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photoshop1234 [79]

3 years ago

14

okay so there is a wheel and it shows it when its high, in the middle, and below. my question is when is its potenial energy the

greatest?

1 answer:

earnstyle [38]3 years ago

4 0

At the highest point

cheers

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You have a potential difference of 8 v. how much work is done to transfer 0.32 c of charge through it? answer in units of j.

MatroZZZ [7]

It would take about half a minute

8 0

3 years ago

What is the amount of heat, in calories, given off from a 5 g piece of aluminum when it cools from 80°C to 20°C? The specific he

vagabundo [1.1K]

Answer:

Q (heat) = S * m * (T2 - T1) where Q is heat gained (loss), S the specific heat capacity of the substance and T2, T1 are the final and initial temps

Q = .215 cal / (g deg C) * 5 g * (20 - 80) deg C = -64.5 cal

Since the question specifies the heat emitted, then 64.5 cal is the

heat emitted (loss).

8 0

2 years ago

5) help plz I need help<br> I don't know if I am right so if I am not can you plz help me

seraphim [82]

the answer is waning Gibbous

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7 0

2 years ago

Read 2 more answers

If Star A is magnitude 1.0 and Star B is magnitude 9.6 , which is brighter and by what factor?

ludmilkaskok [199]

Answer:

Star A is brighter than Star B by a factor of 2754.22

Explanation:

Lets assume,

the magnitude of star A = m₁ = 1

the magnitude of star B = m₂ = 9.6

the apparent brightness of star A and star B are b₁ and b₂ respectively

Then, relation between the difference of magnitudes and apparent brightness of two stars are related as give below: $(m_{2} - m_{1}) = 2.5\log_{10}(b_{1}/b_{2})$

The current magnitude scale followed was formalized by Sir Norman Pogson in 1856. On this scale a magnitude 1 star is 2.512 times brighter than magnitude 2 star. A magnitude 2 star is 2.512 time brighter than a magnitude 3 star. That means a magnitude 1 star is (2.512x2.512) brighter than magnitude 3 bright star.

We need to find the factor by which star A is brighter than star B. Using the equation given above,

$(9.6 - 1) = 2.5\log_{10}(b_{1}/b_{2})$

$\frac{8.6}{2.5} = \log_{10}(b_{1}/b_{2})$

$\log_{10}(b_{1}/b_{2}) = 3.44$

Thus,

$(b_{1}/b_{2}) = 2754.22$

It means star A is 2754.22 time brighter than Star B.

3 0

3 years ago

A 15.0 kg turntable with a radius of 25 cm is covered with a uniform layer of dry ice that has a mass of 9.0 kg. The angular spe

liubo4ka [24]

Answer:

ω₂=1.20

Explanation:

Given that

mass of the turn table ,M= 15 kg

mass of the ice ,m= 9 kg

radius ,r= 25 cm

Initial angular speed ,ω₁ = 0.75 rad/s

Initial mass moment of inertia

$I_1=\dfrac{M+m}{2}r^2$

$I_1=\dfrac{15+9}{2}\times 0.25^2\ kg.m^2$

$I_1=0.75\ kg.m^2$

Final mass moment of inertia

$I_2=\dfrac{M}{2}r^2$

$I_2=\dfrac{15}{2}\times 0.25^2\ kg.m^2$

$I_2=0.468\ kg.m^2$

Lets take final speed of the turn table after ice evaporated =ω₂ rad/s

Now by conservation angular momentum

I₁ ω₁ =ω₂ I₂

$\omega_2=\dfrac{0.75\times 0.75}{0.468}\ rad/s$

ω₂=1.20

7 0

2 years ago