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vodka [1.7K]
3 years ago
11

A 445-N force is used to hold up a Papa Johns sign in place. What is the mass of the sign?

Physics
1 answer:
nikitadnepr [17]3 years ago
3 0

Answer:

The mass of the sign is 44.5 kg.

Explanation:

Given that,

Force acting on the sign, F = 445 N

We need to find the mass of the sign.

Net force acting on it is given by :

F = mg

Where g is the acceleration due to gravity

m=\dfrac{F}{g}\\\\m=\dfrac{445\ N}{10\ m/s^2}\\\\m=44.5\ kg

So, the mass of the sign is 44.5 kg.

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Use Newton's laws to explain why a falling object dropped from a 57m tower accelerates initially but then reaches constant veloc
snow_lady [41]

Answer:

At the point of dropping the object, by Newton's first law due to gravitational force F_g = m × g, accelerates

By Newton's Second law the object reaches impacts on the air with the gravitational force resulting in changing momentum of m×(Final Velocity - Initial Velocity)

As the velocity increases, the rate of change of momentum becomes equivalent to the gravitational force and by Newton's third law, the action action and reaction are equal and opposite hence they cancel each other out

The body then moves at a constant uniform motion down according to Newton's first law

Explanation:

At the point the object of mass, m, is dropped from the height of the tower, the only force acting on the object is the gravitational force such that the object has an acceleration which is the acceleration due to gravity, g, and the gravitational force is therefore = m × g

As the speed of the object increases while the object is falling with the gravitational acceleration the rate at which the object cuts through layers of air which (by Newton's first law of motion, are at rest ) has some buoyancy effect also increases therefore, the object is constantly increasingly changing the momentum of the air which by Newton's second law results, at an high enough velocity, and by Newton's third law, in a force equal to the applied gravitational force

Therefore, the force of the air drag becomes equal to the gravitational force, cancelling each other out and the object then moves according to Newton;s first law, in uniform motion of a constant speed while still falling down.

5 0
3 years ago
What force acts on all objects, all the time on Earth?
TiliK225 [7]

The force that acts on all objects, all the time on Earth is gravitational force.

The force that surface exert on an object perpendicularly is normal reaction.

<h3>What force acts on all objects, all the time on Earth?</h3>
  • Force due to gravity is gravitational pull on objects due to its position on earth's surface.

The force due to gravity on object's is calculated by applying Newton's second law of motion as follows;

F = mg

where;

  • m is the mass of the object
  • g is acceleration due to gravity

The force that surface exert on an object perpendicularly is normal reaction.

Thus, the force that acts on all objects, all the time on Earth is gravitational force.

Learn more about force of gravity here: brainly.com/question/2537310

7 0
3 years ago
All about Henry Mosléy?<br>​
Anni [7]

Answer:

henry moseley was an english physicist, whose contribution to the science of physics was the justification from physical laws of the previous chemical concept of the atomic number. one of his developments were of moseley's law in x-ray spectra.

Explanation:

7 0
3 years ago
A jet transport has a weight of 2.25 x 106 N and is at rest on the runway. The two rear wheels are 16.0 m behind the front wheel
Rudik [331]

Answer:

Explanation:

Given that,

Weight of jet

W = 2.25 × 10^6 N

It is at rest on the run way.

Two rear wheels are 16m behind the front wheel

Center of gravity of plane 10.6m behind the front wheel

A. Normal force entered on the ground by front wheel.

Taking moment about the the about the real wheel.

Check attachment for better understanding

So,

Clock wise moment = anti-clockwise moment

W × 5.4 = N × 16

2.25 × 10^6 × 5.4 = 16•N

N = 2.25 × 10^6 × 5.4 / 16

N = 7.594 × 10^5 N

B. Normal force on each of the rear two wheels.

Using the second principle of equilibrium body.

Let the rear wheel normal be Nr and note, the are two real wheels, then, there will be two normal forces

ΣFy = 0

Nr + Nr + N — W = 0

2•Nr = W—N

2•Nr = 2.25 × 10^6 — 7.594 × 10^5

2•Nr = 1.491 × 10^6

Nr = 1.491 × 10^6 / 2

Nr = 7.453 × 10^5 N

6 0
4 years ago
A transformer is intended to decrease the value of the alternating current from 500 amperes to 25 amperes. The primary coil cont
EastWind [94]

Answer:

The number of turns in secondary coil is 4000

Explanation:

Given:

Current in primary coil I_{P} = 500 A

Current in secondary coil I_{S} = 25 A

Number of turns in primary coil N_{P} = 200

In case of transformer the relation between current and number of turns is given by,

     \frac{N_{S} }{N_{P}  } = \frac{I_{P} }{I_{S} }

For finding number of turns in secondary coil,

     N_{S} = \frac{I_{P} }{I_{S} }  N_{P}

     N_{S} = \frac{500}{25} \times 200

     N_{S} = 4000

Therefore, the number of turns in secondary coil is 4000

5 0
3 years ago
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