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Colt1911 [192]
3 years ago
10

Determine the mechanical energy of this object: 1-kg ball rolls on the ground at 2 m/s

Physics
1 answer:
Fiesta28 [93]3 years ago
5 0
The potential energy would be zero. Only kinetic energy is present in this case. To find out what the answer is we do the equation: mv^2/2 soo...

KE =mv^2/2 
KE= 1(2^2)/2 which the answer will come up by 2 Joules.
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sunlight, wind, and running water are essentially "free." Yet renewable energy sources are a very small part of our energy consu
dimulka [17.4K]
The heat and energy of coal burning is more easier and more efficient to turn into electricity than any method using renewable resources. 
7 0
3 years ago
Read 2 more answers
A shoe and a shirt are released from the same height. They take different amounts of time to fall to the ground. How can this be
Law Incorporation [45]

The best explanation for the difference in time is: A. The difference in weight doesn't affect the time, but they are affected differently by air resistance.

<h3>What is weight?</h3>

Weight can be defined as the force acting on an object or a physical body due to the effect of gravity. Also, the weight of an object (body) is typically measured in Newton.

<h3>The factors that affect weight.</h3>

Some of the factors that affect the weight that is possessed by an object or a physical body include the following:

  • Mass
  • Distance
  • Air resistance

In conclusion, the weight possessed by the shoe and shirt has no effect on time but would be affected differently by air resistance.

Read more on weight here: brainly.com/question/13833323

4 0
2 years ago
Sí un auto viaja a 8m/s determine Tiempo en llegar a 200km de distancia Distancia que recorre en 40 minutos
GrogVix [38]

Answer:

a. Tiempo = 25000 segundos

b. Distancia = 19200 metros

Explanation:

Dados los siguientes datos;

Velocidad = 8 m/s

Distancia = 200 km a metros = 200 * 1000 = 200,000

Para encontrar el tiempo para cubrir la distancia anterior;

Tiempo = distancia/velocidad

Tiempo = 200000/8

Tiempo = 25000 segundos

b. Para encontrar la distancia recorrida en 40 minutos;

Tiempo = 40 minutos a segundos = 40 * 60 = 2400 segundos

Distancia = velocidad * tiempo

Distancia = 8 * 2400

Distancia = 19200 metros

7 0
3 years ago
When carbon bonds with oxygen,
Ierofanga [76]

Answer:

Carbon dioxide

Explination:

I remember it from biology.

I hope this helps ^-^

6 0
3 years ago
A sailboat moves north for a distance of 15.00 km when blown by a wind from the exact southeast with a force of 3.00 x 10^-4 N.
Zolol [24]
These are actually 4 different exercises:

ex 1) The sailboat moves north, while the wind moves from southeast. This means the angle between the direction of the boat and the wind is 45^{\circ}.

Calling F the force of the wind, and d=15~km=15000~m the distance covered by the boat, the work done by the wind is:
W=Fdcos{\theta}=3\cdot10^{-4}~N \cdot 15000~m\cdot cos 45^{\circ}=3.18~J

The total time of the motion is t=1~h=3600~s and therefore the power of the wind is
P= \frac{W}{t} = \frac{3.18~J}{3600~s}=8.8\cdot10^{-4}~W

ex 2) First of all, let's calculate the length of the ramp. Given the two sizes 2.00 m and 6.00 m, we have
d= \sqrt{(2~m)^2+(6~m)^2}= 6.32~m

The mechanical advantage (MA) of the ramp is the ratio between the output load (W) and the input force (F). The output load is the weight of the load, mg, therefore:
MA= \frac{W}{F}= \frac{mg}{F}= \frac{195~Kg\cdot 9.81~m/s^2}{750~N}=2.55

Finally, the efficiency \epsilon of the ramp is the ratio between the output energy and the work done. The output energy is simply the potential energy (Ep) of the load, which is mgh, where h is the height of the ramp. The work done W is the product between the input force, F, and the displacement of the load, which is the length of the ramp: Fd. Therefore:
\epsilon =  \frac{E_p}{W}= \frac{mgh}{Fd}= \frac{195~Kg \cdot 9.81~m/s^2\cdot 2~m}{750~N\cdot6.32~m}=0.81

ex 3) the graph is missing

ex 4) We know that the power is the ratio between the work done W and the time t:
P= \frac{W}{t}
But we can rewrite the work as
W=Fdcos\theta
where F is the force applied, d the displacement of rock and \theta=60^{\circ] is the angle between the direction of the force and the displacement (3 m). 
Therefore we can rewrite the power as
P= \frac{W}{t} = \frac{F d cos\theta}{t}=F v cos\theta
where v=d/t=5~m/s is the velocity, Using the data of the exercise, we can then find the force, F:
F= \frac{P}{v cos\theta} =   \frac{250~W}{5~m/s \cdot cos 60^{\circ}}=100~N

and now we can also calculate the work, which is 
W=Fdcos 60^{\circ}=100~N\cdot 3~m \cos60^{\circ}=150~J
3 0
3 years ago
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