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otez555 [7]
1 year ago
9

A person is hauling their taco stand and it takes 3,500 Joules of work to stop the taco stand. What force was exerted on the tac

o stand if it took 1.5 meters to stop the motion?

Physics
1 answer:
Lapatulllka [165]1 year ago
3 0

Given the work done, the force exerted on the taco stand through the given distance is 2.3 × 10³ Newtons.

<h3>What is Work done?</h3>

Work done is simply defined as the energy transfer that takes place when an object is either pushed or pulled over a certain distance by an external force. It is expressed as;

W = f × d

Where f is force applied and d is distance travelled.

Given that;

  • Work done W = 3500J = 3500kgm²/s²
  • Distance covered d = 1.5m
  • Force applied F = ?

W = f × d

3500kgm²/s² = f × 1.5m

f = 3500kgm²/s² ÷ 1.5m

f = 2.3 × 10³ kgm/s²

f = 2.3 × 10³ N

Given the work done, the force exerted on the taco stand through the given distance is 2.3 × 10³ Newtons.

Learn more about work done here: brainly.com/question/26115962

#SPJ1

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A painter on a ladder, painting the ceiling of a room comments. “It is hotter up here by the ceiling than it is down on the floo
ira [324]

Answer:

1. Hot air is less dense and has moved upward allowing cool air to move downward which is less dens

(Convection current)

2. The ceiling has transferred heat to him by radiation.

Explanation:

During the day when air is heated as a result of the ceiling transferring heat to it, it becomes less dens and it gains energy, which make it lighter in weight than cooler air, this hot air moves upward allowing cooler air to move downward.

4 0
3 years ago
Two forces Upper FSubscript Upper A Baseline Overscript right-arrow EndScripts and Upper FSubscript Upper B Baseline Overscript
Pavel [41]

Answer:

Part a)

F_A = 4.59 N

Part B)

F_B = 1.28 N

Explanation:

As we know that when both the forces are acting on the object in same direction then we will have

F_A + F_B = ma

as we know that

a = 0.554 m/s^2

m = 10.6 kg

now we will have

F_A + F_B = 10.6(0.554)

F_A + F_B = 5.87 N

Now two forces are in opposite direction then we have

F_A - F_B = 10.6(0.313)

F_A - F_B = 3.32 N

Part A)

Now we will have from above two equation

F_A = 4.59 N

Part B)

Similarly for other force we have

F_B = 1.28 N

5 0
2 years ago
A 25N force is acting on a body moving on a straight line with Initial momentum 20 kam's. Find the final momentum after 4 second
Nezavi [6.7K]

The final momentum of the body is equal to 120 Kg.m/s.

<h3>What is momentum?</h3>

Momentum can be described as the multiplication of the mass and velocity of an object. Momentum is a vector quantity as it carries magnitude and direction.

If m is an object's mass and v is its velocity then the object's momentum p is: {\displaystyle \mathbf {p} =m\mathbf {v} . The S.I. unit of measurement of momentum is kg⋅m/s, which is equivalent to the N.s.

Given the initial momentum of the body = Pi = 20 Kg.m/s

The force acting on the body, Pf = 25 N

The time, Δt = 4-0 = 4s

The Force is equal to the change in momentum: F ×Δt = ΔP

25 × 4 = P - 20

100 = P - 20

P = 100 + 20 = 120  Kg.m/s

Therefore, the final momentum of a body is 120 Kg.m/s.

Learn more about momentum, here:

brainly.com/question/4956182

#SPJ1

5 0
9 months ago
Can someone help??
Paul [167]

Answer:

F = 2520 N

Explanation:

We have,

The maximum acceleration of a fist in a karate punch is 4200 m/s². The mass of the fist is 0.6 kg.

It is required to find the force that the wood place on the fist. Force is given by the product of mass and its acceleration such that,

F = ma

F=0.6\times 4200\\\\F=2520\ N

So, the force of 2520 N is acting on the wood.

8 0
3 years ago
What is the specific heat of an object if it takes 1200 J to raise the temperature of a 20 kg object by 6 degrees C?
ch4aika [34]

Answer:

c=10\ J/kg^{\circ} C

Explanation:

Given that,

Heat required, Q = 1200 J

Mass of the object, m = 20 kg

The increase in temperature, \Delta T=6^{\circ} C

We need to find the specific heat of the object. The heat required to raise the temperature is given by :

Q=mc\Delta T\\\\c=\dfrac{Q}{m\Delta T}\\\\c=\dfrac{1200}{20\times 6}\\\\c=10\ J/kg^{\circ} C

So, the specific heat of the object is 10\ J/kg^{\circ} C.

5 0
2 years ago
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