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stiv31 [10]
3 years ago
7

A bullet has a speed of 500 m/s as it leaves a rifle. If it is fired horizontally from a cliff 12m above a lake, how far does th

e bullet travel before striking the water?
Physics
1 answer:
dem82 [27]3 years ago
3 0
Because you don’t have it
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A miner finds a small mineral fragment with a volume of 5.74 cm^3 and a mass of 28.7 g. What is the density of that mineral frag
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P=M÷V
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Tapping the surface of a pan of water generates 17.5 waves per second. If the wavelength of each wave is 45 cm, what is the spee
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Answer:

Speed of the wave is 7.87 m/s.

Explanation:

It is given that, tapping the surface of a pan of water generates 17.5 waves per second.

We know that the number of waves per second is called the frequency of a wave.

So, f = 17.5 Hz

Wavelength of each wave, \lambda=45\ cm=0.45\ m

Speed of the wave is given by :

v=f\lambda

v=17.5\times 0.45

v = 7.87 m/s

So, the speed of the wave is 7.87 m/s. Hence, this is the required solution.

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3 years ago
Difference between discrete and continuous charge distribution?
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Discrete systems are those systems in which are made up of finite component particles a which are non-homogeneously arranged such that no smooth variation exists. It is such that all constituent particles have properties which vary randomly. They are direct opposite to continuous systems, which are smooth arrangement of particles which cannot be individually taken into consideration.
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When buying a car, you tell the salesman that you want to make sure the car can go from 0 miles per hour to 60 miles per hour in
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If the magnitude of the magnetic field is 2.50 mT at a distance of 12.6 cm from a long straight current carrying wire, what is t
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Answer:

The magnetic field at a distance of 19.8 cm from the wire is 1.591 mT

Explanation:

Given;

first magnetic field at first distance, B₁ = 2.50 mT

first distance, r₁ = 12.6 cm = 0.126 m

Second magnetic field at Second distance, B₂ = ?

Second distance, r₂ = ?

Magnetic field for a straight wire is given as;

B = \frac{\mu I}{2 \pi r}

Where:

μ is permeability

B is magnetic field

I is current flowing in the wire

r distance to the wire

Let \ \frac{\mu I}{2\pi}  \ be \ constant; = K\\\\B = \frac{K}{r} \\\\K = Br\\\\B_1r_1 = B_2r_2\\\\B_2 =\frac{B_1r_1}{r_2} \\\\B_2 = \frac{2.5*10^{-3} *0.126}{0.198} \\\\B_2 = 1.591 *10^{-3}\ T\\\\B_2 = 1.591 \ mT

Therefore, the magnetic field at a distance of 19.8 cm from the wire is 1.591 mT

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