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S_A_V [24]
2 years ago
11

A machine does 1,500 joules of work in 30 seconds. What is the power of this machine?

Physics
1 answer:
aniked [119]2 years ago
3 0
The power of the machine is 50
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1. Susan pushes her dad, David, on an ice rink with a force of 30 N. She weighs 45 kg and her dad weighs 100 kg. What are the ac
mamaluj [8]

Explanation:

Given parameters:

Force = 30N

Weight Susan = 45kg

Weight of Dad = 100kg

Unknown:

Acceleration of Susan = ?

Acceleration of Dad = ?

Solution:

                    Force = mass x acceleration

  Acceleration = \frac{force}{mass}

 Acceleration of Susan = \frac{30}{45} = 0.67m/s²

Acceleration of Dad =  \frac{30}{100}  = 0.3m/s²

Learn more:

Acceleration brainly.com/question/3820012

#learnwithBrainly

8 0
3 years ago
What is the area of a rectangle whose sides are 3.2 m and 8.01 m? Answer in units of m²
alexgriva [62]

Answer:

25.632 m²

Explanation:

area = a*b

area = 3.2m * 8.01m

area = 25.632 m²

4 0
3 years ago
Read 2 more answers
7. Two people are pushing a 40.0kg table across the floor. Person 1 pushes with a force of 490N
artcher [175]

Answer:

20.4 m/s^{2}

Explanation:

To start doing this problem, first draw a free body diagram of the table. My teacher always tells us to do this, and I find that it is very helpful. I have attached a free body diagram to this answer- take a look at it.

First, let us see if Net force = MA. To do that, we need to determine whether the object is at equilibrium horizontally. For an object to be at equilibrium, it either needs to be moving at a constant velocity or not moving at all. Also, if an object is at equilibrium, there will not be any acceleration. But we know that there IS acceleration horizontally, so it cannot be in equilibrium. If it is not in equilibrium, we can use the formula ∑F= ma.

Let us determine the net force. Since the object is moving horizontally, we can ignore the weight and normal force, because they are vertical forces. The only horizontal forces we need to worry about are the applied force and force of friction.

Applied force = 1055 N (490 + 565)

Friction force= Unknown

To find the friction force, use the kinetic friction formula, Friction = μkN

μk is the coefficient, which the problem includes- it is 0.613.

N is the normal force, which we have to find.

*To find the normal force, we have to determine if the object is at equilibrium VERTICALLY. Since it has no acceleration vertically (it's not moving up/down), it is at equilibrium. Now, when an object is at equilibrium in one direction, it means that all the forces in that direction are equal. What are our vertical forces? Weight (mg) and Normal force (N). So it means that the Normal force is equal to the Weight.

Weight = mg = (40)(9.8) = 392 N

Normal force = 392 N

Now, plug it back into the formula (μkN): (0.613)(392) = 240.296 N

Friction = 240.296 N

Now that we know the friction, we can find the horizontal net force. Just subtract the friction force, 240.296 from the applied force, 1055 N

Horizontal Net Force: 814.704 N

Now that we know the net force, plug in the numbers for the formula

∑F= ma.

814.704 = (40.0)(a)

*Divide on both sides)

a = 20.3676 m/s^2

Round it to 3 significant figures, to get:

20.4 m/s^{2}

7 0
3 years ago
What is the nucleation mechanisms in which nuclei of the solid phase form in the interior of the liquid as atoms cluster togethe
pshichka [43]
I  think its q because i suck on toes 


8 0
2 years ago
What is the difference between kinetic and gravitacional energy?
kondaur [170]

Answer:

In physics, the kinetic energy (KE) of an object is the energy that it possesses due to its motion

In classical mechanics, the gravitational potential at a location is equal to the work (energy transferred) per unit mass that would be needed to move an object to that location from a fixed reference location. It is analogous to the electric potential with mass playing the role of charge. The reference location, where the potential is zero, is by convention infinitely far away from any mass, resulting in a negative potential at any finite distance.

In mathematics, the gravitational potential is also known as the Newtonian potential and is fundamental in the study of potential theory. It may also be used for solving the electrostatic and magnetostatic fields generated by uniformly charged or polarized ellipsoidal bodies

8 0
2 years ago
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