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3 years ago

(Select all that apply)

Physics

2 answers:

loris [4]3 years ago

5 0

Answer: a. air pollution

c. hazardous wastes

d. potential reactor accident

e. water pollution

A nuclear energy is produced in a thermal power plant. A nuclear energy is produced in a nuclear reactor. In nuclear reactor nuclear fission reactions takes place in which an atoms absorbs energy from radiations and undergo fission and produces energy in the form of high intensity radiations along with heat. Although the fission reactions takes place in a nuclear reactor in a controlled way so that the radiation may not leak out from the reactor. The accidentally leak out radiations or explosion or bursting of the reactor due to uncontrolled thermal energy production can result in air pollution as the leak out air will cause bursting effects which will contaminate the air.

The nuclear waste are radioactive and are non-biodegradable these wastes are disposed off deep in geospheres and in water. They have potential to contaminate both land and water. Radioactive wastes can cause mutations in the genome of the organisms exposed to these wastes which generate deadly diseases and disorders. Therefore, these wastes are hazardous.

Triss [41]3 years ago

5 0

Answer: Hazardous wastes and potential reactor accident.

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Question 30

stellarik [79]

Answer: $0.69\°$

Explanation:

The angular diameter $\delta$ of a spherical object is given by the following formula:

$\delta=2 sin^{-1}(\frac{d}{2D})$

Where:

$d=16 m$ is the actual diameter

$D=1338 m$ is the distance to the spherical object

Hence:

$\delta=2 sin^{-1}(\frac{16 m}{2(1338 m)})$

$\delta=0.685\° \approx 0.69\°$ This is the angular diameter

3 0

3 years ago

On the way to school, the bus speeds up from 20 m/s to 36 m/s in 4 seconds. What distance

Tresset [83]

Answer B. 112 m

Step-by-Step Explanation

initial velocity u = 20 m /s
final velocity v = 36 m /s
time taken t = 4 s
acceleration = (v - U) / t
= (36 - 20) / 4
a=4m/s2
from the formula
7-u2=2as , sis distance covered
putting the values
362-202=2×4×s
1296 - 400 = 8 x S
S= 112 m

4 0

2 years ago

In February 1955, a paratrooper fell 370 m from an airplane without being able to open his chute but happened to land in snow, s

nevsk [136]

a) 0.94 m

The work done by the snow to decelerate the paratrooper is equal to the change in kinetic energy of the man:

$W=\Delta K\\-F d = \frac{1}{2}mv^2 - \frac{1}{2}mu^2$

where:

$F=1.1 \cdot 10^5 N$ is the force applied by the snow

d is the displacement of the man in the snow, so it is the depth of the snow that stopped him

m = 68 kg is the man's mass

v = 0 is the final speed of the man

u = 55 m/s is the initial speed of the man (when it touches the ground)

and where the negative sign in the work is due to the fact that the force exerted by the snow on the man (upward) is opposite to the displacement of the man (downward)

Solving the equation for d, we find:

$d=\frac{1}{2F}mu^2 = \frac{(68 kg)(55 m/s)^2}{2(1.1\cdot 10^5 N)}=0.94 m$

b) -3740 kg m/s

The magnitude of the impulse exerted by the snow on the man is equal to the variation of momentum of the man:

$I=\Delta p = m \Delta v$

where

m = 68 kg is the mass of the man

$\Delta v = 0-55 m/s = -55 m/s$ is the change in velocity of the man

Substituting,

$I=(68 kg)(-55 m/s)=-3740 kg m/s$

7 0

3 years ago

A wheel that is rotating at 33.3 rad/s is given an angular acceleration of 2.15 rad/s 2. Through what angle has the wheel turned

Neporo4naja [7]

Answer:

The angle through which the wheel turned is 947.7 rad.

Explanation:

initial angular velocity, $\omega _i$ = 33.3 rad/s

angular acceleration, α = 2.15 rad/s²

final angular velocity, $\omega_f$ = 72 rad/s

angle the wheel turned, θ = ?

The angle through which the wheel turned can be calculated by applying the following kinematic equation;

$\omega_f^2 = \omega_i^2 + 2\alpha \theta\\\\\theta = \frac{\omega_f^2\ -\ \omega_i^2}{2\alpha } \\\\\theta = \frac{(72)^2\ -\ (33.3)^2}{2(2.15)}\\\\\theta = 947.7 \ rad$

Therefore, the angle through which the wheel turned is 947.7 rad.

5 0

3 years ago

already did the work, i just need someone to see if i did the tangent line right? the line has to touch the point 0.6! thank you

Dmitrij [34]

The tangent looks good.

The curve is a bit crooked, at the 0.9 and 1.

But overall, cool graph.

5 0

3 years ago