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2 years ago

1. A jogger jogs 25 m East, stops to rest, then continues for 10 more meters

Physics

1 answer:

Komok [63]2 years ago

8 0

Answer:

35m

Explanation:

25m + 10m = 35m east.

*(adding since it's in the same direction)

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A 2.60 kg lion runs at a speed of 5.00 m/s until he sees his prey. The lion then speeds up 8.00 m/s to catch it. How much work d

ELEN [110]

Answer: 50.7 J

Explanation:

Given

mass of lion is $m=2.6\ kg$

The initial speed of the lion is $v_1=5\ m/s$

increased speed of lion is $v_2=8\ m/s$

Initially, its kinetic energy is $K_1=\frac{1}{2}mv_1^2$

Final kinetic energy $k_2=\frac{1}{2}mv_2^2$

work did by lion after speed up is $k_2-k_1$

$\Rightarrow W=\dfrac{1}{2}\times 2.6[8^2-5^2]\\\\\Rightarrow W=1.3\times [39]=50.7\ J$

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2 years ago

High-level clouds usually occur above 20,000 ft. Which is the best description of the clouds’ appearance?

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D. Thin and composed of Ice crystals

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3 years ago

Locate the element calcium (Ca) on the periodic table and click on the square. Read about the properties of calcium. Why might c

Ann [662]

In the periodic table, calcium is located at the 2nd column of the 4th row starting from the hydrogen gas.

Calcium is very essential in the human body is it helps maintain a healthy teeth and all other bones. Other calcium are also found in nerves, tissues, and all other parts. Calcium also helps in clotting blood and normalizing a person's heart beat. This may apply not only to humans but to all living things.

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2 years ago

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A bus approaches red traffic light, and slows down from a speed of 15

Keith_Richards [23]

Answer:

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rhtjnh65y7uy6trey56u6yh5ttrdjyj56ejjyjg

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3 years ago

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There is a current of 192 pA when a certain potential is applied across a certain resistor. When that same potential is applied

dalvyx [7]

Answer:

$\frac{d_2}{d_1}=0.089$

Explanation:

The equation that relates voltage, current and resistance is V=RI, and we can calculate this resistance in terms of the resistivity of a material $\rho$ , its length L and cross-section (area) A with:

$R=\frac{\rho L}{A}$

Putting both equations together we have:

$\frac{V}{I}=\frac{\rho L}{A}$

Which will be useful to write as:

$\frac{V}{\rho}=\frac{I L}{A}$

since each term on the left will be the same for both situations, thus making the term on the right the same for both situations (which we will call 1 and 2), so we have:

$\frac{I_1 L_1}{A_1}=\frac{I_2 L_2}{A_2}$

Which we want to write as (since the ratio of the diameter is asked):

$\frac{A_2}{A_1}=\frac{I_2 L_2}{I_1 L_1}$

Since the areas appear to be that of a circle, we can write them as $A=\pi r^2=\frac{\pi d^2}{4}$ where r is the radius of that circle, and d the diameter. Substituting this in our previous equation we then have:

$\frac{A_2}{A_1}=\frac{\frac{\pi d_2^2}{4}}{\frac{\pi d_1^2}{4}}=\frac{d_2^2}{d_1^2}=(\frac{d_2}{d_1})^2=\frac{I_2 L_2}{I_1 L_1}$

Which means:

$\frac{d_2}{d_1}=\sqrt{\frac{I_2 L_2}{I_1 L_1}}$

We know that $I_1=192pA$ , $I_2=0.054pA$ and $L_2=28L_1$ , so substituting our values we have:

$\frac{d_2}{d_1}=\sqrt{\frac{(0.054pA)(28L_1)}{(192pA)(L_1)}}=0.089$

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3 years ago