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olasank [31]
3 years ago
12

A bullet of mass 0.107 kg traveling horizontally at a speed of 300 m/s embeds itself in a block of mass 3 kg that is sitting at

rest on a nearly frictionless surface. What was the transfer of energy (microscopic work) from the surroundings into the block+bullet system during the collision? (Remember that represents energy transfer due to a temperature difference between a system and its surroundings.)
Physics
1 answer:
melisa1 [442]3 years ago
5 0

Answer:

165.77J

Explanation:

M₁ = 0.107kg

u₁ = 300m/s

m₂ = 3kg

u₂ = 0

v =

m₁u₁ + m₂u₂ = (m₁ + m₂)V

(0.107*300) + 0 = (0.107 + 3)V

V = 32.1 / 3.107 = 10.33m/s

kinetic energy of the system after collision =

½m1v² + ½m2v²

K.E = ½(m1 + m2)v²

K.E = ½(0.107+3) * 10.33²

K.E = 165.77J

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The electric field strength is inversely related to the square of the distance.so  the strength of the electric field is given by

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Here,  \frac{1}{4\pi \epsilon _{0}  } = k  is constant depend upon medium and its value is 9.0 \times10^{9} \ N m^2/C^2 and q is charge and  r is the distance.

Given  r = 0.90 mm = 9.0 \times 10^{-4} m and we know the charge of proton, q = 1.6\times 10^{-19} \ C.

Therefore,

E=\frac{9.0 \times10^{9} \ N m^2/C^2 \times1.6\times 10^{-19} \ C  }{(9.0 \times 10^{-4} m)^2} = 0.177 \times 10^{-2} \ N/C \\\\  E= 1.77 \times 10^{-3}  N/C

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When turned on, a fan requires 5.0 seconds to get up to its final operating rotational speed of 1200 rpm. a) How large is the fi
vova2212 [387]

Answer:

a)

125.6 rad/s

b)

25.12 rad/s²

Explanation:

a)

t = time required by the fan to get up to final operating speed = 5 sec

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w = 1200\frac{rev}{min}\frac{2\pi rad}{1 rev}\frac{1 min}{60 sec}

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