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solniwko [45]
3 years ago
5

two campers pull a canoe. If the force of friction acting on the canoe is 84 N, then how much work does the two campers need to

do to keep the canoe in the middle of the river for a displacement of 50 m?
Physics
1 answer:
sladkih [1.3K]3 years ago
8 0
W=Fs 
just multiply them all
84*50 = 4200 J
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What is the energy of a 4kg apple that is sitting on a 2 m high tree branch?
miv72 [106K]

78.4 joules is the energy of a 4 kg apple that is sitting on a 2 m high tree branch.

<u>Explanation: </u>

When an apple falls to the ground from a tree, its positional energy (stored as potential gravitational energy) turns into kinetic energy, during a fall. Chemical potential energy is chemical energy because it is food and potential energy as it can still have ability to move. So, in the given case, kinetic energy is zero.

To find potential energy, the formula would be

             \text { potential energy }(P . E)=m \times g \times h

Where, given

m – Mass – 4 kg

g-\text { Earth'sgravity }-9.8 \mathrm{m} / \mathrm{s}^{2} (Known value)

h – Height -  2 m

Substitute these values, we get

                  P . E=4 \times 9.8 \times 2=78.4 \text { joules }

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A group of atoms with aligned magnetic poles are known as which of the following?
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3 years ago
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A wheel of mass 4kg is pulled up a plane inclined at 30° to the horizontal by a force of 45N applied to the axle and parallel to
Len [333]

Answer:

v = 10 m/s

Explanation:

Let's assume the wheel does not slip as it accelerates.

Energy theory is more straightforward than kinematics in my opinion.

Work done on the wheel

W = Fd = 45(12) = 540 J

Some is converted to potential energy

PE = mgh = 4(9.8)12sin30 = 235.2 J

As there is no friction mentioned, the remainder is kinetic energy

KE = 540 - 235.2 = 304.8 J

KE = ½mv² + ½Iω²

ω = v/R

KE = ½mv² + ½I(v/R)² = ½(m + I/R²)v²

v = √(2KE / (m + I/R²))

v = √(2(304.8) / (4 + 0.5/0.5²)) = √101.6

v = 10.07968...

5 0
3 years ago
When the mass of the cylinder increased by a factor of 3, from 1.0 kg to 3.0 kg, what happened to the cylinder’s gravitational p
const2013 [10]

Answer: Fourth option. It increased by a factor of 3.

Solution:

m1=1.0 kg

Cylinder's gravitational potential energy: Ep=m*g*h

Ep1=(1.0 kg)*g*h

Ep1=g*h

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Ep2=(3.0 kg)*g*h

Ep2=3*g*h

Replacing g*h by Ep1 in the equation above:

Ep2=3*Ep1

Then, the cylinder's gravitational potential energy increased by a factor of 3.

3 0
4 years ago
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