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Viefleur [7K]
4 years ago
8

ListenA person weighing 6.0 × 102 newtons rides an elevator upward at an average speed of 3.0 meters per second for 5.0 seconds.

How much does this person’s gravitational potential energy increase as a result of this ride? a) 3.6 × 102 J b) 1.8 × 103 J c) 3.0 × 103 J d) 9.0 × 103 J.
Physics
1 answer:
pychu [463]4 years ago
4 0

Answer:

(d) 9 × 10^{3] J

Explanation:

from the question we are given the following:

weight of the man = 6.0 × 10 ^ {2} N

average speed (v) = 3 m/s

time (t) = 3 s

potential energy (U) = ?

We can calculate the increase in potential energy of the man by applying the formula below

increase in potential energy = P₂ - P₁

where

  • P₁ is the initial potential energy
  • P₂ is the final potential energy
  • Potential energy = mass x acceleration due to gravity x height

From physic we know that weight = mass x acceleration due to gravity

  • We should take note that the distance in this case is also our height, and we can get it from the formula distance = velocity x time
  • Distance = 3 x 5 = 15 meters
  • Initial potential energy P₁ is zero because the person was initially in motion and potential energy is the energy at rest.

therefore

potential energy = 6.0 × 10 ^ {2} × 15 = 9 × 10^{3] J

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Answer:

ac= 15.07 m/s²

Explanation:

The Wheel rotates with a constant angular acceleration.:

Centripetal acceleration is calculated as follows:

ac =ω² *R   Formula (1)

ac =v² / R    Formula (2)

Kinematics of the wheel

We apply the equations of circular motion uniformly accelerated :

ωf²= ω₀² + 2αθ  Formula (3)

v = ω* R Formula (4)

Where:

θ : angle that the wheel has rotated in a given time interval (rad)

α : angular acceleration (rad/s²)

ω₀ : initial angular speed ( rad/s)

ωf : final angular speed   ( rad/s)

v: tangential velocity of a point on the rim ( m/s)

R : radius of  wheel (m)

ac: centripetal acceleration, (m/s²)

Data:

D = 40.0 cm : diameter of the wheel

R = D/2= 40.0 cm/ 2 = 20 cm = 0.2m

α = 3.00 rad/s^2

ω₀ = 0

n = 2 revolutions : number of revolutions

θ =2πn (rad) = 2π*2 (rad) = 4π rad

Calculate of the ωf

We replace data in the formula (3)

ωf²= ω₀² + 2αθ

ωf²= 0 + 2(3)(4π)

ωf²= 24π

w_{f} = \sqrt{24\pi }

ωf = 8.68 rad/s

Calculate of the v

We replace data in the formula (4)

v = ω*R

v = (8.68)*(0.2)

v = 1.736 m/s

Calculate of the ac

We replace data in the formula (1)

ac = ( ω)²*(R)

ac = (8.68)²*(0.2)

ac = 15.07  m/s²

We replace data in the formula (2)

ac = v²/ R

ac = ( 1.736  )²/(0.2)

ac = 15.07  m/s²

7 0
4 years ago
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blagie [28]

the football player has speed

8 0
3 years ago
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What is the mass of a truck if it is accelerating at a rate of 5 m/s2 and hits a parked car with a
sineoko [7]

Answer:

<h3>The answer is 2800 kg</h3>

Explanation:

The mass of the truck can be found by using the formula

m =  \frac{f}{a}  \\

f is the force

a is the acceleration

From the question we have

m =  \frac{14000}{5}  \\

We have the final answer as

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5 0
3 years ago
A shell is shot with an initial velocity v with arrow0 of 18 m/s, at an angle of θ0 = 60° with the horizontal. At the top of the
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Answer:

D = 43 m

Explanation:

given,

initial velocity = 18 m/s

angle θ = 60°

total horizontal distance covered by the shell is

R = \dfrac{v_0^2sin 2\theta}{g}

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R' = v t

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total distance traveled by the shell is

D = \dfrac{R}{2}+R'

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   = 1.5\dfrac{v_0^2sin 2\theta}{g}

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Explanation:

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