How much is 100 kg in Newtons?

A satellite is in orbit 36000km above the surface of the earth. Its angular velocity is 7.27*10^-5 rad/s. What is the velocity o

bagirrra123 [75]

Answer:

v = 3.08 km/s

Explanation:

Given that,

The angular velocity of the satellite = $\omega=7.27\times 10^{-5} rad/s$

A satellite is in orbit 36000km above the surface of the earth.

The radius of the earth is 6400 km

Let v is the velocity of the satellite. It can be calculated as :

$v=r\omega\\\\v=(36000\times 10^3+6400\times 10^3)\times 7.27\times 10^{-5}\\\\v=3082.48\ m/s\\\\v=3.08\ km/s$

So, the velocity of the satellite is 3.08 km/s.

6 0

3 years ago

A 50.0 kg driver is riding at 35.0 m/s in her red sports car when she must suddenly slam on the brakes to avoid

egoroff_w [7]

Answer:

3500N

Explanation:

Given parameters:

Mass of driver = 50kg

Speed = 35m/s

Time = 0.5s

Unknown:

Average force the seat belt exerts on her = ?

Solution:

The average force the seat belt exerts on her can be deduced from Newton's second law of motion.

F = mass x acceleration

So;

F = mass x $\frac{change in velocity }{time}$

F = 50 x $\frac{35}{0.5}$ = 3500N

8 0

3 years ago

You drop a ball from a height of 2.0 m, and it bounces back to a height of 1.5 m. (a) what fraction of its initial energy is los

Scilla [17]

a) At a position of 2.0m, the Initial energy is all made up of the potential energy=m*g*hi
and meanwhile at 1.5 all its energy is also potential energy=m*g*hf

The percentage of energy remaining is E=m*g*hi/m*g*hf x 100

and since mass and gravity are constant so it leaves us with just E=hi/hf
which 1.5/2.0 x100= 75% so we see that we lost 25% of the energy or 0.25 in fraction

b) Here use the equation vf^2=vi^2+2gd

where g is gravity, vf is the final velocity and vi is the initial velocity while d is the distance travelled

so in here we are looking for the vi so let us isolate that variable
we know that at maximum height or peak, the velocity is 0 so vf is 0

therefore,

vi =sqrt(-2gd) 
vi =sqrt(-2x-9.81x1.5) 
vi =5.4 m/s

c) The energy was converted to heat due to friction with the air and the ground.

6 0

3 years ago

A change in position with respect to a reference point is called?

kumpel [21]

A change in position with respect to a reference point is called motion

hope it helps...

4 0

3 years ago

Sort the forces as producing a torque of positive, negative, or zero magnitude about the rotational axis identified in part

Fantom [35]

a) Angular acceleration: $17.0 rad/s^2$

b) Weight: conterclockwise torque, reaction force: zero torque

Explanation:

a)

In this problem, you are holding the pencil at its end: this means that the pencil will rotate about this point.

The only force producing a torque on the pencil is the weight of the pencil, of magnitude

$W=mg$

where m is the mass of the pencil and g the acceleration of gravity.

However, when the pencil is rotating around its end, only the component of the weight tangential to its circular trajectory will cause an angular acceleration. This component of the weight is:

$W_p =mg sin \theta$

where $\theta$ is the angle of the rod with respect to the vertical.

The weight act at the center of mass of the pencil, which is located at the middle of the pencil. So the torque produced is

$\tau = W_p \frac{L}{2}=mg\frac{L}{2} cos \theta$

where L is the length of the pencil.

The relationship between torque and angular acceleration $\alpha$ is

$\tau = I \alpha$ (1)

where

$I=\frac{1}{3}mL^2$

is the moment of inertia of the pencil with respect to its end.

Substituting into (1) and solving for $\alpha$ , we find:

$\alpha = \frac{\tau}{I}=\frac{mg\frac{L}{2}sin \theta}{\frac{1}{3}mL^2}=\frac{3 g sin \theta}{2L}$

And assuming that the length of the pencil is L = 15 cm = 0.15 m, the angular acceleration when $\theta=10^{\circ}$ is

$\alpha = \frac{3(9.8)(sin 10^{\circ})}{2(0.15)}=17.0 rad/s^2$

b)

There are only two forces acting on the pencil here:

- The weight of the pencil, of magnitude $mg$

- The normal reaction of the hand on the pencil, R

The torque exerted by each force is given by

$\tau = Fd$

where F is the magnitude of the force and d the distance between the force and the pivot point.

For the weight, we saw in part a) that the torque is

$\tau =mg\frac{L}{2} cos \theta$

For the reaction force, the torque is zero: this is because the reaction force is applied exctly at the pivot point, so d = 0, and therefore the torque is zero.

Therefore:

- Weight: counterclockwise torque (I have assumed that the pencil is held at its right end)

- Reaction force: zero torque

8 0

3 years ago