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3 years ago

10. Which of the following is an acceleration? a. 12 m/s2 down b. 5 m/s up c. 8N West

Physics

1 answer:

nadya68 [22]3 years ago

3 0

Answer:

a. 12 m/s² down

Explanation:

Acceleration has units of length per time squared. Acceleration is a vector, so it also has a direction.

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if you have a mass of 55 kg and you are standing 3 meters away from your car, which has a mass of 1234 kg, how strong is the for

bagirrra123 [75]

Gravitational force between two masses is given by formula

$F = \frac{Gm_1m_2}{r^2}$

here we know that

$m_1 = 55 kg$

$m_2 = 1234 kg$

$r = 3 m$

$G = 6.67 \times 10^{-11} Nm^2/kg^2$

now from the above equation we will have

$F = \frac{(6.67 \times 10^{-11})(55)(1234)}{3^2}$

$F = 5.03 \times 10^{-7}N$

so above is the gravitational force between car and the person

5 0

3 years ago

Answer True or Flase1-Electric potential due to a uniform E field doesn’t change with location.2-The equipotential surfaces asso

TEA [102]

Answer:

1. False

2. True

3. True

Explanation:

1- False —> The relation between electric potential and electric field is given such that

$-\int\limits^a_b \vec{E}d\vec{l} = V_{ab}$

Therefore, for a uniform E field, electric potential is linearly proportional to the distance.

2- True —> The electric field lines always cross the equipotential lines perpendicularly.

3- True —> In order to be a potential difference, one source of electric field is enough. The electric potential will decrease radially according to the following formula:

$V = \frac{1}{4\pi\epsilon_0}\frac{q}{r^2}$

There is no test charge in the formula, only the source charge. Even when there is no test charge, the potential difference between points in space can exist.

3 0

3 years ago

A person wants to make a metronome for music practice. He uses a 35-g object attached to a spring to serve as the time standard.

alukav5142 [94]

To develop this problem it will be necessary to apply the concepts related to the frequency of a spring mass system, for which it is necessary that its mathematical function is described as

$f = \frac{1}{2\pi} \sqrt{\frac{k}{m}}$

Here,

k = Spring constant

m = Mass

Our values are given as,

$m = 35g = 35*10^{-3}kg$

$f = 1 Hz$

Rearranging to find the spring constant we have that,

$k = (2\pi f \sqrt{m})^2$

$k = 4\pi^2 f^2 m$

$k = (4) (\pi)^2 (1) (35*10^{-3})$

$k = 1.38N/m$

Therefore the spring constant is 1.38N/m

7 0

3 years ago

When an average force F is exerted over a certain distance on a shopping cart of mass m, its kinetic energy increases by 12mv2.

VMariaS [17]

Answer:

A) $d=\dfrac{1}{2F}mv^2$

B) $\Delta KE'=2\times \dfrac{1}{2}mv^2$

Explanation:

Given that

Force = F

Increase in Kinetic energy = $\dfrac{1}{2}mv^2$

$\Delta KE=\dfrac{1}{2}mv^2$

we know that

Work done by all the forces =change in the kinetic energy

Lets distance = d

We know work done by force F

W= F .d

F.d=ΔKE

$F.d=\dfrac{1}{2}mv^2$

$d=\dfrac{1}{2F}mv^2$

If the force become twice

F' = 2 F

F'.d=ΔKE'

2 F .d = ΔKE' ( F.d =Δ KE)

2ΔKE = ΔKE'

$\Delta KE'=2\times \dfrac{1}{2}mv^2$

Therefore the final kinetic energy will become the twice if the force become twice.

8 0

3 years ago

If you double the velocity of a moving object, how is it's momentum affected?

Allushta [10]

Well momentum is = to Mass*Velocity so let's use an example to figure this out

If I weighed 50kg and I was jogging at 3m/s then I broke into a run at 6m/s how will me momentum be affected?
3m/s*50kg=150
6m/s*50kg=300

So as you can see by doubling the velocity you also double the momentum

8 0

3 years ago