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2 years ago

What is inelastic deformation?

Physics

1 answer:

EleoNora [17]2 years ago

7 0

The answer is A. Deformation is where the object doesnt return to its original shape.

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A resistor is connected in series with a power supply of 23.90 V. The current measure is 0.60 A. What is the resistance (in Ω) o

Yuliya22 [10]

The resistance in this circuit is 39.8 ohms.

Explanation:

Any circuit having resistor, battery and ammeter connected in series will obey the ohm's law in basic case. So according to the Ohm's law, the current flowing in the circuit through the ammeter will be equal to the voltage shown in the voltmeter or battery and resistor is the proportionality constant. So with this law

V = IR

So, Resistance R = V/I

As the voltage is given as 23.90 V and the current is given as 0.6 A, then resistance is

R = 23.90/0.6 = 39.8 ohms.

So, the resistance in this circuit is 39.8 ohms.

8 0

3 years ago

What is Gravity? PLEASE ANSWER

jolli1 [7]

Answer:

Gravity is the force by which a planet or other body draws objects toward its center. The force of gravity keeps all of the planets in orbit around the sun.

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7 0

3 years ago

Read 2 more answers

An old light bulb draws only 54.3 W, rather than its original 60.0 W, due to evaporative thinning of its filament. By what facto

Lemur [1.5K]

Answer:

The factor of the diameter is 0.95.

Explanation:

Given that,

Power of old light bulb = 54.3 W

Power = 60 W

We know that,

The resistance is inversely proportional to the diameter.

$R\propto\dfrac{1}{D}$

The power is inversely proportional to the resistance.

$P\propto\dfrac{1}{R}$

$P\propto D^2$

We need to calculate the factor of the diameter of the filament reduced

Using relation of power and diameter

$\dfrac{P_{i}}{P_{f}}=\dfrac{D_{i}^2}{D_{f}^2}$

Put the value into the formula

$\dfrac{D_{i}^2}{D_{f}^2}=\dfrac{54.3}{60}$

$\dfrac{D_{i}}{D_{f}}=0.95$

$D_{i}=0.95 D_{f}$

Hence, The factor of the diameter is 0.95.

7 0

3 years ago

Read 2 more answers

A nonconducting sphere of diameter 10.0 cm carries charge distributed uniformly inside with charge density of +5.50 µC/m3 . A pr

VLD [36.1K]

Answer:

t = 2.58*10^-6 s

Explanation:

For a nonconducting sphere you have that the value of the electric field, depends of the region:

$rR:\\\\E=k\frac{Q}{r^2}$

k: Coulomb's constant = 8.98*10^9 Nm^2/C^2

R: radius of the sphere = 10.0/2 = 5.0cm=0.005m

In this case you can assume that the proton is in the region for r > R. Furthermore you use the secon Newton law in order to find the acceleration of the proton produced by the force:

$F=m_pa\\\\qE=m_pa\\\\k\frac{qQ}{r^2}=m_pa\\\\a=k\frac{qQ}{m_pr^2}$

Due to the proton is just outside the surface you can use r=R and calculate the acceleration. Also, you take into account the charge density of the sphere in order to compute the total charge:

$Q=\rho V=(5.5*10^{-6}C/m^3)(\frac{4}{3}\pi(0.05m)^3)=2.87*10^{-9}C\\\\a=(8.98*10^9Nm^2/C^2)\frac{(1.6*10^{-19}C)(2.87*10^{-9}C)}{(1.67*10^{-27}kg)(0.05m)^2}=9.87*10^{11}\frac{m}{s^2}$