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Illusion [34]
3 years ago
7

If 3600 j of work is done in 3.0 s what is the power0.00083W1200W3600W11000W

Physics
2 answers:
ivanzaharov [21]3 years ago
8 0

Answer : Power, P = 1200 Watts.

Explanation :

It is given that,

Work done, W = 3600 J

Time taken to complete the work, t = 3 s

The power is defined as the work done per unit time. The SI unit of power is watts.

Mathematically, it can be written as :

P=\dfrac{W}{t}

P=\dfrac{3600\ J}{3\ s}

P = 1200 Watts.

So, the correct option is (B) " 1200 W ".

Hence, this is the required solution.

Viktor [21]3 years ago
5 0

Answer:

1200 W

Explanation:

Power is given by the ratio between work done and time taken:

P=\frac{W}{t}

where W is the work done and t the time taken.

In this problem, W = 3600 J and t = 3.0 s. Therefore, the power in this exercise is

P=\frac{3600 J}{3.0 s}=1200 W

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A 35.8 kg box initially at rest is pushed 2.38 m along a rough, horizontal floor with a constant applied horizontal force of 108
tiny-mole [99]

Answer:

The work done by the applied force is 259.22 J.

Explanation:

The work done by the applied force is given by:

W = F*d

Where:

F: is the applied horizontal force = 108.915 N

d: is the distance = 2.38 m  

Hence, the work is:

W = F*d = 108.915 N*2.38 m = 259.22 J

Therefore, the work done by the applied force is 259.22 J.

I hope it helps you!                                                

6 0
3 years ago
A ‘thermal tap’ used in a certain apparatus consists of a silica rod which
abruzzese [7]

Correct question is;

A thermal tap used in a certain apparatus consists of a silica rod which fits tightly inside an aluminium tube whose internal diameter is 8mm at 0°C.When the temperature is raised ,the fits is no longer exact. Calculate what change in temperature is necessary to produce a channel whose cross-sectional is equal to that of the tube of 1mm. (linear expansivity of silica = 8 × 10^(-6) /K and linear expansivity of aluminium = 26 × 10^(-6) /K).

Answer:

ΔT = 268.67K

Explanation:

We are given;

d1 = 8mm

d2 = 1mm

At standard temperature and pressure conditions, the temperature is 273K.

Thus; Initial temperature; T1 = 273K,

Using the combined gas law, we have;

P1×V1/T1 = P2×V2/T2

The pressure is constant and so P1 = P2. They will cancel out in the combined gas law to give:

V1/T1 = V2/T2

Now, volume of the tube is given by the formula;V = Area × height = Ah

Thus;

V1 = (πd1²/4)h

V2 = (π(d2)²/4)h

Thus;

(πd1²/4)h/T1 = (π(d2)²/4)h/T2

π, h and 4 will cancel out to give;

d1²/T1 = (d2)²/T2

T2 = ((d2)² × T1)/d1²

T2 = (1² × T1)/8²

T2 = 273/64

T2 = 4.23K

Therefore, Change in temperature is; ΔT = T2 - T1

ΔT = 273 - 4.23

ΔT = 268.67K

Thus, the temperature decreased to 268.67K

6 0
2 years ago
Question:
exis [7]

Answer:

She can swing 1.0 m high.

Explanation:

Hi there!

The mechanical energy of Jane (ME) can be calculated by adding her gravitational potential (PE) plus her kinetic energy (KE).

The kinetic energy is calculated as follows:

KE = 1/2 · m · v²

And the potential energy:

PE = m · g · h

Where:

m = mass of Jane.

v = velocity.

g = acceleration due to gravity (9.8 m/s²).

h = height.

Then:

ME = KE + PE

Initially, Jane is running on the surface on which we assume that the gravitational potential energy of Jane is zero (the height is zero). Then:

ME = KE + PE      (PE = 0)

ME = KE

ME = 1/2 · m · (4.5 m/s)²

ME = m · 10.125 m²/s²

When Jane reaches the maximum height, its velocity is zero (all the kinetic energy was converted into potential energy). Then, the mechanical energy will be:

ME = KE + PE      (KE = 0)

ME = PE

ME = m · 9.8 m/s² · h

Then, equallizing both expressions of ME and solving for h:

m · 10.125 m²/s² =  m · 9.8 m/s² · h

10.125 m²/s² / 9.8 m/s²  = h

h = 1.0 m

She can swing 1.0 m high (if we neglect dissipative forces such as air resistance).

6 0
3 years ago
What do you mean by Parallel Universe ?​
BartSMP [9]

Answer:

Parallel universe, or alternate reality, is a hypothetical self-contained plane of existence, co-existing with one's own

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iogann1982 [59]

Answer:

Explanation:

Momentum conservation

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6 0
3 years ago
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