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3 years ago

Please help me with Part B & help me correct the rest of the questions

Physics

1 answer:

AleksAgata [21]3 years ago

5 0

Answer:

Magnetic activity and sunspots are in a proportional relationship.

Explanation:

From the graph, as one increases, so does the other, thus we describe the relationship as proportional.

Other than that, all of your other answers seem correct.

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A 9 cm tall object is placed 16 cm from a converging lens forming an image at 13 cm. What is the height of the image?

Tresset [83]

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2 years ago

Coal burns in a furnace, producing light and heat. This reaction is

olasank [31]

The reaction is Exothermic

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2 years ago

Read 2 more answers

A block is pulled across a flat surface at a constant speed using a force of 50 newtons at an angle of 60 degrees above the hori

vladimir2022 [97]

The magnitude of the friction force is 25 N

Explanation:

To solve this problem, we just have to analyze the forces acting on the block along the horizontal direction. We have:

The horizontal component of the pulling force, $F cos \theta$ , where F = 50 N is the magnitude and $\theta=60^{\circ}$ is the angle between the direction of the force and the horizontal; this force acts in the forward direction
The force of friction, $F_f$ , acting in the backward direction

According to Newton's second law, the net force acting on the block in the horizontal direction must be equal to the product between the mass of the block and its acceleration:

$\sum F_x = ma_x$

where

m is the mass of the block

$a_x$ is the horizontal acceleration

However, the block is moving at constant speed, so the acceleration is zero:

$a_x = 0$

So the equation becomes

$\sum F_x = 0$ (1)

The net force here is given by

$\sum F_x = F cos \theta - F_f$ (2)

And so, by combining (1) and (2), we find the magnitude of the friction force:

$F cos \theta - F_f = 0\\F_f = F cos \theta = (50)(cos 60^{\circ})=25 N$

Learn more about force of friction:

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4 0

3 years ago

1) A force of 20 Newton acts on a bar having a cross sectional area of 0.8m^2 and length 10cm.calculate the stress developed in

Elanso [62]

Answer:25N/M^2

Explanation:

Force=20N Area=0.8M^2

Stress=force/area

Stress=20/0.8

Stress=25N/M^2

4 0

3 years ago

A fisherman notices that his boat is moving up and down periodically without any horizontal motion, owing to waves on the surfac

matrenka [14]

Answer:

a) 6.1 m

b) 4.6 s

c) 1.326 m/s

d) 0.325 m

Explanation:

a) The wave length is the distance between 2 crests λ = 6.1m

b) The period of the wave is the time it takes from the lowest point to the next lowest point, which is twice the time it takes from the lowest point to the highest point = 2*2.3 = 4.6 s

c) The speed of the wave is the distance per unit of time, or wave length over period = 6.1 / 4.6 = 1.326 m/s

d)The amplitude A is half the distance from the highest point to the lowest point = 0.65 / 2 = 0.325 m

6 0

3 years ago