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oksano4ka [1.4K]
4 years ago
5

Due in a few minutes plzzzz help! Will mark brainliest

Physics
1 answer:
nadya68 [22]4 years ago
5 0

Answer:

1. 960

2. 16.6666

3. 0.41666666666

4. 10

Explanation:

A*V=W

A*120=50

V=I*R

R being resistance

(I'm not a 100% sure on 2 & 3)

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I have an object that I want to know when it was made. I calculate that the object contains 0.12g of Iodine-131 but started with
Sav [38]

Answer: 71.72 days

Explanation:

This problem can be solved using the <u>Radioactive Half Life Formula: </u>

A=A_{o}.2^{\frac{-t}{h}} (1)  

Where:  

A=0.12 g is the final amount of Iodine-131

A_{o}=60 g is the initial amount of Iodine-131

t is the time elapsed  

h=8 days is the half life of Iodine-131

Knowing this, let's substitute the values and find t from (1):

0.12 g=(60 g)2^{\frac{-t}{8 days}} (2)  

\frac{0.12 g}{60g}=2^{\frac{-t}{8 days}} (3)  

Applying natural logarithm in both sides:

ln(\frac{0.12 g}{60g})=ln(2^{\frac{-t}{8 days}}) (4)  

-6.21=-\frac{t}{8 days}ln(2) (5)

Finding t:

t=71.72 days

7 0
4 years ago
Which of the following is prohibited by the 2nd law of thermodynamics?
vitfil [10]

(A) A device that converts heat into work with 100% efficiency

It clearly violates the second law of thermodynamics because it warns that while all work can be turned into heat, not all heat can be turned into work. Therefore, despite the innumerable efforts, the efficiencies of the bodies have only been able to reach 60% at present.

6 0
3 years ago
Where is the sun's energy most concentrated
Maru [420]

B. At the equator

Explanation:

The energy coming from the Sun hits the Earth's surface at different angles, depending on the latitude of the place. The more perpendicular the ray of lights hit the surface, the more the energy transmitted to the Earth's surface, the warmer the location.

The angle at which the ray of lights hit the Earth is related to the latitude: in particular, the ray of lights arrive perpendicular at the equator (0^{\circ}), they arrive at larger angle in the United States (which is located at intermediate latitudes) and they arrive at the largest angles at the poles. For this reason, the sun's most energy is concentrated at the equator.

5 0
3 years ago
Which of these examples displays CONDUCTION based on the definition?
dangina [55]
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3 0
3 years ago
A tank, shaped like a cone has height 12 meter and base radius 1 meter. It is placed so that the circular part is upward. It is
Temka [501]

Answer:

376966.991 Joules

Explanation:

Given that :

the height = 12  m

Let assume the tank have a thickness  = dh

The radius of the tank by using the concept of similar triangle is :

\dfrac{1}{r} = \dfrac{12}{h}

r = \dfrac{h}{12}

The area of the tank = \mathbf{\pi r^2}

The area of the tank = \mathbf{\pi( \dfrac{h}{12})^2}

The area of the tank = \mathbf{ \dfrac{\pi}{144}h^2}

The volume of the tank is  = area × thickness

= \mathbf{ \dfrac{\pi}{144}h^2 \  dh}

Weight of the element = \rho_ g * volume

where;

\rho_g = density of water ; which is given as 10000 N/m³

So;

Weight of the element = \mathbf{ 10000 *\dfrac{\pi}{144}h^2 \  dh}

Weight of the element = \mathbf{69.44 \ \pi  \ h^2 \  dh}

However; the work required to pump this water = weight × height  rise

where the height rise = 12 - h

the work required to pump this water  = \mathbf{69.44 \ \pi  \ h^2 \  dh}(12 - h)

the work required to pump this water  = \mathbf{69.44 \pi (12h^2-h^3)dh}

We can determine the total workdone by integrating the work required to pump this water

SO;

Workdone = \mathbf{\int\limits^{12}_0 {69.44 \pi(12h^2-h^3)} dh}

= \mathbf{ 69.44 \pi \int\limits^{12}_0 {(12h^2-h^3)} dh}

=  \mathbf{ 69.44 \pi[ \frac{12h^3}{3}-  \frac{h^4}{4}]^{12}}_0} }

= \mathbf{69.44 \pi [ \frac{12^4}{3}-\frac{12^4}{4}]}

= \mathbf{69.44 \pi*12^4 [ \frac{4-3}{12}]}

= \mathbf{69.44 \pi*12^4 *\frac{1}{12}}

= 376966.991 Joules

6 0
3 years ago
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