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Simora [160]
3 years ago
9

A constant force of 11.8 N in the positive x direction acts on a 4.7-kg object as it moves from the origin to the point (1.6i –

4.6j) m. How much work is done by the given force during this displacement?
Physics
1 answer:
zhenek [66]3 years ago
5 0

Answer:

W = 18.88 J

Explanation:

Given that,

Constant force, F = 11.8 N (in +x direction)

Mass of an object, m = 4.7 kg

The object moves from the origin to the point (1.6i – 4.6j) m

We need to find the work is done by the given force during this displacement. The work done by an object is given by the formula as follows :

W=F{\cdot} d\\\\W=(11.8i){\cdot} (1.6i-4.6j)\\\\=11.8\times 1.6\\\\=18.88\ J

So, the work done by the given force is 18.88 J.

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What is one watt ? Write the relation of watt with kilowatt , megawatt , and horsepower .​
alina1380 [7]

Answer:

See the explanation below

Explanation:

The watt (the power) is equal to the relationship between the work and the time in which that work is performed.

P = W/t

where:

W = work [J] (units of Joules)

t = time [s].

Now 1000 [W] are equal to 1 [kW]

And 1000000 [W] are equal to 1 [MW]

The horsepower is the unit of power in the imperial system of units.

And 745.7 [W] are equal to 1 [Hp]

3 0
3 years ago
Two forces act on a 6.00-kg object. One of the forces is 10.0 N. If the object accelerates at 2.00 m/s2
liubo4ka [24]

Given :

Two forces act on a 6.00-kg object. One of the forces is 10.0 N.

Acceleration of object 2 m/s².

To Find :

The greatest possible magnitude of the other force.\

Solution :

Let, other force is f.

So, net force, F = 10 + f.

Now, acceleration is given by :

a=\dfrac{F}{mass}\\\\a= \dfrac{10+f}{6}\\\\\dfrac{10+f}{6}=2\\\\f = 12 - 10\\\\f = 2 \ N

Therefore, the greatest possible magnitude of the other force is 2 N.

Hence, this is the required solution.

7 0
3 years ago
A car traveling north with a velocity of 33 m/sec slows down to a velocity of 12 m/sec north within 10 sec. What is the car’s de
strojnjashka [21]

a=v/t

v= 33-12 = 21

21÷10=2.1

a= 2.1

4 0
3 years ago
Read 2 more answers
A spring is stretched from x=0 to x=d, where x=0 is the equilibrium position of the spring. It is then compressed from x=0 to x=
VARVARA [1.3K]

Answer:

The same amount of energy is required to either stretch or compress the spring.

Explanation:

The amount of energy required to stretch or compress a spring is equal to the elastic potential energy stored by the spring:

U=\frac{1}{2}k (\Delta x)^2

where

k is the spring constant

\Delta x is the stretch/compression of the spring

In the first case, the spring is stretched from x=0 to x=d, so

\Delta x = d-0=d

and the amount of energy required is

U=\frac{1}{2}k d^2

In the second case, the spring is compressed from x=0 to x=-d, so

\Delta x = -d -0 = -d

and the amount of energy required is

U=\frac{1}{2}k (-d)^2= \frac{1}{2}kd^2

so we see that the amount of energy required is the same.

7 0
4 years ago
Question<br> In order for work to be done, what three things are necessary
Misha Larkins [42]
Perseverance, good mind set, and work ethic
5 0
4 years ago
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