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1 year ago

When a 10 V battery is connected to a resistor, 5 A of current flows through the resistor. What is the resistor's value?

Physics

1 answer:

SashulF [63]1 year ago

7 0

Given data

*The value of battery voltage is V = 10 V

*The current flows through the resistor is I = 5 A

The formula for the resistor is given by the Ohm's law as

$R=\frac{V}{I}$

Substitute the values in the above expression as

$\begin{gathered} R=\frac{10}{5} \\ =2\text{ ohm} \end{gathered}$

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A vaccum unlike sound,light can travel through any matter including a great vacuum of nothing (space)

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3 years ago

3. The car's mass is 400 kg. It moves at a velocity of 20 m/s. Calculate the car's momentum. *

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Answer:

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3 years ago

A river flows due east at 1.60 m/s. A boat crosses the river from the south shore to the north shore by maintaining a constant v

Citrus2011 [14]

Answer:

part (a) $v\ =\ 10.42\ at\ 81.17^o$ towards north east direction.

part (b) s = 46.60 m

Explanation:

Given,

velocity of the river due to east = $v_r\ =\ 1.60\ m/s.$
velocity of the boat due to the north = $v_b\ =\ 10.3\ m/s.$

part (a)

River is flowing due to east and the boat is moving in the north, therefore both the velocities are perpendicular to each other and,

Hence the resultant velocity i,e, the velocity of the boat relative to the shore is in the North east direction. velocities are the vector quantities, Hence the resultant velocity is the vector addition of these two velocities and the angle between both the velocities are $90^o$

Let 'v' be the velocity of the boat relative to the shore.

$\therefore v\ =\ \sqrt{v_r^2\ +\ v_b^2}\\\Rightarrow v\ =\ \sqrt{1.60^2\ +\ 10.3^2}\\\Rightarrow v\ =\ 10.42\ m/s.$

Let $\theta$ be the angle of the velocity of the boat relative to the shore with the horizontal axis.

Direction of the velocity of the boat relative to the shore. $\therefore Tan\theta\ =\ \dfrac{v_b}{v_r}\\\Rightarrow Tan\theta\ =\ \dfrac{10.3}{1.60}\\\Rightarrow \theta\ =\ Tan^{-1}\left (\dfrac{10.3}{1.60}\ \right )\\\Rightarrow \theta\ =\ 81.17^o$

part (b)

Width of the shore = w = 300m

total distance traveled in the north direction by the boat is equal to the product of the velocity of the boat in north direction and total time taken

Let 't' be the total time taken by the boat to cross the width of the river. $\therefore w\ =\ v_bt\\\Rightarrow t\ =\ \dfrac{w}{v_b}\\\Rightarrow t\ =\ \dfrac{300}{10.3}\\\Rightarrow t\ =\ 29.12 s$

Therefore the total distance traveled in the direction of downstream by the boat is equal to the product of the total time taken and the velocity of the river $\therefore s\ =\ u_rt\\\Rightarrow s\ =\ 1.60\times 29.12\\\Rightarrow s\ =\ 46.60\ m$

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3 years ago

Les propriétés de l’air?

crimeas [40]

Answer:

Lorsque l'on détend l'air son volume augmente et sa pression diminue. L'air qui est un mélange de gaz est compressible et expansible. – Lorsque l'on comprime l'air, son volume diminue et sa pression augmente. – Lorsque l'on détend l'air, son volume augmente et sa pression diminue.

4 0

3 years ago

Options are: a)4Cn b)5Cn c)6 Cn d)3 Cn

nasty-shy [4]

Answer:

Option B. 5 nC

Explanation:

From the question given above, the following data were obtained:

Capicitance (C) = 100 pF

Potential difference (V) = 50 V

Quantity of charge (Q) =?

Next, we shall convert 100 pF to Farad (F). This can be obtained as follow:

1 pF = 1×10¯¹² F

Therefore,

100 pF = 100 pF × 1×10¯¹² F / 1 pF

100 pF = 1×10¯¹⁰ F

Next, we shall determine the quantity of charge. This can be obtained as follow:

Capicitance (C) = 1×10¯¹⁰ F

Potential difference (V) = 50 V

Quantity of charge (Q) =?

Q = CV

Q = 1×10¯¹⁰ × 50

Q = 5×10¯⁹ C

Finally, we shall convert 5×10¯⁹ C to nano coulomb (nC). This can be obtained as follow:

1 C = 1×10⁹ nC

Therefore,

5×10¯⁹ C = 5×10¯⁹ C × 1×10⁹ nC / 1 C

5×10¯⁹ C = 5 nC

Thus, the quantity of charge is 5 nC

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3 years ago