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3241004551 [841]
3 years ago
13

A "8" grams box is pushed with a force of 100 N for 1m whereas opposing force is 10 N. A) Find the net work done on the box. B)

Find the total work done by considering work done by each force. "
Physics
1 answer:
Debora [2.8K]3 years ago
8 0

Answer:

A)   Wn = 90J

B)   W1 = 100J

     W2 = 10J

Explanation:

A) The net work done on the box is calculated by using the following formula:

W_n=(F_1-F_2)d     (1)

F1 = 100N

F2 = 10N

d: distance on which the box is moved = 1m

You replace the values of the parameters in the equation (1):

W_n=(100N-10N)(1m)=90J

The net work on the box is 90J

B) The total work done by each force is:

W_1=F_1d=(100N)(1m)=100J\\\\W_2=F_2d=(10N)(1m)=10J

The first box does a work of 100J and the second force does a force of 10J

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What is the speed of a 48-kilogram dog running across<br>a lawn with 216 joules of kinetic energy?​
pishuonlain [190]

Answer:

3m/s

Explanation:

K.E= (1/2)mv^2

216j= (1/2)48kg • v^2

216J=24kg•v^2

v^2 = (216J)/(24kg)

v^2= 9m^2/s^2

/sqrt{v^2} = /sqrt{9m^2/s^2}

V =3m/s

8 0
3 years ago
A mirror forms an erect image 40cm from the object and one third its height where must the mirror be situated ​
iragen [17]

The mirror is located 30 cm from the object

Explanation:

To solve the problem, we can use the magnification equation, which states that:

M=\frac{y'}{y}=-\frac{q}{p}

where

M is the magnification

y' is the size of the image

y is the size of the object

q is the distance of the image  from the mirror

p is the distance of the object from the mirror

In this problem, we notice that the image formed by the mirror is erect and diminished: this means that this is a convex mirror, so the image is virtual, and this means that the image and the object are located on opposite sides of the mirror. Therefore,

p-q=40 cm (distance of the image from the object is 40 cm, but since the image is virtual, q is the negative)

The size of the image is 1/3 that of the object, so

y'=\frac{1}{3}y

Solving the equation for p, we find the distance between the object and the mirror:

\frac{y'}{y}=-\frac{p-40}{p}\\\\p=\frac{40}{1+\frac{y'}{y}}=\frac{40}{4/3}=30 cm

So, the mirror is 30 cm from the object.

#LearnwithBrainly

5 0
3 years ago
A 2.0kg mass is attached to a horizontal spring having a spring constant of 0.05Nm.
Alex Ar [27]

Good.  You can do some very interesting experiments with that equipment.

3 0
3 years ago
7. How do Newton’s laws of motion explain why it is important to keep the ice smooth on a hockey rink so that players can pass a
Arisa [49]
The first law states that “objects at rest and objects in motion remain in motion in a straight line unless acted upon by an unbalanced force”. Keeping the ice smooth will make sure there is not friction, friction would slow the puck down
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3 years ago
Read 2 more answers
Two small stereo speakers are driven in step by the same variable-frequency oscillator. Their sound is picked up by a microphone
anygoal [31]

Answer:

The fundamental frequency at which the sound of speakers at the microphone produce constructive interference is 801.076458 Hz

Explanation:

For a given arrangement having constructive interference, we have;

R₁ - R₂ = 2·x = 0 + n·λ

The distance from one speaker to the microphone. R₁ = 4.50 m

The distance between the two speakers = 2.00 m

The angle formed between the direction from the microphone to the speaker closest and the directional path between the speakers = 90°

Therefore, by Pythagoras's theorem, the distance from the speaker furthest from the microphone, to the microphone, 'R₂' is given as follows;

R₂ = √(4.50² + 2.00²) = √(24.25) = 4.9244289009 ≈ 4.924

∴ R₂ ≈ 4.9244289 m

R₂ - R₁ = 4.9244289 m - 4.5 m = 0.4244289 m

For constructive interference, R₂ - R₁ =0.4244289 m = n·λ

For n = 1, we have;

R₂ - R₁ =0.4244289 m = n·λ = 1 × λ = λ

λ = 0.4244589 m

f = v/λ = 340 m/sec/(0.4244289 m) ≈ 801.076458 Hz

Therefore the lowest possible fundamental frequency at which the speakers produce constructive interference, f = 801.076458 Hz

7 0
3 years ago
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