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Marat540 [252]
3 years ago
15

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:

B

Explanation:

V=IR    I= curren V=volt R=resistor

8=2.R   8/2=R R=4

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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
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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}

<u>We consider merry go-round as a ring:</u>

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}

<u>Moment of inertia of the person considering as a point mass:</u>

I_p=m.R^2

I_p=36\times 2^2

I_p=144\ kg.m^2

<u>Now according to the conservation of angular momentum:</u>

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:-}}}}}

⠀

⠀

<u>The</u><u> </u><u>horizontal</u><u> </u><u>range</u><u> </u><u>of</u><u> </u><u>projectile</u><u> </u><u>at</u><u> </u><u>x</u><u>.</u><u> </u>

⠀

\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><u>Henceforth</u><u>,</u><u> </u><u>the</u><u> </u><u>distance</u><u> </u><u>at</u><u> </u><u>horizon</u><u> </u><u>is</u><u> </u><u>6</u><u>8</u><u>8</u><u>.</u><u>3</u><u>2</u><u>m</u><u> </u><u>and</u><u> </u><u>at</u><u> </u><u>vertical</u><u> </u><u>is</u><u> </u><u>2</u><u>7</u><u>7</u><u>.</u><u>4</u><u>4</u><u>m</u><u>.</u></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} \\

It is necessary to clarify the value of each of the respective variables below

y = - 1.45 [m] "It is negative because this point is below the water outlet"

yo = 0

vo = 0 "The velocity is zero because the component of the speed on the Y-axis does not exist"

therefore:

-1.45=0.5*(-9.81)*t^{2} \\t = \sqrt{\frac{1.45}{0.5*9.81} } \\t = 0.543[s]

The next step is to determine the velocity in component x, knowing the time.

v=\frac{x}{t} \\v=\frac{0.579}{.543} \\v = 1.06[m/s]

Now using torricelli's law we can find the elevation.

v=\sqrt{2*g*h} \\h=\frac{v^{2} }{2*g} \\h=\frac{1.06^{2} }{2*9.81} \\h= 0.057[m] = 57.95[mm]

3 0
3 years ago
explain how you can use a speedometer and a clock to tell how far you've traveled in a car if the cars odometer is not working
astraxan [27]
You multiply your average speed with the time for which you have been traveling 
5 0
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