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Alexandra [31]
3 years ago
12

A 1000 kg car collides with a 1 mg mosquito. By what factor will the acceleration of the mosquito be greater than the accelerati

on of the car when they collide?
Physics
1 answer:
rodikova [14]3 years ago
3 0

Answer:

The acceleration of the mosquito is 1 million (10⁶) times that of the car

Explanation:

From Newton’s Third Law, every action has an equal and opposite reaction. Therefore, when the car and the mosquito collide, the force the car exerts on the mosquito equals the force the mosquito exerts on the car.

F₁ = F₂

Also, from Newton’s Second Law, F₁ = ma₁

Force of the car, F₁ = 1000 * a₁

And force exerted by the mosquito, F₂ = 0.001 * a₂

Equating both sides

1000a₁ = 0.001a₂

Taking the ratio of the acceleration of the mosquito and that of the car, a₂/a₁:

a₂/a₁ = 1000/0.001

a₂/a₁ = 1000000

Therefore, the acceleration of the mosquito is 1000000 (10⁶) times that of the car

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If the distance to a point source of sound is doubled, by what multiplicative factor does the intensity change?
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If the distance to a point source of sound is doubled, by a multiplicative factor of 4, the intensity changes.

Intensity of sound is the sound which is perpendicular to sound wave propogation per unit area. It is dependent on the Surface of source sound.

Intensity is the Power per unit area. Its SI unit is Watt/m².

As we move away from a source of sound, the sound starts to diminish. This is due to the decreasing sound intensity with distance.

It can also be understood by the fact that on increasing distance, the Power radiated by the source spreads over a larger area. Hence, the Intensity decreases gradually.

Since, Intensity is proportional to the square of the distance.

Hence, on doubling the distance, Intensity reduces to one fourth of the initial intensity or reduces by a multiplicative factor of 4.

Learn more about Intensity here, brainly.com/question/17583145

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8 0
2 years ago
In odd nuclei, what determines the final spin of the nucleus?
Katen [24]

Answer:

In odd nuclei, the left out proton or neutron will contribute to the spin of the nucleus.

Explanation:

The meaning of odd nuclei is atomic mass is odd.

A=odd number.

A=Z+n

Here, Z is proton either it will odd or n will odd which is neutron.

Now according to the shell model the left out proton or neutron will contribute to the spin and parity.

For example,

Take the case of isotope of nitrogen-15.

Here Z is 7, and n is 8 will not contribute in spin.

Now, for Z=7.

1S^{2} _{\frac{1}{2} }, 1P^{4} _{\frac{3}{2} }, 1P^{1} _{\frac{1}{2} }

Here,

j=\frac{1}{2}

and, L=1.

Fort parity,

(-1)^{L}

Put the value of L.

Parity will be -1.

Now, spin will be

S=(\frac{1}{2} )^{-1}.

7 0
3 years ago
A boy on a bicycle drags a wagon full of newspapers at 0.80 m/s for 30 min (1800 s) using a force of 40 N. How much work has the
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N133A

Explanation:black

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2 years ago
A T-bar tow is required to pull 85 skiers up a 600 m slope inclined at 15° above horizontal at a speed of 2.5 m/s. The coefficie
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Answer:

P = 1.45 hp or 1.94kW

Explanation:

Given:

v = 2.5m/s

uk = 0.06

m = 60kg

Fk = uk*m*g*fsin(15)

Fk = 0.06*85*9.81*60*sin(15)

Fk = 776.15 N

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P = F*v

P = 776.15N*2.5m/s

P = 1940.36 W

1 Horsepower = 0.7457 Kilowatts

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5 0
3 years ago
A particle initially located at the origin has an acceleration of a=3jm/s^2 and an initial velocity of Vi=5im/s.
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Given the particle's acceleration is

\vec a(t) = \left(3\dfrac{\rm m}{\mathrm s^2}\right)\vec\jmath

with initial velocity

\vec v(0) = \left(5\dfrac{\rm m}{\rm s}\right)\,\vec\imath

and starting at the origin, so that

\vec r(0) = \vec 0

you can compute the velocity and position functions by applying the fundamental theorem of calculus:

\vec v(t) = \vec v(0) + \displaystyle \int_0^t \vec a(u)\,\mathrm du

\vec r(t) = \vec r(0) + \displaystyle \int_0^t \vec v(u)\,\mathrm du

We have

• velocity at time <em>t</em> :

\vec v(t) = \left(5\dfrac{\rm m}{\rm s}\right)\,\vec\imath + \displaystyle \int_0^t \left(3\dfrac{\rm m}{\mathrm s^2}\right)\,\vec\jmath\,\mathrm du \\\\ \vec v(t) = \left(5\dfrac{\rm m}{\rm s}\right)\,\vec\imath + \left(3\dfrac{\rm m}{\mathrm s^2}\right)t\,\vec\jmath \\\\ \boxed{\vec v(t) = \left(5\dfrac{\rm m}{\rm s}\right)\,\vec\imath + \left(3\dfrac{\rm m}{\mathrm s^2}\right)t\,\vec\jmath}

• position at time <em>t</em> :

\vec r(t) = \displaystyle \int_0^t \left(\left(5\dfrac{\rm m}{\rm s}\right)\,\vec\imath + \left(3\dfrac{\rm m}{\mathrm s^2}\right)u\,\vec\jmath\right) \,\mathrm du \\\\ \boxed{\vec r(t) = \left(5\dfrac{\rm m}{\rm s}\right)t\,\vec\imath + \frac12 \left(3\frac{\rm m}{\mathrm s^2}\right)t^2\,\vec\jmath}

8 0
3 years ago
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