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4 years ago

If an object travels 100 m in 5 seconds, how fast is it going on average?

Physics

1 answer:

stealth61 [152]4 years ago

6 0

Answer:

20 m/s

Explanation:

The relevant relationship is ...

speed = distance/time

speed = (100 m)/(5 s) = (100/5) m/s

speed = 20 m/s

The average speed is 20 meters per second.

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What is the value of vEp? Select One of the Following (a) 345 mph (b) -345 mph (c) 255 mph (d) 255 mph

serious [3.7K]

We're not completely sure what vEp stands for.

On Etsy, the "value of Eggs from pheasants" is $7.98 a dozen. Shipping is probably additional.

7 0

4 years ago

If you run the simulation a few times, you should see that for this particular pair of block and disc the block's position has t

user100 [1]

Explanation:

C) is incorrect. The mass is equivalent to the moment of inertia in an angular formula.

The others appear all correct.

T * theta = F * r * theta = F * (r theta) = fx = work

This is the only one that needed explanation. The others should be obvious.

7 0

3 years ago

Look at the vector r plotted below

solong [7]

Answer: option A. r = 3x+2y

Explanation:

Vector r is plotted on the graph. On x-axis each small division corresponds to 1 unit. Similarly on y-axis, each small division corresponds to one unit.

The vector is the resultant of addition of its x and y components. we would draw perpendicular to the x-axis and y-axis from the head of vector r.

On x-axis, $r_x$ = +3 units

on y-axis, $r_y$ = +2 units

$r = r_x+r_y$

Hence, vector r can be written as: r = 3x + 2y . Correct option is A.

7 0

4 years ago

A string is 3.00 m long with a mass of 5.00 g. The string is held taut with a tension of 500.00 N applied to the string. A pulse

NISA [10]

Answer:

The time interval is $t = 5.48 *10^{-3} \ s$

Explanation:

From the question we are told that

The length of the string is $l = 3.00 \ m$

The mass of the string is $m = 5.00 \ g = 5.0 *10^{-3}\ kg$

The tension on the string is $T = 500 \ N$

The velocity of the pulse is mathematically represented as

$v = \sqrt{ \frac{T}{\mu } }$

Where $\mu$ is the linear density which is mathematically evaluated as

$\mu = \frac{m}{l}$

substituting values

$\mu = \frac{5.0 *10^{-3}}{3}$

$\mu = 1.67 *10^{-3} \ kg /m$

Thus

$v = \sqrt{\frac{500}{1.67 *10^{-3}} }$

$v = 547.7 m/s$

The time taken is evaluated as

$t = \frac{d}{v}$

substituting values

$t = \frac{3}{547.7}$

$t = 5.48 *10^{-3} \ s$

5 0

4 years ago

If a ball leaves the ground with a vertical velocity of 5.46 m/s, how long does it take

kifflom [539]

Answer:

0.557 s

Explanation:

Given:

v₀ = 5.46 m/s

v = 0 m/s

a = -9.8 m/s²

Find: t

v = at + v₀

0 m/s = (-9.8 m/s²) t + 5.46 m/s

t = 0.557 s

8 0

3 years ago