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3 years ago

An impulse of 12.2kg m/s is delivered to an object whose initial momentum is 4.5kgm/s. What is the object's final momentum?

Physics

1 answer:

AleksandrR [38]3 years ago

5 0

Answer:

Object's final momentum = 16.7 kg-m/s

Explanation:

Given that,

Impulse, J = 12.2 kg-m/s

Initial momentum, $P_i=4.5\ kg-m/s$

We need to find the object's final momentum. We know that the impulse is equal to the change in momentum. So,

Impulse = change in momentum

$J=P_f-P_i\\\\P_f=J+P_i\\\\P_f=12.2+4.5\\\\P_f=16.7\ kg-m/s$

So, the object's final momentum is 16.7 kg-m/s.

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If you whirl a tin can on the end of a string and the string suddenly breaks, in what direction will the can go?

Contact [7]

Answer:

fly off, tangent to its circular path.

Explanation:

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3 years ago

A straight 1.20 m long conductor has a 2.00 A current travelling toward the East. Earth's magnetic field in this location is 4.9

KATRIN_1 [288]

Answer:

Force's magnitude $=1.176\,10^{-4}\,N$

Direction: down (towards the center of the Earth)

Explanation:

Recall that the magnetic force on a conductor of length L carrying a current I in a magnetic field B is given by the equation: $F=I\,L\,B$ in the case the magnetic field B and the direction of the current are at 90 degrees from each other (which is our case). The direction of the force will be given by the "right hand rule" associated with the vector product that defines this force.

Since the current is moving East, and the magnetic field of the Earth goes from North to South, the resultant Force vector will be pointing towards the Earth (and perpendicular to the plane defined by the current's direction and the magnetic field B)

The magnitude of the force, is given by the formula above, and given that all quantities to be considered are is SI units, it will result in Newtons (N):

$F=I\,L\,B\\F=2\,*\,1.2\,*\,4.9\,10^{-5}\,N\\F=1.176\,10^{-4}\,N$

8 0

4 years ago

The electric field 2.5 mm from a uniform sheet of charge is σ=800, NC. How much charge is contained in a 5.0x5.0 cm section of t

slamgirl [31]

Answer:

The charge is $2.75\times10^{-13}\ C$

Explanation:

Given that,

Distance = 2.5 mm

Electric field = 800 NC

Length $L=5.0\times5.0\times10^{-4}\ m$

We need to calculate the linear charge density

Using formula of linear charge density

$E=\dfrac{2k\lambda}{r}$

$\lambda=\dfrac{Er}{2k}$

Put the value into the formula

$\lambda=\dfrac{800\times2.5\times10^{-3}}{2\times9\times10^{9}}$

$\lambda=1.1\times10^{-10}\ C/m$

We need to calculate the charge

Using formula of charge

$Q=\lambda\timesL$

Put the value into the formula

$Q=1.1\times10^{-10}\times(5.0\times5.0\times10^{-4})$

$Q=2.75\times10^{-13}\ C$

Hence, The charge is $2.75\times10^{-13}\ C$

5 0

3 years ago

Please please help me out I’m lost

disa [49]

Answer:

Option B

The correct answer is 2

3 0

3 years ago

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There are devices to put in a light socket that control the current through a lightbulb, thereby increasing its lifetime. Which

Dmitrij [34]

Answer: B

Explanation:

Limiting the maximum current through the bulb. This will help in preserving or improving the bulb's lifetime and also this won't have an effect on the brightness of the bulb as brightness is affected by the average value. Although brightness is a factor of current, reducing the maximum current won't have any bearing on the average current the bulb is getting.

4 0

3 years ago