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Svetlanka [38]
2 years ago
15

Two objects, labeled A and B, are the same size. Object A has a density of 1.21 g/cm3. Object B has a density of 1.37 g/cm3. Bot

h are placed in a beaker of water. Which will float higher in the water?
Physics
2 answers:
Sedaia [141]2 years ago
4 0

Neither object will float in water. 

They both have densities greater than 1.0 g/cm³,
so both will sink to the bottom of the beaker.

Like rocks.

deff fn [24]2 years ago
4 0

Answer:

None of the object float in water

Explanation:

When an object float in water than we can say that buoyancy force due to water will counter balance the weight of the object

As we can say that

W_g = F_b

so here if object is just floating or we can say that if object is in the state of sinking then we have

\rho_{object} V g = \rho_{water} Vg

so we have

\rho_{object} = \rho_{water}

so from above relation we can say that

i)\rho_{object} > \rho_{water} then it will sink inside water

ii)\rho_{object} < \rho_{water} then it will float in water

since we know density of water is 1 gm/cm^3 so here given both objects are having density more than this

so both the objects will sink inside water

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Naddika [18.5K]
Because of gravity and friction. 
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2 years ago
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In Paul Hewitt's book, he poses this question: "If the forces that act on a bullet and the recoiling gun from which it is fired
Sauron [17]
They have different accelerations because of their masses. According to Newton's Second Law, an objects acceleration is inversely proportional to its mass. Therefore the object with the larger mass, in this case the gun, will have a smaller acceleration. In the same way, the less massive object, being the bullet, will have a higher acceleration.

Hope this helps :)
4 0
2 years ago
A wooden disk of mass m and radius r has a string of negligible mass is wrapped around it. If the disk is allowed to fall and th
Tju [1.3M]

Answer:

a = \frac{2}{3}g

T = \frac{mg}{3}

Explanation:

As the disc is unrolling from the thread then at any moment of the time

We have force equation as

mg - T = ma

also by torque equation we can say

TR = I\alpha

TR = \frac{1}{2}mR^2(\frac{a}{R})

T = \frac{1}{2}ma

Now we have

mg - \frac{1}{2}ma = ma

mg = \frac{3}{2}ma

a = \frac{2}{3}g

Also from above equation the tension force in the string is

T = \frac{1}{2}ma

T = \frac{mg}{3}

7 0
2 years ago
Scenario A: 120 J of work is done in 6 s. Scenario B: 160 J of work is done in 8 s. Scenario C: 200 J of work is done in 10 s. W
hodyreva [135]
Using the equation P = W/t to solve your problem . 

Thus the answer is all of them use the same amount of power. 20 J.  
8 0
1 year ago
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An object attached to an ideal massless spring is pulled across a frictionless surface. If the spring constant is 45 N/m and the
Nataly [62]

Answer:

The mass of the object is 49.5kg which is approximately 50kg

Explanation:

Given that

Spring constant (k)=45N/m

The extension (e)=0.88m

Also given that the acceleration is 0.8m/s²

Force by the spring is given as

Using hooke's law

According to Hooke's law which states that the extension of an elastic material is directly proportional to the applied force provided that the elastic limit is not exceeded. Mathematically,

F = ke where

F is the applied force

k is the spring constant

e is the extension

From the formula k = F/e

F=ke

m is the mass of the block = ?

a is the acceleration = 0.8m/s²

e is the extension of the spring =  0.88m

k is the spring constant = 45N/m

F=45×0.88

F=39.6N

Now this force will set the object in motion, now using newton second law of motion

F=ma

Then, m=F/a

m=39.6/0.8

m=49.5kg

The mass of the object is 49.5kg which is approximately 50kg

6 0
2 years ago
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