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ExtremeBDS [4]
2 years ago
6

Suppose you want to calculate how much work it

Physics
1 answer:
Jobisdone [24]2 years ago
6 0

Mass is not necessary the main thing we need is Displacement

As

  • W=FD

Or

  • Work done=Force×Displacement

The amount of work done is calculated as the force exerted to remove a object to D distance .

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Outside a spherically symmetric charge distribution of net charge Q, Gauss's law can be used to show that the electric field at
Sergio039 [100]

Answer:

Q at the center of the distribution.

Explanation:

  • The Gauss's law is the law that relates to the distribution of electrical charges to the resulting electrical field. It states that a flux of electricity outside the arabatory closed surface is proportional to the electricitical harg enclosed by the surface.
3 0
2 years ago
1. A truck with a mass of 8, 000 kg is traveling at 26.8 m/s when it hits the brakes. A.)What is the momentum of the truck befor
NikAS [45]

Answer:

1. A.) The moment of the truck before it hits the brakes is 214,400 kg·m/s

B.) The force it takes to stop the truck is approximately 17,290.4 N

Explanation:

1. A.) The given parameters are;

The mass of the truck, m = 8,000 kg

The velocity of the truck when it hits the brakes, u = 26.8 m/s

Momentum = Mass × Velocity

The moment of the truck = The mass of the truck × The velocity of the truck

Therefore;

The moment of the truck before it hits the brakes = 8,000 kg × 26.8 m/s = 214,400 kg·m/s

B.) The amount of momentum lost when the truck comes to a stop = The initial momentum of the truck

The time it takes the truck to come to a complete stop, t = 12.4 s

The deceleration, "a" of the truck is given by the following kinematic equation of motion

v = u - a·t

Where;

v = The final velocity of the truck = 0 m/s

u = The initial velocity = 26.8 m/s

a = the deceleration of the truck

t = The time of deceleration of the truck = 12.4 s

Substituting the known values gives;

0 = 26.8 - a × 12.4

Therefore;

26.8 = a × 12.4

a = 26.8/12.4 ≈ 2.1613

The deceleration (negative acceleration) of the truck, a ≈ 2.1613 m/s²

Force = Mass × Acceleration

The force required to stop the truck = The mass pf the truck × The deceleration (negative acceleration) given to the truck

∴ The force it takes to stop the truck = 8,000 kg × 2.1613 m/s² ≈ 17,290.4 N.

8 0
3 years ago
A 150 kg uniform beam is attached to a vertical wall at one end and is supported by a cable at the other end. Calculate the magn
xenn [34]

Answer:

T = 2010 N

Explanation:

m = mass of the uniform beam = 150 kg

Force of gravity acting on the beam at its center is given as

W = mg

W = 150 x 9.8

W = 1470 N

T = Tension force in the wire

θ = angle made by the wire with the horizontal =  47° deg

L = length of the beam

From the figure,

AC = L

BC = L/2

From the figure, using equilibrium of torque about point C

T (AC) Sin47 = W (BC)

T L Sin47 = W (L/2)

T Sin47 = W/2

T Sin47 = 1470

T = 2010 N

6 0
3 years ago
Which has more KE, a large truck moving at 50m/s or a bicycle moving at 10m/s?
klio [65]
The truck has more KE than the bike
8 0
3 years ago
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A bond between two copper atoms would be a<br> bond? *
mihalych1998 [28]
A metallic bond would be formed
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2 years ago
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