What property of equality can be used to solve the equation. n

Far from any other masses, two masses, m1 and m2, are interacting gravitationally. The value for the mass of m1 suddenly doubles

NNADVOKAT [17]

<span>It also doubles The gravitational force between two masses is expressed as: F = G*m1*m2/r^2 where F = Force between the two masses m1 = Mass of object 1 m2 = Mass of object 2 r = distance between centers of object 1 and object 2 G = Gravitational constant The exact values of G, m1, m2, and r don't matter since all except for m1 is held constant. And when m1 suddenly doubles, the force attracting the two object to each other also doubles.</span>

5 0

4 years ago

HELPPPPPPPPPPPPPPPPPPPPPPPPPP

aev [14]

Answer:

C

Explanation:

through the desk....here desk is the student's medium to hear the sound. its oblivious because when he lifts his head away from the desk he hears nothing else

6 0

3 years ago

An infinite conducting cylindrical shell of outer radius r1 = 0.10 m and inner radius r2 = 0.08 m initially carries a surface ch

irinina [24]

Answer:

a) $\sigma_{\rm in} = -2.18~{\rm \mu C/m^2}$

b) $\sigma_{\rm out}= 1.12~{\rm \mu C/m^2}$

c) $E = \frac{\sigma(r_1 + r_2)}{\epsilon_0 r}$

Explanation:

Before the wire is inserted, the total charge on the inner and outer surface of the cylindrical shell is as follows:

$Q_{\rm in} = \sigma A_{\rm in} = \sigma(2\pi r_1 h) = (-0.35)(2\pi (0.08) h) = -0.175h~{\rm \mu C}$

$Q_{\rm out} = \sigma A_{\rm out} = \sigma(2\pi r_2 h) = (-0.35)(2\pi (0.1) h) = -0.22h~{\rm \mu C}$

Here, 'h' denotes the length of the cylinder. The total charge of the cylindrical shell is -0.395h μC.

When the thin wire is inserted, the positive charge of the wire attracts the same amount of negative charge on the inner surface of the shell.

$Q_{\rm wire} = \lambda h = 1.1h~{\rm \mu C}$

a) The new charge on the inner shell is -1.1h μC. Therefore, the new surface charge density of the inner shell can be calculated as follows:

$\sigma_2 = \frac{Q_{\rm in}}{2\pi r_1h} = \frac{-1.1h}{2\pi r_1 h} = \frac{-1.1}{2\pi(0.08)} = -2.18~{\rm \mu C/m^2}$

b) The new charge on the outer shell is equal to the total charge minus the inner charge. Therefore, the new charge on the outer shell is +0.705 μC.

The new surface charge density can be calculated as follows:

$\sigma_{\rm out}= \frac{Q_{\rm out}}{2\pi r_2h} = \frac{0.705h}{2\pi r_2 h} = \frac{0.705}{2\pi(0.1)} = 1.12~{\rm \mu C/m^2}$

c) The electric field outside the cylinder can be found by Gauss' Law:

$\int{\vec{E}d\vec{a} = \frac{Q_{enc}}{\epsilon_0}$

We will draw an imaginary cylindrical shell with radius r > r2. The integral in the left-hand side will be equal to the area of the imaginary surface multiplied by the E-field.

$E(2\pi r h) = \frac{Q_{\rm enc}}{\epsilon_0}\\E2\pi rh = \frac{\sigma 2\pi (r_1 + r_2)h}{\epsilon_0}\\E = \frac{\sigma(r_1 + r_2)}{\epsilon_0 r}$

4 0

3 years ago

A lump of clay whose rest mass is 4 kg is travelling at three-fifths the speed of light when it collides head-on with an identic

Harman [31]

Answer:

mass of the composite lump is 10 kg

Explanation:

given data

mass = 4 kg

to find out

mass of composite lump

solution

we know energy is conserved so

so m1 = m2 = m0 that is 4kg

and

E(1) release+ E(2) release = E(1,2) rest

so γ(1)m(1)c² + γ(2)m(2)c² = Mc² ..........................1

that why here

|v(1)| = |v(2)| = 3/5 c ......................2

and

γ = 1 / √(1 − v²/c²) .......................3

put here v = 3 and c is 5

γ = 1 /√(1 − 9/25)

γ = 5/4

so

γ(1) = γ(2) = γ = 5/4

so from equation 1

γ(1)m(1)c² + γ(2)m(2)c² = Mc²

M = 2γm0

M = 2(5/4 )(4)

M = 10 kg

so mass of the composite lump is 10 kg

7 0

4 years ago

Shaking a long metal chain up and down can create a _____ wave in the chain.

Aloiza [94]

I think Transverse.

8 0

3 years ago

Read 2 more answers

What property of equality can be used to solve the equation. n – 14 = 25