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lys-0071 [83]
3 years ago
10

An iron block of density rhoFe and of volume l 3 is immersed in a fluid of density rhofluid. The block hangs from a scale which

reads W as the weight. The top of the block is a height h below the surface of the fluid. The correct equation for the reading of the scale is
Physics
1 answer:
PolarNik [594]3 years ago
6 0

Answer:

R=m*g-∀fl*g*l3

Explanation:

<em>An iron block of density rhoFe and of volume l 3 is immersed in a fluid of density rhofluid. The block hangs from a scale which reads W as the weight. The top of the block is a height h below the surface of the fluid. The correct equation for the reading of the scale is</em>

From Archimedes' principle we know that a body when immersed in a fluid, fully or partially, experiences an the upward buoyant force equal to the weight of the fluid displaced. As the body is fully submerged in water, volume of water displaced  

density of iron =mass/ volume

rho=m/l3

mass=rhol3

weight fluid=rhofluid*g*Volume

weight of fluid=rhofluid*g*l3

F=∀fl*g*l3

Downward force is weight of iron

w=m*g

Reading on the spring scale

R=w-F

R=m*g-∀fl*g*l3

m=mass of iron

g=acceleration due to ravity

rhfld=density of fluid

l3=volume of fluid displaced

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Yes, because a crumpled piece of paper has more center mass, allowing it to fall faster.
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Why do seat belts help protect passengers when a car stops quickly? Explain your answer in terms of forces and motions.
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Answer:

If the driver is wearing a seat belt, the seat belt rather than the windshield applies the unbalanced force that stops the driver's forward motion. The force from the seat belt is applied over a longer time, so the force causes less damage.

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6 0
3 years ago
A hockey puck is hit on a frozen lake and starts moving with a speed of 13.60 m/s. Exactly 6.2 s later, its speed is 7.20 m/s. (
stellarik [79]

Answer:

-1.03 m/s²

Explanation:

Acceleration: This can be defined as the rate of change of velocity. The S. I unit of acceleration is m/s².

Mathematically, acceleration is expressed as

a = (v-u)/t ........................ Equation 1

Where a = acceleration, v = final velocity, u = initial velocity, t  = time.

Given: u = 13.60 m/s, v = 7.20 m/s t = 6.2 s.

Substituting into equation 2

a = (7.20-13.60)/6.2

a = -6.4/6.2

a = -1.03 m/s²

Note: a is negative because, the hockey puck is decelerating.

Hence the average acceleration = -1.03 m/s²

3 0
3 years ago
A 100 kg roller coaster comes over the first hill at 2 m/sec (vo). The height of the first hill (h) is 20 meters. See roller dia
aleksandr82 [10.1K]

For the 100 kg roller coaster that comes over the first hill of height 20 meters at 2 m/s, we have:

1) The total energy for the roller coaster at the <u>initial point</u> is 19820 J

2) The potential energy at <u>point A</u> is 19620 J

3) The kinetic energy at <u>point B</u> is 10010 J

4) The potential energy at <u>point C</u> is zero

5) The kinetic energy at <u>point C</u> is 19820 J

6) The velocity of the roller coaster at <u>point C</u> is 19.91 m/s

1) The total energy for the roller coaster at the <u>initial point</u> can be found as follows:

E_{t} = KE_{i} + PE_{i}

Where:

KE: is the kinetic energy = (1/2)mv₀²

m: is the mass of the roller coaster = 100 kg

v₀: is the initial velocity = 2 m/s

PE: is the potential energy = mgh

g: is the acceleration due to gravity = 9.81 m/s²

h: is the height = 20 m

The<em> total energy</em> is:

E_{t} = KE_{i} + PE_{i} = \frac{1}{2}mv_{0}^{2} + mgh = \frac{1}{2}*100 kg*(2 m/s)^{2} + 100 kg*9.81 m/s^{2}*20 m = 19820 J

Hence, the total energy for the roller coaster at the <u>initial point</u> is 19820 J.

   

2) The <em>potential energy</em> at point A is:

PE_{A} = mgh_{A} = 100 kg*9.81 m/s^{2}*20 m = 19620 J

Then, the potential energy at <u>point A</u> is 19620 J.

3) The <em>kinetic energy</em> at point B is the following:

KE_{A} + PE_{A} = KE_{B} + PE_{B}

KE_{B} = KE_{A} + PE_{A} - PE_{B}

Since

KE_{A} + PE_{A} = KE_{i} + PE_{i}

we have:

KE_{B} = KE_{i} + PE_{i} - PE_{B} =  19820 J - mgh_{B} = 19820 J - 100kg*9.81m/s^{2}*10 m = 10010 J

Hence, the kinetic energy at <u>point B</u> is 10010 J.

4) The <em>potential energy</em> at <u>point C</u> is zero because h = 0 meters.

PE_{C} = mgh = 100 kg*9.81 m/s^{2}*0 m = 0 J

5) The <em>kinetic energy</em> of the roller coaster at point C is:

KE_{i} + PE_{i} = KE_{C} + PE_{C}            

KE_{C} = KE_{i} + PE_{i} = 19820 J      

Therefore, the kinetic energy at <u>point C</u> is 19820 J.

6) The <em>velocity</em> of the roller coaster at point C is given by:

KE_{C} = \frac{1}{2}mv_{C}^{2}

v_{C} = \sqrt{\frac{2KE_{C}}{m}} = \sqrt{\frac{2*19820 J}{100 kg}} = 19.91 m/s

Hence, the velocity of the roller coaster at <u>point C</u> is 19.91 m/s.

Read more here:

brainly.com/question/21288807?referrer=searchResults

I hope it helps you!

3 0
3 years ago
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