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3 years ago

A person of weight mg, is standing on a scale in an elevator. What will the

Physics

1 answer:

sergejj [24]3 years ago

8 0

Answer:

B. less than mg

Explanation:

A person's weight changes as he rides an elevator at a constant velocity. If the elevator is moving downward, the person's weight becomes lesser. He is considered to be in "free fall" due to gravity. However, the opposite happens when the elevator goes upward. The person will feel heavier because he is going against gravity. This means his weight is more than his actual weight. If the person will stand on an elevator that doesn't move, his weight will just be the same. All of these is due to Newton's Second Law of Motion where an object's acceleration relies on its force and mass.

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What is the momentum of a 546,540 kg train that is travelling at 7.8 m/s

lara [203]

p=mv so wouldn't u multiply them?

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3 years ago

A 0.07 kg tennis ball, initially at rest, leaves a racket with a speed of 56 m/s. If the ball is in contact with

Naddika [18.5K]

The average force on the ball by the racket is 98 N. The correct option is the third option - 98 N

From the question, we are to determine the average force on the ball by the racket.

From the formula,

$F = \frac{mv}{t}$

Where F is the force

m is the mass

v is the velocity

and t is the time

From the given information

m = 0.07 kg

v = 56 m/s

t = 0.04 s

Putting the parameters into the formula,

we get

$F = \frac{0.07 \times 56}{0.04}$

$F = \frac{3.92}{0.04}$

F = 98 N

Hence, the average force on the ball by the racket is 98 N. The correct option is the third option - 98 N

Learn more on calculating force exerted on an object here: brainly.com/question/13590154

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3 years ago

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svetlana [45]

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3 years ago

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Identify the temperature that is equivalent to 95°F. Use this formula to convert the temperature

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4 years ago

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When cars are equipped with flexible bumpers, they will bounce off each other during low-speed collisions, thus causing less dam

Len [333]

Answer:

Explanation:

Given that,

Mass of the heavier car m_1 = 1750 kg

Mass of the lighter car m_2 = 1350 kg

The speed of the lighter car just after collision can be represented as follows

$m_1u_1+m_2u_2=m_1v_1+m_2v_2\\\\v_2=\frac{m_1u_1+m_2u_2-m_1v_1}{m_2}$

$v_2=\frac{(1850)(1.4)+(1450)(-1.10)-(1850)(0.250)}{1450} \\\\=\frac{2590+(-1595)-(462.5)}{1450} \\\\=\frac{2590-1595-462.5}{1450} \\\\=\frac{532.5}{1450}\\\\=0.367m/s$

b) the change in the combined kinetic energy of the two-car system during this collision

$\Delta K.E=(\frac{1}{2} m_1v_1^2+\frac{1}{2} m_2v_2^2)-(\frac{1}{2} m_1u_1^2+\frac{1}{2} m_2u_2^2)\\\\=\frac{1}{2} (m_1(v_1^2-u_1^2)+m_2(v_2^2-u_2^2))$

substitute the value in the equation above

$=\frac{1}{2} (1850((0.250)^2-(1.4)^2)+(1450((0.3670)^2-(-1.10)^2)\\\\=\frac{1}{2}(11850(0.0625-1.96)+(1450(0.1347)-(1.21))\\\\= \frac{1}{2}(11850(-1.8975))+(1450(-1.0753))\\\\=\frac{1}{2} (-3510.375+(-1559.185)\\\\=\frac{1}{2} (-5069.56)\\\\=-2534.78J$

Hence, the change in combine kinetic energy is -2534.78J

8 0

3 years ago