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STatiana [176]
3 years ago
13

a sprinter with a mass of 80kg accelerates uniformly from 0 m/s to 9 m/s in 3 s. a.)what is the runners acceleration? b.) what i

s the net force on the runner? c.) how far does the sprinter travel during the 3 s?
Physics
1 answer:
Katyanochek1 [597]3 years ago
3 0
Part A: 
Acceleration can be calculated by dividing the difference of the initial and final velocities by the given time. That is,
             a = (Vf - Vi) / t
where a is acceleration,
Vf is final velocity,
Vi is initial velocity, and
t is time

Substituting,
            a = (9 m/s - 0 m/s) / 3 s = 3 m/s²
<em>ANSWER: 3 m/s²</em>

Part B: 
From Newton's second law of motion, the net force is equal to the product of the mass and acceleration,
               F = m x a
where F is force,
m is mass, and 
a is acceleration

Substituting,
              F = (80 kg) x (3 m/s²) = 240 kg m/s² = 240 N

<em>ANSWER: 240 N </em>

Part C: 
The distance that the sprinter travel is calculated through the equation,
         d = V₀t + 0.5at²

Substituting,
            d = (0 m/s)(3 s) + 0.5(3 m/s²)(3 s)²
             d = 13.5 m

<em>ANSWER: d = 13.5 m</em>

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Explanation:

Given that,

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V=\dfrac{12}{0.0001}

V=12\times 10^4\ Volt

(b) Similarly, same electric field does 24 J of work on a 0.0002-C charge. The electric potential difference is given by :

V=\dfrac{W}{q}

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3 years ago
I need help with this
fredd [130]
We have here what is known as parallel combination of resistors.

Using the relation:

\frac{1}{ r_{eff} } = \frac{1}{ r_{1} } + \frac{1}{ r_{2} } + \frac{1}{ r_{3} }.. . + \frac{1}{ r_{n} } \\
And then we can turn take the inverse to get the effective resistance.

Where r is the magnitude of the resistance offered by each resistor.

In this case we have,
(every term has an mho in the end)
\frac{1}{10000} + \frac{1}{2000} + \frac{1}{1000} \\ \\ = \frac{1}{1000} ( \frac{1}{10} + \frac{1}{2} + \frac{1}{1} ) \\ \\ = \frac{1}{1000} ( \frac{31}{20}) \\ \\ = \frac{31}{20000}

To ger effective resistance take the inverse:
we get,
\frac{20000}{31} \: ohm \\ = 645 .16 \: ohm

The potential difference is of 9V.

So the current flowing using ohm's law,

V = IR

will be, 0.0139 Amperes.
7 0
3 years ago
A sinusoidal wave of angular frequency 1,203 rad/s and amplitude 3.1 mm is sent along a cord with linear density 3.9 g/m and ten
kobusy [5.1K]

Answer:

18.7842493212 W

Explanation:

T = Tension = 1871 N

\mu = Linear density = 3.9 g/m

y = Amplitude = 3.1 mm

\omega = Angular frequency = 1203 rad/s

Average rate of energy transfer is given by

P=\dfrac{1}{2}\sqrt{T\mu}\omega^2y^2\\\Rightarrow P=\dfrac{1}{2}\sqrt{1871\times 3.9\times 10^{-3}}\times 1203^2\times (3.1\times 10^{-3})^2\\\Rightarrow P=18.7842493212\ W

The average rate at which energy is transported by the wave to the opposite end of the cord is 18.7842493212 W

7 0
3 years ago
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