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3 years ago

Convection can occur in which two substances?

2 answers:

5 0

Convection can't occur in solids because the atoms in a solid objects cannot move from where they are and that's needed in the process of convection, so that rules out wood and ice so the answer is D) air and water.

algol133 years ago

3 0

D) Convection can't occur in solids because the atoms in a solid objects cannot move from where they are and that's needed in the process of convection, so that rules out wood and ice so the answer is D) air and water.

You might be interested in

If izzy mass is 0.3kg he applide 657.9n force what will be the accelration

Sholpan [36]

Answer:

The acceleration of the body, a = 2193 m/s²

Explanation:

Given,

The mass of the body, m = 0.3 kg

The force acting on the body, F = 657.9 N

The force acting on an object is proportional to the product of mass and acceleration of the body.

F = m x a

Therefore, the acceleration of the body is

a = F / m

= 657.9 N / 0.3 kg

= 2193 m/s²

Hence, the acceleration of the body, a = 2193 m/s²

4 0

4 years ago

A rock is stuck deep in the dirt. When you pull on the rock to remove it, what type of force are you exerting?

aleksandrvk [35]

It is Tension as the other 3 answer choices would not make sense. Compression would mean you are pressing the rock on both sides or in this case, pushing it into the dirt. It can't be nuclear force as you are pulling out a rock. Air resistance would not make sense either as there is no air involved in the scenario at all.

4 0

3 years ago

Read 2 more answers

A shell of mass m and speed v explodes into two identical fragments. If the shell was moving horizontally (the positive x direct

-Dominant- [34]

Answer:

The velocity of the other fragment immediately following the explosion is v .

Explanation:

Given :

Mass of original shell , m .

Velocity of shell , + v .

Now , the particle explodes into two half parts , i.e $\dfrac{m}{2}$ .

Since , no eternal force is applied in the particle .

Therefore , its momentum will be conserved .

So , Final momentum = Initial momentum

$mv=\dfrac{mv}{2}+\dfrac{mu}{2}\\\\u=v$

The velocity of the other fragment immediately following the explosion is v .

4 0

3 years ago

A ball with mass of 0.050 kg is dropped from a height of h1 = 1 .5 m. It collides with the floor, then bounces up to a height of

Iteru [2.4K]

Answer:

Explanation:

Impulse of reaction force of floor = change in momentum

Velocity of impact = √ 2gh₁

= √ 2 x 9.8 x 1.5 = 5.4 m /s.

velocity of rebound = √2gh₂

= √ 2x 9.8 x 1

= 4.427 m / s.

Initial momentum = .050 x 5.4 = .27 kg m/s

Final momentum = .05 x 4.427 = .22 kg.m/s

change in momentum = .27 - .22 = .05 kg m/s

Impulse = .05 kg m /s

Impulse = force x time

force = impulse / time

.05 / .015 = 3.33 N.

kinetic energy = 1/2 m v²

Initial kinetic energy = 1/2 x .05 x 5.4²

= 0.729 J

Final Kinetic Energy =1/2 x .05 x 4.427²

= 0.489 J

Change in Kinetic energy =0 .24 J

Lost kinetic energy is due to conversion of energy into sound light etc.

4 0

3 years ago

A negative charge of 20 x 10-6C and another charge of 15 x 10-6C are separated by as distance of 0.7 m.

denpristay [2]

Answer:

Approximately $5.5\; \rm N$ , assuming that the volume of these two charged objects is negligible.

Explanation:

Assume that the dimensions of these two charged objects is much smaller than the distance between them. Hence, Coulomb's Law would give a good estimate of the electrostatic force between these two objects regardless of their exact shapes.

Let $q_1$ and $q_2$ denote the magnitude of two point charges (where the volume of both charged object is negligible.) In this question, $q_1 = 20 \times 10^{-6}\; \rm C$ and $q_2 = 15 \times 10^{-6}\; \rm C$ .

Let $r$ denote the distance between these two point charges. In this question, $r = 0.7\; \rm m$ .

Let $k$ denote the Coulomb constant. In standard units, $k \approx 8.98755\times 10^{9}\; \rm kg \cdot m^{3}\cdot s^{-2}\cdot C^{-2}$ .

By Coulomb's Law, the magnitude of electrostatic force (electric force) between these two point charges would be:

$\begin{aligned}F &= \frac{k \cdot q_1 \cdot q_2}{r^{2}}\end{aligned}$ .

Substitute in the values and evaluate:

$\begin{aligned}F &= \frac{k \cdot q_1 \cdot q_2}{r^{2}}\\ &\approx 8.98755 \times 10^{9}\; \rm kg \cdot m^{3}\cdot s^{-2}\cdot C^{-2} \\ &\quad \times 20\times 10^{-6}\; \rm C\\ &\quad \times 15\times 10^{-6}\; \rm C \\ &\quad \times \frac{1}{{(0.7\; \rm m)}^{2}}\\ &\approx 5.5\; \rm N \end{aligned}$ .

8 0

3 years ago