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3 years ago

Describe the motion of a skydiver from the time he jumps to the time he lands safely on the ground

Physics

1 answer:

Angelina_Jolie [31]3 years ago

7 0

The skydiver accelerates at the beginning, while later he continues at constant velocity

Explanation:

There are two forces acting on the skydiver during its fall:

The force of gravity, which is constant, of magnitude $mg$ , where m is the mass of the skydiver and $g$ is the acceleration of gravity ( $9.8 m/s^2$ downward)
The air resistance, whose magnitude is proportional to the speed of the skydiver: $F_r\propto v$ , and which acts upward

At the beginning of its fall, the speed of the skydiver is small, therefore at the beginning the air resistance is negligible, and therefore the skydiver is accelerated downward with an acceleration of

$g=9.8 m/s^2$

However, as the skydiver falls, he gains speed; therefore, the magnitude of the air resistance increases. The net force acting on the skydiver is

$F_{net}=mg-F_r$

As the air resistance increases, the net force decreases: therefore, the acceleration of the skydiver decreases. Eventually, he will reach a speed at which the air resistance becomes equal to the force of gravity:

$mg=F_r$

when this occurs, the acceleration of the skydiver will become zero, so he will continue his fall at a constant velocity: this value of the velocity is called terminal velocity.

Learn more about free fall:

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A small boy is playing with a ball on a stationary train. If he places the ball on the floor of the train, when the train starts

expeople1 [14]

A small boy is playing with a ball on a stationary train. If he places the ball on the floor of the train, when the train starts moving the ball moves toward the back of the train. This happened due to inertia

An object at rest remains at rest, or if in motion, remains in motion unless a net external force acts on it .

When a train starts moving forward, the ball placed on the floor tends to fall backward is an example of inertia of rest. Due to the reason that the lower part of the ball is in contact with the surface and rest of the part is not . As the train starts moving, its lower part gets the motion as the floor starts moving but the upper part will remain as it is as it is not in contact with the floor , hence do not attain any motion due to the inertia of rest simultaneously i.e. it tends to remain at the same place.

To learn more about inertia here :

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1 year ago

(1.08×10-3)(9.3×10-4)

ira [324]

First you do the first parenthesis, (1.08 x 10 - 3) and you do it in the order of operations! (parenthesis, exponents, multiplication/division, add/subtract) to get 7.8. Then you take the second parenthesis (9.3 x 10 - 4) and do the same thing to get 89! You then times 7.8 by 89 to get 694.2! If it needs more elaboration just ask ^.^

4 0

3 years ago

A DJ starts up her phonograph player. The turntable accelerates uniformly from rest, and takes t1 = 11.9 seconds to get up to it

Degger [83]

Answer:

a) $\omega_1=8.168\,rad.s^{-1}$

b) $n_1=7.735 \,rev$

c) $\alpha_1 =0.6864\,rad.s^{-2}$

d) $\alpha_2=4.1454\,rad.s^{-2}$

e) $t_2=1.061\,s$

Explanation:

Given that:

initial speed of turntable, $N_0=0\,rpm\Rightarrow \omega_0=0\,rad.s^{-1}$

full speed of rotation, $N_1=78 \,rpm\Rightarrow \omega_1=\frac{78\times 2\pi}{60}=8.168\,rad.s^{-1}$

time taken to reach full speed from rest, $t_1=11.9\,s$

final speed after the change, $N_2=120\,rpm\Rightarrow \omega_2=\frac{120\times 2\pi}{60}=12.5664\,rad.s^{-1}$

no. of revolutions made to reach the new final speed, $n_2=11\,rev$

(a)

∵ 1 rev = 2π radians

∴ angular speed ω:

$\omega=\frac{2\pi.N}{60}\, rad.s^{-1}$

where N = angular speed in rpm.

putting the respective values from case 1 we've

$\omega_1=\frac{2\pi\times 78}{60}\, rad.s^{-1}$

$\omega_1=8.168\,rad.s^{-1}$

(c)

using the equation of motion:

$\omega_1=\omega_0+\alpha . t_1$

here α is the angular acceleration

$78=0+\alpha_1\times 11.9$

$\alpha_1 = \frac{8.168 }{11.9}$

$\alpha_1 =0.6864\,rad.s^{-2}$

(b)

using the equation of motion:

$\omega_1\,^2=\omega_0\,^2+2.\alpha_1 .n_1$

$8.168^2=0^2+2\times 0.6864\times n_1$

$n_1=48.6003\,rad$

$n_1=\frac{48.6003}{2\pi}$

$n_1=7.735\, rev$

(d)

using equation of motion:

$\omega_2\,^2=\omega_1\,^2+2.\alpha_2 .n_2$

$12.5664^2=8.168^2+2\alpha_2\times 11$

$\alpha_2=4.1454\,rad.s^{-2}$

(e)

using the equation of motion:

$\omega_2=\omega_1+\alpha_2 . t_2$

$12.5664=8.168+4.1454\times t_2$

$t_2=1.061\,s$

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3 years ago

A free-falling golf ball strikes the ground and exerts a force on it

ozzi

Yes, it do, for a short time.

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3 years ago

Explain why the atomic mass of an element is weighted average mass

madreJ [45]

Explanation:

The mass written on the periodic table is an average atomic mass taken from all known isotopes of an element. This average is a weighted average, meaning the isotope's relative abundance changes its impact on the final average. The reason this is done is because there is no set mass for an element.

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2 years ago