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3 years ago

A car and driver weighing 7130 N passes a sign stating...?

Physics

1 answer:

IgorLugansk [536]3 years ago

4 0

Answer:

321 280 J

Explanation:

Work done = force * distance

The distance is 32 m

The force can be calculated using the second law of motion

F = ma = (7130 N ÷ 9.8 m/s²) * 13.8 m/s² = 10 040 N

Work done = force * distance

= 10 040 N * 32 m

= 321 280 J

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A helicopter starting on the ground is rising directly into the air at a rate of 25 ft/s. You are running on the ground starting

rusak2 [61]

Answer:

The rate of change of the distance between the helicopter and yourself (in ft/s) after 5 s is $\sqrt{725}$ ft/ sec

Explanation:

Given:

h(t) = 25 ft/sec

x(t) = 10 ft/ sec

h(5) = 25 ft/sec . 5 = 125 ft

x(5) = 10 ft/sec . 5 = 50 ft

Now we can calculate the distance between the person and the helicopter by using the Pythagorean theorem

$D(t) = \sqrt{h^2 + x^2}$

Lets find the derivative of distance with respect to time

$\frac{dD}{dt} (t) = \frac{2h \cdot \frac{dh}{dt} +2x \cdot\frac{dx}{dt}} {2\sqrt{h^2 + x^2}}$

Substituting the values of h(t) and x(t) and simplifying we get,

$\frac{dD}{dt}(t) = \frac{50t \cdot \frac{dh}{dt} + 20 \cdot \frac{dx}dt}{2\sqrt{625\cdot t^2 + 100 \cdot t^2}}$

$\frac{dh}{dt} = 25ft/sec$

$\frac{dx}{dt} = 10 ft/sec$

$\frac{Dd}{dt} (t) = \frac{1250t +200t}{2\sqrt{725}t}$ = $\frac{725}{\sqrt{725}}$ = $\sqrt{725}$ ft / sec

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2 years ago

Compare the catching of two different water balloons. Case A: A 600-mL water balloon moving at 8 m/s is caught and brought to a

Vadim26 [7]

Answer:

jduruhrjrhryyrurjjrjrh4h5ghrrhrhhrhrhrhrrhrhrh sus

Explanation:

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8 0

2 years ago

Read 2 more answers

What is the angle of refraction when a ray of light passes frm ethanol to air

77julia77 [94]

Answer: Your answer is<u> 1.36.</u>

Hope this helps!

5 0

2 years ago

How long does it took for 1000-watt microwave oven to do 20000 joules of work

klio [65]

Answer:

Work = power * time

time = 20000 joules / 1000 joules / sec = 20 sec

4 0

2 years ago

A car slows down uniformly from a speed of 30.0 m/s to rest in 7.20 s

Ede4ka [16]

When acceleration is constant, the average velocity is given by

$\bar v=\dfrac{v+v_0}2$

where $v$ and $v_0$ are the final and initial velocities, respectively. By definition, we also have that the average velocity is given by

$\bar v=\dfrac{\Delta x}{\Delta t}=\dfrac{x-x_0}{t-t_0}$

where $x,x_0$ are the final/initial displacements, and $t,t_0$ are the final/initial times, respectively.

Take the car's starting position to be at $t_0=0\,\mathrm s$ . Then

$\dfrac{v+v_0}2=\dfrac{x-x_0}t\implies x=x_0+\dfrac12(v+v_0)t$

So we have

$x=0\,\mathrm m+\dfrac12\left(0\,\dfrac{\mathrm m}{\mathrm s}+30.0\,\dfrac{\mathrm m}{\mathrm s}\right)(7.20\,\mathrm s)=108\,\mathrm m$

You also could have first found the acceleration using the equation

$v=v_0+at$

then solve for $x$ via

$x=x_0+v_0t+\dfrac12at^2$

but that would have involved a bit more work, and it turns out we didn't need to know the precise value of $a$ anyway.

7 0

2 years ago