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eduard
2 years ago
14

A student connects four AA batteries (1.5 V each) in series to light up a light bulb. The circuit has a resistance of 50 Ω. How

much current flows through the circuit?
Physics
2 answers:
VLD [36.1K]2 years ago
5 0

Answer:

0.3

Explanation:

V = I R

1.5= I (50)

I = 1.5/50

I = 0.3 J

Therefore, 0.3joule current flows through the circuit.

Hope this helped

Phantasy [73]2 years ago
5 0

Answer:

0.3

Explanation:

V = I R

1.5= I (50)

I = 1.5/50

I = 0.3 J

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Which unit of a measurement is better for determining the volume for a drop of water. A milliliter, or a liter?
IceJOKER [234]
A milliliter, because it is a smaller unit of measurement more fitting for a drop of water.
6 0
2 years ago
Five race cars speed toward the finish line at the Jasper County Speedway. The table lists each car’s speed in meters/second. If
nekit [7.7K]
We have: v = d/t
From the expression, we conclude speed is indirectly proportional to speed.
So, the car which will take longer time must have the smallest speed. Among all the options Car C has the smallest speed. So, it would be your answer.

In short, Your Answer would be Option C) Car C will take longer time than any other.

Hope this helps!
6 0
3 years ago
Read 2 more answers
An ideal gas, initially at a pressure of 11.2 atm and a temperature of 299 K, is allowed to expand adiabatically until its volum
Tju [1.3M]

Answer:

The pressure is  P_2  =  4.25 \ a.t.m

Explanation:

From the question we are told that

   The initial pressure is P_1  =  11.2\ a.t.m

   The  temperature is  T_1 =   299 \ K

   

Let the first volume be  V_1 Then the final volume will be  2 V_1

 Generally for a diatomic  gas

           P_1 V_1 ^r  =  P_2  V_2  ^r

Here r is the radius of the molecules which is  mathematically represented as

    r =  \frac{C_p}{C_v}

Where C_p \  and\   C_v are the molar specific heat of a gas at constant pressure and  the molar specific heat of a gas at constant volume with values

     C_p=7 \  and\   C_v=5

=>   r =  \frac{7}{5}

=>  11.2*( V_1 ^{\frac{7}{5} } ) =  P_2  *  (2 V_1 ^{\frac{7}{5} } )

=>   P_2  =  [\frac{1}{2} ]^{\frac{7}{5} } * 11.2

=>  P_2  =  4.25 \ a.t.m

8 0
3 years ago
(a) Calculate the acceleration due to gravity on the surface of the Sun.
Ira Lisetskai [31]
<h2>a)Acceleration due to gravity on the surface of the Sun is 274.21 m/s²</h2><h2>b) Factor of increase in weight is 27.95</h2>

Explanation:

a) Acceleration due to gravity

                      g=\frac{GM}{r^2}

 Here we need to find acceleration due to gravity of Sun,

                G = 6.67259 x 10⁻¹¹ N m²/kg²

    Mass of sun, M = 1.989 × 10³⁰ kg

    Radius of sun, r = 6.957 x 10⁸ m

Substituting,

                g=\frac{6.67259\times 10^{-11}\times 1.989\times 10^{30}}{(6.957\times 10^8)^2}\\\\g=274.21m/s^2

Acceleration due to gravity on the surface of the Sun = 274.21 m/s²

b) Acceleration due to gravity in earth = 9.81 m/s²

   Ratio of gravity = 274.21/9.81 = 27.95

   Weight = mg

  Factor of increase in weight = 27.95

8 0
3 years ago
two cars are moving at constant speeds in a straight line along a major highway. The first is travelling at 20ms^-1 and the seco
Reika [66]

Answer:

t=750s

Explanation:

The two cars are under an uniform linear motion. So, the distance traveled by them is given by:

\Delta x=vt\\x_f-x_0=vt\\x_f=x_0+vt

x_f is the same for both cars when the second one catches up with the first. If we take as reference point the initial position of the second car, we have:

x_0_1=6km\\x_0_2=0

We have x_f_1=x_f_2. Thus, solving for t:

x_0_1+v_1t=x_0_2+v_2t\\x_0_1=t(v_2-v_1)\\t=\frac{x_0_1}{v_2-v_1}\\t=\frac{6*10^3m}{28\frac{m}{s}-20\frac{m}{s}}\\t=750s

8 0
2 years ago
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