Ask question

Ask question

All categories

3241004551 [841]

2 years ago

15

What is an example of convection currents? a. marshmallows toasting over a campfire b. a pot being heated by an electric burner,

feet getting hot when stepping across sand, a radiator emitting warm air and drawing in cool airs Only one choice on edge

2 answers:

Svetlanka [38]2 years ago

8 0

Answer: B

Explanation:

Saw it on multiple sources and i got it right on Edge 2020

SCORPION-xisa [38]2 years ago

7 0

Answer:

B

Explanation:

because, convection is the transfer of heat between fluid substances/materials

You might be interested in

A child is riding a bike at a speed of 6m/s with a total kinetic energy of 1224J. If the mass of the child is 30kg, what is the

UkoKoshka [18]

Answer:

Mass of bike = 38 kg.

Explanation:

Kinetic energy is given by the expression, $KE = \frac{1}{2} mv^2$ , where m is the mass and v is the velocity.

Here speed of child riding bike = 6 m/s

Mass of child = 30 kg

Total kinetic energy = 1224 J

Let the mass of bike be, m kg

So, total mass of child and bike = (m + 30) kg

Substituting,

$1224 = \frac{1}{2}* (m+30)*6^2\\ \\ m+30=68\\ \\ m=38kg$

So, mass of bike = 38 kg.

3 0

2 years ago

Mickey, a daredevil mouse of mass m , m, is attempting to become the world's first "mouse cannonball." He is loaded into a sprin

Sati [7]

Answer:

h = v₀² / 2g , h = k/4g x²

Explanation:

In this exercise we can use the law of conservation of energy at two points, the lowest, before the shot and the highest point that the mouse reaches

Starting point. Lower compressed spring

Em₀ = K = ½ m v²

Final point. Highest on the path

$Em_{f}$ = U = mg h

As or no friction the energy is conserved

Em₀ = Em_{f}

½ m v₀²² = m g h

h = v₀² / 2g

We can also use as initial energy the energy stored in the spring that will later be transferred to the mouse

½ k x² = 2 g h

h = k/4g x²

8 0

2 years ago

Read 2 more answers

What is (Fnet3)x, the x-component of the net force exerted by these two charges on a third charge q3 = 55.0 nC placed between q1

notka56 [123]

Complete Question

Part of the question is shown on the first uploaded image

The rest of the question

What is (Fnet3)x, the x-component of the net force exerted by these two charges on a third charge q3 = 55.0 nC placed between q1 and q2 at x3 = -1.220 m ? Your answer may be positive or negative, depending on the direction of the force. Express your answer numerically in newtons to three significant figures.

Answer:

The net force exerted on the third charge is $F_{net}= 3.22*10^{-5} \ J$

Explanation:

From the question we are told that

The third charge is $q_3 = 55 nC = 55 *10^{-9} C$

The position of the third charge is $x = -1.220 \ m$

The first charge is $q_1 = -16 nC = -16 *10^{-9} \ C$

The position of the first charge is $x_1 = -1.650m$

The second charge is $q_2 = 32 nC = 32 *10^{-9} C$

The position of the second charge is $x_2 = 0 \ m$

The distance between the first and the third charge is

$d_{1-3} = -1.650 -(-1.220)$

$d_{1-3} = -0.43 \ m$

The force exerted on the third charge by the first is

$F_{1-3} = \frac{k q_1 q_3}{d_{1-3}^2}$

Where k is the coulomb's constant with a value $9*10^{9} \ kg\cdot m^3\cdot s^{-4}\cdot A^2.$

substituting values

$F_{1-3} = \frac{9*10^{9}* 16 *10^{-9} * (55*10^{-9})}{(-0.43)^2}$

$F_{1-3} = 4.28 *10^{-5} \ N$

The distance between the second and the third charge is

$d_{2-3} = 0- (-1.22)$

$d_{2-3} =1.220 \ m$

The force exerted on the third charge by the first is mathematically evaluated as

$F_{2-3} = \frac{k q_2 q_3}{d_{2-3}^2}$

substituting values

$F_{2-3} = \frac{9*10^{9} * (32*10^{-9}) *(55*10^{-9})}{(1.220)^2}$

$F_{2-3} = 1.06*10^{-5} N$

The net force is

$F_{net} = F_{1-3} -F_{2-3}$

substituting values

$F_{net} = 4.28 *10^{-5} - 1.06*10^{-5}$

$F_{net}= 3.22*10^{-5} \ J$

6 0

1 year ago

Physics help, please? :)

Svetllana [295]

Answer:

Explanation:

3. Newton’s third law explains how every action has an equal but opposite reaction, meaning that forces comes in pairs. While the locomotive’s wheels are pushing back against the ground as the action force, the ground is producing a reaction force towards the locomotive, propelling it forward. Another pair of forces that act on the locomotive is gravity and normal force. While gravity is pulling the locomotive towards the ground, the normal force the ground exerts on the locomotive is why the locomotive doesn’t fall through the ground.

4. The force of Earth’s gravity on the Sun is weaker than the force of the Sun’s gravity on Earth. The Sun’s attraction affects the motion of Earth more than the Earth’s attraction affects the Sun’s motion because according to Newton’s second law, force has mass as one of its factors. The Sun has a significantly higher mass than Earth, meaning that its force of gravity would also be significantly higher. Newton’s third law is why the Earth doesn’t get marginally closer to the Sun, stating that every action has an equal and opposite reaction. As the Sun is pulling Earth towards itself, Earth is pulling away from the Sun.

8 0

1 year ago

A mass m slides down a frictionless ramp and approaches a frictionless loop with radius R. There is a section of the track with

Lana71 [14]

Answer:

h = 2 R (1 +μ)

Explanation:

This exercise must be solved in parts, first let us know how fast you must reach the curl to stay in the

let's use the mechanical energy conservation agreement

starting point. Lower, just at the curl

Em₀ = K = ½ m v₁²

final point. Highest point of the curl

$Em_{f}$ = U = m g y

Find the height y = 2R

Em₀ = Em_{f}

½ m v₁² = m g 2R

v₁ = √ 4 gR

Any speed greater than this the body remains in the loop.

In the second part we look for the speed that must have when arriving at the part with friction, we use Newton's second law

X axis

-fr = m a (1)

Y Axis

N - W = 0

N = mg

the friction force has the formula

fr = μ N

fr = μ m g

we substitute 1

- μ mg = m a

a = - μ g

having the acceleration, we can use the kinematic relations

v² = v₀² - 2 a x

v₀² = v² + 2 a x

the length of this zone is x = 2R

let's calculate

v₀ = √ (4 gR + 2 μ g 2R)

v₀ = √4gR( 1 + μ)

this is the speed so you must reach the area with fricticon

finally have the third part we use energy conservation

starting point. Highest on the ramp without rubbing

Em₀ = U = m g h

final point. Just before reaching the area with rubbing

$Em_{f}$ = K = ½ m v₀²

Em₀ = Em_{f}

mgh = ½ m 4gR(1 + μ)

h = ½ 4R (1+ μ)

h = 2 R (1 +μ)

7 0

1 year ago