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svlad2 [7]
2 years ago
8

a man carries a hand bag by hanging on his hand and moves horizontally where the bag does not move up or down. What is the work

done on the bag? The man gets tired after sometime of the movement. Why
Physics
1 answer:
vlada-n [284]2 years ago
3 0

The work that the man does is the scalar product of the force applied by the man and the horizontal displacement of the bag.

<h3>What is the work done?</h3>

In Physics, we define the work done as the product of the of force and distance. Hence, we generally define the work done as that which occurs when the force applied moves a distance in the direction of the force. This implies that the work done is not a vector but a  scalar quantity.

In relation to the man and the bag, the work done is the product of the force that the man applies and the displacement of the bag. As such, the reason for the tiredness of the man is that the internal energy that he possesses is transferred to moving the bag.

Thus, the work that the man does is the scalar product of the force applied by the man and the horizontal displacement of the bag.

Learn more about work done:brainly.com/question/13662169

#SPJ1

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Answer:

explained

Explanation:

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The intensity produces more electron but does not change the maximum kinetic energy of electrons.

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7 0
3 years ago
A velocity selector in a mass spectrometer uses a 0.150 T magnetic field. (a) What electric field strength (in volts per meter)
Alekssandra [29.7K]

Answer:

The electric field strength is 6.6\times10^{5}\ V/m

Explanation:

Given that,

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Using formula of velocity

v=\dfrac{E}{B}

E=v\times B

Where, v = speed

B = magnetic field

Put the value into the formula

E=4.40\times10^{6}\times0.150

E=660000\ V/m

E=6.6\times10^{5}\ V/m

Hence, The electric field strength is 6.6\times10^{5}\ V/m

4 0
3 years ago
Name of a body that changes Chemical energy to electric energy.​
Ratling [72]

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022 (part 1 of 4) 10.0 points A ball is thrown vertically upward with a speed of 24.5 m/s. How high does it rise? The accelerati
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1)

Answer:

Part 1)

H = 30.6 m

Part 2)

t = 2.5 s

Part 3)

t = 2.5 s

Part 4)

v_f = 24.5 m/s

Explanation:

Part 1)

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v_i = 24.5 m/s

so maximum height of the ball is given by

H = \frac{v_i^2}{2g}

H = \frac{24.5^2}{2(9.80)}

H = 30.6 m

Part 2)

As we know that final speed will be zero at maximum height

so we will have

v_f - v_i = at

0 - 24.5 = (-9.8)t

t = 2.5 s

Part 3)

Since the time of ascent of ball is same as time of decent of the ball

so here ball will same time to hit the ground back

so here it is given as

t = 2.5 s

Part 4)

since the acceleration due to earth will be same during its return path as well as the time of the motion is also same

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v_f = 24.5 m/s

2)

Answer:

a = 9.76 m/s/s

Explanation:

As we know that the object is released from rest

so the displacement of the object in vertical direction is given as

y = \frac{1}{2}at^2

4.88 = \frac{1}{2}a(1^2)

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Answer:

v = 29.7 m/s

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acceleration of the rocket is given as

a = 90 m/s^2

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Part 1)

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d = 6.72 m

Explanation:

Part 1)

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y = 25.95 m

Part 2)

Distance traveled by it in horizontal direction is given as

d = v_x t

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