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3 years ago

Find the force exerted by each rope in the sign below

1 answer:

4 0

Answer: Example 1: Consider a crate being pulled along a frictionless floor (while such a floor is very hard to find, this will still help us understand the concept and we can return to this situation later, after considering friction, and solve it more realistically).

Consider a crate being pulled along a horizontal, frictionless floor. A rope is tied around it and a man pulls on the rope with a force of T. T is the tension in the rope. What happens to the crate?

Before we can apply Newton's Second Law,

F = m a

we must find the net force -- the vector sum of all the forces -- acting on the object. In addition to the force T exerted by the rope, what other forces act on the object?

As discussed in class, in Mechanics, we can restrict our attention to "contact" forces and "gravity". That means gravity pulls down on this crate with a force equal to its weight, w. But the floor supports the crate. The floor responds by pushing up on the crate with a force we call the normal force. "Normal" means "perpendicular". We will call this force n; you may also encounter it labeled N or FN.

Explanation:

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Please help it’s easy will give brainlist !! Thank youu worth a lot be serious pls

Genrish500 [490]

Answer:

Part 1

Stationary

Part 2

20 Newtons

Part 3

Force

Part 4

4.0 m/s²

Part 5

Normal

Part 6

Cart m

Part 7

The gravitational force is less than magnetic force

Explanation:

Part 1

The position time graph of the object is an horizontal straight line passing across the top of the position 3 boxes vertically up from the origin

As the time increases by the units of number boxes to the left, the position of the object does not change and remains at the 3 boxes up above the origin, therefore, the object is stationary

Part 2

By Newton's third law of motion, the action action obtained from a force is equal to the reaction given to the force, therefore, we have;

The force exerted by the student on the scale = The force exerted by the scale on the student = 20 N

Part 3

A force is a the directional push on an object or pull from the object as a form of interaction with another object which tends to alter or maintain the motion of the object

Part 4

The given parameters are'

The mass of block A = 1.0 kg

The mass of block B = 2.0 kg

Both blocks, "A" and "B" are initially at rest

The applied horizontal force, F = 12-N

The nature of the surface over which the blocks move = Smooth surface

Force, F = Mass, m × Acceleration, a

F = m × a

The blocks two blocks experience a common acceleration, a

The combined mass of the two blocks, m = 1.0 kg + 2.0 kg = 3.0 kg

m = 3.0 kg

Therefore, a = F/m = 12-N/(3.0 kg) = 4 m/s²

Part 5

A normal force is a force acting perpendicularly to a surface that supports the weight of an object

Part 6

The given parameters are;

M = 2 kg

The mass of the left cart attached to the spring = M = 2 kg

The mass of the right cart attached to the spring = 2·M = 2 × 2 kg = 4 kg

Therefore, given that the force exerted by one cart on the other after the spring is removed, we have;

Force, F = M × a₁ = 2·M × a₂

Where;

a₁ = The average acceleration of the cart with mass, M

a₂ = The average acceleration of the cart with mass, 2·M

M × a₁ = 2·M × a₂

∴ a₁ = 2·a₂

The acceleration, a₁, of car M = 2 × The acceleration, a₂, of car 2·M

The acceleration of cart M is two times the acceleration of cart 2·M

Cart M will experience a greater average acceleration

Part 7

For the top magnet of two magnets placed two magnets arranged so that one is on top the other, to be held in the air with a space between the two magnets, the gravitational force pulling the top magnet down, is less than the repelling magnetic force between the two magnets.

7 0

3 years ago

An aerobatic airplane pilot experiences weightlessness as she passes over the top of a loop-the-loop maneuver. the acceleration

Yanka [14]

The acceleration due to gravity serves as the centripetal acceleration of the objects that orbits the Earth. The centripetal acceleration due to gravity is calculated through the equation,

a = v²/r

where v is the speed and r is the radius. Substituting the known values to the equation,

9.8 m/s² = (420 m/s)² / r

The value of r from the equation is 18000 m or equal to 18 km.

<em>Answer: 18 km</em>

6 0

3 years ago

Your little sister is building a radio from scratch. Plans call for a 500 μH inductor wound on a cardboard tube. She brings you

gayaneshka [121]

Answer:

N = 195 turns

Explanation:

The inductance of the inductor, L = 500 μH = 500 * 10⁻⁶H

The length of the tube, l = 12 cm = 0.12 m

The diameter of the tube, d = 4 cm = 0.04 m

Radius, r = 0.04/2 = 0.02 m

Area of the tube, A = πr² = 0.02²π = 0.0004π m²

$\mu_{0} = 4\pi * 10^{-7}$

The inductance of a solenoid is given by:

$L = \frac{\mu_{0}N^{2} A }{l}$

$500 * 10^{-6} = \frac{4\pi *10^{-7} N^{2} *4\pi *10^{-4} }{0.12}\\500 * 10^{-6} = 0.00000001316N^{2} \\N^{2} = \frac{500 * 10^{-6}}{0.00000001316}\\N^{2} = 37995.44\\N = \sqrt{37995.44} \\N = 194.92 turns$

8 0

3 years ago

At this radius, what is the magnitude of the net force that maintains circular motion exerted on the pilot by the seat belts, th

Ainat [17]

Answer:

Fc=5253 N

Explanation:

Answer:

Fc=5253 N

Explanation:

sequel to the question given, this question would have taken precedence:

"The 86.0 kg pilot does not want the centripetal acceleration to exceed 6.23 times free-fall acceleration. a) Find the minimum radius of the plane’s path. Answer in units of m."

so we derive centripetal acceleration first

ac (centripetal acceleration) = v^2/r

make r the subject of the equation

r= v^2/ac

ac is 6.23*g which is 9.81

v is 101m/s

substituing the parameters into the equation, to get the radius

(101^2)/(6.23*9.81) = 167m

Now for part

( b) there are two forces namely, the centripetal and the weight of the pilot, but the seat is exerting the same force back due to newtons third law.

he net force that maintains circular motion exerted on the pilot by the seat belts, the friction against the seat, and so forth is the centripetal force.

Fc (Centripetal Force) = m*v^2/r

So (86kg* 101^2)/(167) =

Fc=5253 N

4 0

3 years ago

Electric Field of Ring of Charge A ring shaped charge has a radius of 0.250 meters and bears a total charge of +5.00 C What is t

Eduardwww [97]

Answer:

6.44 × 10^10 N/C

Explanation:

Electric field due to the ring on its axis is given by

E = K q r / (r^2 + x^2)^3/2

Where r be the radius of ring and x be the distance of point from the centre of ring and q be the charge on ring.

r = 0.25 m, x = 0.5 m, q = 5 C

K = 9 × 10^9 Nm^2/C^2

E = 9 × 10^9 × 5 × 0.25 / (0.0625 + 0.25)^3/2

E = 6.44 × 10^10 N/C

8 0

4 years ago