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2 years ago

What is the velocity between the times t=3 to t = 4?

Physics

1 answer:

saveliy_v [14]2 years ago

3 0

Answer:

The answer is t=4 + t=3=t=7

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A 20-kg boy slides down a smooth, snow-covered hill on a plastic disk. The hill is at a 10° angle to the horizontal, and the slo

zvonat [6]

Answer:

13 m/s

Explanation:

I assume we are ignoring friction.

The boy's PE will all be converted to KE at the bottom of the hill.

to find PE = mgh we need to know h

h = 50 sin 10 = 8.68 meters

then: PE = 20 * 9.81 * 8.68 =<u> 1703.49</u> j

KE = 1/2 m v^2 = <u>1703 .49</u>

v = 13 m/s

7 0

2 years ago

What would be most useful to help make a simple compass a nonmetal bar,a round metal can,a small iron nail,or a Quartz needle

pav-90 [236]

Iron nail. the rest of those are not iron or some form of magnetic material.

5 0

3 years ago

Read 2 more answers

A delivery truck leaves a warehouse and travels 2.60 km north. The truck makes a left turn and travels 1.25 km west before makin

yulyashka [42]

Answer:

4.19 km and 107.35 degrees north of east

Explanation:

So in the end, the truck is (2.6 + 1.4 = 4km) north and 1.25 km west from the warehouse. We can use the Pythagorean formula to calculate the magnitude and direction α of the truck displacement from the warehouse:

$s = \sqrt{s_n^2 + s_w^2} = \sqrt{4^2 + 1.25^2} = \sqrt{16 + 1.5625} = \sqrt{17.5625} = 4.19$ km

$tan\alpha = \frac{s_n}{s_w} = \frac{4}{1.25} = 3.2$

$\alpha = tan^{-1}3.2 = 1.27 rad \approx 72.65 degrees$ north or west or (180 - 72.65) = 107.35 degrees north of east

3 0

3 years ago

Which of the following are not vector directions?

Svetradugi [14.3K]

Answer:

here north are not vector option b hope ur help

6 0

3 years ago

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Two protons in an atomic nucleus are typically separated by a distance of 2 ✕ 10-15 m. The electric repulsion force between the

castortr0y [4]

Answer:

The magnitude of the electric force between the to protons will be 57.536 N.

Explanation:

We can use Coulomb's law to find out the force, in scalar form, will be:

$F \ = \ \frac{1}{4 \pi \epsilon_0 } \frac{q_1 q_2}{d^2}$ .

Now, making the substitutions

$d \ = \ 2.00 * 10 ^{-15} \ m$ ,

$q_1 = q_2 = 1.60 * 10 ^ {-19} \ C$ ,

$\frac{1}{4\pi\epsilon_0}=8.99 * 10^9 \frac{Nm^2}{C^2}$ ,

we can find:

$F \ = \ 8.99 * 10^9 \frac{Nm^2}{C^2} \frac{(1.60 * 10 ^ {-19} \ C)^2}{(2.00 * 10 ^{-15} \ m)^2}$ .

$F \ = 57.536 N$ .

Not so big for everyday life, but enormous for subatomic particles.

4 0

3 years ago

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