1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Katena32 [7]
3 years ago
12

PLEASE ASAP ILL GIVE BRAINLIEST.

Physics
2 answers:
Sergeu [11.5K]3 years ago
6 0

Answer:

law of gravity

Explanation:

cause the ball was still moving

Debora [2.8K]3 years ago
5 0

Answer: the answer is law of gravity

Explanation:

You might be interested in
As a recreational boat operator, what actions must you take when in a narrow channel?
alexandr402 [8]
One should never anchor in a narrow channel, until unless required very importantly. One should stay to the starboard side, and use a prolonged blast. The announcement must be done to alarm the nearby vessels, about your approach. The vessel should be kept at the outer limit of the starboard side. 
6 0
3 years ago
Read 2 more answers
What is an example of a non contact force
olchik [2.2K]

Answer: Gravitational force

Explanation:

A non contact force can be described as a force applied to an object by another body that is not in direct contact with it.

For example, an object thrown upwards will return back due to the force of gravity acting on it. So, it means Gravitational force is acting on the body without necessarily being in contact with that body.

8 0
4 years ago
A fan blade is rotating with a constant angular acceleration of 11.5 rad/s2. At what point on the blade, as measured from the ax
vichka [17]

Answer:

Alpha = ω^2 R    where R is radius of blade

g = w^2 r      where r is distance from center

ω^2 R = 11.5 ω^2 r

R / r = 11.5 / 9.8 = 1.17

Or r = .852 R

Since the angular acceleration depends on both R and ω it seems that one can only get r as it depends on R

7 0
2 years ago
You stand on a straight desert road at night and observe a vehicle approaching. This vehicle is equipped with two small headligh
g100num [7]

To solve this problem we will apply trigonometric and optical concepts that allow us to obtain the minimum distance required. The resolution of the eye is given under the following condition,

\theta = \frac{1.22\lambda}{D}

Here,

\lambda = Wavelength

D = Diameter

With the values we have that the diameter will be,

\theta = \frac{1.22(534nm)}{5.37mm}

\theta = 1.213*10^{-4}

The relation between the distance of the lights and the distance from the eye to the lamp is given under the function,

sin\theta = \frac{d}{L}

For small angles sin\theta = \theta, then

\theta = \frac{d}{L}

Here,

d = Distance between lights

L = Distance from eye to lamp

1.213*10^{-4} = \frac{0.673m}{L}

L = \frac{0.673m}{1.213*10^{-4}}

L = 5548.22m

Therefore the distance will be 5.5km

5 0
3 years ago
2. Which positions of the pendulum would have the most potential energy?
g100num [7]

Explanation:

im listening to metallica

turn the page

4 0
3 years ago
Other questions:
  • Cosmologist study ____
    12·2 answers
  • Please answer. Thanks
    6·1 answer
  • Why sodium &pottasium are kept under kerosine
    10·1 answer
  • Help please!!!!!!!!!!!!!!!!!
    12·2 answers
  • Walking across a carpet is an example of charge being transferred by
    13·1 answer
  • A string along which waves can travel is 4.29 m long and has a mass of 228 g. The tension in the string is 31.1 N. What must be
    8·1 answer
  • The volume of liquid flowing per second is called the volume flow rate Q and has the dimensions of [L]3/[T]. The flow rate of li
    9·1 answer
  • You are standing next to your friend, ready to race across a 100 meter field. When the go signal is given, you take 5 seconds to
    9·1 answer
  • The attraction between the Sun's gravity and Earth's gravity holds Earth in is ?
    8·1 answer
  • Currents in dc transmission lines can be 100 A or higher. Some people are concerned that the electromagnetic fields from such li
    8·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!