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JulijaS [17]
3 years ago
13

A machine lifts a 1000 N object 40 meters in 15 seconds. How much work is done by the machine

Physics
1 answer:
patriot [66]3 years ago
6 0

Answer:

40000 J or 40 kJ

Explanation:

Work is the net force times the distance.

W = Fd

W = (1000 N) (40 m)

W = 40000 J

Notice that the time doesn't affect the amount of work done.

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Assume this 1.20-mm-radius copper wire is electrically neutral in the Earth reference frame, in which it is at rest and carrying
agasfer [191]

Answer:

The charge density in the system is 4.25*10^4C/m

Explanation:

To solve this problem it is necessary to keep in mind the concepts related to current and voltage through the density of electrons in a given area, considering their respective charge.

Our data given correspond to:

r=1*10^{-3}m\\v = 5.2*10^{-4}m/s\\e= 1.6*10^{-19}C

We need to asume here the number of free electrons in a copper conductor, at which is generally of 8.5 *10^{28}m^{-3}

The equation to find the current is

I = VenA

Where

I =Current

V=Velocity

A = Cross-Section Area

e= Charge for a electron

n= Number of free electrons

Then replacing,

I = (5.2*10^{-4})(1.6*10^{-19})(88.5 *10^{28})(\pi(1*10^{-3})^2)

I= 22.11a

Now to find the linear charge density, we know that

I = \frac{Q}{t} \rightarrow Q = It

Where:

I: current intensity

Q: total electric charges

t: time in which electrical charges circulate through the conductor

And also that the velocity is given in proportion with length and time,

V_d = \frac{l}{t} \rightarrow l = V_d t

The charge density is defined as

\lambda = \frac{Q}{l}\\\lambda = \frac{It}{V_d t}\\\lambda = \frac{I}{V_d}

Replacing our values

\lambda = \frac{22.11}{5.20*10{-4}}

\lambda= 4.25*10^4C/m

Therefore the charge density in the system is 4.25*10^4C/m

5 0
3 years ago
What is tarzan's speed vf just before he reaches jane? express your answer in meters per second to two significant figures?
e-lub [12.9K]
Before swinging, T has only potential energy, (no speed)
Ui = mgh
Where h is the vertical displacement of T
From the laws of geometry,
cos45 = (L-h)/L
cos45 = 1-h/L
h/L = 1-cos45
h = L(1-cos45)

Therefore
Ui = mgL(1-cos45)

Proceeding the same way,
Twill raise to aheight of h' due to swing
h' = L(1-cos30)
The PE of T after swing is
Uf = mgh'
Uf = mgL(1-cos30)

Along with the PE , T has some kinetic energy results due to the moment.
Tf = 0.5*mv^2

According to the law of conservation of energy,
Ui = Uf+Tf
mgL(1-cos45) = mgL(1-cos30) + 0.5*mv^2
gL(co30-cos45) = 0.5*v^2
9.8*20*(co30-cos45) = 0.5*V^2
v = 7.89 m/s

<span>The speed f T after swing is 7.89 m/s</span>
3 0
3 years ago
What is the STANDARD unit of measurement for velocity in Physics?
rusak2 [61]

Answer:

m. {s}^{ - 1}

4 0
3 years ago
Answer please urgent​
motikmotik

Answer:

equal and unlike charges

Explanation:

5 0
3 years ago
Read 2 more answers
A 0.660 kg ball is dropped from rest from the top of a building and falls for 5.65 seconds. How tall was the building ?
kirill115 [55]

Answer:

The height of the building is approximately 156.58 m

Explanation:

The mass of the ball dropped from rest from the building top = 0.660 kg

The time in which the ball falls, t = 5.65 seconds

The height, h, of the building is given from the following equation of motion;

h = u·t + ¹/₂·g·t²

Where;

u = The initial velocity of the ball = 0 m/s

g = The acceleration due to gravity = 9.81 m/s²

Plugging in the values, we have;

h = 0 × 5.65 + ¹/₂ × 9.81 × 5.65² ≈ 156.58 m

The height of the building, h ≈ 156.58 m.

6 0
3 years ago
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