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3 years ago

What's is the power output of an engine that does 60000 J of work in 10 seconds

Physics

1 answer:

ad-work [718]3 years ago

8 0

A: 6,000 watts

Equation for power: P= work / time

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If the displacement of a horizontal mass-spring system was doubled, the elastic potential energy in the system would change by a

bixtya [17]

Answer:

K'=4K

Explanation:

The electric potential energy is given by :

$E=\dfrac{1}{2}kx^2$

Where

k is spring constant

x is compression or extension in the spring

If the displacement of a horizontal mass-spring system is doubled, x'= 2x

New elastic potential energy :

$E'=\dfrac{1}{2}kx'^2\\\\E'=\dfrac{1}{2}k(2x)^2\\\\=\dfrac{1}{2}k\times 4x^2\\\\K'=4\times \dfrac{1}{2}kx^2\\\\K'=4K$

So, new elastic potential energy 4 times the initial elastic potential energy.

3 0

3 years ago

An electron with a charge of -1.6 × 10-19 coulombs experiences a field of 1.4 × 105 newtons/coulomb. What is the magnitude of th

kipiarov [429]

Answer:

$2.24 \times 10^{-14}$ Newtons

Explanation:

Magnitude of charge on the electron = q = $1.6 \times 10^{-19}$ Coulombs

The negative sign in the question statement indicates that the charge is negative.

Magnitude of Electric Field experienced by the electron = E = $1.4 \times 10^{5}$ Newtons/Coulomb

Magnitude of Force on the electron = F = ?

The relation between the charge, electric field and the force on the charge because of electric field is given by:

$E=\frac{F}{q}$

From here we can write:

F = qE

Using the values, we get:

$F = 1.6 \times 10^{-19} \times 1.4 \times 10^{5}\\\\ F = 2.24 \times 10^{-14}$

Thus, the magnitude of the electric force experience by the electron would be $2.24 \times 10^{-14}$ Newtons

8 0

3 years ago

If a 5.5 kg object experiences 15 N of force for .15 seconds what is the speed change

Nady [450]

The speed change : Δv = 0.41 m/s

<h3>Further explanation</h3>

Given

mass = 5.5 kg

Force = 15 N

time = 0.15 s

Required

the speed change

Solution

Newton 2nd's law

Impulse and momentum

F = m.a

F = m . Δv/t

F.t = m.Δv

Input the value :

15 N x 0.15 s = 5.5 kg x Δv

Δv = 0.41 m/s

8 0

3 years ago

Please answer me!!! asappp

Anastasy [175]

Answer:

omg this is hard

Explanation:

Yes it is lol

5 0

3 years ago

Rank the tensions in the ropes, t1, t2, and t3, from smallest to largest, when the boxes are in motion and there is no friction

gizmo_the_mogwai [7]

<span>AS T1,T2,T3 are the tensions in the ropes,assuming that there are Three blocks of mass 3m, 2m, and m.T3 is the string between 3m and 2m,T2 is the string between 2m and m ,T1 is the string attached to m thus T1 pulls the whole set of blocks along, so it must be the largest. T2 pulls the last two masses, but T3 only pulls the last mass, so T3 < T2 < T1.</span>

5 0

3 years ago

What's is the power output of an engine that does 60000 J of work in 10 seconds​

What's is the power output of an engine that does 60000 J of work in 10 seconds