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labwork [276]
3 years ago
11

A child pulls a wagon with a force F at an angle o with respect to the horizontal. Which of the following free body diagrams is

correct?

Physics
1 answer:
NeTakaya3 years ago
5 0

Option d) is the correct free body diagram.

Explanation:

A free-body diagram is a diagram that shows all the forces acting on a body. Each force is represented using an arrow, where:

- The length of the arrow is proportional to the magnitude of the force

- The direction of the arrow corresponds to the direction of the force

For the block in this problem, we have the following forces:

- The force F applied from the child, which acts at an angle with respect to the horizontal --> this rules out option a) and c), where the force acts horizontally

- The force of gravity (the weight of the object), labelled with W, which always acts downward --> this rules out option b), since the weight acts downward.

Therefore, the correct option is d).

(in reality, there should be another force: the normal reaction exerted by the floor on the block, N, acting upward).

Learn more about forces and weight:

brainly.com/question/8459017

brainly.com/question/11292757

brainly.com/question/12978926

#LearnwithBrainly

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Electrons flow from the negative terimal of a
gizmo_the_mogwai [7]

Answer:

Option B

Explanation:

The electrons flow from negative terminal of a battery to the positive terminal because as the charge of electron is negative, it will get repelled by the negative terminal of the battery

Conventional flow actually assumes that the current flows out of the positive terminal, through the circuit and into the negative terminal of the battery

It actually says that direction of flow of current is in opposite direction to the direction of flow of electrons

The direction of current will not change even a resistor is placed in the circuit and it generates a potential difference across the resistor

But the current tries to move in such a way in which there will be less resistance to the flow

3 0
3 years ago
What is the potential energy of an object 20 m in the air with a<br> mass of 600 kg?
Lana71 [14]

Answer:

Ep = 117600 J

Explanation:

Data:

  • Mass (m) = 600 kg
  • Height (h) = 20 m
  • Gravity (g) = 9.8 m/s²
  • Potential Energy (Ep) = ?

Use formula:

  • Ep = m * g * h

Replace:

  • Ep = 600 kg * 9.8 m/s² * 20 m

Multiply operations, and units:

  • Ep = 117600 J

What is the potential energy?

The potential energy is <u>117600 Joules.</u>

7 0
2 years ago
True or false? All waves need a medium in order to travel
Art [367]
That's false.  Mechanical waves (like sound and ocean waves) do
need a medium to travel in, but electromagnetic waves (like radio
and light) don't.
4 0
3 years ago
Read 2 more answers
SOMEONE PLEASE HELP PLEASE PLEASE
Black_prince [1.1K]
Ais the correct answer
4 0
2 years ago
A body which has surface area 5cm² and temperature of 727°C radiates 300J of energy in one minute. Calculate it's emissivity giv
cestrela7 [59]
<h2>Answer: 0.17</h2>

Explanation:

The Stefan-Boltzmann law establishes that a black body (an ideal body that absorbs or emits all the radiation that incides on it) "emits thermal radiation with a total hemispheric emissive power proportional to the fourth power of its temperature":  

P=\sigma A T^{4} (1)  

Where:  

P=300J/min=5J/s=5W is the energy radiated by a blackbody radiator per second, per unit area (in Watts). Knowing 1W=\frac{1Joule}{second}=1\frac{J}{s}

\sigma=5.6703(10)^{-8}\frac{W}{m^{2} K^{4}} is the Stefan-Boltzmann's constant.  

A=5cm^{2}=0.0005m^{2} is the Surface area of the body  

T=727\°C=1000.15K is the effective temperature of the body (its surface absolute temperature) in Kelvin.

However, there is no ideal black body (ideal radiator) although the radiation of stars like our Sun is quite close.  So, in the case of this body, we will use the Stefan-Boltzmann law for real radiator bodies:

P=\sigma A \epsilon T^{4} (2)  

Where \epsilon is the body's emissivity

(the value we want to find)

Isolating \epsilon from (2):

\epsilon=\frac{P}{\sigma A T^{4}} (3)  

Solving:

\epsilon=\frac{5W}{(5.6703(10)^{-8}\frac{W}{m^{2} K^{4}})(0.0005m^{2})(1000.15K)^{4}} (4)  

Finally:

\epsilon=0.17 (5)  This is the body's emissivity

3 0
2 years ago
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