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Rufina [12.5K]
3 years ago
13

A worker pushes horizontally on a large crate with force of 298 N and the crate is moved 4.3 m. How much work was done?

Physics
1 answer:
Katyanochek1 [597]3 years ago
5 0

Answer:

1281.4 J

Explanation:

F = 298 N

d = 4.3 m

the work done is defined as the product of force applied in the direction of displacement and the displacement.

W = force x displacement

Work is a scalar quantity and its SI unit is Joule.

W = F x d

W = 298 x 4.3 = 1281.4 J

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How many feet does a No. 2 pencil last?
Ray Of Light [21]

Answer:45,000

Explanation:

:TIL the average pencil holds enough graphite to draw a line about 35 miles long or to write roughly 45,000 words

6 0
2 years ago
What is the gravitational force that two objects would feel if they are 3.5 meters apart? Object 1 has a mass of 10x10^5 kg and
polet [3.4K]

Answer:

1.63 N

Explanation:

F = GMm/r^2

  = (6.67x10^-11)(10x10^5)(3x10^5) / 3.5^2

  = 1.63 N  ( 3 sig. fig.)

6 0
3 years ago
A uniform metal tube of length 5m and mass 9kg is suspended by two vertical wires attached at 50cm and 150cm respectively from t
Elena-2011 [213]

Answer:

force (tension) of 29.4 N (upward)  in 100 cm

force (tension) of 58.4 N (upward)  in 200 cm

Explanation:

Given:

Length of tube = 5 m (500 cm)

Mass of tube = 9

Suspended vertically from 150 cm and 50 cm.

Computation:

Force = Mass × gravity acceleration.

Force = 9.8 x 9

Force = 88.2 N

So,

Upward forces = Downward forces

D1 = 150 - 50 = 100 cm

D2 = 150 + 50 = 200 cm

And F1 = F2

F1 x D1 = F2 x D2

F1 x 100 = F2 x 200

F = 2F

Total force = Upward forces + Downward forces

3F = 88.2

F = 29.4 and 2F = 58.8 N

force (tension) of 29.4 N (upward)  in 100 cm

force (tension) of 58.4 N (upward)  in 200 cm

4 0
3 years ago
A sample container of carbon monoxide occupies a volume of 435 ml at a pressure of 785 torr and a temperature of 298 k. What wou
strojnjashka [21]

Answer:

181.54 K

Explanation:

From gas laws, we know that v1/t1= v2/t2 where v and t represent volume and temperatures, 1 and 2 for the first and second container. Making t2 the subject of the formula then

T2=v2t1/ v1

Given information

V1 435 ml

V2 265 ml

T1 298K

Substituting the given values then

T2=265*298/435=181.54 K

3 0
3 years ago
2. the dipole moment of a dipole in a 300-n/c electric field is initially perpendicular to the field, but it rotates so it is in
aleksandr82 [10.1K]

Work Done (W) by the field is-6x 107 J,

<h3>What is Electric dipole?</h3>

A pair of opposite, non-coplanar, equally powerful electric charges that are in opposition to one another. An atom is said to have a "induced electric dipole" if the center of the negative cloud of electrons has moved a little bit away from the nucleus due to an external electric field. When the external field is taken away, dipolarity is lost.

Electric field (E) = 300 N/C

Dipole moment (p) = 2 x 10° Cm

Solution:

From the formula we know.

U = -pE cosФ

Here,

p Denotes Dipole moment.

E Denotes Electric field.

Ф Denotes angle b/w them

Now, as given, firstly the dipole is perpendicular to the electric field, so

angle (Ф1) will be 90° and now the dipole is rotated such that they are in same

direction so the angle (Ф2) will be 0°

So, let's find Change in Potential energy which will be equal to the work done

by the electric field.

ΔU = Uf - Ui

ΔU = [-pE*cos Ф2] - [-pE *cos Ф1]

ΔU = [-pE*cos Ф2] + pE *cos Ф1

ΔU = pE * [cos Ф2+ cos Ф1]

Substituting the values,

ΔU = pE * [cos 0° + cos 90°]

ΔU = pE * (-1 +0)

ΔU = -pE

ΔU = -2x 10^-9 × 300

ΔU = 6 x 10^(-9+2)

ΔU = 6 x 10^-7

W = ΔU = -6 x 10^-7

W = - 6 x 10 7 J

Work Done (W) by the field is - 6 x 10-7 J.

To learn more about Work Done visit:

https://brainly.in/question/48222628

#SPJ4

6 0
1 year ago
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