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2 years ago

What is the potential energy of an object that has mass of 410 kg at a height of 12m

1 answer:

8 0

PE = (weight) x (height)

   = (mass) x (gravity) x (height)

   = (410 kg) x (9.8 m/s²) x (12 m)

   = (410 x 9.8 x 12) kg-m²/s²

   = 48,210 joules (rounded)

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A milliliter of very hot water is added to a liter of very cold water. Which of these events will occur? Assume the surrounding

prisoha [69]

Answer:b

Explanation:the water will not be hot nomore because of the cold water

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2 years ago

Read 2 more answers

The intensity level is 65 db at a distance 5.00 m from a barking dog. What would be a reasonable estimate for the intensity leve

BARSIC [14]

Answer:

68 db

Explanation:

Since now instead of one two dogs are barking simultaneously close to each other, therefore we take n =2.

Ignoring interference effects, the barking of two dogs result in a higher level of intensity which is given by,

β(db)=10×㏒(2)

=3 db

So, a reasonable estimate for the raised Intensity Level is: 65db+3db = 68db

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3 years ago

A baseball is thrown through the air. It's initial velocity, described as a vector, is → v ( t = 0 ) = 17.1 ˆ i + 14.7 ˆ j m / s

ludmilkaskok [199]

Answer:

a = - 9.8 j ^ m/s²

Explanation:

This is a projectile launch problem, they give us the initial velocity in the two components

v₀ₓ = 17.1 m / s

$v_{oy}$ = 14.7 m / s

They indicate that the only acceleration that exists is the acceleration of gravity, which acts in the direction towards the center of the Earth, in general in a coordinate system it coincides with the direction of the y axis.

a = - g j ^

a = - 9.8 j ^ m /s²

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2 years ago

Two people are standing on a 1.75-m-long platform, one at each end. The platform floats parallel to the ground on a cushion of a

Fiesta28 [93]

Answer:

0.05312 m

Explanation:

Given:

Length of platform L = 1.75 m

mass of ball m_b = 5.76 kg

mass of (people + platform) m_p = 184 kg

Initial Velocity of ball V_i,b = 0

Initial Velocity of ball (people + platform) V_i,p = 0

Find:

How far does the platform recoils to rest

Solution:

Using the conservation of momentum on ust before and after the ball was thrown P_i = P_f :

Where, P_i = 0 (initially at rest)

P_f = m_p*V_f,p + m_b * V_f,b

0 = m_p*V_f,p + m_b * V_f,b

V_f,p = - (m_b /m_p) * V_f,b

V_f,p = - (5.76 / 184)*V_f,b

V_f,p = - 0.0313*V_f,b ....1

The time the ball is in air:

t = L / (V_f,b - V_f,p) ...2

The distance that the platform moves d:

d = V_f,p *t ....3

Substitute 2 into 3

d = V_f,p*L /(V_f,b - V_f,p) .... 4

Solve 1 and 4 simultaneously :

d = - m_b*L / (m_b + m_p)

d = - 5.76*1.75 / (5.76 + 184)

d = -0.05312 m

The platform moves 0.05312 m to the opposite to which the ball is thrown.

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3 years ago

I NEED HELP ASAP PLEASE!!!<br> (19 pts!)

zhenek [66]

They go in the boxes in this order:
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2 years ago