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2 years ago

7

A 10 ohms, a 7 ohms and a 14 ohms resistor are connected in series with a 24 V battery. Calculate the equivalent resistance. Ans

wer in units of ohms.

1 answer:

Aloiza [94]2 years ago

5 0

Answer:

31ohms

Explanation:

in a series u add all the ohms together

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What is initial velocity

Lapatulllka [165]

Initial Velocity is the velocity at time interval t = 0 and it is represented by u. It is the velocity at which the motion starts. They are four initial velocity formulas: (1) If time, acceleration and final velocity are provided, the initial velocity is articulated as. u = v – at.

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3 years ago

Sherlock Holmes examines a clue by holding his magnifying glass (with

Alborosie

Answer:

Distance: -30.0 cm; image is virtual, upright, enlarged

Explanation:

We can find the distance of the image using the lens equation:

$\frac{1}{f}=\frac{1}{p}+\frac{1}{q}$

where:

f = 15.0 cm is the focal length of the lens (positive for a converging lens)

p = 10.0 cm is the distance of the object from the lens

q is the distance of the image from the lens

Solving for q,

$\frac{1}{q}=\frac{1}{f}-\frac{1}{q}=\frac{1}{15.0}-\frac{1}{10.0}=-0.033\\q=\frac{1}{0.033}=-30.0 cm$

The negative sign tells us that the image is virtual (on the same side of the object, and it cannot be projected on a screen).

The magnification can be found as

$M=-\frac{q}{p}=-\frac{-30}{10}=3$

The magnification gives us the ratio of the size of the image to that of the object: since here |M| = 3, this means that the image is 3 times larger than the object.

Also, the fact that the magnification is positive tells us that the image is upright.

8 0

4 years ago

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 :

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

The X and Y coordinates of the rocket relative of firing

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⠀

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

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

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<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>

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

⠀

⠀

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

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

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The vertical position of projectile at y.

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

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

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

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<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>

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2 years ago

Which two statements offer evidence that the clocks energy is conserved?

Natali [406]

Answer:

C is the best answer because we all know that clock is part of our daily lives but we don't know the about its background

3 0

3 years ago

Read 2 more answers

Saturn is moving with uniform angular speed of 2πf along the circumference of it's orbit around the sun with radius R, having ce

Luda [366]

Answer:Sir haymo knows

Explanation:

he gave as homework

8 0

3 years ago