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ELEN [110]
2 years ago
11

A woman pushes a 3.2kg textbook on a table as she walks forward 3.5 meters. How much work does the vertical component of the for

ce from her hand do on the textbook?in j
Physics
1 answer:
Effectus [21]2 years ago
4 0

Answer:

The done by the vertical component of force is zero.

Explanation:

Given data,

The mass of the textbook the women pushes horizontally, m = 3.2 kg

The displacement of the textbook, S = 3.5 m

Let the force of the women acts on the book horizontally,

Therefore, the horizontal component of force is maximum and the vertical component is zero.

If  F is the force applied by the women, then the horizontal and vertical component of the force is,

                                         F_{x}=F Cos\theta

                                         F_{y}=F Sin\theta

Since the force is acting along with the horizontal x component, the vertical component of the force is zero.

Hence, the done by the vertical component of force is zero.

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Boyle's Law mainly involves _______.
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A basketball with a mass of 20 kg is accelerated with a force of 10 N. If resisting forces are ignored, what is the acceleration
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I’m pretty sure it would be 10/20= 0.5m/s2
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What is the displacement of the particle in the time interval 7 seconds to 8 seconds? A. 0 meters B. 1.5 meters C. 3 meters D. 7
Dmitrij [34]

Answer: B. 1.5 meters

Explanation:

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2 years ago
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A ball is released from the top of a tower of height h meters. it takes T seconds to reach the ground . what is the position of
alekssr [168]

Answer:the

8/9 h

Explanation:

Height  =   1/2 a T^2     now change to T/3

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3 0
2 years ago
A particle of charge 2.0 x 10^-8C experiences an upward force of magnitude 4.0 x10^-6 when it is placed in a particular point in
koban [17]

Answer:

a) The electric field at that point is 2.0\times 10^{2} newtons per coulomb.

b) The electric force is 2.0\times 10^{-6} newtons.

Explanation:

a) Let suppose that electric field is uniform, then the following electric field can be applied:

E = \frac{F_{e}}{q} (1)

Where:

E - Electric field, measured in newtons per coulomb.

F_{e} - Electric force, measured in newtons.

q - Electric charge, measured in coulombs.

If we know that F_{e} = 4.0\times 10^{-6}\,N and q = 2.0\times 10^{-8}\,C, then the electric field at that point is:

E = \frac{4.0\times 10^{-6}\,N}{2.0\times 10^{-8}\,C}

E = 2.0\times 10^{2}\,\frac{N}{C}

The electric field at that point is 2.0\times 10^{2} newtons per coulomb.

b) If we know that E = 2.0\times 10^{2}\,\frac{N}{C} and q = 1.0\times 10^{-8}\,C, then the electric force is:

F_{e} = E\cdot q

F_{e} = \left(2.0\times 10^{2}\,\frac{N}{C} \right)\cdot (1.0\times 10^{-8}\,C)

F_{e} = 2.0\times 10^{-6}\,N

The electric force is 2.0\times 10^{-6} newtons.

7 0
3 years ago
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