Decide which examples are Nuclear Fission reaction.

Physics

2 answers:

MrMuchimi3 years ago

4 0

a,c,d, is your answer I think I hope it helps

Grace [21]3 years ago

4 0

Answer is the letter A

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Find the distance along an arc on the surface of the earth that subtends a central angle of 1 minutes (1 minute = 1/60 degree).

-Dominant- [34]

Answer:

1.152 miles

Explanation:

Given: central angle = 1 minute = $(\frac{1}{60}) ^{o}$

radius of the earth = 3960 miles

The length of an arc = $\frac{\alpha }{360^{o} }$ 2 $\pi$ r

where: $\alpha$ is the central angle, and r is the radius.

Thus,

Distance along the arc = $\frac{\alpha }{360^{o} }$ 2 $\pi$ r

Distance along the arc = $\frac{(\frac{1}{60}) ^{o} }{360^{o} }$ x 2 x $\frac{22}{7}$ x 3960

= $\frac{(\frac{1}{60}) ^{o} }{360^{o} }$ x 24891.4286

= 1.1524

The required distance along an arc is 1.152 miles.

4 0

4 years ago

A house has well-insulated walls. It contains a volume of 105 m3 of air at 305 K.

REY [17]

Answer: $85.46\ kJ$

Explanation:

Given

Volume of air $V=105\ m^3$

Temperature of air $T=305\ K$

Increase in temperature $\Delta T=0.7^{\circ}C$

Specific heat for diatomic gas is $C_p=\dfrac{7R}{2}$

Energy required to increase the temperature is

$\Rightarrow Q=nC_pdT\\\\\Rightarrow Q=n\times \dfrac{7R}{2}\times \Delta T\\\\\Rightarrow Q=\dfrac{7}{2}nR\Delta T\\\\\Rightarrow Q=\dfrac{7}{2}\times \dfrac{PV}{T}\times \Delta T\quad [\text{using PV=nRT}]$

Insert the values

$\Rightarrow Q=\dfrac{7}{2}\times \dfrac{1.01325\times 10^5\times 105}{305}\times 0.7\\ \text{Assuming air pressure to be atmospheric P=}1.01325\times 10^5\ N/m^2\\\\\Rightarrow Q=0.8546\times 10^5\\\Rightarrow Q=85.46\ kJ$