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3 years ago

on cloudless nighs, heat escapes from our planet and goes into empty space A.conduction B.convection C.radiation

2 answers:

6 0

Convention.

On cloudless nighs, heat escapes from our planet and goes into empty space convection.

PIT_PIT [208]3 years ago

5 0

On cloudless nighs, heat escapes from our planet and goes into empty space is an example of convection.

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A 1090 kg car moving +11.0 m/s

stiks02 [169]

The mass of the second car is 1434.21 kg

Explanation:

Using law of conservation of momentum,

$m_{1} u_{1}+m_{2} u_{2}=\left(m_{1}+m_{2}\right) v$

Given:

$m_{1}$ = 1090 kg

$u_{1}$ = 11 m/s

$u_{2}$ = 0

v = 4.75 m/s

We need to find $m_{2}$

When substituting the given values in the above equation, we get

$(1090 \times 11)+\left(m_{2} \times 0\right)=\left(1090+m_{2}\right) 4.75$

$11990=5177.5+4.75 m_{2}$

$4.75 m_{2}=11990-5177.5$

$4.75 m_{2}=6812.5$

$m_{2}=\frac{6812.5}{4.75}=1434.21 \mathrm{kg}$

6 0

3 years ago

PLEASE PLEASE HELP ME!

Vlad1618 [11]

B: heat is transferred as thermal energy by the interaction of moving particles

6 0

3 years ago

A uniformly charged ball of radius a and charge –Q is at the center of a hollowmetal shell with inner radius b and outer radius

vlabodo [156]

Answer:

r < a:

$E = \frac{1}{4\pi \epsilon_0}\frac{Qr}{a^3}$

r = a:

$E = \frac{1}{4\pi a^2}\frac{Q}{\epsilon_0}$

a < r < b:

$E = \frac{1}{4\pi \epsilon_0}\frac{Q}{r^2}$

r = b:

$E = \frac{1}{4\pi b^2}\frac{Q}{\epsilon_0}$

b < r < c:

E = 0

r = c:

$E = \frac{1}{4\pi \epsilon_0}\frac{Q}{c^2}$

r < c:

$E = \frac{1}{4\pi \epsilon_0}\frac{Q}{r^2}$

Explanation:

Gauss' Law will be applied to each region to find the E-field.

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

An imaginary sphere is drawn with radius r, which is equal to the point where the E-field is asked. The area of this imaginary sphere is multiplied by E, and this is equal to the charge enclosed by this imaginary surface divided by ε0.

r<a:

Since the ball is uniformly charged and not hollow, then the enclosed charge can be found by the following method: If the total ball has a charge -Q and volume V, then the enclosed part of the ball has a charge Q_enc and volume V_enc. Then;

$\frac{Q}{V} = \frac{Q_{encl}}{V_{encl}}\\\frac{Q}{\frac{4}{3}\pi a^3} = \frac{Q_{encl}}{\frac{4}{3}\pi r^3}\\Q_{encl} = \frac{Qr^3}{a^3}$

Applying Gauss' Law:

$E4\pi r^2 = \frac{-Qr^3}{\epsilon_0 a^3}\\E = -\frac{1}{4\pi \epsilon_0}\frac{Qr}{a^3}\\E = \frac{r}{4\pi a^3}\frac{Q}{\epsilon_0}$

The minus sign determines the direction of the field, which is towards the center.

At r = a:

$E = \frac{1}{4\pi a^2}\frac{Q}{\epsilon_0}$

At a < r < b:

The imaginary surface is drawn between the inner surface of the metal sphere and the smaller ball. In this case the enclosed charge is equal to the total charge of the ball, -Q.

$E4\pi r^2 = \frac{-Q}{\epsilon_0}\\E = -\frac{1}{4\pi \epsilon_0}\frac{Q}{r^2}$

At r = b:

$E = -\frac{1}{4\pi b^2}\frac{Q}{\epsilon_0}$

Again, the minus sign indicates the direction of the field towards the center.

At b < r < c:

The hollow metal sphere has a net charge of +2Q. Since the sphere is a conductor, all of its charges are distributed across its surface. No charge is present within the sphere. The smaller ball has a net charge of -Q, so the inner surface of the metal sphere must possess a net charge of +Q. Since the net charge of the metal sphere is +2Q, then the outer surface of the metal should possess +Q.

Now, the imaginary surface is drawn inside the metal sphere. The total enclosed charge in this region is zero, since the total charge of the inner surface (+Q) and the smaller ball (-Q) is zero. Therefore, the Electric region in this region is zero.

E = 0.

At r < c:

The imaginary surface is drawn outside of the metal sphere. In this case, the enclosed charge is +Q (The metal (+2Q) plus the smaller ball (-Q)).

$E4\pi r^2 = \frac{Q}{\epsilon_0}\\E = \frac{1}{4\pi \epsilon_0}\frac{Q}{r^2}$

At r = c:

$E = \frac{1}{4\pi \epsilon_0}\frac{Q}{c^2}$

3 0

3 years ago

A 4.0-kg object is supported by an aluminum wire of length 2.0 m and diameter 2.0 mm. How much will the wire stretch?

forsale [732]

Answer:

The extension of the wire is 0.362 mm.

Explanation:

Given;

mass of the object, m = 4.0 kg

length of the aluminum wire, L = 2.0 m

diameter of the wire, d = 2.0 mm

radius of the wire, r = d/2 = 1.0 mm = 0.001 m

The area of the wire is given by;

A = πr²

A = π(0.001)² = 3.142 x 10⁻⁶ m²

The downward force of the object on the wire is given by;

F = mg

F = 4 x 9.8 = 39.2 N

The Young's modulus of aluminum is given by;

$Y = \frac{stress}{strain}\\\\Y = \frac{F/A}{e/L}\\\\Y = \frac{FL}{Ae} \\\\e = \frac{FL}{AY}$

Where;

Young's modulus of elasticity of aluminum = 69 x 10⁹ N/m²

$e = \frac{FL}{AY} \\\\e = \frac{(39.2)(2)}{(3.142*10^{-6})(69*10^9)} \\\\e = 0.000362 \ m\\\\e = 0.362 \ mm$

Therefore, the extension of the wire is 0.362 mm.

8 0

3 years ago

Solved, can someone check it over!

kakasveta [241]

Answer:

its good no need to change anything :))

4 0

3 years ago