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3 years ago

A weightlifter lifts a 125-kg barbell straight up 1.15 m in 2.5 s. What was the power expended by the weightlifter?

Physics

1 answer:

labwork [276]3 years ago

7 0

Answer:

560

Explanation:

Round your answer to two significant figures.

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A ball is shot straight up into the air from the ground with initial velocity of 44ft/sec44ft/sec. assuming that the air resista

solniwko [45]

Convert the given in SI units.

(44 ft/sec)(1 m/ 3.28 ft) = 13.41 m/sec

The distance traveled and the initial velocity can be related through the equation,

d = (Vf)² - (Vi)²/ 2a

where d is the distance, Vf is the final velocity, Vi is the initial velocity, a is the acceleration due to gravity. Substituting the known values from the given above,

d = ((0 m/s)² - (13.41 m/s)²)/ 2(-9.8 m/s²)

The value of d from the equation,

d = 9.17 meters

Convert this to feet,

d = (9.17 m)(3.28 ft / 1 m) = 30 ft

Answer: 30 ft

3 0

3 years ago

RHOOLIOTTO<br> How much mass would be needed to produce 2.7 x 1016 J?

Radda [10]

E = mc^2
m = e/c^2
m = 2.7*10^16/(300000^2)
m = 300000

8 0

3 years ago

escribe the differences between the terrestrial and jovian planets and classify each of the 8 planets to the proper type

Radda [10]

Answer:

*******

Explanation:

is obviously the correct answer they are both so different yet so alike

6 0

2 years ago

Read 2 more answers

An object is moving along a straight line, and the uncertainty in its position is 1.90 m.

just olya [345]

Answer:

$2.78\times 10^{-35}\ \text{kg m/s}$

$6.178\times 10^{-34}\ \text{m/s}$

$0.31\times 10^{-4}\ \text{m/s}$

Explanation:

$\Delta x$ = Uncertainty in position = 1.9 m

$\Delta p$ = Uncertainty in momentum

h = Planck's constant = $6.626\times 10^{-34}\ \text{Js}$

m = Mass of object

From Heisenberg's uncertainty principle we know

$\Delta x\Delta p\geq \dfrac{h}{4\pi}\\\Rightarrow \Delta p\geq \dfrac{h}{4\pi\Delta x}\\\Rightarrow \Delta p\geq \dfrac{6.626\times 10^{-34}}{4\pi\times 1.9}\\\Rightarrow \Delta p\geq 2.78\times 10^{-35}\ \text{kg m/s}$

The minimum uncertainty in the momentum of the object is $2.78\times 10^{-35}\ \text{kg m/s}$

Golf ball minimum uncertainty in the momentum of the object

$m=0.045\ \text{kg}$

Uncertainty in velocity is given by

$\Delta p\geq m\Delta v\geq 2.78\times 10^{-35}\\\Rightarrow \Delta v\geq \dfrac{2.78\times 10^{-35}}{m}\\\Rightarrow \Delta v\geq \dfrac{2.78\times 10^{-35}}{0.045}\\\Rightarrow \Delta v\geq 6.178\times 10^{-34}\ \text{m/s}$

The minimum uncertainty in the object's velocity is $6.178\times 10^{-34}\ \text{m/s}$

Electron

$m=9.11\times 10^{-31}\ \text{kg}$

$\Delta v\geq \dfrac{\Delta p}{m}\\\Rightarrow \Delta v\geq \dfrac{2.78\times 10^{-35}}{9.11\times 10^{-31}}\\\Rightarrow \Delta v\geq 0.31\times 10^{-4}\ \text{m/s}$

The minimum uncertainty in the object's velocity is $0.31\times 10^{-4}\ \text{m/s}$ .

6 0

2 years ago

How much energy is required to move an electron through a potential difference of

Tresset [83]

I think it’s going to be the 2nd one

3 0

2 years ago