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3 years ago

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A force of 80 N is exerted on an object on a frictionless surface for a distance of 4 meters. If the object has a mass of 10 kg,

calculate its velocity. v =
64 m/s
16 m/s
8 m/s
4 m/s

1 answer:

solmaris [256]3 years ago

6 0

There are two ways to find energy. Energy=F*d=mv^2. We can use this relationship to find v:
$Fd=0.5mv^{2} \\ 80*4=0.5*10v^{2} \\ 64=v^{2} \\ v=8 m/s$

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How much time does it take a person to walk 12km north at the velocity at a velocity of 6.5 km/hrs?

alukav5142 [94]

"6.5 km/hr" is not a velocity. It's just a speed, so
we don't know what direction he's walking.

If he happens to be walking north, then it takes him

(12 km) / (6.5 km/hr) = 1.846... hours (rounded) .

If he's walking in any other direction, it takes him longer than that.

If the angle between north and the direction he's walking is
90 degrees or more, then he can never cover any northward
distance, no matter how long he walks.

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3 years ago

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As a box slides down a ramp, friction does 23.0 joules of work. At the bottom of the ramp, the box has 3.8 joules of kinetic ene

tensa zangetsu [6.8K]

Answer:

The high of the ramp is 2.81[m]

Explanation:

This is a problem where it applies energy conservation, that is part of the potential energy as it descends the block is transformed into kinetic energy.

If the bottom of the ramp is taken as a potential energy reference point, this point will have a potential energy value equal to zero.

We can find the mass of the box using the kinetic energy and the speed of the box at the bottom of the ramp.

$E_{k}=0.5*m*v^{2}\\\\where:\\E_{k}=3.8[J]\\v = 2.8[m/s]\\m=\frac{E_{k}}{0.5*v^{2} } \\m=\frac{3.8}{0.5*2.8^{2} } \\m=0.969[kg]$

Now applying the energy conservation theorem which tells us that the initial kinetic energy plus the work done and the potential energy is equal to the final kinetic energy of the body, we propose the following equation.

$E_{p}+W_{f}=E_{k}\\where:\\E_{p}= potential energy [J]\\W_{f}=23[J]\\E_{k}=3.8[J]\\$

And therefore

$m*g*h + W_{f}=3.8\\ 0.969*9.81*h - 23= 3.8\\h = \frac{23+3.8}{0.969*9.81}\\ h = 2.81[m]$

8 0

3 years ago

How much work is done when I apply 200 N of force to a brick wall

nalin [4]

W=Force*Distance
If you just apply force but do not move the brick, no work has been done.

8 0

3 years ago

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The sound source of a ship’s sonar system operates at a frequency of 22.0 kHz . The speed of sound in water (assumed to be at a

Dafna1 [17]

Answer:

$\Delta f=f_{Lr}-f_{Se}$

147.45 Hz

Explanation:

v = Speed of sound in water = 1482 m/s

$v_w$ = Speed of whale = 4.95 m/s

Frequency of the wave in stationary condition

$f_{Lr}=f\dfrac{v+v_w}{v-v_w}$

Ship's frequency which is reflected back

$f_{Se}=f\dfrac{v}{v-v_w}$

The difference in frequency is given by

$\Delta f=f\dfrac{v+v_w}{v-v_w}-f\dfrac{v}{v-v_w}\\\Rightarrow \Delta f=f_{Lr}-f_{Se}$

$\mathbf{\Delta f=f_{Lr}-f_{Se}}$

$f_{Lr}=22\times \dfrac{1482+4.95}{1482-4.95}\\\Rightarrow f_{Lr}=22.14745\ kHz$

$f_{Se}=22\ kHz$

$\Delta f=f_{Lr}-f_{Se}\\\Rightarrow \Delta f=22.14745-22\\\Rightarrow \Delta f=0.14745\ kHz\\\Rightarrow \Delta f=147.45\ Hz$

The difference in wavelength is 147.45 Hz

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3 years ago

A 591 μF capacitor is discharged through a resistor, whereby its potential difference decreases from its initial value of 88.5 V

Stels [109]

Answer:

2.6 kilo Ohm

Explanation:

Capacitance, C = 591 μF = 591 x 10^-6 F

Vo = 88.5 V

V = 11.9 V

t = 3.09 s

Let the resistance is R.

$V = V_{0}e^{\frac{-t}{RC}}$

$\frac{11.9}{88.5} = e^{\frac{-t}{RC}}$

$0.135 = e^{\frac{-t}{RC}}$

Take natural log on oth the sides

ln 0.135 = - 3.09 / RC

RC = 1.545

R = 1.545 / ( 591 x 10^-6)

R = 2614.2 ohm

R = 2.6 kilo Ohm

Thus the resistance is 2.6 kilo Ohm.

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3 years ago