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alina1380 [7]
3 years ago
6

Does a force cause motion or a change in motion

Physics
1 answer:
Brrunno [24]3 years ago
3 0
It changes the motion as Newton's second law of motion states that a force, acting on an object, will change its velocity by changing either its speed or its direction or both. If your basketball goes rolling into the street and is hit by a bike, either the ball will change direction or its speed or both.
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What holds the moon in orbit around the earth?<br> A. The sun's gravity<br> B. The Earth's gravity
irina [24]

Answer:

Earth's gravity

Explanation:

hope this helps

7 0
3 years ago
Read 2 more answers
For a satellite of mass mS in a circular orbit of radius rS around the Earth, determine its kinetic energy, K . Express your ans
agasfer [191]

Since it is asking you to find the kinetic energy in relation to the mass, radius, mechanical energy (total energy), and constants, you will need to setup an equation first to "find" the Mechanical Energy, so that we can then solve for the kinetic energy, as from my experience with high school physics, there is only the graviational potential energy equation and force in relation to celestial bodies.

Knowing the ME is the total energy, we add up the energies of the system. Since it is being influenced by the Earth, as per the problem stating the satellite has circular orbit around the Earth, we know there is gravitational potential. Since it is orbiting, we can assume some type of velocity. Nothing else that we need to worry about should be occuring at this level of physics, leaving you with

ME= Ug+K

from here we solve for K, as plugging in could get confusing and messy at the moment.

ME-Ug=K

now using the equations presumably given in class, if not then using this equation, we can find the Ug

Ug=(-(Gm*M)/r)     note that M is the mass of the Earth and m is the satellite

this should give us

ME-(-(GmM)/r)=K

since there is a negative being subracted, we can change that to

ME+(GmM)/r=K

I believe this should be fine, as the Earth's mass is constant, but if not, then all you need to figure left is how to get rid of the M in the equation, as the rest of the terms and constants are for sure within the requirements.

8 0
3 years ago
As a laudably skeptical physics student, you want to test Coulomb's law. For this purpose, you set up a measurement in which a p
Oksana_A [137]

Answer:

a.Attractive

2.31531\times 10^{-16}\ N

Explanation:

When it comes to charges, the charges which are alike repel each other and the charges which are different will attract each other.

Here, there is a proton and electron which are different particles hence, they will attract each other.

q_1=q_2 = Charge of electron and proton = 1.6\times 10^{-19}\ C

r = Distance between them = 997 nm

k = Coulomb constant = 8.99\times 10^{9}\ Nm^2/C^2

Force is given by

F=\dfrac{kq_1q_2}{r^2}\\\Rightarrow F=\dfrac{8.99\times 10^9\times 1.6\times 10^{-19}\times 1.6\times 10^{-19}}{(997\times 10^{-9})^2}\\\Rightarrow F=2.31531\times 10^{-16}\ N

The force of attraction between the particles will be 2.31531\times 10^{-16}\ N

8 0
3 years ago
What is happening in the picture?
Veseljchak [2.6K]
The gravity is pushing rhe boat down
3 0
3 years ago
Read 2 more answers
how fast will and in what direction will a 20kg object accelerate if one force pushes at a 30 degree angle and another pushes at
DiKsa [7]

Answer:

|a|=2.83\ m/s^2

\theta=75^o

Explanation:

<u>Net Force And Acceleration </u>

The Newton's second law relates the net force applied on an object of mass m and the acceleration it aquires by

\vec F_n=m\vec a

The net force is the vector sum of all forces. In this problem, we are not given the magnitude of each force, only their angles. For the sake of solving the problem and giving a good guide on how to proceed with similar problems, we'll assume both forces have equal magnitudes of F=40 N

The components of the first force are

\vec F_1=

\vec F_1=\ N

The components of the second force are

\vec F_2=

\vec F_2=\ N

The net force is

\vec F_n=

\vec F_n=\ N

The magnitude of the net force is

|F_n|=\sqrt{14.64^2+54.64^2}

|F_n|=\sqrt{3200}=56.57\ N

The acceleration has a magnitude of

\displaystyle |a|=\frac{|F_n|}{m}

\displaystyle |a|=\frac{56.57}{20}

|a|=2.83\ m/s^2

The direction of the acceleration is the same as the net force:

\displaystyle tan\theta=\frac{54.64}{14.64}

\theta=75^o

5 0
3 years ago
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