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goblinko [34]
3 years ago
5

You are carrying a 7.0-kg bag of groceries as you walk at constant velocity along the sidewalk. You walk a distance of 82 meters

and hold the bag at a height of 1.5 m above the sidewalk. How much work against gravity do you do on the bag?
100 J

0 J

5600 J

82 J
Physics
1 answer:
MAVERICK [17]3 years ago
7 0

Answer:

0J

Option: B

Explanation:

Work is done when something is moved by the force in the direction of the force. That is the force (e.g., the weight) and the direction the object moves must be aligned for work to be done. In this given condition, the direction is horizontal and the force is downward as its gravity force. That 90° between the two vectors.

The work function is W = m × g ×h × cosθ

\text { Where, in this case theta }=90^{\circ}

Hence,  

Work done = 7 × 9.8 × 1.5 × cos(90)  

Work done = 0 (cos90^{\circ} = 0)

Work done = 0

Therefore work done is 0 J.

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Paul [167]

Answer:

F = 2520 N

Explanation:

We have,

The maximum acceleration of a fist in a karate punch is 4200 m/s². The mass of the fist is 0.6 kg.

It is required to find the force that the wood place on the fist. Force is given by the product of mass and its acceleration such that,

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F=0.6\times 4200\\\\F=2520\ N

So, the force of 2520 N is acting on the wood.

8 0
3 years ago
You are standing 2.5m directly in front of one of the two loudspeakers. They are 3.0m apart and both are playing a 686Hz tone in
ahrayia [7]

Answer:

distance from speaker is 17.87 m

Explanation:

given data

distance from loudspeaker = 2.5 m

distance between loudspeaker = 3.0 m

room temperature = 20c

wavelength f  = 686Hz

to find out

what distances from the speaker

solution

we know sound velocity c = 331.5  + 0.6 × 20c = 343.5

so wavelength of sound  λ = c / f  

wavelength = 343.5 /  686 = 0.5 m

when the difference in distance of speaker destructive interference will be

d = λ/2 × (2n-1)

for n = 1, 2 3 4 ..

d = 0.5/2 × (2n-1)

d = 0.250 , 0.75 , 1.25 , 1.750............   for n = 1, 2 3 .............

so

for d = 0.250

side of triangle by hypotenuse of triangle are

\sqrt{3^{2}+(2..5+x)^{2} } - (2.5 + x1) = 0.250

0.5 x1 = 7.6875

x1 = 15.375 m

for d = 0.75

side of triangle by hypotenuse of triangle are

\sqrt{3^{2}+(2..5+x)^{2} } - (2.5 + x2) = 0.75

1.5 x2 = 4.6875

x2 = 3.125 m

for d = 1.250

side of triangle by hypotenuse of triangle are

\sqrt{3^{2}+(2..5+x)^{2} } - (2.5 + x3) = 1.250

2.5 x2 = 1.1875

x3 = 0.475 m

for d = 1.750

x4 will be negative so we stop here

so the distance from speaker here is given below

distance = 2.5 + x

here x = 0.475 , 3.125 and 15.375 so

distance 1 = 2.5 + 0.475  = 2.975 m

distance 2 = 2.5 + 3.125  = 5.625 m

distance 3 = 2.5 + 15.375 = 17.875 m

final distance from speaker is 17.87 m

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