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2 years ago

Use F = 1/T as your basis:

Physics

1 answer:

skelet666 [1.2K]2 years ago

6 0

Time Period=T=2.5×10^{-3}s

Now

Frequency:-

$\\ \rm\rightarrowtail \nu=\dfrac{1}{T}$

$\\ \rm\rightarrowtail \nu=\dfrac{1}{2.5\times 10^{-3}}$

$\\ \rm\rightarrowtail \nu=0.4\times 10^{3}$

$\\ \rm\rightarrowtail \nu=400Hz$

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How Many Negative (-) Electrons are there

makvit [3.9K]

Answer:

All electrons are negative(-) charged

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3 years ago

A stationary police car emits a sound of frequency 1240 HzHz that bounces off of a car on the highway and returns with a frequen

Tju [1.3M]

Answer

given,

frequency from Police car= 1240 Hz

frequency of sound after return = 1275 Hz

Calculating the speed of the car = ?

Using Doppler's effect formula

Frequency received by the other car

$f_1 = \dfrac{f_0(u + v)}{u}$ ..........(1)

u is the speed of sound = 340 m/s

v is the speed of the car

Frequency of the police car received

$f_2= \dfrac{f_1(u)}{u-v}$

now, inserting the value of equation (1)

$f_2= f_0\dfrac{u+v}{u-v}$

$1275=1240\times \dfrac{340+v}{340-v}$

1.02822(340 - v) = 340 + v

2.02822 v = 340 x 0.028822

2.02822 v = 9.799

v = 4.83 m/s

hence, the speed of the car is equal to v = 4.83 m/s

5 0

3 years ago

Two cars collide at an intersection. Car A , with a mass of 2000kg , is going from west to east, while car B , of mass 1400kg ,

ahrayia [7]

Complete Question:

Two cars collide at an intersection. Car A , with a mass of 2000 kg, is going from west to east, while car B, of mass 1500 kg, is going from north to south at 15 m/s. As a result, the two cars become enmeshed and move as one. As an expert witness, you inspect the scene and determine that, after the collision, the enmeshed cars moved at an angle of 65∘ south of east from the point of impact. (a) How fast were the enmeshed cars moving just after the collision? (b) How fast was car A going just before the collision?

Answer:

a) 6.36 m/s b) 4.57 m/s

Explanation:

a) Assuming no external forces acting during the collision, total momentum must be conserved.

As momentum is a vector, we can decompose it along two directions perpendicular each other.

Just for convenience, we choose as our x-axis to the W-E direction, and as our y-axis, the direction N-S.

If we know that total momentum must be conserved, same must be true for both components, px and py.

Applying the information provided (both cars become enmeshed after the collision, moving at an angle of 65º south of east from the point of impact), we have:

px = ma * va = (ma+mb) * vab * cos 65º (1)

py = mb * vb = (ma + mb) * vab * sin 65º (2)

Replacing by the values of ma, mb, and sin 65º, we can solve for vab, as follows:

vab = 1,400 kg* 14.0 m/s / (3,400 Kg * sin 65º) = 6.36 m/s

b) Replacing vab from above in (1), and solving for va, we have:

va = 3,400 kg* 6.36 m/s* cos 65º / 2,000 Kg = 4.57 m/s

7 0

3 years ago

The elementary particle called a muon is unstable and decays in about 2.20μs2.20μs , as observed in its rest frame, into an elec

vredina [299]

Answer:

5.865 μs

Explanation:

t₀ = Time taken to decay a muon = 2.20 μs

c = Speed of Light in vacuum = 3×10⁸ m/s

v = Velocity of muon = 0.927 c

t = Lifetime observed

Time dilation

$t=\frac{t_0}{\sqrt{1-\frac{v^2}{c^2}}}\\\Rightarrow t=\frac{2.2\times 10^{-6}}{\sqrt{1-\frac{(0.927c)^2}{c^2}}}\\\Rightarrow t=\frac{2.2\times 10^{-6}}{\sqrt{1-0.927^2}}\\\Rightarrow t=\frac{2.2\times 10^{-6}}{\sqrt{0.140671}}\\\Rightarrow t=\frac{2.2\times 10^{-6}}{0.3750}\\\Rightarrow t=5.865\times 10^{-6}\ seconds$

∴Lifetime observed for muons approaching at 0.927 the speed of light is 5.865 μs

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3 years ago

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Kamila [148]

Answer:

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Explanation:

6 0

3 years ago