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Keith_Richards [23]
3 years ago
7

You need to determine the density of a ceramic statue. if you suspend it from a spring scale, the scale reads 32.4 n . if you th

en lower the statue into a tub of water, so that it is completely submerged, the scale reads 17.0 n . part a what is the density?
Physics
1 answer:
Mandarinka [93]3 years ago
4 0
Well, the density of the water is
{1000 \frac{kg}{ {m}^{3} } }
so i believe that is what the question is asking for :)
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Is it meaning kinetic energy ?
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A rigid tank having a volume of 0.100 m3 contains helium gas at 150 atm. how many balloons can be inflated by opening the valve
GalinKa [24]
The initial value of P*V = 0.100*150 atm-m³ 
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</span><span> 1.20*[N*(4/3)*π*r³ + 0.100] 
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4 0
3 years ago
What is magnetic declination?
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5 0
3 years ago
Read 2 more answers
answer A vertical spring stretches 3.4 cm when a 8-g object is hung from it. The object is replaced with a block of mass 26 g th
victus00 [196]

Answer:

0.695s

Explanation:

From Hooke's law, the restoring force is given has

F = -ky .......1

Where F is the force, y is the spring displacement and k force constant of the spring.

Also recall,

F=mg ............ 2

Where m is the mass of object, g is the acceleration due to gravity.

Equating 1 and 2

Ky = mg

Given that g=9.8m/s2 , y is 3.4cm and g is 8g

K×3.4/100m =8/1000kg × 9.8m/s2

K= ( 0.008kg × 9.8m/s2 ) ÷ 0.034

K= 0.0784÷0.035

K=2.24N/m

Mass ofvthe second object is 25g =0.025kg

Period of oscillation T

T=2π√m/k

T=2×3.142√0.025/2.24

T=6.284√0.0111

T=0.659seconds

8 0
3 years ago
Read 2 more answers
A pot on the stove contains 200 g of water at 20°C. An unknown mass of ice that is originally at −10°C is placed in an identical
Mumz [18]

Answer:

a) The mass of the ice is smaller than the mass of the water

b) The ice reaches first 80°C ,

Explanation:

Since the heat Q that should be provided to ice

Q = sensible heat to equilibrium temperature (as ice) + latent heat + sensible heat until final temperature ( as water)

m ice * c ice * ( T equil -T initial  ) + m ice* L + m ice* c water * ( T final - T equil)

and the heat Q that should be provided to water is

Q= m water * c water * ( T final - T equil )

since the rate of heat addition q = constant and the time t taken to reach the final temperature is the same , then the heat absorbed Q=q*t is the same for both, therefore

m water * c water *  ( T final - T equil ) = m ice* [c ice *( T equil -T initial  ) + L + c water * ( T final - T equil)]

m water/ m ice =  [c ice * ( T equil -T initial  )  + L + c water * ( T final - T equil)]/ [ c water * ( T final - T equil)]

m water/ m ice = [c ice * ( T equil -T initial  )  + L ]/[c water * ( T final - T equil) ] + 1

since  [c ice * ( T equil -T initial  )  + L ]/[c water * ( T final - T equil) ] >0 , then

m water/ m ice > 1

m water > m ice

so the mass of ice is smaller that the mass of water

b) Since the heat Q that should be provided to the ice, starting from 55°C mass would be

Q ice= m ice * c water * ( T final2 - T final1 )

and for the water mass

Q water = m water * c water * ( T final2 - T final1 )

dividing both equations

Q water / Q ice = m water / m ice >1

thus

Q water > Q ice

since the heat addition rate is constant

Q water = q* t water and Q ice=q* t ice

therefore

q* t water > q* t ice

t water >  t ice

so the time that takes to reach 80°C is higher for water , thus the ice mass reaches it first.

5 0
3 years ago
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