explanation:

a) The linear acceleration of the watermelon seed is:

a_{r} = \omega^{2}\cdot r

a_{r} = \left[\left(33\,\frac{rev}{min} \right)\cdot \left(2\pi\,\frac{rad}{rev} \right)\cdot \left(\frac{1}{60}\,\frac{min}{s} \right)\right]^{2}\cdot (0.023\,m)

a_{r} = 0.275\,\frac{m}{s^{2}}

b) The watermelon seed is experimenting a centrifugal acceleration. The coefficient of static friction between the seed and the turntable is calculated by the Newton's Laws:

\Sigma F = \mu_{s}\cdot m\cdot g = m\cdot a

a = \mu_{s}\cdot g

\mu_{s} = \frac{a}{g}

\mu_{s} = \frac{0.275\,\frac{m}{s^{2}} }{9.807\,\frac{m}{s^{2}} }

\mu_{s} = 0.028

c) Angular acceleration experimented by the turntable is:

\alpha = \frac{\omega-\omega_{o}}{\Delta t}

\alpha = \frac{3.456\,\frac{rad}{s}-0\,\frac{rad}{s} }{0.36\,s}

\alpha = 9.6\,\frac{rad}{s^{2}}

The tangential acceleration experimented by the watermelon seed is:

a_{t} = \left(9.6\,\frac{rad}{s^{2}} \right)\cdot (0.023\,m)

a_{t} = 0.221\,\frac{m}{s^{2}}

The linear acceleration experimented by the watermelon seed is:

a = \sqrt{a_{t}^{2}+a_{r}^{2}}

a = \sqrt{\left(0.221\,\frac{m}{s^{2}} \right)^{2}+\left(0.275\,\frac{m}{s^{2}} \right)^{2}}

a = 0.353\,\frac{m}{s^{2}}

The minimum coefficient of static friction is:

\mu_{s} = \frac{0.353\,\frac{m}{s^{2}} }{9.807\,\frac{m}{s^{2}} }

\mu_{s} = 0.036

4 0
3 years ago
The period of a 400 hertz sound wave is ___ second
Oksi-84 [34.3K]

Answer:

0.0025 sec

Explanation:

Period = 1 / frequency = 1/400 = 0.0025 sec

3 0
2 years ago
Read 2 more answers
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