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posledela
3 years ago
6

A 1.0 kg ball falls from rest a distance of 15m. what was its change in potential energy

Physics
1 answer:
expeople1 [14]3 years ago
8 0

Answer:

112.5 J

Explanation:

I calculated it by K/G BY M/S TO POTENTIAL ENERGY.

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Two identical spheres a and b carry charges of +0.6 coulomb and -0.2 coulomb, respectively. If these spheres touch, the resultin
Slav-nsk [51]
The answer is D because when the positive charged side touches the negative charged side it nullifys part of the positively charged side, basically subtraction from my understanding?
3 0
3 years ago
Electron grops could be considered which of the following
Maurinko [17]

Answer:

Electron groups could be considered as  Lone pair electrons and bonded pairs of electrons.

Answer: Option D & B

Explanation:

The two or more electrons can be bonded by single bond, double bond, covalent bond of electrons can simply be lone pair of electrons. Unshared pair of electrons are generally termed as lone pair of electrons in an atom which are generally present in the outermost shell of atoms. Hence electron groups can be determined by bonded pairs and lone pairs of electrons.

I got this from another brainly user

5 0
2 years ago
Even when shut down after a period of normal use, a large commercial nuclear reactor transfers thermal energy at the rate of 150
Sever21 [200]

Answer:2.89\approx 2.9^{\circ}C/s

Explanation:

Given

Power\left ( P\right )=150 MW

mass of core\left ( m\right )=1.60\times 10^5 kg

Average specific heat \left ( C\right )=0.3349 KJ/kg^{\circ}C

And rate of increase of temperature =\frac{\mathrm{d}T}{\mathrm{d} t}

Now

P=mc\frac{\mathrm{d}T}{\mathrm{d} t}

150\times 10^6=1.60\times 10^5\times 0.3349\times \frac{\mathrm{d}T}{\mathrm{d} t}

Thus \frac{\mathrm{d}T}{\mathrm{d} t}=[tex]\frac{1.60\times 10^5\times 0.3349}{150\times 10^6}

\frac{\mathrm{d}T}{\mathrm{d} t}=2.89\approx 2.9^{\circ}C/s

6 0
3 years ago
A spaceship leaves a space station and emits a flash of light that travels at a speed of 300,000 km/s. The spaceship starts out
balandron [24]
The answer is number D
7 0
3 years ago
Read 2 more answers
A 1 m long wire of diameter 1mm is submerged in an oil bath of temperature 25-degC. The wire has an electrical resistance per un
Zarrin [17]

Answer:

steady state temperature =88.7deg C

t=time within  1 deg C of it steady state is 8.31s

Explanation:

A 1 m long wire of diameter 1mm is submerged in an oil bath of temperature 25-degC. The wire has an electrical resistance per unit length of 0.01 Ω/m. If a current of 100 A flows through the wire and the convection coefficient is 500W/m2K, what is the steady state temperature of the wire? From the time the current is applied, how long does it take for the wire to reach a temperature within 1-degC of the steady state value? The density of the wire is 8,000kg/m3, its heat capacity is 500 J/kgK and its thermal condu

The diameter of the wire is known to be=1mm

properties=

The density of the wire is 8,000 kg/m3,

heat capacity is 500 J/kgK

themal conductivity is 20W/m.K

electrical resistance per unit length of 0.01 Ω/m

from lump capavity method

B_{i} =\frac{hr/2}{k}

500*(2.5*10^-4)/20

0.006<0.1

we know also, to find steady state temperature

\piDh(T-Tinf)=I^{2} R_{e}

make T the subject of the equation , we have

T=25+\frac{100^2*0.01}{\pi*0.001*500 }

T=88.7 degC

rate of chnage in temperature

dT/dt=\frac{I^2*Re}{rho*c*\pi*D^2/4 } -\frac{4h}{rho*c*D} (T-Tinf)

at t=o and integrating both sides\frac{T-Tinf-(I^2*Re/\pi*Dh) }{Ti-Tinf-(I^2*Re/\pi*Dh } =exp\frac{-4ht}{rho*c*D}

we have

\frac{87.7-25-63.7}{25-25-63.7} =exp\frac{4*500t}{8000*500*0.001}

t=8.31s

steady state temperature =88.7deg C

t=time within  1 degC of it steady stae is 8.31s

7 0
3 years ago
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