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yan [13]
1 year ago
5

how much work did the movers do (horizontally) pushing a 41.0- kg crate 10.3 m across a rough floor without acceleration, if the

effective coefficient of friction was 0.60?
Physics
1 answer:
11111nata11111 [884]1 year ago
7 0

W=2485.65 J ,work did the movers do friction force a 41.0- kg crate 10.3 m across a rough floor without acceleration

<h3>What is a basic friction force?</h3>

Two surfaces that come into contact and slide against one another produce a force known as frictional force. Several aspects that influence the frictional force include: The surface texture as well as the amount of force attracting them together have the most effects on these forces.

<h3>What outcomes does friction produce?</h3>

It generates heat, which is useful for warming our bodies or specific areas of any object. Power is also lost as a result. It makes noise during every operation. We are able to walk, run, dance, etc. due to friction.

To know  more about friction force visit:

brainly.com/question/1389727

#SPJ4

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Define malleability?<br>​
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Answer:

Explanation:

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Suppose a certain car supplies a constant deceleration of A meter per second per second. If it is traveling at 90km/hr. When. th
aksik [14]

Answer:

i)-6.25m/s

ii)18 metres

iii)26.5 m/s or 95.4 km/hr

Explanation:

Firstly convert 90km/hr to m/s

90 × 1000/3600 = 25m/s

(i) Apply v^2 = u^2 + 2As...where v(0m/s) is the final speed and u(25m/s) is initial speed and also s is the distance moved through(50 metres)

0 = (25)^2 + 2A(50)

0 = 625 + 100A....then moved the other value to one

-625 = 100A

Hence A = -6.25m/s^2(where the negative just tells us that its deceleration)

(ii) Firstly convert 54km/hr to m/s

In which this is 54 × 1000/3600 = 15m/s

then apply the same formula as that in (i)

0 = (15)^2 + 2(-6.25)s

-225 = -12.5s

Hence the stopping distance = 18metres

(iii) Apply the same formula and always remember that the deceleration values is the same throughout this question

0 = u^2 + 2(-6.25)(56)

u^2 = 700

Hence the speed that the car was travelling at is the,square root of 700 = 26.5m/s

In km/hr....26.5 × 3600/1000 = 95.4 km/hr

3 0
3 years ago
How much energy is produced if you have a current carrying a charge of 0.0006 C running 8.0 V?
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Answer:

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= 1/2x0.0006 x 8

=2.4x10-³J.

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3 years ago
A delivery truck travels with a constant velocity up an 8 slope. A 60 kg box sits on the floor of the truck and, because of sta
Blababa [14]

Explanation:

Given that,

The slope of the ramp, \theta=8^{\circ}

Mass of the box, m = 60 kg

(a) Distance covered by the truck up the slope, d = 300 m

Initially the truck moves with a constant velocity. We know that the net work done on the box is equal to 0 as per work energy theorem as :

W=\dfrac{1}{2}m(v^2-u^2)=0

u and v are the initial and the final velocity of the truck

(b) The work done on the box by the force of gravity is given by :

W=Fd\ cos\theta

Here, \theta=90+8=98^{\circ}

W=mgd\ cos\theta

W=60\times 9.8\times 300\ cos(98)

W = -24550.13 J

(c) What is the work done on the box by the normal force is equal to 0 as the angle between the force and the displacement is 90 degrees.

(d) The work done by friction is given by :

W_f=-W

W_f=24550.13\ J

Hence, this is the required solution.

6 0
3 years ago
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