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Ostrovityanka [42]

3 years ago

7

Why is it inaccurate to use mgy to calculate the potential energy of a satellite orbiting earth at a height one earth radius abo

ve the earth’s surface

1 answer:

Alex3 years ago

4 0

Because the gravitational constant is not same as on earth

You might be interested in

consider the free-body diagram. if you want the box to move, the force applied while dragging must be greater than the

NeTakaya

Answer:

Force of static friction between the two surfaces

Explanation:

When two surfaces come into contact, they exert a force that resist the sliding of the two surfaces. This force is called static friction.

This force is given by the relation

$F_{s}=\mu_{s}\eta$

Where,

μ - coefficient of static friction

η - normal force acting on the body

When a force acts on a body placed on a rough surface, it doesn't do any work if the applied force was less than the force of static friction.

So, in order to move the body, the applied force should be greater than the force of static friction.

6 0

2 years ago

You set out to design a car that uses the energy stored in a flywheel consisting of a uniform 101-kg cylinder of radius r that h

Ket [755]

Ok, assuming "mj" in the question is Megajoules MJ) you need a total amount of rotational kinetic energy in the fly wheel at the beginning of the trip that equals
(2.4e6 J/km)x(300 km)=7.2e8 J
The expression for rotational kinetic energy is

E = (1/2)Iω²

where I is the moment of inertia of the fly wheel and ω is the angular velocity.
So this comes down to finding the value of I that gives the required energy. We know the mass is 101kg. The formula for a solid cylinder's moment of inertia is

I = (1/2)mR²

We want (1/2)Iω² = 7.2e8 J and we know ω is limited to 470 revs/sec. However, ω must be in radians per second so multiply it by 2π to get
ω = 2953.1 rad/s
Now let's use this to solve the energy equation, E = (1/2)Iω², for I:
I = 2(7.2e8 J)/(2953.1 rad/s)² = 165.12 kg·m²

Now find the radius R,

165.12 kg·m² = (1/2)(101)R²,
√(2·165/101) = 1.807m

R = 1.807m

8 0

3 years ago

the metal wire in an incandescent lightbulb glows when the light is switched on and stops glowing when it is switched off. the s

Travka [436]

When the metal wire in an incandescent lightbulb glows when the light is switched on and stops glowing when it is switched off, this is an example of resistance, which provides light and heat.

3 0

2 years ago

A 0.50 kg toy is attached to the end of a 1.0 m very light string. The toy is whirled in a horizontal circular path on a frictio

xenn [34]

Answer:

The maximum speed will be 26.475 m/sec

Explanation:

We have given mass of the toy m = 0.50 kg

radius of the light string r = 1 m

Tension on the string T = 350 N

We have to find the maximum speed without breaking the string

For without breaking the string tension must be equal to the centripetal force

So $T=\frac{mv^2}{r}$

So $350=\frac{0.5\times v^2}{1}$

$v^2=700$

v = 26.475 m /sec

So the maximum speed will be 26.475 m/sec

6 0

2 years ago

Calculate the gravitational attraction between two objects which masses are: mass A: 2.5kg, mass B: 5kg. The distance between th

jonny [76]

From the calculation, the gravitational force of attraction is 1.33 * 10^-14 N.

<h3>What is the gravitational force?</h3>

The gravitational force is an attractive force that acts between any two masses.

It is given by;

F = Gm1m2/r^2

F = 6.67 * × 10−11 * 2.5 * 5/(250)^2

F = 83.4 × 10−11 /62500

F= 1.33 * 10^-14 N

Learn more about gravitational force:brainly.com/question/12528243

#SPJ1

4 0

1 year ago