Ask question

Ask question

All categories

3 years ago

7

If you and your friend are riding your bikes side by side and your reference points are each other, are you in motion? Explain y

es or no

1 answer:

geniusboy [140]3 years ago

6 0

If your ref points are each other and not the surrounding environment, at the same speed you would both appear to be stationary

You might be interested in

Which celestial body would have the strongest gravitational pull on a satellite orbiting 100 km above its surface?

8090 [49]

According to the Law of Universal Gravitation, the gravitational force is directly proportional to the mass, and inversely proportional to the distance. In this problem, let's assume the celestial bodies to be restricted to the planets and the Sun. Since the distance is specified, the other factor would be the mass. Among all the celestial bodies, the Sun is the most massive. So, the Sun would cause the strongest gravitational pull to the satellite.

3 0

3 years ago

Read 2 more answers

The first and second coils have the same length, and the third and fourth coils have the same length. They differ only in the cr

stealth61 [152]

Answer:

$\frac{R_2}{R_1}=\frac{A_1}{A_2}\\\frac{R_4}{R_3}=\frac{A_3}{A_4}$

Explanation:

The resistance of a conductor is directly proportional to its length and is inversely proportional to its cross-sectional area, this dependence is given by:

$R=\frac{\rho L}{A}$

$\rho$ is the material's resistance, L is the legth and A is the cross-sectional area.

For the first and second coils, we have:

$R_1=\frac{\rho L}{A_1}\\R_2=\frac{\rho L}{A_2}\\\rho L=R_1A_1\\\rho L=R_2A_2\\R_1A_1=R_2A_2\\\frac{R_2}{R_1}=\frac{A_1}{A_2}$

For the third and fourth coils, we have:

$R_3=\frac{\rho L'}{A_3}\\R_4=\frac{\rho L'}{A_4}\\\rho L'=R_3A_3\\\rho L'=R_4A_4\\R_3A_3=R_4A_4\\\frac{R_4}{R_3}=\frac{A_3}{A_4}$

6 0

3 years ago

When work is done on an object, where does the energy used to do the work go?

borishaifa [10]

-- If the work is done to make the object move faster, then
the work done becomes kinetic energy of the object.

-- If work is done on the object but it doesn't move any faster,
then there must be friction holding it back. In that case, the work
that's done just to keep the object moving becomes heat, in the
places where the friction acts on it.

3 0

3 years ago

You have a two-wheel trailer that you pull behind your ATV. Two children with a combined mass of 77.2 kg hop on board for a ride

lions [1.4K]

To solve this problem it is necessary to apply the kinematic equations of motion and Hook's law.

By Hook's law we know that force is defined as,

$F= kx$

Where,

k = spring constant

x = Displacement change

PART A) For the case of the spring constant we can use the above equation and clear k so that

$k= \frac{F}{x}$

$k = \frac{mg}{x}$

$k= \frac{77.2*9.8}{0.0637}$

$k = 11876.92N/m$

Therefore the spring constant for each one is 11876.92/2 = 5933.46N/m

PART B) In the case of speed we can obtain it through the period, which is given by

$T = \frac{2\pi}{\omega}$

Re-arrange to find \omega,

$\omega = \frac{2\pi}{T}$

$\omega = \frac{2\pi}{2.14}$

$\omega = 2.93rad/s$

Then through angular kinematic equations where angular velocity is given as a function of mass and spring constant we have to

$\omega^2 = \frac{k}{m}$

$m = \frac{k}{\omega^2}$

$m = \frac{ 11876.92}{2.93}$

$m = 4093.55Kg$

Therefore the mass of the trailer is 4093.55Kg

PART C) The frequency by definition is inversely to the period therefore

$f = \frac{1}{T}$

$f = \frac{1}{2.14}$

$f = 0.4672 Hz$

Therefore the frequency of the oscillation is 0.4672 Hz

PART D) The time it takes to make the route 10 times would be 10 times the period, that is

$t_T = 10*T$

$t_T = 10 *2.14s$

$t_T = 21.4s$

Therefore the total time it takes for the trailer to bounce up and down 10 times is 21.4s

5 0

3 years ago

What weather would create the weakest surfing waves?

ella [17]

Its B hope I could help

5 0

2 years ago