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V125BC [204]
4 years ago
8

If several resistors are connected in series in

Physics
1 answer:
Nostrana [21]4 years ago
6 0

Answer: varies directly with its resistance

Explanation:

In a serious circuit, the potential difference across each resistor varies directly with its resistance. The sum of the potential drops across all the resistors equals the applied potential

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Yanka [14]
Is there information in the previous question which relates to this one?
8 0
3 years ago
A closed box is filled with dry ice at a temperature of -87.1°C, while the outside temperature is 17.6°C. The box is cubical, me
natta225 [31]

Answer:

K=24.17 x 10⁻² J s⁻¹c⁻¹m⁻¹

Explanation:

Rate of flow of heat through a material is given by the following expression

\frac{Q}{t} =\frac{KA\delta T}{d}

where Q is amount of heat flowing in time t through area A and  a medium of thickness d having two faces at temperature difference δT . K is thermal conductivity of the medium .

Here Q = 3.34 x 10⁶/6 , t = 24 x 60 x 60 = 86400 s , A = .332 X .332 = .0110224 m² ,  δT = 104.7

Put these values here

\frac{3.34\times10^6}{6\times86400}= \frac{k\times.011224\times104.7}{4.41\times10^{-2}}

K=\frac{3.34\times4.41\times10^4}{6\times86400\times.011224\times104.7}

K=24.17 x 10⁻² J s⁻¹c⁻¹m⁻¹

4 0
3 years ago
Use the kinetic theory of matter to explain thermal expansion in liquids​
blagie [28]
When a substance is heated the average kinetic energy of molecules increases and they start moving with an increased speed.As a result between mean separation between the molecules also increases

6 0
3 years ago
2. A force of 50 N is applied to the object for a distance of 2.0 m. Assume that object(the mass of the object is 3kg)
Gwar [14]

Answer:

2. A force of 50 N is applied to the object for a distance of 2.0 m. Assume that object(the mass of the object is 3kg)

was at rest at the beginning, what speed did it achieved because of the work done on it? (Hint:

Calculate the works performed by the force first.)

I figured that is 8.2m/S,I am just not sure can anyone help me i much appreciate it.

4 0
3 years ago
An 80 kg astronaut has gone outside his space capsule to do some repair work. Unfortunately, he forgot to lock his safety tether
Anna [14]

Answer:

the time taken by the astronaut to reach safety = 9.8 hr

Explanation:

The equation for intensity can be written as :

I = \frac{P}{A}

where :

\frac{I}{c}= \frac{F}{A}

Replacing that into the above previous equation; we have:

\frac{P}{Ac}=\frac{F}{A}

F = \frac{P}{c}

However ; the force needed to push the astronaut is as follows:

F = ma

where ;

m = mass of the  astronaut and a = its acceleration

we as well say;

\frac{P}{c} = ma

a = \frac{P}{mc}

Replacing P with 1000 W ; m with 80 kg and 3*10^{8} \  m/s for  c

Then; a = \frac{1000 \ W}{(80)(3.0*10^8)}

a = 4.2*10^{-8} \ m/s

It is also known that the battery will run for one hour and after which the battery on the laser will run out

Then to determine the change in the position after the first hour ; we have:

\Delta x_1 = \frac{1}{2}*4.2*10^{-8} \ m/s^2 ) (1.0 \ h)^2

\Delta x_1 = \frac{1}{2}*4.2*10^{-8} \ m/s^2 ) (1.0 *3600 s)^2

= 0.27 m

Furthermore, the final velocity of the astronaut is determined as:

v_1 = at_1

where ;

v_1 = final \ velocity

replacing t_1 = 1.0 \ h and a =  4.2*10^{-8} \ m/s; Then:

v_1 = (4.2*10^8 \ m/s * 1.0 \ h * \frac{ 3600\ s}{1.0 \ h})

v_1 =  1.51 *10^{-4} \ m/s

Also; when he drifted 5.0 m away from the capsule; the distance is far short of the 5 m but he still have 9 hours left of oxygen . In addition to that, he acceleration is also zero and the final velocity remains the same, so:

To find the final distance traveled by the astronaut ;we have:

\Delta x_2 = d - \Delta x_1

where;

\Delta x_2 = the final distance

d = total distance

So;

\Delta x_2 = 5 m - 0.27 m \\ \\ \Delta x_2 = 4.73 \ m

The time taken to reach the final distance can be calculated as:

t_2 = \frac{\Delta x_2 }{v_1}

where;

t_2 = is the  time to  reach the final distance

Replacing 4.73 for {\Delta x_2 } and  1.51*10^{-4} m/s for v_1

t_2 = \frac{4.73 \ m }{1.51*10^{-4} \ m/s}

t_2 = 31500 \ s (\frac{1.0 \ h}{3600 \ s} )

t_2 = 8.8 \ h

We knew the laser was operated for 1 hour; thus the total time taken by the astronaut to  reach the final distance is the sum of the time taken to reach the final distance and the operated time of the laser.

Hence ; the time taken by the astronaut to reach safety = 9.8 hr

8 0
3 years ago
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