All categories

3 years ago

Find acceleration given the following:

Physics

1 answer:

Lorico [155]3 years ago

5 0

Answer:

Explanation:df=378

Di=0

Vi=0

T=6s

Formula

Df=Di+Vit+1/2(at)2

378=0+0+6+1/2×a×t2

378=6+1/2×a×36

378=6+18a

18a=378+6=384

a=385/18=21.33(cm/s)/s

You might be interested in

The phases of the moon depend on how much of the lighted side of the moon can be seen from earth.is this true or false

Sindrei [870]

The answer to this question is going to be False

6 0

3 years ago

15 PTS! HELP! BRAINLIEST!

vodka [1.7K]

I cant do all but I can do number 4
When force moves ur leg u start skatin and when that happenes ur using a lot of force to move

3 0

3 years ago

Space-faring astronauts cannot use standard weight scales (since they are constantly in free fall) so instead they determine the

valentinak56 [21]

Answer:

ma = 48.48kg

Explanation:

To find the mass of the astronaut, you first calculate the mass of the chair by using the information about the period of oscillation of the empty chair and the spring constant. You use the following formula:

$T=2\pi\sqrt{\frac{m_c}{k}}$ (1)

mc: mass of the chair

k: spring constant = 600N/m

T: period of oscillation of the chair = 0.9s

You solve the equation (1) for mc, and then you replace the values of the other parameters:

$m_c=\frac{T^2k}{4\pi^2}=\frac{(0.9s)^2(600N/m)}{4\pi^2}=12.31kg$ (2)

Next, you calculate the mass of the chair and astronaut by using the information about the period of the chair when the astronaut is sitting on the chair:

T': period of chair when the astronaut is sitting = 2.0s

M: mass of the astronaut plus mass of the chair = ?

$T'=2\pi\sqrt{\frac{M}{k}}\\\\M=\frac{T'^2k}{4\pi^2}=\frac{(2.0s)^2(600N/m)}{4\pi^2}\\\\M=60.79kg$ (3)

Finally, the mass of the astronaut is the difference between M and mc (results from (2) and (3)) :

$m_a=M-m_c=60.79kg-12.31kg=48.48kg$

The mass of the astronaut is 48.48 kg

3 0

3 years ago

Which is likely to be more common in our Galaxy: white dwarfs or black holes? Why?

Nookie1986 [14]

Answer:

White dwarfs are likely to be much more common. The number of stars decreases with increasing mass, and only the most massive stars are likely to complete their lives as black holes. There are many more stars of the masses appropriate for evolution to a white dwarf.

4 0

2 years ago

A metal rod of length (L) moves with velocity (v), perpendicular to its length, in a magnetic field B, which is perpendicular to

Alborosie

Explanation:

If a metal rod of length L moves with velocity v is moving perpendicular to its length, in a magnetic field B, the induced emf is given by :

$\epsilon=Blv$

The electric field in the conductor is given by :

$E=\dfrac{\epsilon}{l}\\\\E=\dfrac{Blv}{l}\\\\E=Bv$

It is clear that the electric field is independent of the length of the rod. If the length of the rod is doubled, the electric field in the rod remains the same.

6 0

2 years ago