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pantera1 [17]
2 years ago
12

How much power does it take to do 500J of work in 10 sec

Physics
2 answers:
tatuchka [14]2 years ago
8 0
P=w/t
so the answer is 500/10=50
aliya0001 [1]2 years ago
3 0

Answer: Power = 50J/s

Explanation: Power ,P is the rate of doing work and it is a scalar quantity with no direction.

P= Workdone/ time

P= 500/10

P= 50J/s

Therefore the power requires is 50J/s

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a student is pushing a 50 kilogram cart with a force of 500 newtons another students measures the speed of the cart and finds th
Irina-Kira [14]

The force of friction is 300 N

Explanation:

We can solve the problem by applying Newton's second law of motion: in fact, the net force acting on an object is equal to the product between the mass of the object and its acceleration. So we can write

\sum F = ma

where

\sum F is the net force acting on the object

m is its mass

a is its acceleration

For the cart in this problem, we have two forces acting on it:

- The force of push, F = 500 N, forward

- The force of friction, F_f, backward

So Newton's second law can be rewritten as

F-F_f = ma

where

m = 50 kg

a=4 m/s^2 is the acceleration of the cart

And solving for F_f, we find the force of friction:

F_f = F-ma=500-(50)(4)=300 N

Learn more about force of friction:

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4 0
2 years ago
How do scientist use information about an object or event
Arlecino [84]
Many ways for example they look up related things and study them or they can test it and see what happens
4 0
3 years ago
A stone is dropped from a tower 100 meters above the ground. The stone falls past ground level and into a well. It hits the wate
lilavasa [31]

Take the stone's position at ground level to be the origin, and the downward direction to be negative. Then its position in the air y at time t is given by

y=100\,\mathrm m-\dfrac g2t^2

Let d be the depth of the well. The stone hits the bottom of the well after 5.00 s, so that

-d=100\,\mathrm m-\dfrac g2(5.00\,\mathrm s)^2\implies d=\boxed{22.6\,\mathrm m}

7 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

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6 0
1 year ago
un futbolista patea una pelota que se encuentra en el pasto con un angulo de 30° (medido desde la horizontal) con la intención d
Rus_ich [418]

Answer:

i dont really know what it is

8 0
3 years ago
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