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alexdok [17]
3 years ago
13

A merchant in Katmandu sells you a solid gold 1.00-kg statue for a very reasonable price. When you get home, you wonder whether

or not you got a bargain, so you lower the statue into a container of water and measure the volume of displaced water.
A) Find the volume of water that will be displaced for pure gold.
B) For my answer I found 51813 kg/m^3. It marked me down saying enter your answer using dimension of volume.... i thought kg/m^3 was dimesnsion of volume?
Physics
1 answer:
White raven [17]3 years ago
7 0

Answer:

a) 51.8 cm³

b) kg/m³ is a dimension of density (mass/volume). The regular unitys for volume are m³, cm³, L, gallons.

Explanation:

a) The density of pure gold is 19.3 g/cm³. When put in water, the piece of gold will occupy a volume, so that the volume of water will be displaced. To know the volume, we must divide the mass for the density (mass must be in grams because of the units of the density)

V = 1000/19.3

V = 51.8 cm³

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The kinetic energy K of a moving object varies jointly with its mass m and the square of its velocity v. If an object weighing 4
nata0808 [166]

Answer:

K' = 1777.777  J

Explanation:

Given that

m = 40 kg

v= 15 m/s

K=1000

Given that kinetic energy(K) varies with mass(m) and velocity(v)

 K= C(mv²)

Where

C= Constant

m=mass

v=velocity

When

m = 40 kg ,v= 15 m/s ,K=1000

 K= C(mv²)

1000 = C( 40 x 15²)

C=0.111111

When m = 40 kg and v= 20 m/s

K' = C(mv²)

K= 0.1111 x (40 x 20²)

K' = 1777.777  J

5 0
3 years ago
Read 2 more answers
A child bounces a 60 g superball on the sidewalk. The velocity change of the superball is from 22 m/s downward to 15 m/s upward.
fgiga [73]

complete question:

A child bounces a 60 g superball on the sidewalk. The velocity change of the superball is from 22 m/s downward to 15 m/s upward. If the contact time with the sidewalk is 1/800 s, what is the magnitude of the average force exerted on the superball by the sidewalk

Answer:

F = 1776  N

Explanation:

mass of ball = 60 g = 0.06 kg

velocity of downward direction = 22 m/s = v1

velocity of upward direction = 15 m/s = v2

Δt = 1/800 = 0.00125 s

Linear momentum of a particle with mass and velocity is the product of the mass and it velocity.

p = mv

When a particle move freely and interact with another system within a period of time and again move freely like in this scenario it has a definite change in momentum. This change is defined as Impulse .

I = pf − pi = ∆p

F =  ∆p/∆t  =  I/∆t

let the upward velocity be the positive

Δp =  mv2 - m(-v1)

Δp =  mv2 - m(-v1)

Δp = m (v2 + v1)

Δp = 0.06( 15 + 22)

Δp = 0.06(37)

Δp = 2.22 kg m/s

∆t  = 0.00125

F =  ∆p/∆t

F =  2.22/0.00125

F = 1776  N

4 0
3 years ago
A ball is thrown upward.
valentina_108 [34]

Answer:

It is 10.75

Explanation:

8 0
3 years ago
If you were to mix 3 liters of 20 degrees celsius water with 4 liters of 30 degrees water at 7 liters of water what temperature
Debora [2.8K]
Lets say final temperature is @ celsius
3c(@-20)= 4c([email protected])
[email protected]=180
@=180/7
3 0
3 years ago
During a football game, the running back changes direction from running 4.5 m/s [N] to 4.5 m/s [S] over 8 s. What is the acceler
Nimfa-mama [501]

Explanation:

Take north to be positive and south to be negative.

a = (v − v₀) / t

a = (-4.5 m/s − 4.5 m/s) / 8 s

a = -1.125 m/s²

The acceleration is 1.125 m/s² south.

8 0
2 years ago
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