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4 years ago

7

Which best describes what forms in nuclear fission?A. two smaller, more stable nucleiB. two larger, less stable nucleiC. one sma

ller, less stable nucleusD. one larger, more stable nucleus

2 answers:

Step2247 [10]4 years ago

4 0

Answer:B. two larger, less stable nuclei

Explanation: They collied and don't combine

quester [9]4 years ago

4 0

Answer:

B) two larger, less stable nuclei

Explanation:

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The micrometer (1 μm) is often called the micron. (a) How many microns make up 3.0 km? (b) How many centimeters equal 3.0 μm? (c

Tomtit [17]

Answer:

3 x 10⁻⁹km

3 x 10⁻⁴cm

2.73 x 10⁶μm

Explanation:

A micron is a subunit of measurement usually for length dimensions.

1μm = 1 x 10⁻⁶m

a. How many microns make up 3km;

Now convert to meter first;

1000m = 1km

So, 3km will be made up of 3000m

So;

1 x 10⁻⁶m = 1μm

3000m = $\frac{3000}{1 x 10^{-6} }$ = $\frac{3 x 10^{3} }{ 1 x 10^{-6} }$ = 3 x 10⁻⁹km

b. How many centimeters equal 3.0 μm?

Since;

1μm = 1 x 10⁻⁶m

3μm = 3 x 1 x 10⁻⁶ = 3 x 10⁻⁶m

So;

100cm = 1m;

1m = 100cm

3 x 10⁻⁶m = 3 x 10⁻⁶ x 10² = 3 x 10⁻⁴cm

c. How many microns are in 3.0 yd?

1yd = 0.91m

3yd = 3 x 0.91 = 2.73m

So;

1 x 10⁻⁶m = 1μm

2.73m will give $\frac{2.73}{1 x 10^{-6} }$ = 2.73 x 10⁶μm

3 0

3 years ago

A 7.25 kg7.25 kg block is sent up a ramp inclined at an angle ????=28.5°θ=28.5° from the horizontal. It is given an initial velo

Free_Kalibri [48]

Answer:

15.03 m

Explanation:

Given:

mass of the block, m = 7.25 kg

Angle, Θ = 28.5°

Initial speed of the block, v₀ = 15 m/s

let the distance traveled by the block be 's'

Now, applying the work energy theorem,

we have

$(m\times g\times\sin(\theta)\times s) + \mu_k\times mg\times s\times cos(\theta) = \frac{1}{2}\times m\times v^2$

on substituting the values in the above equation, we get

$(7.25\times 9.8\times\sin(28.5^o)\times s) + 0.326\times 7.25\times9.8\times s\times cos(28.5^o) = \frac{1}{2}\times 7.25\times 15^2$

or

$33.902\times s) +20.35\times s = 815.625$

or

$54.252\times s = 815.625$

s = 15.03 m

Hence, the block will travel 15.03 m up the ramp

6 0

3 years ago

A student with a mass of 65.5 kg runs up a flight of 18 stairs. Each of the steps has a height of 16.5 cm. The student was able

Helga [31]

The power developed by the student is 756.9 J/s and remains the same if the student takes the same time to climb the stairs when climbing it in two's and three's.

<h3>What is power?</h3>

Power is the rate at work is done.

Power = work done/time
work done = mass × acceleration due to gravity × height

Work done = 65.5 × 10 × (18 × 0.165) = 1945.35 J

Power = 1945.35/2.57 = 756.9 J/s

If the student climbed the steps in two or three at a time, the power does not change if the time remains the same.

Time required = Energy/ power

The time required to convert the Big Mac meal from McDonalds = 4 853 440/756.9

Time required = 6412.26 seconds

Therefore, from the power developed by the student, it will take him 6412.26 seconds to convert all the energy in a Big Mac meal.

Learn more about power at: brainly.com/question/1634438

#SPJ1

5 0

2 years ago

A car accelerates from rest at a rate of 3 m/s² for 4 seconds. How far did the car travel?

Gnom [1K]

Distance= distance(initial)+v(initial)*t+0.5at^2
D=0+0*4+0.5*3*4^2
D=24 m

4 0

3 years ago

Read 2 more answers

The exit nozzle in a jet engine receives air at 1200 K, 150 kPa with negligible kinetic energy. The exit pressure is 80 kPa, and

nikitadnepr [17]

Complete question:

The exit nozzle in a jet engine receives air at 1200 K, 150 kPa with negligible kinetic energy. The exit pressure is 80 kPa, and the process is reversible and adiabatic. Use constant specific heat at 300 K to find the exit velocity.

Answer:

The exit velocity is 629.41 m/s

Explanation:

Given;

initial temperature, T₁ = 1200K

initial pressure, P₁ = 150 kPa

final pressure, P₂ = 80 kPa

specific heat at 300 K, Cp = 1004 J/kgK

k = 1.4

Calculate final temperature;

$T_2 = T_1(\frac{P_2}{P_1})^{\frac{k-1 }{k}$

k = 1.4

$T_2 = T_1(\frac{P_2}{P_1})^{\frac{k-1 }{k}}\\\\T_2 = 1200(\frac{80}{150})^{\frac{1.4-1 }{1.4}}\\\\T_2 = 1002.714K$

Work done is given as;

$W = \frac{1}{2} *m*(v_i^2 - v_e^2)$

inlet velocity is negligible;

$v_e = \sqrt{\frac{2W}{m} } = \sqrt{2*C_p(T_1-T_2)} \\\\v_e = \sqrt{2*1004(1200-1002.714)}\\\\v_e = \sqrt{396150.288} \\\\v_e = 629.41 \ m/s$

Therefore, the exit velocity is 629.41 m/s

6 0

3 years ago