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3 years ago

A motorcycle accelerated from 10 metre per second to 30 metre per second in 6 seconds. How far did it travel in this time ?

1 answer:

Ber [7]3 years ago

7 0

U=10 m/s
v=30 m/s
t=6 sec

therefore, a=(v-u)/t
=(30-10)/6
=(10/3) ms^-2

now, displacement=ut+0.5*a*t^2
=60+ 0.5*(10/3)*36
=120 m
And you can solve it in another way:

v^2=u^2+2as
or, s=(v^2-u^2)/2a
=(900-100)/6.6666666.......
=120 m

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A 1050 W carbon-dioxide laser emits light with a wavelength of 10μm into a 3.0-mm-diameter laser beam. What force does the laser

avanturin [10]

The force exerted by the laser beam on a completely absorbing target is $3.5 \times 10^{-6} \ N$ .

The given parameters;

power of the laser light, P = 1050 W
wavelength of the emitted light, λ = 10 μm

The speed of the emitted laser light is given as;

v = 3 x 10⁸ m/s

The force exerted by the laser beam on a completely absorbing target is calculated as follows;

P = Fv

$F = \frac{P}{v} \\\\F = \frac{1050}{3\times 10^8} \\\\F = 3.5 \times 10^{-6} \ N$

Thus, the force exerted by the laser beam on a completely absorbing target is $3.5 \times 10^{-6} \ N$ .

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2 years ago

A skateboarder starting from rest accelerates down a ramp at 2 m/s for 2 s. What is the final speed of the skateboarder?

AlekseyPX

Answer: 4m/s

Explanation:

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3 years ago

A solid sphere has a radius of 0.200 m and a mass of 150.0 kg. how much work is required to get the sphere rolling with an angul

Allisa [31]

Here in this case we can use work energy theorem

As per work energy theorem

Work done by all forces = Change in kinetic Energy of the object

Total kinetic energy of the solid sphere is ZERO initially as it is given at rest.

Final total kinetic energy is sum of rotational kinetic energy and translational kinetic energy

$KE = \frac{1}{2}Iw^2 +\frac{1}{2} mv^2$

also we know that

$I = \frac{2}{5}mR^2$

$w= \frac{v}{R}$

Now kinetic energy is given by

$KE = \frac{1}{2}(\frac{2}{5}mR^2)w^2 +\frac{1}{2} m(Rw)^2$

$KE = \frac{1}{5}mR^2w^2 +\frac{1}{2} mR^2w^2$

$KE = \frac{7}{10}mR^2w^2$

$KE = \frac{7}{10}*150*(0.200)^2(50)^2$

$KE = 10500 J$

Now by work energy theorem

Work done = 10500 - 0 = 10500 J

So in the above case work done on sphere is 10500 J

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3 years ago

You drop a ball from a height of 2.0 m, and it bounces back to a height of 1.5 m. a) What fraction of its initial energy is lost

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3 years ago

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