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Maurinko [17]
3 years ago
9

To determine power, it is necessary to know the A. energy required and the time it takes. B. amount of work and the distance cov

ered. C. amount of work in a given time. D. distance and time it takes to do work.
Physics
2 answers:
maxonik [38]3 years ago
7 0

Answer:

Option (A)

Explanation:

The rate of doing work is called power.

Power is given by the formula

Power = Work / time

Work done is equivalent to energy.

So, Energy required to the time taken is called power. Its SI unit is Watt and it is a scalar quantity.

Y_Kistochka [10]3 years ago
5 0
The answer for this would be A. since power is Joules/seconds and energy is rated in Joules
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Do electric fields point towards or away from positive charges?
juin [17]

Answer:

it points towards positive charges

3 0
3 years ago
A positive point charge q1 = +5.00 × 10−4C is held at a fixed position. A small object with mass 4.00×10−3kg and charge q2 = −3.
Lelechka [254]

Answer:

Therefore the speed of q₂ is 1961.19 m/s when it is 0.200 m from from q₁.

Explanation:

Energy conservation law: In isolated system the amount of total energy remains constant.

The types of energy are

  1. Kinetic energy.
  2. Potential energy.

Kinetic energy =\frac{1}{2} mv^2

Potential energy =\frac{Kq_1q_2}{d}

Here, q₁= +5.00×10⁻⁴C

q₂=-3.00×10⁻⁴C

d= distance = 4.00 m

V = velocity = 800 m/s

Total energy(E) =Kinetic energy+Potential energy

                      =\frac{1}{2} mv^2+ \frac{Kq_1q_2}{d}

                     =\frac{1}{2} \times 4.00\times 10^{-3}\times(800)^2 +\frac{9\times10^9\times 5\times10^{-4}\times(-3\times10^{-4})}{4}

                    =(1280-337.5)J

                    =942.5 J

Total energy of a system remains constant.

Therefore,

E =\frac{1}{2} mv^2 + \frac{Kq_1q_2}{d}

\Rightarrow  942.5 = \frac{1}{2} \times 4 \times10^{-3} \times V^2 +\frac{9\times10^{9}\times5\times 10^{-4}\times(-3\times 10^{-4})}{0.2}

\Rightarrow 942.5 = 2\times10^{-3}v^2 -6750

\Rightarrow 2 \times10^{-3}\times v^2= 942.5+6750

\Rightarrow v^2 = \frac{7692.5}{2\times 10^{-3}}

\Rightarrow v= 1961.19   m/s

Therefore the speed of q₂ is 1961.19 m/s when it is 0.200 m from from q₁.

5 0
3 years ago
A driver drives for 30.0 minutes at 80.0 km/h, then 45.0 minutes at 100 km/h. She then stops 30 minutes for lunch. She then trav
bija089 [108]

Answer:

b) 68,9 km/h a) picture

Explanation:

In this problem, since velocity is expressed in km/h and time in minutes, we have to convert either time to hours or velocity to km/min. It is easier to use hours.

Using this formula we pass time to hours:

t_{hours}=t_{min}*\frac{1 h}{60 min}\\30min*\frac{1 h}{60 min}=0,5h\\45min*\frac{1 h}{60 min}=0,75h

Now we can plot speed vs time (image 1). The problem says that the driver uses constant speed, so all lines have to be horizontal.

Using the values of the speed we calculate the distance in each interval

d=v*t\\80km/h*0.5h=40km\\100km/h*0.75h=75km

Using these values and the fact that she was having lunch in the third one (therefore stayed in the same position), we plot position vs time, using initial position zero (image 2, distance is in km, not meters).

Finally, we compute the average speed with the distance over time:

v_{average}=\frac{155km}{2.25h}=68.9km/h

6 0
3 years ago
A 5000kg freight car moving at 2 m/s East collides with a 10,000kg freight car at rest. Upon collision, they got stuck and moved
mr_godi [17]

Answer:

A

Explanation:

6 0
2 years ago
What is the value of work done on an object when a
tino4ka555 [31]
W = F * s
Here, F = 50 N
s = 15 m

Substitute their values, 
W = 50 * 15
W = 750 J

In short, Your Answer would be 750 Joules

Hope this helps!
7 0
3 years ago
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