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3 years ago

9

Newton's universal law of gravitation says that every object exerts a force on another object. The shuttle has a gravitational f

orce with Earth, the moon, and whatever planet it is near. This force prevents the shuttle from drifting.
Question 1 options:
True
False

1 answer:

Gemiola [76]3 years ago

7 0

The correct answer is true

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How does approaching a temperature of absolute zero affect kinetic energy.?

Alexus [3.1K]

First of all the kinetic energy is when the particles move in continuous random motion.

If the temperature is high the colliding particles will collide more. and if the temperature is low the colliding particles will collide less.

Low temperature result in low kinetic energy
High temperature result in high kinetic energy

Absolute zero is the point where where all molecules have no kinetic energy. It is a theoretical value (it has never been reached).

The Kelvin temperature scale is based on absolute zero being the lowest possible temperature that could theoretically be reached. That is why there is no such thing as a negative Kelvin temperature value.

7 0

3 years ago

Ann walked 1.5 miles south to her house in 0.5 hours. what is Ann's speed? what is Ann's velocity

ipn [44]

To calculate the speed and velocity of the Ann`s we use the formula,

$v= \frac{d}{t}$

Here, d is distance and t is time and v if we take it with direction then it is called velocity and if we take it without the direction then it is called speed.

Given $d=1.5 miles$ and $t=0.5 hours$ .

Substituting these values in above equation we get

$v=\frac{1.5mi }{0.5 h} = 3 mi/h$

As Ann walked towards south direction therefore, Ann`s velocity is 3 mi/h south and her speed is 3 mi/h .

7 0

3 years ago

A small object of mass 3.82 g and charge -16.5 µC is suspended motionless above the ground when immersed in a uniform electric f

horrorfan [7]

Answer:

2271.16N/C upward

Explanation:

The diagram well illustrate all the forces acting on the mass. The weight is acting downward and the force is acting upward in other to balance the weight.since the question says it is motionless, then indeed the forces are balanced.

First we determine the downward weight using

$W=mg\\g=9.81m/s^{2}$

Hence for a mass of 3.82g 0r 0.00382kg we have the weight to be

$W=0.00382kg*9.81m/s^{2}$

$W=0.0375N$

To calculate the electric field,

$E=f/q\\E=0.0375/16.5*10^{-6} \\E=2271.16N/C$

Since the charge on the mass is negative, in order to generate upward force, there must be a like charge below it that is repelling it, Hebce we can conclude that the electric field lines are upward.

Hence the magnitude of the electric force is 2271.16N/C and the direction is upward

4 0

3 years ago

A boy notices that 45 complete waves travel past him in ninety seconds. What is the

san4es73 [151]

Answer:

0.5 Hz

Hope you find this helpful. Please mark me as brainliest!

5 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