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3 years ago

Sean applies a force of 100N to move a box 5 meters. How much work did he do?

Physics

1 answer:

Grace [21]3 years ago

6 0

Answer:

500 Joules

Explanation:

W(work)= force * distance

w = 100*5

W= 500

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How much meters is a mile

elena-s [515]

Roughly 1609 meters in one mile

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3 years ago

7. What can be done to prevent such devastating destruction as seen in the earthquake in Haiti? correct me if im wrong on the su

Alex787 [66]

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3 0

3 years ago

A Van de Graaff generator is one of the original particle accelerators and can be used to accelerate charged particles like prot

Rzqust [24]

Answer:

595391.482946 m/s

$3.21875\times 10^{6}$

Explanation:

E = Energy = 1.85 keV

I = Current = 5.15 mA

e = Charge of electron = $1.6\times 10^{-19}\ C$

t = Time taken = 1 second

m = Mass of proton = $1.67\times 10^{-27}\ kg$

Velocity of proton is given by

$v=\sqrt{\dfrac{2E}{m}}\\\Rightarrow v=\sqrt{\dfrac{2\times 1.85\times 10^3\times 1.6\times 10^{-19}}{1.67\times 10^{-27}}}\\\Rightarrow v=595391.482946\ m/s$

The speed of the proton is 595391.482946 m/s

Current is given by

$I=\dfrac{\Delta Q}{t}\\\Rightarrow \Delta Q=It\\\Rightarrow \Delta Q=5.15\times 10^{-3}\times (1\ sec)\\\Rightarrow Q=5.15\times 10^{-3}\ C$

Number of protons is

$n=\dfrac{Q}{e}\\\Rightarrow n=\dfrac{5.15\times 10^{-3}}{1.6\times 10^{-19}}\\\Rightarrow n=3.21875\times 10^{6}\ protons$

The number of protons is $3.21875\times 10^{6}$

3 0

4 years ago

In which of these examples is the greatest movement occurring?

Elenna [48]

Answer:

not clear pic...but it's definitely not A)

7 0

3 years ago

At a certain instant a particle is moving in the +x direction with momentum +8 kg m/s. During the next 0.13 seconds a constant f

jeka94

Answer:

The momentum of the particle at the end of the 0.13 s time interval is 7.12 kg m/s

Explanation:

The momentum of the particle is related to force by the following equation:

Δp = F · Δt

Where:

Δp = change in momentum = final momentum - initial momentum

F = constant force.

Δt = time interval.

Let´s calculate the x-component of the momentum after the 0.13 s:

final momentum - 8 kg m/s = -7 N · 0.13 s

final momentum = -7 kg m/s² · 0.13 s + 8 kg m/s

final momentum = 7.09 kg m/s

Now let´s calculate the y-component of the momentum vector after the 0.13 s. Since the particle wasn´t moving in the y-direction, the initial momentum in this direction is zero:

final momentum = 5 kg m/s² · 0.13 s

final momentum = 0.65 kg m/s

Then, the mometum vector will be as follows:

p = (7.09 kg m/s, 0.65 kg m/s)

The magnitude of this vector is calculated as follows:

$|p| = \sqrt{(7.09 kg m/s)^{2} + (0.65 kg m/s)^{2}} = 7.12 kg m/s$

The momentum of the particle at the end of the 0.13 s time interval is 7.12 kg m/s

4 0

3 years ago