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ikadub [295]
3 years ago
11

A sports car accelerates at 2.80 m/s2 along a straight road. At t1 = 5.00s and t2 = 5.50 s it passes two marks that are 35.0 m a

part. What is the car's velocity at t0 = 0 s?​
Physics
1 answer:
harina [27]3 years ago
7 0

Explanation:

Using kinematics,

s = ut + 0.5at².

We have s = 35.0m, t = 0.50s, a = 2.80m/s².

Substitute in the variables, we have u = 69.3m/s.

This u is only the initial velocity at t1. To find the velocity at t0, we apply a different kinematics equation with different values:

We have t = 5.00s, a = 2.80m/s², v = 69.3m/s.

Using kinematics,

v = u + at.

Substitute in the variables, we have u = 55.3m/s.

Hence the car's initial velocity is 55.3m/s.

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A boy standing throws a penny horizontally at 7.25 m/s out of the window of his apparent buliding. If the window is 10.0 m above
Ksivusya [100]

Answer:

Explanation:

This is a 2D problem (parabolic) so we have to think that way. We have to split up the problem into its 2 dimensions to solve it. Think "y-stuff" and "x-stuff".

In the y-stuff category:

v₀ = 0 (initial upwards velocity is 0 since we are told the penny is thrown horizontally)

Δx = -10.0 m (this displacement is negative because the penny lands 10.0 m below the point from which it was thrown)

a = -9.8 m/s/s

t = ? (we need to find the time in this dimension so we can use it in the x dimension to find the displacement, our unknown)

In the "x-stuff" category:

v₀ = 7.25 m/s (this is given)

Δx = ???

a = 0 (acceleration in this dimension is ALWAYS 0)

t = (we will solve for this in the y-dimension and plug it in here).

In the y dimension:

Δx = v₀t + \frac{1}{2}at^2 and plugging in from the y-dimension info:

-10.0=0t+\frac{1}{2}(-9.8)t^2 which simplifies to

-10.0=-4.9t^2 so

t=\sqrt{\frac{-10.0}{-4.9} } which, to 2 significant digits is

t = 1.4 seconds

Now we will do the same in the x-dimension, using t = 1.4:

Δx = v₀t + \frac{1}{2}at^2 and filling in the x-stuff:

Δx = 7.25(1.4)+\frac{1}{2}(0)(1.4)^2 Notice that the stuff after the + sign goes to 0 cuz of the multiplication of 0, so what we are left with is another form of the d = rt equation:

Δx = 7.25(1.4) + 0 so

Δx = 1.0 × 10¹ m (That's rounded correctly to 2 sig dig's: 10 m from the base of the building).

6 0
3 years ago
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Nutka1998 [239]

Answer:

Explanation:

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7 0
3 years ago
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How much energy does the electron have initially in the n=4 excited state?what is the change in energy if the electron from part
Ierofanga [76]
What it looks to be that you found in A was the "initial"...b/c the question asks: 
<span>"how much energy does the electron have 'initially' in the n=4 excited state?" </span>

<span>"final" would be where it 'finally' ends up at, ie. its last stop...as for this question...the 'ground state' as in its lowest energy level. </span>

The answer comes to: <span>−1.36×10^−19 J</span>


You use the same equation for the second part as for part a. 
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A ball is tied to a string and whirled in a horizontal circle at a constant 4 revolutions per second. Categorize the following a
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The velocity of the ball is constantly changing because its direction is changing.
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svetlana [45]

A < B < C


C will increase faster than B which is faster than A

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