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3 years ago

During a fission each time an atom divides ___.

2 answers:

7 0

It releases a lot of energy so I would say, the answer IS NOT half. More energy than what is released though? I would say the Third option

garri49 [273]3 years ago

3 0

“More energy is released”

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The "lead" in pencils is a graphite composition with a Young’s modulus of about 1×1010N/m21×1010⁢N/m2. Calculate the change in l

Tanya [424]

Answer:

b) 0.1 mm

Explanation:

Given that

E= 1 x 10¹⁰ N/m²

F= 4 N

d= 0.5 mm

L = 60 mm

We know that elongation due to force F given as

$\Delta L=\dfrac{FL}{AE}$

$\Delta L=\dfrac{FL}{\dfrac{\pi d^2}{4}\times E}$

$\Delta L=\dfrac{4\times 60}{\dfrac{\pi \times 0.5^2}{4}\times 10^4}$

ΔL = 0.12 mm

Therefore the answer is -

b) 0.1 mm

6 0

3 years ago

A 1.8 kg uniform rod with a length of 90 cm is attached at one end to a frictionless pivot. It is free to rotate about the pivot

Leokris [45]

Answer:

a. 32.67 rad/s² b. 29.4 m/s²

Explanation:

a. The initial angular acceleration of the rod

Since torque τ = Iα = WL (since the weight of the rod W is the only force acting on the rod , so it gives it a torque, τ at distance L from the pivot )where I = rotational inertia of uniform rod about pivot = mL²/3 (moment of inertia about an axis through one end of the rod), α = initial angular acceleration, W = weight of rod = mg where m = mass of rod = 1.8 kg and g = acceleration due to gravity = 9.8 m/s² and L = length of rod = 90 cm = 0.9 m.

So, Iα = WL

mL²α/3 = mgL

dividing through by mL, we have

Lα/3 = g

multiplying both sides by 3, we have

Lα = 3g

dividing both sides by L, we have

α = 3g/L

Substituting the values of the variables, we have

α = 3g/L

= 3 × 9.8 m/s²/0.9 m

= 29.4/0.9 rad/s²

= 32.67 rad/s²

b. The initial linear acceleration of the right end of the rod?

The linear acceleration at the initial point is tangential, so a = Lα = 0.9 m × 32.67 rad/s² = 29.4 m/s²

5 0

3 years ago

A 4.00 kg block is suspended from a spring with k 500 N/m. A 50.0 g bullet is fired into the block from directly below with a sp

Yanka [14]

Answer:

a. A = 0.1656 m

b. % E = 1.219

Explanation:

Given

mB = 4.0 kg , mb = 50.0 g = 0.05 kg , u₁ = 150 m/s , k = 500 N / m

To find the amplitude of the resulting SHM using conserver energy

ΔKe + ΔUg + ΔUs = 0

¹/₂ * m * v² - ¹/₂ * k * A² = 0

A = √ mB * vₓ² / k

vₓ = mb * u₁ / mb + mB

vₓ = 0.05 kg * 150 m / s / [0.050 + 4.0 ] kg = 1.8518

A = √ 4.0 kg * (1.852 m/s)² / (500 N / m)

A = 0.1656 m

The percentage of kinetic energy

%E = Es / Ek

Es = ¹/₂ * k * A² = 500 N / m * 0.1656²m = 13.72 N*0.5

Ek = ¹/₂ * mb * v² = 0.05 kg * 150² m/s = 1125 N

% E = 13.72 / 1125 = 0.01219 *100

% E = 1.219

6 0

3 years ago

The wattage of a bulb is 24 W when it is connected to a 12 V battery. Calculate its effective wattage if it operates on a 6 V ba

tresset_1 [31]

It would be 12W because: 6v is half of 12v so half of 24w would be 12w

3 0

3 years ago

Read 2 more answers

A baseball player leads off the game and hits a long home run. The ball leaves the bat at an angle of 70.0 from the horizontal w

professor190 [17]

Answer: 211.059 m

Explanation:

We have the following data:

$\theta=70\°$ The angle at which the ball leaves the bat

$V_{o}=55 m/s$ The initial velocity of the ball

$g=-9.8 m/s^{2}$ The acceleration due gravity

We need to find how far (horizontally) the ball travels in the air: $x$

Firstly we need to know this velocity has two components:

<u>Horizontally:</u>

$V_{ox}=V_{o}cos \theta$ (1)

$V_{ox}=55 m/s cos(70\°)=18.811 m/s$ (2)

<u>Vertically:</u>

$V_{oy}=V_{o}sin \theta$ (3)

$V_{oy}=55 m/s sin(70\°)=51.683 m/s$ (4)

On the other hand, when we talk about parabolic movement (as in this situation) the ball reaches its maximum height just in the middle of this parabola, when $V=0$ and the time $t$ is half the time it takes the complete parabolic path.

So, if we use the following equation, we will find $t$ :

$V=V_{o}+gt=0$ (5)

Isolating $t$ :

$t=\frac{-V_{o}}{g}$ (6)

$t=\frac{-55 m/s}{-9.8 m/s^{2}}$ (7)

$t=5.61 s$ (8)

Now that we have the time it takes to the ball to travel half of is path, we can find the total time $T$ it takes the complete parabolic path, which is twice $t$ :

$T=2t=2(5.61 s)=11.22 s$ (9)

With this result in mind, we can finally calculate how far the ball travels in the air:

$x=V_{ox}T$ (10)

Substituting (2) and (9) in (10):

$x=(18.811 m/s)(11.22 s)$ (11)

Finally:

$x=211.059 m$

8 0

3 years ago