2 years ago

The lift took 5 seconds to arrive at the second floor. What was the power of lift?

Physics

2 answers:

Lubov Fominskaja [6]2 years ago

8 0

Answer:

Explanation:

Sana maka tulong huhuhuhuh

gtnhenbr [62]2 years ago

7 0

Answer:

It should be D. 3,700 W

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2. A ball is dropped from rest. The acceleration due to gravity is 10m/s? and the time it

d1i1m1o1n [39]

Answer:

STEP BY STEP

V = S/T

V= 10/5

V = 2 m/s

3 0

3 years ago

How to solve the problem

erica [24]

If you're looking for distance, you have to multiply the time and speed.

0.75 × 45 = 33.75

The rabbit hopped 33.75 m.

5 0

3 years ago

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A 60 kg skier starts from rest at the top of a frictionless slope of height of 35 meters. The velocity at the bottom is v. If a

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Answer:

Explanation:

Speed of skier without parachute

= √ 2gh

= √ 2 x 9.8 x 35

= 26.2 m / s

Speed of skier with parachute

net force downwards

mg - 200

= 60 x 9.8 -200

= 388 N

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a = 6.47 m / s

v = √ 2ah

= √ 2 x 6.47 x 35

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3 years ago

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A racing car travels on a circular track of radius 158 m, moving with a constant linear speed of 19.1 m/s. Find its angular spee

SOVA2 [1]

Answer:

$\omega=0.12\frac{rad}{s}$

Explanation:

In a uniform circular motion, since a complete revolution represents 2π radians, the angular velocity, which is defined as the angle rotated by a unit of time, is given by:

$\omega=\frac{2\pi}{T}(1)$

Here T is the period, that is, the time taken to complete onee revolution:

$T=\frac{2\pi r}{v}(2)$

Replacing (2) in (1):

$\omega=\frac{2\pi}{\frac{2\pi r}{v}}=\frac{v}{r}\\\omega=\frac{19.1\frac{m}{s}}{158m}\\\omega=0.12\frac{rad}{s}$

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3 years ago

Work done depends on

natima [27]

Answer:

C. Both force and displacement

Explanation:

Hope this helps

3 0

2 years ago