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lorasvet [3.4K]
3 years ago
15

A moving car initially has kinetic energy K. The car then moves in the opposite direction with four times its initial speed. Wha

t is the kinetic energy now
Physics
1 answer:
Liula [17]3 years ago
6 0

If initial speed was v, Kinetic energy was K = 1/2mv^2

When speed will be 4v, KE will be:

= 1/2 m (4v)^2

=1/2m 16 v^2

=16K

Now kinetic energy will become <em>1</em><em>6</em><em> </em><em>times</em><em> </em><em>of</em><em> </em><em>initial</em><em> </em><em>Kinetic</em><em> </em><em>Energy</em>

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A student constructed a series circuit consisting of a 12.0-volt battery, a 10.0-ohm lamp, and a
Stels [109]
The power dissipated across a component can be calculated through the formula P=I^2xR

Substituting the values in we get P=(0.5)^2x10=2.5W
4 0
3 years ago
How are frequency and wave period related?
dedylja [7]
-- The unit of frequency is "per second"  (Hz), which is [reciprocal time].

-- The unit of period is "second", which is [time].

Do you see where this is going ?

'Frequency' and 'period' are reciprocals of each other.

For any wave ...

Period  =  (1) / (frequency) .

Frequency  =  (1) / (period) .
6 0
3 years ago
A thin double convex glass lens with an index of 1.56 while surrounded by air has a 10 cm focal length. If it is placed under wa
bearhunter [10]

Explanation:

Formula which holds true for a leans with radii R_{1} and R_{2} and index refraction n is given as follows.

          \frac{1}{f} = (n - 1) [\frac{1}{R_{1}} - \frac{1}{R_{2}}]

Since, the lens is immersed in liquid with index of refraction n_{1}. Therefore, focal length obeys the following.  

            \frac{1}{f_{1}} = \frac{n - n_{1}}{n_{1}} [\frac{1}{R_{1}} - \frac{1}{R_{2}}]  

             \frac{1}{f(n - 1)} = [\frac{1}{R_{1}} - \frac{1}{R_{2}}]

and,       \frac{n_{1}}{f(n - n_{1})} = \frac{1}{R_{1}} - \frac{1}{R_{2}}

or,          f_{1} = \frac{fn_{1}(n - 1)}{(n - n_{1})}

              f_{w} = \frac{10 \times 1.33 \times (1.56 - 1)}{(1.56 - 1.33)}

                          = 32.4 cm

Using thin lens equation, we will find the focal length as follows.

             \frac{1}{f} = \frac{1}{s_{o}} + \frac{1}{s_{i}}

Hence, image distance can be calculated as follows.

       \frac{1}{s_{i}} = \frac{1}{f} - \frac{1}{s_{o}} = \frac{s_{o} - f}{fs_{o}}

              s_{i} = \frac{fs_{o}}{s_{o} - f}

             s_{i} = \frac{32.4 \times 100}{100 - 32.4}

                       = 47.9 cm

Therefore, we can conclude that the focal length of the lens in water is 47.9 cm.

4 0
3 years ago
What is zero on the kelvin scale
elena55 [62]
The point at which all motion stops.
8 0
3 years ago
Read 2 more answers
A ball is thrown straight up from ground level. It passes a 2-m-high window. The bottom of the window is 7.5 m off the ground. T
slamgirl [31]

Answer:

u=14.48m/s

Explanation:

From the question we are told that:

Height of window h=2m

Height of window off the ground h_g=7.5m

Time to fall and drop t=1.3s

 

Generally the Newton's equation motion  is mathematically given by

 s=ut+\frac{1}{2}at^2

Where

h=ut+\frac{1}{2}at^2

2=u1.3-\frac{1}{2}*9.8*1.3^2

2=u1.3-8.281

u=7.91m/s^2  

Generally the Newton's equation motion  is mathematically given by

2as=v^2-u^2

Where

-2gh_g=v^2-u^2

-2*9.8*7.5=(7.91)^2-u^2

-147=62.5681-u^2

u=\sqrt{209.5681}

u=14.48m/s

Therefore the  ball’s initial speed

u=14.48m/s

8 0
3 years ago
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