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3 years ago

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A 833 kg automobile is sliding on an icy street. It collides with a parked car which has a mass of 561 kg. The two cars lock up

and slide together with a speed of 16.8 km/h. What was the speed of the first car just before the collision?

1 answer:

V125BC [204]3 years ago

5 0

Answer:

The speed of the first car just before the collision is 7.841 m/s.

Explanation:

Given that,

Mass of the automobile, m = 833 kg

Mass of the parked car, m' = 561 kg

The final speed of the cars lock up and slide together with a speed of 16.8 km/h, V = 16.8 km/h = 4.67 m/s

If two cars lock up and slide together, then it is the case of inelastic collision. The momentum will remains conserved such as :

$mv+m'v'=(m+m')V$

v is the speed of the first car

v' is the speed of the second car that is at rest

So,

$mv=(m+m')V$

$v=\dfrac{(m+m')V}{m}$

$v=\dfrac{(833+561)\times 4.67}{833}$

$v=7.81\ m/s$

So, the speed of the first car just before the collision is 7.841 m/s. Hence, this is the required solution.

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A 70.0-kilogram man is walking at a speed if 2.0 m/s. What is the kinetic energy?

Paraphin [41]

Answer:

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Explanation:

From the the question, the mass of the man =70.0 kg and the speed at which the man is walking =2.0 m/s.

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where K.E = the kinetic energy, m=mass and v= speed.

By substitution,

$K.E = \frac{1}{2} \times 70 \times {2}^{2}$

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$\implies K.E =140$

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The closest stars are 4 light years away from us. How far away must you be from a 854 kHz radio station with power 50.0 kW for t

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Answer:

The distance from the radio station is 0.28 light years away.

Solution:

As per the question:

Distance, d = 4 ly

Frequency of the radio station, f = 854 kHz = $854\times 10^{3}\ Hz$

Power, P = 50 kW = $50\times 10^{3}\ W$

$I_{p} = 1\ photon/s/m^{2}$

Now,

From the relation:

P = nhf

where

n = no. of photons/second

h = Planck's constant

f = frequency

Now,

$n = \frac{P}{hf} = \frac{50\times 10^{3}}{6.626\times 10^{- 34}\times 854\times 10^{3}} = 8.836\times 10^{31}\ photons/s$

Area of the sphere, A = $4\pi r^{2}$

Now,

Suppose the distance from the radio station be 'r' from where the intensity of the photon is $1\ photon/s/m^{2}$

$I_{p} = \frac{n}{A} = \frac{n}{4\pi r^{2}}$

$1 = \frac{8.836\times 10^{31}}{4\pi r^{2}}$

$r = \sqrt{\frac{8.836\times 10^{31}}{4\pi}} = 2.65\times 10^{15}\ m$

Now,

We know that:

1 ly = $9.4607\times 10^{15}\ m$

Thus

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3 years ago

In a single wire, how much current would be required to generate 1 Tesla magnetic field at a 2 meter distance away from the wire

aev [14]

Answer:

12.56 A.

Explanation:

The magnetic field of a conductor carrying current is give as

H = I/2πr ............................... Equation 1

Where H = Magnetic Field, I = current, r = distance, and π = pie

Making I the subject of the equation,

I = 2πrH............... Equation 2

Given: H = 1 T, r = 2 m.

Constant: π = 3.14

Substitute into equation 2

I = 2×3.14×2×1

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A pilot is upside down at the top of an inverted loop of radius 3.20 x 103 m. At the top of the loop his normal force is only on

n200080 [17]

Answer:

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Explanation:

Given that:

Radius, r

R = 3.20 * 10^3.

Normal force = 0.5 * normal weight

Normal force = Fn ; Normal weight = Fg

Fn = 0.5Fg

Recall:

mv² / R = Fn + Fg

Fn = 0.5Fg

mv² / R = 0.5Fg + Fg

mv² /R = 1.5Fg

mv² = 1.5Fg * R

F = mg

mv² = 1.5* mg * R

v² = 1.5gR

v = sqrt(1.5gR)

V = sqrt(1.5 * 9.8 * 3.2 * 10^3)

V = sqrt(47.04^3)

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