The field around a solenoid is similar to the field around a bar magnet.<br><br> true or false?

Sarah weighs 392 N. She pushes a 56 kg rock with a force of 65 N. What force does the rock exert on Sarah?

Leona [35]

Answer:

The force the rock exerts on Sarah =#is 65 N

Explanation:

The given parameters are;

Sarah's weight = 392 N

The force with which Sarah pushes the rock = 65 N

The mass of the rock = 56 kg

The weight of the rock = The mass of the rock × Acceleration due to gravity

∴ The weight of the rock = 56 kg ×9.81 m/s² = 549.36 N

Given that the force Sarah applies to push the rock = 65 N, then by Newton's third law of motion which states that action and reaction are equal and opposite, the force that the rock exert on Sarah is equal an opposite to the force Sarah is applying

Therefore, the force the rock exerts on Sarah = 65 N (in the opposite direction).

6 0

3 years ago

A Man is pulling a trolley on a horizontal road with a force of 200N making 30° with the road . Find the horizontal and vertical

artcher [175]

The horizontal component of the force is 173.2 N.

The vertical component of the force is 100 N.

The given parameters:

Applied force, F = 200 N
Angle of inclination, Ф = 30⁰

<h3>Horizontal component of the force</h3>

The horizontal component of the force is calculated as follows;

$F_x = Fcos(\theta)\\\\
F_x = 200 \times cos(30)\\\\
F_x = 173.2 \ N$

<h3>Vertical component of the force</h3>

The vertical component of the force is calculated as follows;

$F_y =F sin(\theta)\\\\
F_y= 200 \times sin(30)\\\\
F_y = 100 \ N$

Learn more about components of forces here: brainly.com/question/8106035

6 0

2 years ago

Read 2 more answers

In April 1974, Steve Prefontaine completed a 10.0 km race in a time of 27 min , 43.6 s . Suppose "Pre" was at the 7.43 km mark a

Novay_Z [31]

Answer:

Acceleration, a = 0.101 m/s²

Explanation:

Average speed = total distance / total time.

At the 7.43km mark, total distance = 7.43km or 7430m

Total time = 25 * 60 s = 1500s

Average speed = 7430m/1500s = 4.95m/s

He then covers (10 - 7.43)km = 2.57 km = 2570 m

in t = 27m43.6s - 25min = 2m43.6s = 163.6 s

Then he accelerates for 60 s, and maintains this velocity V, for the remaining (163.6 - 60)s = 103.6 s.

From V = u + at; V = 4.95m/s + a *60s

Distance covered while accelerating is

s = ut + ½at² = 4.95m/s * 60s + ½ a *(60s)² = 297m + a*1800s²

Distance covered while at constant velocity, v after accelerating is

D = velocity * time

Where v = 4.95m/s + a*60s

D = (4.95m/s + a*60s) * 103.6s = 512.82m + a*6216s²

Total distance covered after initial 7.43 km, S + D = 2570 m, so

2570 m = 297m + a*1800s² + 512.82m + a*6216s²

2570 = 809.82 + a*8016

a = 809.82m / 8016s² = 0.101 m/s²

8 0

3 years ago

How do I solve this?

prohojiy [21]

unlock your inner brain

3 0

2 years ago

A research vessel is mapping the bottom of the ocean using sonar. It emits a short sound pulse called "ping" downward. The frequ

Elena L [17]

Answer:

The ocean is 6485.6m deep when measured from the vessel

Explanation:

v=1474m/s

t=8.88s

let d represent distance from the vessel to the ocean bottom.

an echo travels a distance equivalent to 2d, that is to and fro after it reflects from the obstacle.

$velocity=\frac{distance}{time}\\ v=\frac{2d}{t} \\vt=2d\\d=\frac{vt}{2}$

$d=\frac{1474*8.8}{2}$

d= 6485.6m

7 0

3 years ago

The field around a solenoid is similar to the field around a bar magnet. true or false?