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3 years ago

1. The picture below shows Jamal pushing with a 100-Newton (N) force on a large box. Neither

Physics

2 answers:

posledela3 years ago

5 0

Answer:

<h2>100 newtons</h2>

Explanation:

r-ruslan [8.4K]3 years ago

4 0

Answer:

100 newtons

Explanation:

Given,

Jamal pushing a large box by a force, F = 100 N

Work done on the large box is, W = 0

It is because the applied force is less than the force of the friction between the two surfaces.

Yet, there will be a force that is exerted by the large box on Jamal.

According to newton's third law of motion, every action has an equal and opposite reaction. The reaction force is in the direction opposite to the force of action. But, their magnitude remains the same.

$F_{a} =-F_{r}$

Hence, If the action force is 100 N, then the reaction force should be in 100 N

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3 years ago

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Ray creates an energy transfer diagram for a hair dryer. However, the diagram contains an error that could be corrected in sever

Tresset [83]

If the energy transfer diagram created by Ray has error in it, it can be corrected by applying the principle of energy conservation.

<h3>What is energy diagram?</h3>

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4 0

2 years ago

Please help with this problem. Thank you.

4vir4ik [10]

Answer:

Explanation:

F = mω²R

F = 15(2π/8.5)²(7.8)

F = 63.93044788...

F = 63.9 N

answer a) is the closest. No idea how they got a value that low unless they used a poor approximation for π.

4 0

2 years ago

A charge of 50 µC is placed on the y axis at y = 3.0 cm and a 77-µC charge is placed on the x axis at x = 4.0 cm. If both charge

zheka24 [161]

Answer:

The acceleration of an electron is $1.2\times10^{20}\ m/s^2$

Explanation:

Given that,

One Charge = 50 μC

Distance on y axis = 3.0 cm

Second charge = 77 μC

Distance on x axis = 4.0 cm

We need to calculate the force on electron due to q₁

Using formula of force

$F_{1}=\dfrac{kq_{1}q}{r^2}$

Here, q = charge of electron

Put the value into the formula

$F_{1}=\dfrac{9\times10^{9}\times50\times10^{-6}\times1.6\times10^{-19}}{(3\times10^{-2})^2}$

$F_{1}=8\times10^{-11}j\ \ N$

We need to calculate the force on electron due to q₂

Using formula of force

$F_{2}=\dfrac{kq_{2}q}{r^2}$

Here, q = charge of electron

Put the value into the formula

$F_{2}=\dfrac{9\times10^{9}\times77\times10^{-6}\times1.6\times10^{-19}}{(4\times10^{-2})^2}$

$F_{2}=6.93\times10^{-11}i\ \ N$

We need to calculate the net force

Using formula of net force

$F=F_{1}+F_{2}$

Put the value into the formula

$F=8\times10^{-11}j+6.93\times10^{-11}i$

The magnitude of the net force

$F=\sqrt{(8\times10^{-11})^2+(6.93\times10^{-11})^2}$

$F=1.058\times10^{-10}\ N$

We need to calculate the acceleration of an electron

Using formula of force

$F = ma$

$a=\dfrac{F}{m}$

Put the value into the formula

$a=\dfrac{1.058\times10^{-10}}{9.1\times10^{-31}}$

$a=1.2\times10^{20}\ m/s^2$

Hence, The acceleration of an electron is $1.2\times10^{20}\ m/s^2$

3 0

3 years ago