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Oksana_A [137]
3 years ago
12

30 points please help a girl out! 

Physics
1 answer:
ziro4ka [17]3 years ago
5 0

the answer would have to be A tell me if im wrong



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A 0.11 kg bullet traveling at speed hits a 18.3 kg block of wood and stays in the wood. The block with the bullet imbedded in it
arlik [135]

Explanation:

The energy of the system before the collision must equal the energy after the collision.

After the collision the bullet and the block have a total mass of 18.41 kg and they move at a speed of 8.8 m/s. The kinetic energy after the collision is

\frac{18.41 kg (8.8 m/s)^2}{2} = 713 J

Before the collision only the bullet has kinetic energy.

So we can now determine the speed of the bullet using

\frac{0.11kg (v^2)}{2} = 713 J\\v = 114 m/s

5 0
2 years ago
Help me with this please
pochemuha

Answer:

1. 31.5

3. 3.5

Explanation:

4 0
3 years ago
How much work (in joules) is done in moving a charge of 2.5 μc a distance of 35 cm along an equipotential at 12 v? do not includ
crimeas [40]

The work done for moving a charge is due to the change in potential between the points. That is the work done is given by the expression W = qΔV. Where q is the charge and ΔV is the change in potential between two points.

In case of equipotential lines the value of potential is same at all the points.

So the value of ΔV when moving along a equipotential line is zero.

So W = qΔV = q*0=0

So no work is needed to move charge along an equipotential line.

Work done = 0

3 0
3 years ago
Helium gas contained in a piston cylinder assembly undergoes an isentropic polytropic process from the given initial state with
jeyben [28]

Answer:

Heat transfer during the process = 0

Work done during the process = - 371.87 KJ

Explanation:

Initial pressure P_{1} = 0.02 bar

Initial temperature T_{1} = 200 K

Final pressure P_{2} = 0.14 bar

Gas constant for helium R = 2.077 \frac{KJ}{kg k}

This is an isentropic polytropic process so temperature - pressure relationship is given by the following formula,

\frac{T_{2} }{T_{1} } = [\frac{P_{2} }{P_{1} } ]^{\frac{\gamma - 1}{\gamma} }

Put all the values in above formula we get,

⇒ \frac{T_{2} }{200} = [\frac{0.14 }{0.02 } ]^{\frac{1.4 - 1}{1.4} }

⇒  \frac{T_{2} }{200} = 1.74

⇒ T_{2} = 348.72 K

This is the final temperature of helium.

For isentropic polytropic process heat transfer to the system is zero.

⇒ ΔQ = 0

Work done W = m × ( T_{1} - T_{2} ) × \frac{R}{\gamma - 1}

⇒ W = 1 × ( 200 - 348.72 ) × \frac{2.077}{1.4 - 1}

⇒ W = 371.87 KJ

This is the work done in this process. here negative sign shows that work is done on the gas in the compression of gas.

3 0
3 years ago
5 single loop of copper wire, lying flat in a plane, has an area of 8.60cm2 and a resistance of 3.00. A uniform magnetic field p
Finger [1]

Answer:

0.652 mA

Explanation:

According to Faraday's Law :

Emf = -  \frac{d \phi}{dt}

= \frac{- \delta \phi }{\delta \ t}

|E| = \frac{A ( \delta B)}{\delta t}

where ;

A = 8.60*10^{-4}

\delta B = 3.00 T - 0.5 T = 2.5 T

|E| = \frac{8.60*10^{-4}*2.5}{1.10}

|E| = 1.96*10^{-3}

Induced current  I = \frac{|E|}{R}

= \frac{1.96*10^{-3}}{3.0}

= 6.52*10^{-4} A

= 0.652 mA

Thus, the induced current in the loop of wire over this time = 0.652 A

4 0
3 years ago
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