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3 years ago

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90. An uranium nucleus (at rest) undergoes fission and splits into two fragments, one heavy and the other light. Which fragment

is moving faster after the fission?

1 answer:

Mumz [18]3 years ago

3 0

Answer:

Explanation:

No external force is acting on uranium nucleus so we can apply law of conservation of momentum . If M and m be the mass of bigger and smaller fragments respectively and V and v be their velocity after fission , Total momentum after fission = total momentum before fission

MV - mv = 0 , since the nucleus was at rest , its momentum before fission is zero . negative sign of mv is taken because smaller fragment will move in opposite direction to that of bigger fragment .

MV = mv

M / m = v / V

M > m

Hence,

v > V

So velocity of smaller fragment will be higher .

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I need it due in 10 mins

ExtremeBDS [4]

Answer:

B. 14.4 N

Rotational speed (Angular Velocity) = 2

The Radius of the circle = 1.2 m

Velocity = Angular velocity × radius = 2×1.2 = 2.4 m/s

Centripetal force= mv²/r = 3 × 2.4×2.4/1.2 = 3 × 2.4 × 2

= 14.4 N

7 0

3 years ago

A spring has a relaxed length of 35 cm (0.35 m) and its spring stiffness is 12 N/m. You glue a 66 gram block (0.066 kg) to the t

Sonbull [250]

We have that the value for y is

$y=0.12376m$

From the Question we are told that

Relaxed length of 35 cm (0.35 m)

Spring stiffness is 12 N/m.

Glue a 66 gram block (0.066 kg)

Total length is l_t=16 cm

Block during a 0.24-second interval

Generally the equation for the Force of spring of block is mathematically given as

$F_{spring}=kx\\\\\ F_{spring}=12*(0.35-0.16)\\\\ F_{spring}=2.28N$

Generally

$F_{elastic}=-mg\\\\F_{elastic}=-0.066*9.8\\\\F_{elastic}=-0.6468N$

Generally,Net Force

$F_{net}=F_{spring}-F_{elastic}\\\\F_{net}=2.28N-(-0.6468N)\\\\F_{net}=2.9268N$

Generally,Velocity of block

$V_y=\frac{py}{m}$

Where

$Py= F_{net}*dt\\\\Py= 2.9268*0.08\\\\Py=0.234144$

Therefore

$V_y=\frac{0.234144}{0.066}\\\\V_y=3.547m/s$

Generally the equation for the Velocity of block is mathematically given as

$Vy=\frac{yf-yt}{t-t_0}\\\\y=(3.547(0.08-0))-0.16$

$y=0.12376m$

For more information on this visit

brainly.com/question/22271063?referrer=searchResults

4 0

3 years ago

A car travels at a constant speed around a circular trackwhose

Mashutka [201]

Answer:

a = 0.77 m/s^2

Explanation:

The centripetal acceleration is given as

$a = \frac{v^2}{R}$

The velocity of the car can be calculated using the circumference of the track and the period of the car.

$2\pi R = v T\\2\pi 2.6 = v 360\\v = 0.045 ~km/s = 45~m/s$

So, the acceleration is

$a = \frac{45^2}{2.6\times 10^3} = 0.77~m/s^2$

8 0

4 years ago

A pendulum that moves through its equilibrium position once every 1.000 s is sometimes called a “seconds pendulum.” what is the

horrorfan [7]

The equilibrium position for a pendulum is straight down. If it moves through that position every second then its period is actually 2 seconds. This is because the period is how long it takes to go from one extreme and back again. It will pass through the equilibrium point twice when doing this. Once on the way down and again on the way back.

5 0

3 years ago

If position vector r = bt^2i + ct^3j, where b and c are positive constants, when does the velocity vector make an angle of 450 w

vazorg [7]

$\vec r(t)=bt^2\,\vec\imath+ct^3\,\vec\jmath$

The velocity at time $t$ is

$\dfrac{\mathrm d\vec r(t)}{\mathrm dt}=2bt\,\vec\imath+3ct^2\,\vec\jmath$

Take two vectors that point in the positive $x$ and positive $y$ directions, such as $\vec\imath$ and $\vec\jmath$ . The dot products of the velocity vector with $\vec\imath$ and $\vec\jmath$ are

$\dfrac{\mathrm d\vec r(t)}{\mathrm dt}\cdot\vec\imath=2bt=\sqrt{4b^2t^2+9c^2t^4}\cos\theta$

and

$\dfrac{\mathrm d\vec r(t)}{\mathrm dt}\cdot\vec\jmath=3ct^2=\sqrt{4b^2t^2+9c^2t^4}\cos\theta$

We want the angles between these vectors to be 45º, for which we have $\cos45^\circ=\frac1{\sqrt2}$ . So

$\begin{cases}2\sqrt2\,bt=\sqrt{4b^2t^2+9c^2t^4}\\3\sqrt2\,ct^2=\sqrt{4b^2t^2+9c^2t^4}\end{cases}\implies3\sqrt2\,ct^2-2\sqrt2\,bt=0$

$\implies t(3ct-2b)=0$

$\implies t=0\text{ or }t=\dfrac{2b}{3c}$

When $t=0$ , the velocity vector is equal to the zero vector, which technically has no direction/doesn't make an angle with any other vector. So the only time this happens is for

$\boxed{t=\dfrac{2b}{3c}}$

7 0

3 years ago