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mamaluj [8]
2 years ago
14

A bicycle racer rides from a starting marker to a turnaround marker at 10 m/s. She then rides back along the same route from the

turnaround marker to the starting marker at 16 m/s. What is her average speed for the whole race?
Physics
1 answer:
vitfil [10]2 years ago
8 0

Answer:

12.31 m/s

Explanation:

If we recall from the previous knowledge we had about speed,

we will know that:

speed = distance/ time.

As such:

The average speed of the rider bicycle is

average speed = total distance/ total time

Mathematically, it can be computed as:

v_{avg} = \dfrac{d+d}{\dfrac{d}{v_1}+ \dfrac{d}{v_2}}

v_{avg} = \dfrac{2d}{\dfrac{d}{10 \ m/s}+ \dfrac{d}{16 \ m/s}}

v_{avg} = \dfrac{2}{\dfrac{1}{10 \ m/s}+ \dfrac{1}{16 \ m/s}}

v_{avg} = \dfrac{2}{\dfrac{13}{80 \ m/s}}

\mathbf{v_{avg} =12.31 \ m/s}

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Answer:

Suppose the micrometeoroid weighed 1 g = .001 kg

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Usually, the smaller particle would be moving but for simplicity suppose that it were stationary wrt the ground

v = 18000 miles / hr * 1500 m/mile / 3600 sec/hr = 7500 m/s

KE = 1/2 * .001 kg * (7500 m)^2 = 28,125 Joules

One can see that 28000 Joules could be damaging amount of energy

7 0
2 years ago
A small package is attached to a helium-filled balloon rising at 2 m/s. The package drops from the balloon when it is 14 meters
scZoUnD [109]
I would assume air resistance is negligible and so the acceleration of the package would be approximately 9.81 m/s².

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4 0
3 years ago
A box of mass 14 kg sits on an inclined surface with an angle of 52degrees. What is the component of the weight of the box along
kolezko [41]
First, we calculate for the weight of the object by multiplying the given mass by the acceleration due to gravity which is equal to 9.8 m/s²
                       Weight = (14 kg)(9.8 m/s²)
                          Weight = 137.2 N
The component of the weight that is along the surface of the inclined plane is equal to this weight times the sine of the given angle. 
                         Weight = (137.2 N)(sin 52°)
                               weight = 108.1 N
5 0
3 years ago
Read 2 more answers
PLEASEEEEEE HEEEEEEEEEEEEEEELPPPPPPPPPPPPPPPPP I NED HELP WHIT THIS!!!!! D:
aleksley [76]
The meters per second
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6 0
2 years ago
Three point charges, two positive and one negative, each having a magnitude of 20 C are placed at the vertices of an equilateral
Daniel [21]

The resultant force on the positive charge  is mathematically given as

X=40N

<h3>What is the magnitude of the electrostatic force on the negative charge?</h3>

Question Parameters:

Three-point charges, two positive and one negative, each having a magnitude of 20

Generally, the -ve charge   is mathematically given as

Q+=\sqrt{x^2+x^2+2x.xcos120}\\\\Q+=\sqrt{2x^2+2x*(1/2)}

Q+=X

Therefore

x=\frac{Kq1q2}{r2}\\\\x=\frac{9*10^9*20*10^{-6}*20*10^{-6}}{(30*10^-2)^2}

X=40N

For more information on Force

brainly.com/question/26115859

5 0
2 years ago
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