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worty [1.4K]
3 years ago
11

Suppose a person whose mass is m is being held up against the wall with a constant tangential velocity v greater than the minimu

m necessary. Find the magnitude of the frictional force between the person and the wall
Physics
1 answer:
Likurg_2 [28]3 years ago
7 0

Answer:

Explanation:

The man is moving on a circular path . The reaction of wall is providing centripetal force .so

R = mv² / r

Frictional force = μR where  μ is coefficient of friction

= μ x mv² / r

Frictional force = μ x mv² / r

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Alana is skateboarding at 19 km/h and throws a tennis ball at 11 km/h to her friend Oliver who is behind her leaning against a w
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The speed of the tennis ball is 8 km/h when Alana throws the ball behind at the speed of 11 km/h.

<h3>What is the frame of reference?</h3>

It is a point from where the motion is observed. The speed changes with the frame of reference.

For the person standstill on the sidewalk, If  Alana throw the ball in the front direction, The speed will be,

19+11 = 30 km/h

It Alana just leave the ball, the speed,

19 +0 = 19 km/h

If Alana throws the ball behind her at the speed of 19 km/h. The speed of the ball for a sidewalk observer,

19 +(-19 ) =  0 km/h

Thus, when Aana throw the ball behind the velocity will be negative,

So,

19 + (-11 ) = 8 km/h

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Learn more about Frame of reference:

brainly.com/question/9820962

8 0
2 years ago
After driving a portion of the route, the taptap is fully loaded with a total of 25 people including the driver, with an average
alexira [117]

Answer:

Incomplete Question

Explanation:

String constant is not stated

I'll solve this in a general way.

At the end of my explanation, I'll assume any value for string constant.

The total weight on the taptap : Weight of people + Weight of goats + Weight of chickens + weight of banana

= (25 * 66) + (3 * 15) + (5 * 3) + 25 = 1735 kg

The gravitational force exerted on the taptap

Fig = m*g

where m = the total mass 1735kg

and g = 9.8m/s²

Hooke's law of Elasticity states that

Fs = Kx

Where Fs = spring force

k = spring constant

x = spring stretch or compression

Since Fs = Fg

So, Kx = mg

Make x the subject of the formula

x = mg/k --------------- This is the general way of solving for spring compressor in cases like this

I'll replace m and g with their values, respectively.

Because, the spring constant (k) is omitted in the question, I'll assume it to be 30000

x = 1735 * 9.8/30000

x = 17003/30000

x = 0.566767 metres

x = 0.57 metres ------------- Approximated

In conclusion, the spring compression is 0.57 metres if the spring constant is 30000

6 0
3 years ago
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