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laiz [17]
3 years ago
6

1. A 1000 kg equipment lift travels up a 200 m shaft in 45 seconds. Assuming the speed is constant, what is the

Physics
1 answer:
ANTONII [103]3 years ago
8 0

Answer:

43,555 W

Explanation:

Since the speed is constant, this means that the force applied to lift the equipment is equal to the weight of the equipment. So we can write:

F=mg

where

m = 1000 kg is the mass

g=9.8 m/s^2 is the acceleration due to gravity

So,

F=(1000)(9.8)=9800 N

Then the work done in lifting the equipment is:

W=Fd

where

d = 200 m is the displacement of the equpment

Substituting,

W=(9800)(200)=1.96\cdot 10^6  J

Finally, the power used to lift the equipment is the ratio between the work done and time taken:

P=\frac{W}{t}

where

W=1.96\cdot 10^6 J

t = 45 s is the time taken

Solving,

P=\frac{1.96\cdot 10^6}{45}=43,555 W

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Calculate the force of gravity between a 2.50 kg newborn baby and a 80.0 kg doctor standing 0.250 m away. G = 6.67 E -11 N*m2/kg
Alex787 [66]

Answer: 2.13 × 10⁻⁷ N

Explanation:

Gravitational force exists between any two bodies having mass.

Force of gravity is given by:

F =G\frac{Mm}{r^2}

It is given that, mass of newborn baby is M = 2.50 kg

Mass of the doctor, m = 80.0 kg

Distance between the two, r = 0.250 m

Gravitational constant, G = 6.67 × 10⁻¹¹ N m²/kg²

⇒F = (6.67 × 10⁻¹¹ N m²/kg² × 2.50 kg × 80.0 kg )÷ (0.250 m)² = 2.13 × 10⁻⁷ N

Thus, the force of gravity between new born baby and doctor is 2.13 × 10⁻⁷ N.

7 0
3 years ago
What’s the voltage of a battery in a circuit with resistance of 3 ohms and current of 5 amps?
alexira [117]

Answer: The correct answer is-15 Volts.

Explanation-

Voltage of a battery can be defined as the difference in electric potential that lies between the positive and negative terminals of a battery.

It can be calculated using Ohm's law, which states that the electric potential difference between two points on a circuit is equal to the product of the current that flows between the two points (I) and the total resistance that sis present between the two points. It can be mathematically depicted as-

ΔV = I • R  

Putting the value of 'I' and 'R', we get-

ΔV = 5 X 3

    =  15 V

8 0
3 years ago
The human ear canal is, on average, 2.5cm long and aids in hearing by acting like a resonant cavity that is closed on one end an
Troyanec [42]

Answer:

3400 Hz

Explanation:

We know that

1 cm = 0.01 m

L = Length of the human ear canal = 2.5 cm = 0.025 m

V = Speed of sound = 340 ms⁻¹

f = First resonant frequency

The human ear canal behaves as a closed pipe and for a closed pipe, nth resonant frequency is given as

f = \frac{(2n - 1)V}{4L}

for first resonant frequency, we have n = 1

Inserting the values

f = \frac{(2(1) - 1) 340}{4(0.025)}

f = \frac{340}{4(0.025)}

f = 3400 Hz

4 0
3 years ago
An oscillator consists of a block attached to a spring (k = 427 N/m). At some time t, the position (measured from the system's e
White raven [17]

Answer:

a) 4.49Hz

b) 0.536kg

c) 2.57s

Explanation:

This problem can be solved by using the equation for he position and velocity of an object in a mass-string system:

x=Acos(\omega t)\\\\v=-\omega Asin(\omega t)\\\\a=-\omega^2Acos(\omega t)

for some time t you have:

x=0.134m

v=-12.1m/s

a=-107m/s^2

If you divide the first equation and the third equation, you can calculate w:

\frac{x}{a}=\frac{Acos(\omega t)}{-\omega^2 Acos(\omega t)}\\\\\omega=\sqrt{-\frac{a}{x}}=\sqrt{-\frac{-107m/s^2}{0.134m}}=28.25\frac{rad}{s}

with this value you can compute the frequency:

a)

f=\frac{\omega}{2\pi}=\frac{28.25rad/s}{2\pi}=4.49Hz

b)

the mass of the block is given by the formula:

f=\frac{1}{2\pi}\sqrt{\frac{k}{m}}\\\\m=\frac{k}{4\pi^2f^2}=\frac{427N/m}{(4\pi^2)(4.49Hz)^2}=0.536kg

c) to find the amplitude of the motion you need to know the time t. This can computed by dividing the equation for v with the equation for x and taking the arctan:

\frac{v}{x}=-\omega tan(\omega t)\\\\t=\frac{1}{\omega}arctan(-\frac{v}{x\omega })=\frac{1}{28.25rad/s}arctan(-\frac{-12.1m/s}{(0.134m)(28.25rad/s)})=2.57s

Finally, the amplitude is:

x=Acos(\omega t)\\\\A=\frac{0.134m}{cos(28.25rad/s*2.57s )}=0.45m

5 0
3 years ago
What would be the current of the same wire if it was moved up towards the magnet instead of through the magnetic field?
Aleks04 [339]

Answer:

0 v

Explanation:

3 0
3 years ago
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