Matt swims one hundred meters in fifty seconds. What is his speed

Calculate the speed of a satellite moving in a stable circular orbit about the earth at a height of 4800 km

forsale [732]

The speed of satellite or orbital speed is given by below formula.

V = sqrt (GM/R)

where

G = is the gravitational force

M = earth's mass

R = radius of its orbit.

Solving for the Orbital Speed,

First add the earth's radius to the height of satellite wrt to the earths surface, we have R = 6.59x10^6 m + 4.8 x 10 6 m = 11.39x10^6

Now

V = sqrt ((6.673 x 10-11*5.98x1024 kg)/11.39) =5919 m/s

6 0

4 years ago

If the child pulls with a force of 20 N for 12.0 m , and the handle of the wagon is inclined at an angle of 25 ∘ above the horiz

lilavasa [31]

Answer: 217.52 N

Explanation: The applied force is 20 N, the distance covered is 12.0 m and the angle is 25° above the horizontal.

Hence the formulae that defines work done is given by

W = Force × distance

But since the force has been inclined at an angle θ above the horizontal, the horizontal component of force is neccesary to produce the required motion to make the child do work on the wagon.

Hence

Work done = (horizontal component of force) × distance

Work done = F cos θ × distance

Work done = 20 cos 25 × 12 = 217.52 N

4 0

3 years ago

A textbook of mass 2.09kg rests on a frictionless, horizontal surface. A cord attached to the book passes over a pulley whose di

nydimaria [60]

Answer:

(A) 9.7 N

(B) 15.4 N

(C) I = 0.0045 kg m^{2}

Explanation:

mass of text book (M1) = 2.09 kg

mass of book (M2) = 2.99 kg

diameter of the pulley (d) = 0.12 m

radius (r) = 0.06 m

distance moved (s) = 1.30 m

time (t) = 0.75 s

acceleration due to gravity (g) = 9.8 m/s^[2}

(a) what is the tension in the part of the cord attached to the text book?

the text book is moving horizontally, so the tension in this case becomes

tension = mass x acceleration

we can get the acceleration from s = ut + 0.5 at^{2}

since the books are initially at rest u = 0

s = 0.5 at^{2}

1.3 = 0.5 x a x 0.75^{2}

a = 4.643 m/s^[2}

tension (T1) = 2.09 x 4.643 = 9.7 N

(b) what is the tension in the part of the cord attached to the book?

the book is hanging vertically, so the tension in this case becomes

tension = m x ( g - a )

(g-a) is the net acceleration of the first book

tension (T2) = 2.99 x (9.8 - 4.643) = 15.4 N

(c) What is the moment of inertia of the pulley?

if the books were to move in the direction of the book, it will cause the pulley to rotate clockwise and if they were to move in the direction of the text book on the table the pulley will rotate in an anticlockwise direction. Taking clockwise rotation of the pulley to be negative while anticlockwise to be positive, we can say ( T2 - T1 )r = I∝

where ∝ is the angular acceleration of the pulley relative to its radial

acceleration, ∝ = \frac{a}{r}

( T2 - T1 )r = I\frac{a}{r}

I = \frac{(T2 - T1)r^{2}}{a}

I = \frac{(15.5 - 9.7)0.060^{2}}{4.643}

I = 0.0045 kg m^{2}

6 0

3 years ago

A car travelling at 15 m/s comes to rest in a distance of 14 m when the brakes are applied.

stira [4]

Answer:

-8.04 m/s2