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lesantik [10]
2 years ago
14

Q1.

Physics
1 answer:
adelina 88 [10]2 years ago
3 0

Answer:Because if the shape gets changed it will move faster without to much weight

Explanation:

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A box is pushed 40 m by a mover. The amount of work done was 2,240 j. How much force was exerted on the box
Georgia [21]

The force exerted on the box is 56 N

Explanation:

The work done by a force on an object is given by

W=Fd cos \theta

where

F is the magnitude of the force

d is the displacement of the object

\theta is the angle between the direction of the force and of the displacement

For the box in this problem, we have:

W = 2240 J is the work done

d = 40 m is the displacement of the box

Assuming that the  force is parallel to the displacement, \theta=0

Solving the equation for F, we find the force exerted on the box:

F=\frac{W}{d cos \theta}=\frac{2240}{(40)(cos 0)}=56 N

Learn more about work:

brainly.com/question/6763771

brainly.com/question/6443626

#LearnwithBrainly

3 0
3 years ago
in loading a long lorry a man lifts boxes each of weight 100N and height of 1.5M.how much energy is transfered when one box is l
gizmo_the_mogwai [7]

Explanation:

energy=100×1.5=150 joules

4 0
3 years ago
Sarah observed that different kinds and amounts of fossils were present in a cliff behind her house. She wondered if changes in
MAXImum [283]
This is the answer: Fossil's found in Susan's yard are from prehistoric times.
6 0
3 years ago
A 0.29 kg particle moves in an xy plane according to x(t) = - 19 + 1 t - 3 t3 and y(t) = 20 + 7 t - 9 t2, with x and y in meters
Artist 52 [7]

Answer:

Part a)

F = 7.76 N

Part b)

\theta = -137.7 degree

Part c)

\theta = -127.7 degree

Explanation:

As we know that acceleration is rate of change in velocity of the object

So here we know that

x = -19 + t - 3t^3

y = 20 + 7t - 9t^2

Part a)

differentiate x and y two times with respect to time to find the acceleration

a_x = \frac{d^2}{dt^2}(-19 + t - 3t^3)

a_x = \frac{d}{dt}(0 +1 - 9t^2)

a_x = -18t

a_y = \frac{d^2}{dt^2}(20 + 7t - 9t^2)

a_y = \frac{d}{dt}(0 +7 - 18t)

a_y = -18

Now the acceleration of the object is given as

\vec a = (-18t)\hat i + (-18)\hat j

at t= 1.1 s we have

\vec a = -19.8 \hat i - 18 \hat j

now the net force of the object is given as

\vec F = m\vec a

\vec F = (0.29 kg)(-19.8 \hat i - 18 \hat j)

\vec F = -5.74 \hat i - 5.22 \hat j

now magnitude of the force will be

F = \sqrt{5.74^2 + 5.22^2} = 7.76 N

Part b)

Direction of the force is given as

tan\theta = \frac{F_y}{F_x}

tan\theta = \frac{-5.22}{-5.74}

\theta = -137.7 degree

Part c)

For velocity of the particle we have

v_x = \frac{dx}[dt}

v_x = (0 +1 - 9t^2)

v_y = \frac{dy}{dt}

v_y = (0 +7 - 18t)

now at t = 1.1 s

\vec v = -9.89\hat i - 12.8 \hat j

now the direction of the velocity is given as

\theta = tan^{-1}(\frac{v_y}{v_x})

\theta = tan^{-1}(\frac{-12.8}{-9.89})

\theta = -127.7 degree

7 0
3 years ago
Determine the kinetic energy of 1000-kg roller coaster car that is moving with speed of 20.0m/s
nevsk [136]
B, i got the same question
7 0
3 years ago
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