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2 years ago

A body displaces 50cm^3 of water. What is the up-thrust on the body?

Physics

1 answer:

SVEN [57.7K]2 years ago

3 0

Answer:

Force of buoyancy is given by

Fb = Weight of displaced water

= mg

But, m=density x volume

= ρV g

where ρ = density of the water

V = volume of displaced water = volume of block

g = acceleration due to gravity

Given that,

V = 50 cm3

g = 981 cm/s2

ρ = 1 g/cm3

Fb = 1*50*981

= 49050 dyne

= 0.49 N

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A 125kg bumper car going 12m/s hits a 235kg bumper car going -13m/s.if the first car bounces back at -12.5m/s what is the veloci

vovikov84 [41]

According to the law of conservation of momentum:

$m_{1}v_{1}+m_{2}v_{2}=m_{1}v_{1}'+m_{2}v_{2}'$

m1 = mass of first object
m2 = mass of second object
v1 = Velocity of the first object before the collision
v2 = Velocity of the second object before the collision
v'1 = Velocity of the first object after the collision
v'2 = Velocity of the second object after the collision

Now how do you solve for the velocity of the second car after the collision? First thing you do is get your given and fill in what you know in the equation and solve for what you do not know.

m1 = 125 kg v1 = 12m/s v'1 = -12.5m/s
m2 = 235kg v2 = -13m/s v'2 = ?

$m_{1}v_{1}+m_{2}v_{2}=m_{1}v_{1}'+m_{2}v_{2}'$
$(125kg)(12m/s)+(235kg)(-13m/s)=(125kg)(-12.5m/s)+(235kg)(v_{2}'$
$1,500kg.m/s+(-3055kg.m/s)=(-1562.5kg.m/s)+(235kg)(v_{2}')$
$-1,555kg.m/s=(-1562.5kg.m/s)+(235kg)(v_{2}')$

Transpose everything on the side of the unknown to isolate the unknown. Do not forget to do the opposite operation.

$-1,555kg.m/s + 1562.5kg.m/s=(235kg)(v_{2}')$
$7.5kg.m/s=(235kg)(v_{2}')$
$(7.5kg.m/s)/(235kg)=(v_{2}')$
$0.03m/s=(v_{2}')$

The velocity of the 2nd car after the collision is 0.03m/s.

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3 years ago

A small plastic bead has been charged to -15nC.

marishachu [46]

Answer:

Explanation:

q = - 15 n C = - 15 x 10^-9 c

(a) d = 0.9 cm = 0.009 m

electric field on proton, E = Kq/d²

E = 9 x 10^9 x 15 x 10^-9 / (0.009)²

E = 1.67 x 10^6 N/C

Force on proton, F = charge on proton x Electric field

F = 1.6 x 10^-19 x 1.67 x 10^6 = 2.67 x 10^-13 N

acceleration of proton, a = F / m

a = (2.67 x 10^-13) / (1.67 x 10^-27)

a = 1.6 x 10^14 m/s²

(B) the direction of acceleration is towards the bead, as the force is attractive.

electric field on proton, E = Kq/d²

E = 9 x 10^9 x 15 x 10^-9 / (0.009)²

E = 1.67 x 10^6 N/C

Force on electron, F = charge on electron x Electric field

F = 1.6 x 10^-19 x 1.67 x 10^6 = 2.67 x 10^-13 N

acceleration of proton, a = F / m

a = (2.67 x 10^-13) / (9.1 x 10^-31)

a = 3 x 10^17 m/s²

(D) the direction of acceleration is away from the bead, as the force is repulsive.

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3 years ago

If the Earth rotated in the opposite direction, how would that affect the wind? *

Alik [6]

Answer:

The wind would still blow, but it would curve and spin in the opposite direction.

Explanation:

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If a cable insulation is described as RHW, you would be able to use the cable for what maximum temperature?

BartSMP [9]

167•F which is c

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3 years ago

HELPPP<br> why is the united states considered energy rich, while japan is considered energy poor

Temka [501]

Answer:

Energy resources can be measured. They will include the fossil fuels, geothermal and hydroelectric potential, and increasingly the renewable resources. When the US list is compared to the World it is considered energy Rich. When Japan's list is compared to the world standard it considered energy poor.

A changing technology like nuclear fusion could substantially change the assessment.

Japan does not have any substantial, oil, coal, gas, deposits, while the US does.

Explanation:

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3 years ago

A body displaces 50cm^3 of water. What is the up-thrust on the body? ​

A body displaces 50cm^3 of water. What is the up-thrust on the body?