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3 years ago

12

What is the speed of light in air ?

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ty! ~

2 answers:

VikaD [51]3 years ago

6 0

Answer:

the speed of light in air is 3*10⁸m/s.

hichkok12 [17]3 years ago

4 0

Answer:

the speed of light in air is about 299,000,000 and 3×10⁸ m/s

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What is the momentum of an object with a mass of 60 kg and velocity of 10 m/s?<br><br> heeelp

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Answer:

Linear momentum= Mass*Velocity

P=mv

P=60*10

P=600Kg m/s

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3 years ago

Every 4 years we have a leap year on our calander to make up the extra.......

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3 years ago

Two identical steel balls, each of mass 67.8 g, are moving in opposite directions at 4.80 m/s.They collide head-on and bounce ap

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Answer:

Explanation:

To detrmine the time interval at which the balls are in contact.

<u>Given information</u>

The mass of each steal ball 67.8g. The speed of ball towards each other is 4.80 m/s. The exerted force by each jaw 15.9 kN and the forece reduce the diameter by 0.130 mm.

Expression for the effective spring constant ball is shown below.

K = |F|/|x|

Here,

k is a spring constant

F is the force exerted on the ball

x is dispalcement due force

substitute 15.9 kN for F and 0.130 mm in above equation

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3 years ago

Why did donald hings invent the walkie talkie ?

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3 years ago

A 0.42 kg mass is attached to a light spring with a force constant of 34.9 N/m and set into oscillation on a horizontal friction

Whitepunk [10]

(a) 0.456 m/s

The maximum speed of the oscillating mass can be found by using the conservation of energy. In fact:

- At the point of maximum displacement, the mechanical energy of the system is just elastic potential energy:

$E=U=\frac{1}{2}kA^2$ (1)

where

k = 34.9 N/m is the spring constant

A = 5.0 cm = 0.05 m is the amplitude of the oscillation

- At the point of equilibrium, the displacement is zero, so all the mechanical energy of the system is just kinetic energy:

$E=K=\frac{1}{2}mv_{max}^2$ (2)

where

m = 0.42 kg is the mass

vmax is the maximum speed, which is maximum when the mass passes the equilibrium position

Since the mechanical energy is conserved, we can write (1) = (2):

$\frac{1}{2}kA^2=\frac{1}{2}mv_{max}^2\\v_{max}=\sqrt{\frac{kA^2}{m}}=\sqrt{\frac{(34.9 N/m)(0.05 m)^2}{0.42 kg}}=0.456 m/s$

(b) 0.437 m/s

When the spring is compressed by x = 1.5 cm = 0.015 m, the equation for the conservation of energy becomes:

$E=\frac{1}{2}kx^2+\frac{1}{2}mv^2$ (3)

where the total mechanical energy can be calculated at the point where the displacement is maximum (x = A = 0.05 m):

$E=\frac{1}{2}kA^2=\frac{1}{2}(34.9 N/m)(0.05 m)^2=0.044 J$

So, solving (3) for v, we find the speed when x=1.5 cm:

$v=\sqrt{\frac{2E-kx^2}{m}}=\sqrt{\frac{2(0.044 J)-(34.9 N/m)(0.015 m)^2}{0.42 kg}}=0.437 m/s$

(c) 0.437 m/s

This part of the problem is exactly identical to part b), since the displacement of the mass is still

x = 1.5 cm = 0.015 m

So, the speed when this is the displacement is

$v=\sqrt{\frac{2E-kx^2}{m}}=\sqrt{\frac{2(0.044 J)-(34.9 N/m)(0.015 m)^2}{0.42 kg}}=0.437 m/s$

(d) 4.4 cm

In this case, we have that the speed of the mass is 1/2 of the maximum value, so:

$v=\frac{v_{max}}{2}=\frac{0.456 m/s}{2}=0.228 m/s$

And by using the conservation of energy again, we can find the corresponding value of the displacement x:

$E=\frac{1}{2}kx^2+\frac{1}{2}mv^2\\x=\sqrt{\frac{2E-mv^2}{k}}=\sqrt{\frac{2(0.044 J)-(0.42 kg)(0.228 m/s)^2}{34.9 N/m}}=0.044 m=4.4 cm$

4 0

3 years ago