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alisha [4.7K]
2 years ago
15

Bartek w czasie 2minut przesunął ruchem jednostajnym kamień na odległość 0,5m.Oblicz moc mięśni Bartka jeśli siła oporów wynsiła

640N Prosze szybką odpowiedz daje naj
Physics
1 answer:
Ipatiy [6.2K]2 years ago
4 0

Answer:

Bartek's muscle power = 2.667 W

Moc mięśni Bartka = 2.667 W

Explanation:

English Translation

Bartek within 2 minutes moved the stone

steadily at a distance of 0.5 m. Calculate Bartek's muscle power , if the resistance force was 640N.

The power, P, expended or generated by moving a body to move a velocity, v, against a resistive force, F is given as

P = Fv

P = ?

F = 640 N

v = velocity = (0.5) ÷ (2×60) = 0.00416667 m/s

P = 640 × 0.0041666667 = 2.667 W

In Polish/Po polsku

Moc P, zużyta lub wytworzona przez poruszanie ciałem w celu przemieszczenia prędkości v przeciwko sile oporowej, F jest podawana jako

P = Fv

P = ?

F = 640 N

v = velocity = (0.5) ÷ (2×60) = 0.00416667 m/s

P = 640 × 0.0041666667 = 2.667 W

Hope this Helps!!!

Mam nadzieję że to pomoże!!

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<span>According to its definition, friction is generated when atoms interfere with each other on sliding surfaces.</span>
5 0
3 years ago
An electron is released from rest in a uniform electric field and accelerates to the north at a rate of 115 m/s squared. What ar
Iteru [2.4K]
The direction of the electric field would be south. 

qE/m = 115 
<span>       E = 115*m/q </span>
<span>           = 115 * 9.1 * 10^(-31) / 1.67*10^(-19) </span>
<span>           = 762.87 * 10^(-12) </span>
<span>           = 6.27 x 10^-10 N/C
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Hope this answers the question. Have a nice day. Feel free to ask more questions.
6 0
2 years ago
The question and options are in the photo
inessss [21]

Answer:

Option C

Explanation:

Metal conducts heat much better than wood does.

Since Azam's hand is hotter than room temperature, both the metal and the wood conduct heat away from it.

4 0
3 years ago
Calculate the magnitude of the electric field 2.00 m from a point charge of 5.00 mC (such as found on the terminal of a Van de G
MakcuM [25]

Answer:

The magnitude of the electric field is 1.124 X 10⁷ N/C

Explanation:

Magnitude of electric field is given as;

E = \frac{kq}{r^2} , N/C

where;

E is the magnitude of the electric field, N/C

q is the point charge, C

k is coulomb's constant, Nm²/C²

r is the distance of the point charge, m

Given;

q = 5mC = 5×10⁻³ C

r = 2m

k = 8.99 × 10⁹ Nm²/C²

Substitute these values and solve for magnitude of electric field E

E = \frac{kq}{r^2}  = \frac{(8.99 X10^9)(5X10^{-3})}{2^2}  = 1.124 X10^7 N/C

Therefore, the magnitude of the electric field is 1.124 X 10⁷ N/C

7 0
2 years ago
A constant unbalanced force is applied to an object for a period of time. Which graph best represents the acceleration of the ob
balu736 [363]

Answer:

Here is an image attached with similar questions.

The correct answer is D where acceleration is function of time.

Explanation:

The force acting on the object is constant.

There is no change in the application of forces.

And we know that Force is the product of acceleration and masses.

Newtons second law: F=m\times a

Regarding mass we know that it can neither be created nor destroyed.

So in F=m \times a we have two constant terms, constant divided by constant will give the same result as a=\frac{Force}{mass}

There is no change in acceleration (a) with respect to time (t).

So the most appropriate graph where time (t) is changing on x-axis but acceleration doesn't changes is D.

The graph will be similar to y=any\ constant and will be horizontal to the x-axis.

Option D depicts the same.

5 0
2 years ago
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