All categories

3 years ago

When an 8 V battery is connected to a resistor, a 2 A current flows in the resistor. What is the resistor's value?

Physics

1 answer:

Vinvika [58]3 years ago

5 0

Answer:

Explanation:

V=IR I= curren V=volt R=resistor

8=2.R 8/2=R R=4

You might be interested in

Two iron balls of different mass are heated to 100°C and dropped in water. If the same amount of heat is lost by the two balls t

luda_lava [24]

The correct answer for this question is this one:"D) The heavier ball will have a higher temperature because the change of temperature is inversely proportional to mass."
Hope this helps answer your question and have a nice day ahead.

3 0

3 years ago

Read 2 more answers

Joey, whose mass is m = 36 kg, stands at rest at the outer edge of a frictionless merry-go-round with the mass M = 300 kg and th

Evgen [1.6K]

Answer:

$\omega=0.24\ rad.s^{-1}$

Explanation:

Given:

mass of person, $m=36\ kg$

mass of merry go-round, $M=300\ kg$

radius of merry go-round, $R=2\ m$

velocity of the person running, $v=4\ m.s^{-1}$

We consider merry go-round as a ring:

Now the moment of inertial of the ring is given as,

$I=M.R^2$

$I=300\times 2^2$

$I=1200\ kg.m^{-2}$

Moment of inertia of the person considering as a point mass:

$I_p=m.R^2$

$I_p=36\times 2^2$

$I_p=144\ kg.m^2$

Now according to the conservation of angular momentum:

$I.\omega=I_p.\omega_p$

where:

$\omega =$ angular velocity of the merry-go-round

$\omega_p=$ angular velocity of the person running

$1200\times \omega=144\times \frac{v}{R}$

$\omega=\frac{144}{1200} \times \frac{4}{2}$

$\omega=0.24\ rad.s^{-1}$

4 0

4 years ago

Read 2 more answers

A rocket is fired with an initial VELOCITY OF 100m/s at an angle of 55° above the horizontal, It explodes On the mountain Side 1

GuDViN [60]

Answer

688.32m and 277.44m

Explanation :

⠀

$\large{\maltese{\textsf{\underline{To find :-}}}}$

The X and Y coordinates of the rocket relative of firing

⠀

$\large{\maltese{\textsf{\underline{Given :-}}}} \\ \\ \sf velocity (v_i) = 100m{s}^{-1} \\ \sf angle ({\theta}_{1}) = 55.0{\degree} \\ \sf time (t) = 12s$

⠀

$\Large{\maltese{\textsf{\underline{\underline{Step by Step Solution:-}}}}}$

⠀

The horizontal range of projectile at x.

⠀

$\sf \large{x = v_{xi}} \times t \\ \\ \sf \large{x = v_i \times \cos {\theta}_{i} \times t}$

⠀

$\large\textsf{\underline{Now substituting the required values}}$

⠀

$\sf x = 300 \times \cos 55{\degree} \times 12 \\ \\ \sf x = 100 \times 0.5756 \times 12 \\ \\ {\underline{\boxed{\bold{ x = 688.32m}}}}$

⠀

The vertical position of projectile at y.

⠀

$\sf \large y = v_{yi} \times t - (\frac{1}{2} \times g \times {t}^{2}) \\ \\ \sf \large y = v_i \times \cos \theta \times t - \frac{1}{2} g {t}^{2}$

⠀

$\textsf{ \large {\underline{Now substituting the required values}} }$

⠀

$\sf y = 100 \times \cos55{ \degree} \times 12 - \frac{1}{2} \times 9.80 \times {12}^{2} \\ \\ \sf y = 100 \times 0.8192 \times 12 - 0.5 \times 9.8 \times 144 \\ \\ \sf y = 983.04 - 705.6 \\ \\ \underline{ \boxed{ \bold{y = 277.44m}}}$

⠀

<h3>Henceforth, the distance at horizon is 688.32m and at vertical is 277.44m.</h3>

8 0

2 years ago

A jet of water squirts out horizontally from a hole on the side of the tank as shown below. 0.579 m 1.45 m h If the hole has a d

klemol [59]

Answer:

The height of the water above the hole in the tank is 58 mm

Explanation:

In order to solve this problem we need to draw a sketch of the dimensions that include the input variables of the problem.

Where:

x = 0.579[m]

y = 1.45 [m]

Using the following kinematic equation we can find the time that takes the water to hit the ground, and then with this time, we can find the velocity of the water in the x-component.

$y = y_{o} +v_{o} *t+\frac{1}{2} *g*t^{2} \\$