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2 years ago

Is 5m.whatIn si system, the length of a body isdoes it mean?

1 answer:

4 0

Answer: Some conversions from one system of units to another need to be exact, without increasing or decreasing the precision of the first measurement. This is sometimes called soft conversion. It does not involve changing the physical configuration of the item being measured.

Explanation:

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A weightlifter lifts a 1250-N barbell 2 m in 3 s.How much power was used to lift the barbell?

Vlad [161]

The power is 833.3 W

Explanation:

First of all, we need to calculate the work done in lifting the barbell, which is equal to the change in gravitational potential energy of the barbell:

$W=(mg)h$

where

mg = 1250 N is the weight of the barbell

h = 2 m is the change in height

Substituting,

$W=(1250)(2)=2500 J$

Now we can calculate the power, which is equal to the work done per unit time:

$P=\frac{W}{t}$

where

W = 2500 J is the work done

t = 3 s is the time taken

Substituting,

$P=\frac{2500}{3}=833.3 W$

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2 years ago

A ball rolls from rest at the top of a 15.5 m hill, downward to the right and back upward onto a 8.25 m hill. How fast is the ba

lana66690 [7]

Answer:

Explanation:

6 0

2 years ago

A 0.041-kg bullet has a kinetic energy of 600j. What is the velocity of the bullet?

Paladinen [302]

171.0798 M/S

In classical mechanics, kinetic energy (KE) is equal to half of an object's mass (1/2*m) multiplied by the velocity squared. For example, if a an object with a mass of 10 kg (m = 10 kg) is moving at a velocity of 5 meters per second (v = 5 m/s), the kinetic energy is equal to 125 Joules, or (1/2 * 10 kg) * 5 m/s2.

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5 0

2 years ago

A hiker determines the length of a lake by listening for the echo of her shout reflected by a cliff at the far end of the lake.

ArbitrLikvidat [17]

Answer:

L = 499 m

Explanation:

If we assume that the speed of sound is constant, that travels along a straight line, and that the echo is instantaneous, we can find the total distance travelled by the sound, as follows, just applying the definition of average velocity:

$\Delta x = v_{s} * t = 343 m/s* 2.91 s = 998 m$

If we assume that the time needed to reach to the cliff, is the same used for the return travel, the length of the lake will be exactly half of the total distance calculated:

$l_{lake} = \frac{\Delta x}{2} = \frac{998m}{2} = 499 m$