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ryzh [129]
3 years ago
12

3. The mass of Aimi is 47kg. She wears a pair of high-heeled shoes, each with a surface area of contact of 1.5cm with the floor.

What is the pressure exerted by Aimi on the floor?
4. The mass of a car is 1000kg. If the area of contact between the tyres and the road surface is 25cm^2,calculate the pressure exerted by each tyre on the surface of the road.​
Physics
1 answer:
Svetach [21]3 years ago
5 0

question 3 is 3133333.33N/m^2 and question 4 is 4000000N^m2

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G a magnetic field perpendicular to the plane of a wire loop is uniform in space but changes with time t in the region of the lo
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e = Δφ / Δt     induced emf is proportional to enclosed flux

Also φ  = B * A      flux is proportional to area and enclosed field

If the induced emf e increases with time than the flux and hence the magnetic field is increasing with time  (replace B with G)

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What is the difference between revolution and rotation
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3 years ago
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A knight moves on a chessboard two squares up, down, left, or right followedby one square in one of the two directions perpendic
Contact [7]

Answer:

Explanation:

Check attachment for solution.

Generally the movement of the knight is L, i.e (2,1), (1,2),(-1,2),(1,-2) etc.

So using Pythagoras theorem

x^2+y^2=1^2+2^2

x^2+y^2=5

Then, the knight will make one movement when the displacement is √5.

So let take a look at other positions

(1,0),(0,1),(-1,0), (0,-1).

Then, for us to have this kind of movement, the knight has to make 3 movements.

When the displacement is 1, then it will make 3 movement.

Let examine other positions

(2,2) or(-2,-2) or (2,-2) or (-2,2)

When the displacement is √8

Then, the movement of the knight is 4.

Let examine other points

(2,0) or (0,2) or (-2,0) or (0,-2)

When the displacement is 2.

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5 0
3 years ago
If a 0.50-kg block initially at rest on a frictionless, horizontal surface is acted upon by a force of 3.7 N for a distance of 6
mel-nik [20]

The block's velocity is determined as 10.03 m/s.

<u>Explanation:</u>

According to work energy theorem, the work done on an object is equal to the change in kinetic energy of the object.

So, work done = Kinetic energy

\text {Force } \times \text { displacement }=\frac{1}{2} \times m \times v^{2}

Thus, the velocity can be determined as

\text {velocity}=\sqrt{\frac{2 \times \text {Force} \times \text {displacement}}{m}}

\text {velocity}=\sqrt{\frac{2 \times 3.7 \times 6.8}{0.50}}=\sqrt{\frac{50.32}{0.50}}=\sqrt{100.64}

Velocity = 10.03 m/s.

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How do you add this vector graphically
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