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Nana76 [90]
3 years ago
9

A 40kg load is raised to a height of 25m. If the operation requires 1 min, find the power required​

Physics
1 answer:
OlgaM077 [116]3 years ago
8 0

Answer:

163.33 Watts

Explanation:

From the question given above, the following data were obtained:

Mass (m) = 40 Kg

Height (h) = 25 m

Time (t) = 1 min

Power (P) =..?

Next, we shall determine the energy. This can be obtained as follow:

Mass (m) = 40 Kg

Height (h) = 25 m

Acceleration due to gravity (g) = 9.8 m/s²

Energy (E) =?

E = mgh

E = 40 × 9.8 × 255

E = 9800 J

Finally, we shall determine the power. This can be obtained as illustrated below:

Time (t) = 1 min = 60 s

Energy (E) = 9800 J

Power (P) =?

P = E/t

P = 9800 / 60

P = 163.33 Watts

Thus, the power required is 163.33 Watts

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For a concave mirror, when the object is at focus the image is formed at
Sergeeva-Olga [200]

Answer:

When the object is placed at the focus the image is formed at infinity.

Explanation:

When a ray passes through focus and incident on a concave mirror then it will travel parallel to principal axis after reflection.Hence the image is formed at infinity.

8 0
4 years ago
Assume that at sea-level the air pressure is 1.0 atm and the air density is 1.3 kg/m3.
scoundrel [369]

Answer

Pressure, P = 1 atm

air density, ρ = 1.3 kg/m³

a) height of the atmosphere when the density is constant

   Pressure at sea level = 1 atm = 101300 Pa

   we know

   P = ρ g h

   h = \dfrac{P}{\rho\ g}

   h = \dfrac{101300}{1.3\times 9.8}

          h = 7951.33 m

height of the atmosphere will be equal to 7951.33 m

b) when air density decreased linearly to zero.

  at x = 0  air density = 0

  at x= h   ρ_l = ρ_sl

 assuming density is zero at x - distance

 \rho_x = \dfrac{\rho_{sl}}{h}\times x

now, Pressure at depth x

dP = \rho_x g dx

dP = \dfrac{\rho_{sl}}{h}\times x g dx

integrating both side

P = g\dfrac{\rho_{sl}}{h}\times \int_0^h x dx

P =\dfrac{\rho_{sl}\times g h}{2}

 now,

h=\dfrac{2P}{\rho_{sl}\times g}

h=\dfrac{2\times 101300}{1.3\times 9.8}

  h = 15902.67 m

height of the atmosphere is equal to 15902.67 m.

6 0
3 years ago
99 POINTS
antoniya [11.8K]

Answer:

Send a pic of it so i can mark it.

Explanation:

8 0
3 years ago
A healthy astronaut's heart rate is 60 beats/min. Flight doctors on Earth can monitor an astronaut's vital signs remotely while
Anton [14]
GIVEN:
   60 beats per minute
   21 beats per minute
   find x= how fast would an astronaut be flying away
 1            x
-----   *  ------ = (60x = 21)  -------> 60x = 21    ------------>  x= 0.35 
60         21                                  -------   -----
                                                     60      60

The answer is 0.35 seconds which refers to how fast would an astronaut be flying away from the earth if he has a heart rate of 21 beats/min. 
5 0
3 years ago
Read 2 more answers
In a certain electrolysis experiment involving Al3+ ions, 60.2 g of Al is recovered when a current of 0.352 A is used. How many
yawa3891 [41]

Answer:

30.5 x 10³ min.

Explanation:

Atomic weight of aluminium = 27

For the reaction

Al⁺³ + 3e  = Al

3 mole of electron is required by 1 mole of aluminium

3 x 96500 C of charge is required for 27 gram of aluminium

60.2 g aluminium will require

\frac{3\times96500}{27} \times60.2

645.47 x 10³ C

So charge required = 645.47 x 10³ C

Now Charge  = Current x time

Current ( given ) = 0.352A

Time = charge / current = 645.47 x 10³ / .352

=1833 x 10³ s

30.5 x 10³ minutes

6 0
4 years ago
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