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sergiy2304 [10]
3 years ago
10

Which statements accurately describe the waste generated by nuclear reactions? Check all that apply. It is hazardous for hundred

s to thousands of years. Some waste can be recycled and used again. It is hazardous to living organisms forever. Some waste can be completely transformed into energy. It is not hazardous after a couple of years.
Physics
2 answers:
yarga [219]3 years ago
8 0
A and B is correct..
Gwar [14]3 years ago
4 0
The correct answers are A and B.
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Your roommate drops your wallet down to you from the third-floor window of your apartment, which is 11.5 m from the ground. What
Aneli [31]

Answer:

15 m/s

Explanation:

Using the law of conservation of energy, potential energy equals kinetic energy hence

mgh=0.5mv^{2}

Therefore

v=\sqrt{2gh}

where g is the acceleration due to gravity, m is the mass of the object, h is the height and v is the speed of the wallet

Taking g as 9.81 then

v=\sqrt{2\times 9.8\times 11.5}=15.02098532  m/s\approx 15 m/s

6 0
3 years ago
Two fishing boats depart a harbor at the same time, one traveling east, the other south. the eastbound boat travels at a speed 1
Novosadov [1.4K]
<span>they are travelling at right angles to each other.
 At any given instant they form a right triangle with their starting point
 </span>South bound <span>= x  [mi/h]
</span> East bound <span> = x+1 [mi/h]
 after five hours they will be
 d=5x
 and
 d=5(x+1)
 miles away from the starting point 
 (5x)^2+(5(x+1))^2=625
 25x^2+(5x+5)^2=625
 25x^2+25x^2+50x+25=625
 50</span>x^2+50x-600=0
<span> x^2+ x - 12=0
 (x+4)(x-3)=0
 take the postive value
 x= 3 mph the speed of south bound
 4mph east bound </span>
6 0
3 years ago
What is the difference between a spring and a stream?
Lena [83]
<span>A spring is water coming from under the ground to the surface of the earth and a stream is water that is running along the ground through a trench like place on earth down a hill or steep a area.</span>
3 0
3 years ago
The intensity of light from a star (its brightness) is the power it outputs divided by the surface area over which it’s spread:
kow [346]

Answer:

\frac{d_{1}}{d_{2}}=0.36

Explanation:

1. We can find the temperature of each star using the Wien's Law. This law is given by:

\lambda_{max}=\frac{b}{T}=\frac{2.9x10^{-3}[mK]}{T[K]} (1)

So, the temperature of the first and the second star will be:

T_{1}=3866.7 K

T_{2}=6444.4 K

Now the relation between the absolute luminosity and apparent brightness  is given:

L=l\cdot 4\pi r^{2} (2)

Where:

  • L is the absolute luminosity
  • l is the apparent brightness
  • r is the distance from us in light years

Now, we know that two stars have the same apparent brightness, in other words l₁ = l₂

If we use the equation (2) we have:

\frac{L_{1}}{4\pi r_{1}^2}=\frac{L_{2}}{4\pi r_{2}^2}

So the relative distance between both stars will be:

\left(\frac{d_{1}}{d_{2}}\right)^{2}=\frac{L_{1}}{L_{2}} (3)

The Boltzmann Law says, L=A\sigma T^{4} (4)

  • σ is the Boltzmann constant
  • A is the area
  • T is the temperature
  • L is the absolute luminosity

Let's put (4) in (3) for each star.

\left(\frac{d_{1}}{d_{2}}\right)^{2}=\frac{A_{1}\sigma T_{1}^{4}}{A_{2}\sigma T_{2}^{4}}

As we know both stars have the same size we can canceled out the areas.

\left(\frac{d_{1}}{d_{2}}\right)^{2}=\frac{T_{1}^{4}}{T_{2}^{4}}

\frac{d_{1}}{d_{2}}=\sqrt{\frac{T_{1}^{4}}{T_{2}^{4}}}

\frac{d_{1}}{d_{2}}=\sqrt{\frac{T_{1}^{4}}{T_{2}^{4}}}

\frac{d_{1}}{d_{2}}=0.36

I hope it helps!

5 0
3 years ago
A basketball player makes a jump shot. The 0.599 kg ball is released at a height of 2.18 m above the floor with a speed of 7.05
7nadin3 [17]

Answer:

W_{drag} = 4.223\,J

Explanation:

The situation can be described by the Principle of Energy Conservation and the Work-Energy Theorem:

U_{g,A}+K_{A} = U_{g,B} + K_{B} + W_{drag}

The work done on the ball due to drag is:

W_{drag} = (U_{g,A}-U_{g,B})+(K_{A}-K_{B})

W_{drag} = m\cdot g\cdot (h_{A}-h_{B})+ \frac{1}{2}\cdot m \cdot (v_{A}^{2}-v_{B}^{2})

W_{drag} = (0.599\,kg)\cdot (9.807\,\frac{m}{s^{2}} )\cdot (2.18\,m-3.10\,m)+\frac{1}{2}\cdot (0.599\,kg)\cdot [(7.05\,\frac{m}{s} )^{2}-(4.19\,\frac{m}{s} )^{2}]

W_{drag} = 4.223\,J

7 0
3 years ago
Read 2 more answers
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