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3 years ago

Given equal time periods, which statement is correct?

Physics

2 answers:

Liono4ka [1.6K]3 years ago

4 0

Answer:

B. Nuclear fusion generates less radioactive waste than nuclear fission.

Explanation:

There are basically two types of nuclear reaction nuclear fusion and nuclear fission. In nuclear fission one molecule breaks into two atoms and generate enormous energy and in nuclear fusion two atoms combined and form a molecule. In nuclear fusion energy released is more than nuclear fission. And waste is generated more in case of nuclear fission.

Mekhanik [1.2K]3 years ago

3 0

The correct option is B.
Nuclear fission and fusion are two different types of nuclear reactions, through which energy may be obtained. Nuclear fission involves the splitting of a molecule into two different part in order to generate energy while nuclear fusion reaction involves the joining together of two elements in other to form one product. Nuclear fission generate much radioactive waste than nuclear fusion. The radioactive waste that is obtainable during nuclear fusion is less than 1% of that produce by nuclear fission.

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A pirate ship shoots a cannon towards an oncoming ship. The cannon ball has a mass of 25 kg and the ship has a mass of 2500 kg.

mars1129 [50]

Answer:

25 m/s in the opposite direction with the ship recoil velocity.

Explanation:

Assume the ship recoil velocity and velocity of the cannon ball aligns. By the law of momentum conservation, the momentum is conserved before and after the shooting. Before the shooting, the total momentum is 0 due to system is at rest. Therefore, the total momentum after the shooting must also be 0:

$m_sv_s + m_bv_b = 0$

where $m_s = 2500 kg, m_b = 25 kg$ are masses of the ship and ball respectively. $v_s = 0.25 m/s, v_b$ are the velocities of the ship and ball respectively, after the shooting.

$2500*0.25 + 25*v_b = 0$

$25v_b = -2500*0.25$

$v_b = -2500*0.25/25 = -25 m/s$

So the cannon ball has a velocity of 25 m/s in the opposite direction with the ship recoil velocity.

3 0

3 years ago

A physics student stands on a cliff overlooking a lake and decides to throw a softball to her friends in the water below. She th

aliina [53]

Answer:

58.5 m

Explanation:

First of all, we need to find the total time the ball takes to reach the water. This can be done by looking at the vertical motion only.

The initial vertical velocity of the ball is

$u_y = u sin \theta$

where

u = 21.5 m/s is the initial speed

$\theta=33.5^{\circ}$ is the angle

Substituting,

$u_y = (21.5) sin 33.5^{\circ} =11.9 m/s$

The vertical position of the ball at time t is given by

$y = h + u_y t + \frac{1}{2}gt^2$

where

h = 13.5 m is the initial heigth

$g = -9.8 m/s^2$ is the acceleration of gravity (negative sign because it points downward)

The ball reaches the water when y = 0, so

$0 = h + u_yt +\frac{1}{2}gt^2\\0 = 13.5 +11.9 t - 4.9t^2$

Which gives two solutions: t = 3.27 s and t = -0.84 s. We discard the negative solution since it is meaningless.

The horizontal velocity of the ball is

$u_y = u cos \theta = (21.5) cos 33.5^{\circ} =17.9 m/s$

And since the motion along the horizontal direction is a uniform motion, we can find the horizontal distance travelled by the ball as follows:

$d= u_x t = (17.9)(3.27)=58.5 m$

3 0

3 years ago

A train that has wheels with a diameter of 91.44 cm (36 inches used for 100 ton capacity cars) slows down from 82.5 km/h to 32.5

podryga [215]

Answer:

The answer is below

Explanation:

The initial velocity = u = 82.5 km/h = 22.92 m/s, the final velocity = 32.5 km/h = 9.03 m/s, diameter = 91.55 cm = 0.9144 cm

radius (r) = diameter / 2 = 0.9144 / 2= 0.4572 m

a) Initial angular velocity ( $\omega_o$ ) = u /r = 22.92 / 0.4572 = 50.13 rad/s, final velocity (ω) = v / r = 9.03 / 0.4592 = 19.67 rad / s

θ = 95 rev * 2πr = 95 * 2π * 0.4572= 272.9 rad

angular acceleration (α) is:

$\omega^2=\omega_o^2+2\alpha \theta\\\\19.67^2-50.13^2=2\alpha(272.9)\\\\19.67^2=50.13^2+2\alpha(272.9)\\\\2\alpha(272.9)=-2126.108\\\\\alpha=-3.89\ rad/s^2\\\\$

$\omega=\omega_o+\alpha t\\\\19.67=50.13+(-3.89t)\\\\3.89t=50.13-19..67\\\\3.89t=30.46\\\\t=7.83\ s$

c) θ = 95 rev * 2πr = 95 * 2π * 0.4572= 272.9 rad

a) When it stops, the final angular velocity is 0. Hence:

$\omega^2=\omega_o^2+2\alpha \theta\\\\0=50.13^2+2(-3.89)\theta\\\\2(3.89)\theta=50.13^2\\\\2(3.89)\theta=2513\\\\\theta=323\ rad\\\\revolutions=\frac{\theta}{2\pi r}=\frac{323}{2\pi(0.4572)} =112.4\ rev$