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Dafna11 [192]
3 years ago
9

A charge of 4.4 10-6 C is located in a uniform electric field of intensity 3.9 105 N/C. How much work (in Joules) is required to

move this charge 50 cm along a path making an angle of 40
Physics
1 answer:
schepotkina [342]3 years ago
4 0

Answer:

The work required to move this charge is 0.657 J

Explanation:

Given;

magnitude of charge, q = 4.4 x 10⁻⁶ C

Electric field strength, E =  3.9 x 10⁵ N/C

distance moved by the charge, d = 50 cm = 0.5m

angle of the path, θ = 40°

Work done is given as;

W = Fd

W = FdCosθ

where;

F is the force on the charge;

According the coulomb's law;

F = Eq

F = 3.9 x 10⁵ x 4.4 x 10⁻⁶  = 1.716 N

W = FdCosθ

W = 1.716 x 0.5 x Cos40

W = 0.657 J

Therefore, the work required to move this charge is 0.657 J

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A 0.2 kg hockey puck moving at 80 m/s is caught by a 40 kg goalie at rest. With what speed does the goalie slide on the slide?
Lady_Fox [76]

Answer:

V = 0.39 m/s

Explanation:

Given that,

Mass of hockey puck, m = 0.2 kg

Mass of goalie = 40 kg

Speed of hockey puck, v = 80 m/s

We need to find the speed with which the goalie slide on the slide. Let V be the speed. Using the conservation of momentum as follows :

mv=(M+m)V\\\\V=\dfrac{mv}{M+m}\\\\V=\dfrac{0.2\times 80}{0.2+40}\\\\V=0.39\ m/s

So, the required speed is 0.39 m/s.

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2 years ago
10. A satellites is in a circular orbit around the earth at a height of 360 km above the earth’s surface. What is its time perio
Afina-wow [57]

Answer:

Orbital speed=8102.39m/s

Time period=2935.98seconds

Explanation:

For the satellite to be in a stable orbit at a height, h, its centripetal acceleration V2R+h must equal the acceleration due to gravity at that distance from the center of the earth g(R2(R+h)2)

V2R+h=g(R2(R+h)2)

V=√g(R2R+h)

V= sqrt(9.8 × (6371000)^2/(6371000+360000)

V= sqrt(9.8× (4.059×10^13/6731000)

V=sqrt(65648789.18)

V= 8102.39m/s

Time period ,T= sqrt(4× pi×R^3)/(G× Mcentral)

T= sqrt(4×3.142×(6.47×10^6)^3/(6.673×10^-11)×(5.98×10^24)

T=sqrt(3.40×10^21)/ (3.99×10^14)

T= sqrt(0.862×10^7)

T= 2935.98seconds

4 0
2 years ago
A truck with 0.410 m radius tires travels at 25.0 m/s. What is the angular velocity of the rotating tires in radians per second?
Eduardwww [97]

relation between linear velocity and angular velocity is given as

v = R\omega

here

v = linear speed

R = radius

\omega = angular speed

now plug in all data in the equation

25.0 = 0.410 \omega

\omega  = \frac{25}{0.410}

\omega = 60.9 rad/s

so rotating speed is 60.9 rad/s

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2 years ago
What is the distance from one crest to the next crest?
ella [17]

<span>The distance between wave crests is called wavelength. It is a characteristic shared by waves of all kinds, including ocean waves and sound waves. Wavelength is measured from the highest point, or summit, of one wave's crest to the summit of the next wave's <span>crest</span></span>

<span><span>hope this helps</span></span>

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2 years ago
A test piolot flies with an acceleration of 5
Colt1911 [192]

On Earth, 1 g = 9.8 m/s² .

5 g = 5 · (9.8 m/s²)

5 g = 49 m/s²

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