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3 years ago

Show that the 1 and 3 laws of motion are collection of 2 law of motion ?

Physics

1 answer:

Leya [2.2K]3 years ago

3 0

Answer:

Yes, va

Explanation:

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A moving car has momentum. if it moves twice as fast its momentum is ____________ as much

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It's momentum is twice as much.

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What type of energy is sunlight referred as?

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The sun's energy is refferd to solor energy

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3 years ago

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A particle of mass 4.00 kg is attached to a spring with a force constant of 100 N/m. It is oscillating on a frictionless, horizo

zloy xaker [14]

Solution :

Given :

Mass attached to the spring = 4 kg

Mass dropped = 6 kg

Force constant = 100 N/m

Initial amplitude = 2 m

Therefore,

a). $v_{initial} = A w$

$= 2 \times \sqrt{\frac{100}{4}}$

= 10 m/s

Final velocity, v at equilibrium position, v = 5 m/s

Now, $\frac{1}{2}(4+4)5^2 = \frac{1}{2} kA'$

A' = amplitude = 1.4142 m

b). $T=2 \pi \sqrt{\frac{m}{k}}$

m' = 2m

Hence, $T'=\sqrt2 T$

c). $\frac{\frac{1}{2}(4+4)5^2 + \frac{1}{2}\times 4 \times 10^2}{\frac{1}{2} \times 4 \times 10^2}$

$=\frac{1}{2}$

Therefore, factor $=\frac{1}{2}$

Thus, the energy will change half times as the result of the collision.

7 0

3 years ago

Near the end of a marathon race, the first two runners are separated by a distance of 45.6 m. The front runner has a velocity of

morpeh [17]

Answer:17.08 s

Explanation:

Given

distance between First and second Runner is 45.6 m

speed of first runner $(v_1)$ =3.1 m/s

speed of second runner $(v_2)$ =4.65 m/s

Distance between first runner and finish line is 250 m

Second runner need to run a distance of 250+45.6=295.6 m

Time required by second runner $t=\frac{295.6}{4.65}=63.56 s$

time required by first runner to reach finish line $=\frac{250}{3.1}=80.64 s$

Thus second runner reach the finish line 80.64-63.56=17.08 s earlier

3 0

3 years ago

How far can a sound wave travel in 90 seconds when the ambient air temperature is 10 C?

Ksju [112]

Answer:

s = 30330.7 m = 30.33 km

Explanation:

First we need to calculate the speed of sound at the given temperature. For this purpose we use the following formula:

v = v₀√[T/273 k]

where,

v = speed of sound at given temperature = ?

v₀ = speed of sound at 0°C = 331 m/s

T = Given Temperature = 10°C + 273 = 283 k

Therefore,

v = (331 m/s)√[283 k/273 k]

v = 337 m/s

Now, we use the following formula to calculate the distance traveled by sound:

s = vt

where,

s = distance traveled = ?

t = time taken = 90 s

Therefore,

s = (337 m/s)(90 s)

6 0

3 years ago