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3 years ago

9

a 0.11kg moving hockey puck is caught by a 80kg goalie who then slides on the frictionless ice at 0.076 m/s. What was the initia

l velocity of the hockey puck

1 answer:

aleksandrvk [35]3 years ago

3 0

Answer:

55mph

Explanation:

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A fisherman notices that his boat is moving up and down periodically without any horizontal motion, owing to waves on the surfac

Y_Kistochka [10]

Answer:

See Explanation

Explanation:

a) We know that;

v = λf

Where;

λ = wavelength of the wave

f = frequency of the wave

v = velocity of the wave

So;

T = 2 * 2.10 s = 4.2 s

Hence f = 1/4.2 s

f = 0.24 Hz

The wavelength = 6.5 m

Hence;

v = 6.5 m * 0.24 Hz

v = 1.56 m/s

b)The amplitude of the wave is;

A = 0.600 m/2 = 0.300 m

c) Since the wave speed does not depend on the amplitude of the wave then the answer in (a) above remains the same

Where d = 0.30 m

A = 0.30 m/2 = 0.15 m

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3 years ago

A 55 newton force applied on an object moves the object 10 meters in the same direction as the force. What is the value of work

kifflom [539]

Answer: Option D: 5.5×10²Joules

Explanation:

Work done is the product of applied force and displacement of the object in the direction of force.

W = F.s = F s cosθ

It is given that the force applied is, F = 55 N

The displacement in the direction of force, s = 10 m

The angle between force and displacement, θ = 0°

Thus, work done on the object:

W = 55 N × 10 m × cos 0° = 550 J = 5.5 × 10² J

Hence, the correct option is D.

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3 years ago

What distance does light travel in water, glass, and diamond during the time that it travels 1.0 m in vacuum? The refractive ind

lidiya [134]

Answer:

refractive index for water,glass,diamond are 0.752m, 0.667m, 0.413m respectively

Explanation:

refractive index (n) = $\frac{velocity of light in air/vacuum}{velocity of light in substance}$

velocity = $\frac{distance}{time}$

The time for travel is kept constant for all mediums.

refractive index (n) = $\frac{\frac{distance in vacuum}{time} }{\frac{distance in medium}{time} }\\ \\=\frac{distance in vacuum}{distance in medium}$

distance in medium = $\frac{distance in vacuum}{refractive index of medium}$

$S_{medium} = \frac{S_{vacuum} }{n_{medium} }$

For water, n= 1.33

$S_{water} = \frac{1 }{1.33}$

$S_{water} = 0.752m$

For glass, n=1.5

$S_{glass}= \frac{1 }{1.5}$

$S_{glass} = 0.667m$

For diamond, n= 2.42

$S_{diamond} = \frac{1 }{2.42}$

$S_{diamond} = 0.413m$

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3 years ago

Air bags are designed to deploy in 10 ms. Given that the air bags expand 20 cm as they deploy, estimate the acceleration of the

joja [24]

As it is given that the air bag deploy in time

$t = 10 ms = 0.010 s$

total distance moved by the front face of the bag

$d = 20 cm = 0.20 m$

Now we will use kinematics to find the acceleration

$d = v_i*t + \frac{1}{2}at^2$

$0.20 = 0 + \frac{1}{2}a*0.010^2$

$0.20 = 5 * 10^{-5}* a$

$a = 4000 m/s^2$

now as we know that

$g = 10 m/s^2$

so we have

$a = 400g$

so the acceleration is 400g for the front surface of balloon

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3 years ago

What are each layer on an atom

Zarrin [17]

Protons, electrons, and neutrons. The nucleus (center) of the atom contains the protons (positively charged) and the neutrons (no charge).

5 0

3 years ago

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