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Montano1993 [528]
3 years ago
7

Difference between force and acceleration

Physics
2 answers:
galina1969 [7]3 years ago
7 0
Force your doing it purposely and acceleration it’s just happening
kramer3 years ago
3 0

Anwer: option B.Velocity is how fast an object moves and acceleration is the rate of change in velocity.


As you know velocity is the change is position divided by time, which is a measure of how fast something is moving. Now acceleration is the change in velocity divided by time which is how fast the velocity changes, this is its rate of change.


Both are rate of changes: velocity is the rate of change of position, and acceleration is the rate of change of velocity.


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I have a picture of the question I need help with,
eduard

Given:

The initial velocity of the object, v=30 m/s

a_t=0

a_c≠0

The time period is Δt.

To find:

The right conclusion among the given choices.

Explanation:

a_t represents the tangential accleration on the object and a_c represents the centripetal acceleration on the object.

The centripetal acceleration is the acceleration that keeps the object in its circular path. The centripetal force only changes the direction of the velocity and not the magnitude.

Thus the magnitude of the velocity of the object remains the same after a time interval of Δt. But the direction of the velocity of the object will be changed and will be unknown after Δt seconds.

Final answer:

Thus the object will be traveling at 30 m/s in some unknown direction.

Therefore, the correct answer is option a.

7 0
1 year ago
The most common isotope of hydrogen contains a proton and an electron 'separated by about -11-27 5.0 x 10 m. The mass of proton
Brrunno [24]

Answer:

A)   F_g = 4.05 10⁻⁴⁷ N, B)   F_e = 9.2 10⁻⁸N, C)    \frac{F_e}{F_g} = 2.3 10³⁹

Explanation:

A) It is asked to find the force of attraction due to the masses of the particles

Let's use the law of universal attraction

            F = G \frac{m_1m_2}{r^2}

let's calculate

            F = 6.67 \ 10^{-11} \ \frac{9.1 \ 10^{-31} \ 1.67 \ 10 ^{-27} }{(5 \ 10^{-11})^2 }

            F_g = 4.05 10⁻⁴⁷ N

B) in this part it is asked to calculate the electric force

Let's use Coulomb's law

            F = k \  \frac{q_1q_2}{r^2}

let's calculate

            F = 9 \ 10^9 \  \frac{(1.6 \ 10^{-19} )^2}{(5 \ 10^{-11})^2}

             F_e = 9.2 10⁻⁸N

C) It is asked to find the relationship between these forces

        \frac{F_e}{F_g} = \frac{9.2 \ 10^{-8} }{4.05 \ 10^{-47} }

        = 2.3 10³⁹

therefore the electric force is much greater than the gravitational force

4 0
3 years ago
HELP WILL MEDAL!!!! HELP
steposvetlana [31]
Here try this. The pic is the answer

8 0
3 years ago
1. Why is it important to use units in any graph?
Scorpion4ik [409]

Answer:

so its easier to understand for the reader

Explanation:

4 0
3 years ago
Read 2 more answers
Please help me promise would mark you brainiest
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B. Please mark me the brilliantest pls
3 0
3 years ago
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