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777dan777 [17]
3 years ago
15

A car of mass m accelerates from speed v1 to speed v2 while going up a slope that makes an angle θ with the horizontal. The coef

ficient of static friction is μs, and the acceleration due to gravity is g. Find the total work W done on the car by the external forces. Express your answer in terms of the given quantities. You may or may not use all of them.
Physics
1 answer:
nlexa [21]3 years ago
7 0

Answer:

The answer is W=\frac{1}{2} m (v_2^{2} -v_1^{2} )

Explanation:

Here you have to take into account the theorem of work and energy.  This theorem says that the total work done by external forces on a body is used to modify the kinetic energy. So the only thing you have to do is determine the kinetic energy in the initial (K_1) and final moment (K_2), then the diference between them is the amount of net work that act over the body.

K_1=\frac{1}{2} m v_1^{2}

K_2=\frac{1}{2} m v_2^{2}

W=K_2-K_1=\frac{1}{2} m v_2^{2} - \frac{1}{2} m v_1^{2}=\frac{1}{2} m (v_2^{2} -v_1^{2} )

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<span>Waves that move matter back and forth are called                 a.transverse waves  <u>b.longitudinal wave</u>     c. Medium wave</span>
3 0
3 years ago
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How much does a 0.15 kg baseball weigh on earth?
Gnesinka [82]
1.472 N

to get weight you multiply an object's mass in kilograms with the acceleration of gravity(9.81m/s) :)
7 0
3 years ago
What would the weight of an astronaut be on Saturn if his mass is 68 kg and acceleration of gravity on Saturn is 10.44 m/s2? Ple
alex41 [277]

Here's the part you need to know:

       (Weight of anything) =

                 (the thing's mass)
times
                 (acceleration of gravity in the place where the thing is) .

                 Weight = (mass ) x (gravity) .

That's always true everywhere.
You should memorize it.

For the astronaut on Saturn . . .

                   Weight = (mass ) x (gravity) .

                
  Weight =  (68 kg) x (10.44 m/s²)

                        
      =    709.92 newtons .
__________________________________

On Earth, gravity is only  9.8 m/s².
So as long as the astronaut is on Earth, his weight is only

                                   (68 kg) x (9.8 m/s²)

                               =    666.4 newtons .

Notice that his mass is his mass ... it doesn't change
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But his weight changes in different places, because
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4 0
3 years ago
What will the stopping distance be for a 2,000-kg car if -2,000 N of force are applied when the car is traveling 20 m/s?
astraxan [27]

Answer is B- 200 m

Given:

m (mass of the car) = 2000 Kg

F = -2000 N

u(initial velocity)= 20 m/s.

v(final velocity)= 0.

Now we know that

<u>F= ma</u>

Where F is the force exerted on the object

m is the mass of the object

a is the acceleration of the object

Substituting the given values

-2000 = 2000 × a

a = -1 m/s∧2

Consider the equation

<u>v=u +at</u>

where v is the initial velocity

u is the initial velocity

a is the acceleration

t is the time

0= 20 -t

t=20 secs


s = ut +1/2(at∧2)

where s is the displacement of the object

u is the initial velocity

t is the time

v is the final velocity

a is the acceleration

s= 20 ×20 +(-1×20×20)/2

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3 0
3 years ago
Read 2 more answers
a projectile is fired in such away that its horizontal range is equal to three times its naximum height.what is the angle of pro
finlep [7]

Answer:

\theta=53.13^o

Explanation:

<u>2-D Projectile Motion</u>

In 2-D motion, there are two separate components of the acceleration, velocity and displacement. The horizontal component has zero acceleration, while the acceleration in the vertical direction is always the acceleration due to gravity. The basic formulas for this type of movement are

V_x=V_{ox}=V_ocos\theta

V_y=V_{oy}-gt=V_osin\theta-gt

x=V_{ox}t

\displaystyle y=y_o+V_{oy}t-\frac{gt^2}{2}

\displaystyle x_{max}=\frac{2V_{ox}V_{oy}}{g}

\displaystyle y_{max}=\frac{V_{oy}^2}{2g}

The projectile is fired in such a way that its horizontal range is equal to three times its maximum height. We need to find the angle \theta at which the object should be launched. The range is the maximum horizontal distance reached by the projectile, so we establish the base condition:

x_{max}=3y_{max}

\displaystyle \frac{2V_{ox}V_{oy}}{g}=3\frac{V_{oy}^2}{2g}

Using the formulas for V_{ox}, V_{oy}:

\displaystyle \frac{2V_{o}cos\theta V_{o}sin\theta}{g}=3\frac{V_{o}^2sin^2\theta}{2g}

Simplifying

4cos\theta sin\theta=3sin^2\theta

Dividing by sin\theta

4cos\theta=3sin\theta

Rearranging

tan\theta=\frac{4}{3}

\theta=arctan\frac{4}{3}

\theta=53.13^o

4 0
3 years ago
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