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DerKrebs [107]
3 years ago
5

Calculate the time it would take a motor with a power of 0.6kW to lift a suitcase of 20kg a distance of 50cm

Physics
1 answer:
True [87]3 years ago
7 0

Answer:

0.163 s

Explanation:

Appying,

P = mgh/t................ Equation 1

Where P = power of the motor, m = mass of the suitcase, h = vertical distance, t = time, g = acceleration due to gravity.

make t the subject of the equation,

t = mgh/P................ Equation 2

Given: m = 20 kg, h = 50 cm = 0.5 m, P = 0.6 kW = 600 W

Constant: g = 9.8 m/s²

Substitute these values into equation 2

t = (20×0.5×9.8)/600

t = 0.163 s

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Volume=mass/density

volume=455.6/19.3

volume=23.6 mL

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3 years ago
A 150g copper bowl contains 220g of water, both at 20.0oC, A very hot 300 g copper cylinder is dropped into the water, causing t
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A typical person's eye is 2.5 cm in diameter and has a near point (the closest an object can be and still be seen in focus) of 2
Paraphin [41]

Answer:

2.27 cm

2.5 cm

Explanation:

u = Object distance =  25 cm

v = Image distance = 2.5 cm

f = Focal length

Lens Equation

\frac{1}{f}=\frac{1}{u}+\frac{1}{v}\\\Rightarrow \frac{1}{f}=\frac{1}{25}+\frac{1}{2.5}\\\Rightarrow \frac{1}{f}=\frac{11}{25}\\\Rightarrow f=\frac{25}{11}=2.27\ cm

The minimum effective focal length of the focusing mechanism of the typical eye is 2.27 cm

when u=\infty

\frac{1}{f}=\frac{1}{u}+\frac{1}{v}\\\Rightarrow \frac{1}{f}=\frac{1}{\infty}+\frac{1}{2.5}\\\Rightarrow \frac{1}{f}=\frac{1}{2.5}\\\Rightarrow f=\frac{2.5}{1}=2.5\ cm

The maximum effective focal length of the focusing mechanism of the typical eye is 2.5 cm

3 0
3 years ago
Give two important reasons to do a rough sketch of the crime scene?
Sergio [31]
Tampered evidence/tampered corpse?
6 0
4 years ago
Question 11 (3 points)
kykrilka [37]

Answer:

С. 10.41 Ampere

Explanation:

Given that,

Power, P = 125 watts

Voltage, V = 12 volts

We need to find current. Power in terms of current and voltage is given by :

P = VI, I = current

I=\dfrac{P}{V}\\\\I=\dfrac{125\ W}{12\ V}\\\\I=10.41\ A

So, the current is 10.41 A.

5 0
3 years ago
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