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HACTEHA [7]
2 years ago
14

What is the factor involved in increasing an object’s inertia?

Physics
1 answer:
Monica [59]2 years ago
7 0
Inertia= inactivity
The factor involved in increasing inertia would be "mass". The more mass something has, the more inertia. 
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According to Dr. paul Narguizian professor of Biology and Science Education at California State University, ______ are generaliz
Mazyrski [523]

Answer:

I believe the correct answer would be A :)

Explanation:

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1 year ago
Most binary systems with an invisible companion contain a large, bright star and a small, dim star hidden by the light of its la
Vlada [557]

Answer:

Brain signals are converted

8 0
2 years ago
An LR circuit contains an ideal 60-V battery, a 42-H inductor having no resistance, a 24-ΩΩ resistor, and a switch S, all in ser
s2008m [1.1K]

Answer:

1.6 s

Explanation:

To find the time in which the potential difference of the inductor reaches 24V you use the following formula:

V_L=V_oe^{-\frac{Rt}{L}}

V_o: initial voltage = 60V

R: resistance = 24-Ω

L: inductance = 42H

V_L: final voltage = 24 V

You first use properties of the logarithms to get time t, next, replace the values of the parameter:

\frac{V_L}{V_o}=e^{-\frac{Rt}{L}}\\\\ln(\frac{V_L}{V_o})=-\frac{Rt}{L}\\\\t=-\frac{L}{R}ln(\frac{V_L}{V_o})\\\\t=-\frac{42H}{24\Omega}ln(\frac{24V}{60V})=1.6s

hence, after 1.6s the inductor will have a potential difference of 24V

3 0
2 years ago
10 kg cart and a 5 kg cart are placed on identical surfaces. The 10 kg cart experiences a net force of 12 N to the left, while t
SashulF [63]
F=ma

For the first (10kg) cart,
12=10a
a=6/5 m/s^2 to the left

For the second (5kg) cart,
8=5a
a=8/5 m/s^2 to the left

Therefore, the lighter (5kg) cart experiences a greater acceleration.
7 0
2 years ago
A real heat engine operates between temperatures tc and th. during a certain time, an amount qc of heat is released to the cold
tino4ka555 [31]

q_{c} = Heat released to cold reservoir

q_{h} = Heat released to hot reservoir

W_{max} = maximum amount of work

t_{c} = temperature of cold reservoir

t_{h} = temperature of hot reservoir

we know that

\frac{q_{c}}{q_{h}}=\frac{t_{c}}{t_{h}}

q_{h} = (\frac{t_{h}}{t_{c}})q_{c}                                eq-1

maximum work is given as

W_{max} = q_{h} - q_{c}

using eq-1

W_{max} =  (\frac{t_{h}}{t_{c}})q_{c} - q_{c}



6 0
2 years ago
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