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Gala2k [10]
2 years ago
7

A 1100-N crate rests on the floor.

Physics
1 answer:
Tom [10]2 years ago
6 0

(a) The work done in moving the crate along the floor is 7,050 J.

(b) The work done in moving the crate vertically is 0 J.

<h3>Work done in moving the crate horizontally</h3>

W = Fdcosθ

where;

  • F is total force to be applied
  • d is displacement of the crate
  • θ is the horizontal angle = 0

W = (1100 + 310) x 5

W = 7,050 J

<h3>Work done in moving the crate vertically</h3>

W  = Fd cos(90)

W = 0

Thus, the work done in moving the crate along the floor is 7,050 J.

the work done in moving the crate vertically is 0 J.

Learn more about work done here: brainly.com/question/8119756

#SPJ1

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elixir [45]
The answer is d.... Knife.

Hope this helped :)
3 0
3 years ago
how fast will and in what direction will a 20kg object accelerate if one force pushes at a 30 degree angle and another pushes at
DiKsa [7]

Answer:

|a|=2.83\ m/s^2

\theta=75^o

Explanation:

<u>Net Force And Acceleration </u>

The Newton's second law relates the net force applied on an object of mass m and the acceleration it aquires by

\vec F_n=m\vec a

The net force is the vector sum of all forces. In this problem, we are not given the magnitude of each force, only their angles. For the sake of solving the problem and giving a good guide on how to proceed with similar problems, we'll assume both forces have equal magnitudes of F=40 N

The components of the first force are

\vec F_1=

\vec F_1=\ N

The components of the second force are

\vec F_2=

\vec F_2=\ N

The net force is

\vec F_n=

\vec F_n=\ N

The magnitude of the net force is

|F_n|=\sqrt{14.64^2+54.64^2}

|F_n|=\sqrt{3200}=56.57\ N

The acceleration has a magnitude of

\displaystyle |a|=\frac{|F_n|}{m}

\displaystyle |a|=\frac{56.57}{20}

|a|=2.83\ m/s^2

The direction of the acceleration is the same as the net force:

\displaystyle tan\theta=\frac{54.64}{14.64}

\theta=75^o

5 0
3 years ago
a force is applied to a box of 10.0 kg for 4.0 s. the box goes from rest to 25 m/s in that time. What is the magnitude of that f
Paul [167]

Given:

m(mass of the box)=10 Kg

t(time of impact)=4 sec

u(initial velocity)=0.(as the body is initially at rest).

v(final velocity)=25m/s

Now we know that

v=u+at

Where v is the final velocity

u is the initial velocity

a is the acceleration acting on the body

t is the time of impact

Substituting these values we get

25=0+a x 4

4a=25

a=6.25m/s^2

Now we also know that

F=mxa

F=10 x6.25

F=62.5N

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4 years ago
A train is moving on a flat, continuous track. label all forces acting upon the object
Firdavs [7]
2nd and third laws of energy
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The fictional rocket ship Adventure is measured to be 65 m long by the ship's captain inside the rocket.When the rocket moves pa
kipiarov [429]

Answer:

1.04 s

Explanation:

The computation is shown below:

As we know that

t = t' × 1 ÷ (√(1 - (v/c)^2)

here

v = 0.5c

t = 1.20 -s

So,

1.20 = t' × 1 ÷ (√(1 - (0.5/c)^2)

1.20 = t' × 1 ÷ (√(1 - (0.5)^2)

1.20 = t' ÷ √0.75

1.20 = t' ÷ 0.866

t' = 0.866 × 1.20

= 1.04 s

The above formula should be applied

8 0
3 years ago
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