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ludmilkaskok [199]
3 years ago
14

PLEASE HELP QUICKLY!!!!!

Physics
1 answer:
Grace [21]3 years ago
3 0

Answer:

D air

Explanation:

it is not found on the periodic table

brainliest plsssssssssssssssss

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A 75-m-long train begins uniform acceleration from rest. the the train has a speed of 23 m/s when it passes a railway worker tan
Lapatulllka [165]

Answer:

acceleration a = 1.04 m/s2

Explanation:

Assume the train has a speed of 23m/s when the last car passes the railway workers. Once this happens the last car would have traveled a total distance of the 180m distance between the railway worker standing 180 m from where the front of the train started plus the 75m distance from the first car to the last car:

s = 75 + 180 = 255 m

We can use the following equation of motion to find out the distance traveled by the car:

v^2 - v_0^2 = 2aswhere v = 23 m/s is the velocity of the car when it passes the worker, v_0 = 0m/s is the initial velocity of the car when it starts, a m/s2 is the acceleration, which we are looking for.

23^2 - 0^2 = 2*a*255

510a = 529

a = 529 / 510 = 1.04 m/s^2

8 0
3 years ago
What likely happens to the hart rate as the cardiac mucle gets stronger
Kaylis [27]
The heart rate will likely decrease. As the cardiac muscle, or heart, gets stronger, it takes less effort to pump more blood. As a result, the heart will probably beat less, decreasing the heart rate. This is why athletes often have lower heart rates than the average person.
5 0
3 years ago
the brightness of stars as they appear from earth is measured by _______ magnitude. a. light b. apparent c. relative d. absolute
xxTIMURxx [149]
How bright a star appears from Earth is the star's apparent magnitude.
8 0
3 years ago
Compare the mass of block A, the mass of block B is:
Alexxx [7]

the answer would be b. it`s half as great

6 0
3 years ago
You use a slingshot to launch a potato horizontally from the edge of a cliff with speed v0. The acceleration due to gravity is g
sergij07 [2.7K]

Answer:

 t = 2 v₀ / g

Explanation:

For this projectile launch exercise we use the displacement equations

        x = vox t

       y = y₀ + v_{oy} t - ½ g t²

As it is launched horizontally the vertical velocity is zero and the point of origin of the coordinate system is here, so y₀ is zero.

       x = v₀ t

       y = ½ g t²

They ask us for the time for which

        x = y

        vo t = ½ g t²

         t = 2 v₀ / g

8 0
3 years ago
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