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Vika [28.1K]
1 year ago
7

how much work www does the motor do on the platform during this process? enter your answer in joules to four significant figures

.
Physics
1 answer:
soldi70 [24.7K]1 year ago
4 0

The motor in the platform works for W=1885J during this process.

<h3>What is joules?</h3>

The amount of energy expended when a force of one newton is applied across a displacement of one meter is known as a joule. A joule is equal to one watt of electricity emitted or lost over the course of one second. The British thermal unit (Btu) is sometimes used to express energy. the work- or energy-measurement unit derived from the SI. The energy required to accelerate a body with a mass of one kilogram over a distance of one meter using one newton of force is equal to one joule. Additionally, one joule is equal to one watt-second.

<h3>What is joules formula?</h3>

The energy required to exert one newton of force through one meter is measured in joules. For instance, the energy required to lift a one-kilogram object one meter is given by E = Fd = (ma)d = mgd = (1 kg)(9.8m/s2)(1 m) = 9.8 kg m/s2 = 9.8 joules.

To know more about Joules visit:

brainly.com/question/25947916

#SPJ4

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Sphinxa [80]
Acceleration occurs when there is a change in speed or direction. If it travels in a straight line, there is no speed or change in direction as it is constant througout, hence 0 acceleration.

hope this helps!! ✨
6 0
3 years ago
Read 2 more answers
the ratio of the energy per second radiated by the filament of a lamp at 250k to that radiated at 2000k, assuming the filament i
Naily [24]

Answer:

(a) \frac{P_{250k}}{P_{2000k}}=2.4\ x\ 10^{-4}

(b) P =  0.816 Watt

Explanation:

(a)

The power radiated from a black body is given by Stefan Boltzman Law:

P = \sigma AT^4

where,

P = Energy Radiated per Second = ?

σ = stefan boltzman constant = 5.67 x 10⁻⁸ W/m².K⁴

T = Absolute Temperature

So the ratio of power at 250 K to the power at 2000 K is given as:

\frac{P_{250k}}{P_{2000k}}=\frac{\sigma A(250)^4}{\sigma A(2000)^4}\\\\\frac{P_{250k}}{P_{2000k}}=2.4\ x\ 10^{-4}

(b)

Now, for 90% radiator blackbody at 2000 K:

P = (0.9)(5.67\ x\ 10^{-8}\ W/m^2.K^4)(1\ x\ 10^{-6}\ m^2)(2000\ K)^4

<u>P =  0.816 Watt</u>

7 0
3 years ago
A spring with a spring constant value of 125 N/m is compressed 12.2 cm by pushing on it with a 215 g block. When the block is re
allsm [11]

Answer:

v = 2.94 m/s

Explanation:

When the spring is compressed, its potential energy is equal to (1/2)kx^2, where k is the spring constant and x is the distance compressed. At this point there is no kinetic energy due to there being no movement, meaning the net energy in the system is (1/2)kx^2.

Once the spring leaves the system, it will be moving at a constant velocity v, if friction is ignored. At this time, its kinetic energy will be (1/2)mv^2. It won't have any spring potential energy, making the net energy (1/2)mv^2.

Because of the conservation of energy, these two values can be set equal to each other, since energy will not be gained or lost while the spring is decompressing. That means

(1/2)kx^2 = (1/2)mv^2

kx^2 = mv^2

v^2 = (kx^2)/m

v = sqrt((kx^2)/m)

v = x * sqrt(k/m)

v = 0.122 * sqrt(125/0.215)        <--- units converted to m and kg

v = 2.94 m/s

3 0
3 years ago
The ancient Greek recommendation of rest in the treatment of abnormal behavior soon gave rise to ______. A. Trephining B. The as
Lilit [14]
C. Is the correct answer
7 0
3 years ago
The coefficient of the restitution of an object is defined as the ratio of its outgoing to incoming speed when the object collid
IgorLugansk [536]

Answer:

48.16 %

Explanation:

coefficient of restitution = 0.72

let the incoming speed be = u

let the outgoing speed be = v

kinetic energy = 0.5 x mass x x velocity^{2}

  • incoming kinetic energy = 0.5 x m x x u^{2}

     

  •  coefficient of restitution =\frac{v}{u}

       0.72 =\frac{v}{u}

       v = 0.72u

        therefore the outgoing kinetic energy = 0.5 x m x (0.72u)^{2}

        outgoing kinetic energy = 0.5 x m x 0.5184 x u^{2}

        outgoing kinetic energy = 0.5184 (0.5 x m x x u^{2})

recall that 0.5 x m x x u^{2} is our incoming kinetic energy, therefore

outgoing kinetic energy = 0.5184 x (incoming kinetic energy)

from the above we can see that the outgoing kinetic energy is 51.84 % of the incoming kinetic energy.

The energy lost would be 100 - 51.84 = 48.16 %

5 0
3 years ago
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