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ruslelena [56]
3 years ago
12

A crane takes 2.0 minutes to lift a load to the top of a building. The change in gravitational potential energy of the load is 3

60kJ.
What is the useful power output of the crane?

A. 3.0kW

B. 180kW

C. 720kW

D. 43,200kW
Physics
2 answers:
weeeeeb [17]3 years ago
8 0

Answer:

The answer is option A

Explanation:

why?

because to get the power, you need this equation. P= Work done/Time taken.

the unit of time are seconds, and the unit of work is kJ. So, the equation is=

360kJ/120s = 3kW

steposvetlana [31]3 years ago
7 0

Answer: 3000 W = 3.0 kW

Explanation: power P = W /t = 360 000 J / 120s

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A cat is chasing a mouse across a 1.3 m tall dining table. The mouse darts to the side and the cat accidentally slides off, land
erastovalidia [21]

Answer:

1) 0.51 seconds.

2) 1.45 m/s.

Explanation:

given, height from which cat falls = 1.3 m

we know that, s = ut + \frac{1}{2}at².

here if we consider cat moment only in downward direction,

intial velocity of cat in downward direction , u = 0.

so, time, t = \sqrt{\frac{2h}{g} }.

⇒ t = \sqrt{\frac{2(1.3)}{9.81} } = 0.51 seconds.

t = 0.51 seconds.

now, consider cat moment only in forward direction

s = ut , since acceleration is zero in forward direction

⇒ u = \frac{s}{t}.

so, u = \frac{0.75}{0.51} = 1.45 m/s .

6 0
3 years ago
A proton is accelerated from rest through a potential difference V0 and gains a speed v0. If it were accelerated instead through
Neporo4naja [7]

Answer:

Explanation:

Let the charge on proton be q .

energy gain by proton in a field having potential difference of V₀

= V₀ q

Due to gain of energy , its kinetic energy becomes 1/2 m v₀²

where m is mass and v₀ is velocity of proton

V₀ q = 1/2 m v₀²

In the second case , gain of energy in electrical field

= 2 V₀q , if v be the velocity gained in the second case

2 V₀q = 1/2 m v²

1/2 m v² = 2 V₀q = 2 x 1/2 m v₀²

mv² = 2  m v₀²

v = √2 v₀

6 0
3 years ago
How is tidal energy obtained
yKpoI14uk [10]
Under water turbans that are placed at the above to middle of the ocean they are used to capture kinetic motion
7 0
4 years ago
Dolphin echolocation is similar to ultrasound. Reflected sound waves
s344n2d4d5 [400]

Answer:

Waves with high frequencies have shorter wavelengths that work better  than low frequency waves for successful echolocation.

Explanation:

To understand why high-frequency waves work better  than low frequency waves for successful echolocation, first we have to understand the relation between frequency and wavelength.

The relation between frequency and wavelength is given by

λ = c/f

Where λ is wavelength, c is the speed of light and f is the frequency.

Since the speed of light is constant, the wavelength and frequency are inversely related.

So that means high frequency waves have shorter wavelengths, which is the very reason for the successful echolocation because waves having shorter wavelength are more likely to reach and hit the target and then reflect back to the dolphin to form an image of the object.

Thus, waves with high frequencies have shorter wavelengths that work better  than low frequency waves for successful echolocation.

3 0
4 years ago
The main purpose of an air bag is to stop a passenger during a car accident in a greater amount of time than if the air bag were
Simora [160]

Answer:

a) 45571 N  

b) 22786 N

c) 4557 N

Explanation:

  • Since the goal of the airbag is helping the person to stop after the collision in a greater time, this means that the change in momentum must finish when this is just zero.
  • In other words, the change in momentum, must be equal to the initial one, but with opposite sign.

       \Delta p = - p_{o} = -m*v = -55 kg*29m/s = -1595 kgm/s (1)

  • Now, just applying the original form of  Newton's 2nd Law, we know that this change in momentum must be equal to the impulse needed to stop the person:

       \Delta p = F* \Delta t  (2)

  • So, as we know the magnitude of Δp from (1) and we have different Δt as givens, we can get the different values of F (in magnitude) required to stop the person for each one of them, as follows:

       F_{1} = \frac{\Delta p}{\Delta t_{1}} = \frac{1595kgm/s}{0.035s} = 45571 N (3)

       F_{2} = \frac{\Delta p}{\Delta t_{2}} = \frac{1595kgm/s}{0.07s} = 22786 N (4)

       F_{3} = \frac{\Delta p}{\Delta t_{3}} = \frac{1595kgm/s}{0.35s} = 4557 N (5)

4 0
3 years ago
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