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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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An object initially at rest experiences an acceleration
lidiya [134]

Answer:

1.4m/s

Explanation:

Average velocity is the total distance covered divided by the total time taken.

 Average velocity  = \frac{total distance }{time }  

 Total time taken  = 5s + 6s  = 11s

The first distance covered  = velocity x time  = 1.4 x 5 = 7m

     second distance covered  = velocity x time  = 1.4 x 6  = 8.4m

So;

  Average velocity  = \frac{7 + 8.4}{11}    = 1.4m/s

5 0
3 years ago
If you weigh 690 N on the earth, what would be your weight on the surface of a neutron star that has the same mass as our sun an
lana [24]

Answer:

Explanation:

Given that,

Mass of star M(star) = 1.99×10^30kg

Gravitational constant G

G = 6.67×10^−11 N⋅m²/kg²

Diameter d = 25km

d = 25,000m

R = d/2 = 25,000/2

R = 12,500m

Weight w = 690N

Then, the person mass which is constant can be determined using

W =mg

m = W/g

m = 690/9.81

m = 70.34kg

The acceleration due to gravity on the surface of the neutron star is can be determined using

g(star) = GM(star)/R²

g(star) = 6.67×10^-11 × 1.99×10^30 / 12500²

g (star) = 8.49 × 10¹¹ m/s²

Then, the person weight on neutron star is

W = mg

Mass is constant, m = 70.34kg

W = 70.34 × 8.49 × 10¹¹

W = 5.98 × 10¹³ N

The weight of the person on neutron star is 5.98 × 10¹³ N

5 0
3 years ago
what distance is a book from the floor if the book contains 196 joules f potential energy and has a mass of 5 kg?
AleksAgata [21]
E=mgh.   196=5kg*9.81m/s^2*h.  So h=196/(5*9.81)=4m
5 0
3 years ago
At a certain instant, coil A is in a 10-T external magnetic field and coil B is in a 1-T external magnetic field. Both coils hav
Neporo4naja [7]

Answer:

A) coil A

Explanation:

According to Faraday, Induced emf is given as;

E.M.F = ΔФ/t

ΔФ = BACosθ

where;

ΔФ  is change in magnetic flux

θ is the angle between the magnetic field, B, and the normal to the loop of area A

A is the area of the loop

B is the magnetic field

From the equation above, induced emf depends on the strength of the magnetic field.

Both coils have the same area and are oriented at right angles to the field.

Coil A has a magnetic field strength of 10-T which is greater than 1 T of coil B, thus, coil A will have a greater emf induced in it.

7 0
3 years ago
Early black-and-white television sets used an electron beam to draw a picture on the screen. The electrons in the beam were acce
oee [108]

Answer:

3.25 × 10^7 m/s

Explanation:

Assuming the electrons start from rest, their final kinetic energy is equal to the electric potential energy lost while moving through the potential difference (ΔV)

Ek = 1/2 mv2 = qΔV .................. 1

Given that V is the electron speed in m/s

Charge of electron = 1.60217662 × 10-19 coulombs

Mass of electron = 9.109×10−31 kilograms

ΔV = 3.0kV = 3000V

Make V the subject of the formula in eqaution 1

V = sqr root 2qΔV/m

V = 2 × 1.60217662 × 10-19 × 3000 / 9.109×10−31

V = 3.25 × 10^7 m/s

3 0
3 years ago
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