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Kruka [31]
3 years ago
13

What material of mass 39g,has a volume of 5cm^3

Physics
1 answer:
insens350 [35]3 years ago
4 0
The material of density 7.8 g/cm³
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Which statement best explains how a planet affects the orbit of a comet as the comet passes by the planet?
mezya [45]
Comets orbit the sun just like planets do. Except a comet usually has a very elongated orbit. Thanks to the laws of gravity comets obey the same laws. A comets orbit takes it very close to the sun and then far away again.
8 0
2 years ago
A 50-gram sample of water is initially at a temperature of 22 °C. The sample is heated until the temperature is 32 °C The specif
forsale [732]

Answer:

500cal

Explanation:

Given parameters:

Mass of water  = 50g

Initial temperature  = 22°C

Final temperature  = 32°C

Specific heat of water  = 1cal/g

Unknown:

Amount of heat absorbed by the water in calories  = ?

Solution:

To solve this problem, we use the expression below:

       H  = m c Ф

H is the amount of heat absorbed

m is the mass

c is the specific heat capacity

Ф is the temperature change

       H  = 50 x 1 x (32  - 22)  = 500cal

5 0
3 years ago
Please help: I don't know how to do these problems
antiseptic1488 [7]
d =2.55.68m and t = 11.36s
In my opinion
3 0
3 years ago
In a series RLC resonance circuit, the resonance frequency f0 = 700 kHz. The resistor R = 10 Ohm. The specified bandwidth (BW) s
sladkih [1.3K]

Answer:

  • quality factor (Q) = 69.99
  • inductor = 1.591 x 10⁻⁴ H
  • capacitor = 3.248 x 10⁻¹⁰ F

Explanation:

Given;

resonance frequency (F₀) = 700 kHz

resistor, R =  10 Ohm

bandwidth (BW) = 10 kHz

bandwidth (BW)  = \frac{R}{2\pi L}

BW = \frac{R}{2\pi L}

make L (inductor) the subject of the formula

L = \frac{R}{2\pi *BW}  =  \frac{10}{2\pi *10,000} =1.591 *10^{-4} \ H = \ 0.1591\ mH

F_o =\frac{1}{2\pi\sqrt{LC} } \\\\\sqrt{LC} = \frac{1}{2\pi F_o} \\\\LC = \frac{1}{4\pi ^2F_o^2}= \frac{1}{4\pi ^2(700,000)^2} = 5.168*10^{-14}

make C (capacitor)  the subject of the formula

C = \frac{5.168*10^{-14}}{1.591*10^{-4}} = 3.248*10^{-10} \ F = \ 3.248*10^{-4} \ \mu F

quality factor (Q) = \frac{1}{R} \sqrt{\frac{L}{C}} \ = \frac{1}{10} \sqrt{\frac{1.591*10^{-4}}{3.248*10^{-10}}}=69.99

quality factor (Q) =  69.99

5 0
4 years ago
A small lab cart and one of larger mass collide and rebound off each other. Which of them has the greater average force on it du
Elza [17]

When a small cart collide with a large mass then during collision they must be in contact with each other for some interval of time

During this contact interval we can say they will exert normal force on each other

This normal force is always equal and opposite on two balls which means this force will follow Newton's III law

It will be same in magnitude but opposite in the direction

So here correct answer would be

<u><em>They both experience the same magnitude of the collision force.</em></u>

4 0
3 years ago
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