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mihalych1998 [28]

2 years ago

5

a 455n gymnast jumps upward a distance of 1.5 meters to reach the uneven parallel bars. how much work did she do before she even

began her routine

1 answer:

Taya2010 [7]2 years ago

3 0

Work = force x distance
Work = 455N (weight force) x 1.5m (distance)
Work done = 682.5J
Hope this helped!

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A 13-kg sled is moving at a speed of 3.0 m/s. At which of the following speeds will the sled have twice as much kinetic energy?

AysviL [449]

K.E. = 1/2 mv²
K.E. is directly proportional to v^2
So, when K.E. increase by 2, K.E. increase by root. 2
v' = 1.41v
original v value was 3 so, final would be:
v' = 1.41*3 = 4.23
After round-off to it's tenth value, it will be:
v' = 4.2

So, option B is your answer!

Hope this helps!

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Assoli18 [71]

Explanation:

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2 years ago

What is impulse? How does this relate to momentum?

lawyer [7]

Impulse is a force acting briefly on a body and producing a finite change of momentum.
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4 0

3 years ago

What is the change in potential energy if the distance separating the electron and proton is increased to 1.0 nm?

Vlada [557]

Answer:

$Ep=-2.3*10^{-19}J$

Explanation:

The change in potential energy can be expressed as:

$Ep=K.\frac{q1.q2}{r}$

where K is a constant with a value of $9*10^{9}\frac{N.m^{2}}{C^{2}}$ , q1 and q2 are the charges of the proton and the electron and r is the distance between them.

The charge for the proton is $+1.6*10^{-19}C$ and the charge for the electron is $-1.6*10^{-19}C$ .

Converting r=1.0nm to m:

$1.0nm*\frac{1*10^{-9}m}{1.0nm}=1*10^{-9}m$

Replacing values:

$Ep=9*10^{9}\frac{N.m^{2}}{C^{2}}.\frac{(+1.6*10^{-19}C).(-1.6*10^{-19}C)}{1*10^{-9}m}$

$Ep=-2.3*10^{-19}J$

5 0

2 years ago

Write a numerical expression for the emissive intensity (in W/m^2.sr) coming out of a tiny hole in an enclosure of surface tempe

stiks02 [169]

Answer:

6.0 × $10^{11}$ W/ $m^{2}$

Explanation:

From Wien's displacement formula;

Q = e A $T^{4}$

Where: Q is the quantity of heat transferred, e is the emissivity of the surface, A is the area, and T is the temperature.

The emissive intensity = $\frac{Q}{A}$ = e $T^{4}$

Given from the question that: e = 0.6 and T = 1000K, thus;

emissive intensity = 0.6 × $(1000)^{4}$

= 0.6 × 1.0 × $10^{12}$

= 6.0 × $10^{11}$ $\frac{W}{m^{2} }$

Therefore, the emissive intensity coming out of the surface is 6.0 × $10^{11}$ W/ $m^{2}$ .

3 0

2 years ago