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3 years ago

A light year is _____.

2 answers:

4 0

A light year is a unit of astronomical distance which is equivalent to the distance a light travels in one year( which is nearly 6 trillion miles)

hope this helps :)

777dan777 [17]3 years ago

3 0

Alight year is the distance that ligth travels in a year.

Answer: C. The distance tat light travels in a year.

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In this lab you will use a cart and track to explore various aspects of motion. You will measure and record the time it takes th

Keith_Richards [23]

It is given that by using track and cart we can record the time and the distance travelled and also the speed of the cart can be recorded. With all this data we can solve questions on the laws of motion.

Like using the first law of motion we can determine the force of gravity acting on the cart that has moved a certain distance and the velocity or the speed of card has already been registered and since time is known putting the values in formula would help us calculate the gravitational pull acting on cart.

3 0

3 years ago

Read 2 more answers

1. A mass is on a level plane, it has a weight of 20N. What is the coefficient of kinetic friction if an applied force

Arturiano [62]

Answer:

0.4

Explanation:

F-Fr=ma where F is applied force, Fr is friction, m is mass and a is acceleration.

Since the mass is moving with a constant velocity, there's no acceleration hence

$F=Fr=\mu N$ where N is the weight of object and \mu is coefficient of kinetic friction.

$F=\mu N and making [tex]\mu$ the subject

$\mu=\frac {F}{N}$

Substituting F for 8 N and N for 20 N

$\mu=\frac {8}{20}=0.4$

Therefore, coefficient of kinetic friction is 0.4

4 0

3 years ago

(a) At what height above Earth’s surface is the energy required to lift a satellite to that height equal to the kinetic energy r

Nikitich [7]

Answer:

Explanation:

Gravitational Potential Energy at earth surface $U_1=\frac{GM_em}{R_e}$

Gravitational Potential Energy at height h is $U_2=\frac{GM_em}{R_e+h}$

Energy required to lift the satellite $E_1=U_1-U_2$

$E_1=\frac{GM_em}{R_e}-\frac{GM_em}{R_e+h}$

Now Energy required to orbit around the earth

$E_2=\frac{1}{2}mv_{orbit}^2=\frac{GM_2m}{2(R_e+h)}$

$\Delta E=E_1-E_2$

$\Delta E=\frac{GM_em}{R_e}-\frac{GM_em}{R_e+h}-\frac{GM_2m}{2(R_e+h)}$

$E_1=E_2$ (given)

$\frac{GM_em}{R_e}-\frac{GM_em}{R_e+h}-\frac{GM_2m}{2(R_e+h)}=0$

$\frac{1}{R_e}-\frac{3}{2(R_e+h)}=0$

$h=\frac{R_e}{2}$

$h=3.19\times 10^6\ m$

(b)For greater height $E_1$ is greater than $E_2$

thus energy to lift the satellite is more than orbiting around earth

4 0

3 years ago

A moving truck has more __ energy than a parked truck

Karo-lina-s [1.5K]

Kinetic energy than parked

7 0

3 years ago

A clay vase on a potter's wheel experiences an angular acceleration of 5.69 rad/s2 due to the application of a 16.0-n m net torq

Digiron [165]

The equivalent of the Newton's second law for rotational motions is:
$\tau = I \alpha$
where
$\tau$ is the net torque acting on the object
$I$ is its moment of inertia
$\alpha$ is the angular acceleration of the object.

Re-arranging the formula, we get
$I= \frac{\tau}{\alpha}$
and since we know the net torque acting on the (vase+potter's wheel) system, $\tau=16.0 Nm$ , and its angular acceleration, $\alpha = 5.69 rad/s^2$ , we can calculate the moment of inertia of the system:
$I= \frac{\tau}{\alpha}= \frac{16.0 Nm}{5.69 rad/s^2} =2.81 kg m^2$

8 0

3 years ago