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forsale [732]
4 years ago
10

A ramp is 2.8 m long and 1.2 m high. How much power is needed to push a box up the ramp in 4.6 s with a force of 96 N?

Physics
1 answer:
tino4ka555 [31]4 years ago
6 0
Power = work /time

and  W = force * dis

W = 96 * 3.046 J = 292.44 J

P = 292.44 / 4.6 = 63.575 watt
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The ventilation fan of the bathroom has a volumetric flow rate of 1,640 L/s. If the air density is 1.28 kg/m3, calculate the mas
liq [111]

Answer:

7557120 kg/hour

Explanation:

Given data;

Volume of air in one second = 1640 L

Density of air = 1.28 kg/L

Mass of air in 1 hour =?

Since mass = density × volume

==> Mass of air in one second = 1.28 ×1640 = 2099.2 kg

==> Mass of air in one minute = 2099.2×60=125952 kg

==> Mass of air in one hour = 125952× 60 = 7557120 kg

So rate of flow of air is 7557120 kg/hour

5 0
4 years ago
During the collision of a big truck with a small passenger carGroup of answer choicesthe force from the truck on the car is alwa
lorasvet [3.4K]

Answer:

The force from the truck on the car is always equal to the force from the car on the truck.

Explanation:

According to Newton's third law; action and reaction are equal and opposite. Hence, when the big truck and small passenger car are involved in a collision, we expect that the force from the truck on the car is always equal to the force from the car on the truck. The forces on the car and the truck are equal in magnitude but opposite in direction.

This follows directly from Newton's third law of motion hence the answer above.

3 0
3 years ago
Calculate the amount of energy produced in joules by 100- watt light bulb lit for 2.5 hours.
9966 [12]
100/2.5 is 40.
40 is the energy that is being produced 
4 0
4 years ago
Newton's first law
Bond [772]

Answer:E. an object will remain in uniform motion unless acted upon by a force

Explanation:

7 0
3 years ago
A sled of mass m is given a kick on a frozen pond. The kick imparts to the sled an initial speed of 2.00 m/s . The coefficient o
kobusy [5.1K]

2.04 meters distance is traveled by the sled before stopping.

Mass of the sled = m

The initial speed of the sled = 2 m/s

Coefficient of kinetic friction between sled and ice = 0.100

Let the distance the sled moves before it stops be d.

Gravity = 9.8 m/ s²

Let the initial kinetic energy sled be

= K _{i}

K_{i} =  \frac{1}{2} mv ^{2}

The work done by the frictional force is,

Work \: done \:  by \: frictional \: force =W_{f}

W _{f} = μ_{k}mgd

Work done by frictional force= Initial kinetic energy of the sled

W_{f} = K_{i}

μ_{k}mgd= \frac{1}{2} mv ^{2}

So, the distance traveled by the sled before stopping is

d= \frac{1mv ^{2} }{2 \:μ_{k}mg}

d= \frac{1v ^{2} }{2 \:μ_{k}g}

d= \frac{2^{2} }{2  \times \:0.100 \times 9.8}

d= 2.04 \: m

Therefore, the distance traveled by the sled before stopping is 2.04 meters.

To know more about work done, refer to the below link:

brainly.com/question/13662169

#SPJ4

5 0
2 years ago
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