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mafiozo [28]
3 years ago
5

An object with height h, mass M, and a uniform cross-sectional area A floats upright in a liquid with density ρ.

Physics
1 answer:
soldi70 [24.7K]3 years ago
7 0
** Missing information: The vertical distance from surface of liquid to bottom of the object is sought in this question, with the condition that the object is at equilibrium **

Ans: The vertical distance = y = M/(ρA)

Explanation:

Support the vertical distance = y

Object's density = M/(A*h) (since A*h = volume)

By applying the condition, 

(M/(Ah))/ρ = y/h

M/(ρAh) = y/h

y = M/(ρA)  

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An IGCSE student thinks it may be possible to identify different rocks (A, B and C) by measuring their
kherson [118]

Answer:

See the answer below

Explanation:

a. The volume in the first measuring cylinder reads 70 cm^3 while that of the second reads 95  cm^3. Hence;

V1 = 70  cm^3

V2 = 95  cm^3

b. <u>An object will always displace its own volume in a liquid</u>. Hence:

Volume V of the rock sample = V2 - V1

   = 95 - 70 = 25  cm^3

c. Mass of A = 102 g

   Volume of A = 25  cm^3

<em>Density = mass/volume</em>

Hence, density of A = 102/25 = 4.08 g/cm^3

8 0
3 years ago
In which phase is the full illuminated face of the Moon visible on Earth? A) New moon Eliminate B) Full moon C) Waning gibbous D
Nitella [24]
Full moon

New cant be seen

Gibbous is 3/4

Crescent is 1/4
4 0
3 years ago
If a machine has an efficiency of 94% and you apply 574J of work, how much work do you get out of the machine
Mariana [72]
<h3>Answer:</h3>

539.56 Joules

<h3>Explanation:</h3>
  • Efficiency of a machine is the ratio of work output to work input expressed as a percentage.
  • Efficiency = (work output/work input) × 100%
  • Efficiency of a machine is not 100% because so energy is lost due to friction of the moving parts and also as heat.

In this case;

Efficiency = 94%

Work input = 574 Joules

Therefore, Assuming work output is x

94% = (x/574 J) × 100%

0.94 = (x/574 J)

<h3>x = 539.56 J</h3>

Thus, you get work of 539.56 J from the machine

7 0
3 years ago
A block of ice is sliding down a ramp of slope 40 to the horizontal. What is the rate of acceleration of the block? Assume the f
uysha [10]

The given problem can be exemplified in the following diagram:

Since there is no friction or any other external force, the only force acting in the direction of the movement is the component of the weight of the block, therefore, applying Newton's second law:

\Sigma F=ma

Replacing the values:

mg\sin 40=ma

We may cancel out the mass:

g\sin 40=a

Using the gravity constant as 9.8 meters per square second:

9.8\frac{m}{s^2}\sin 40=a

Solving the operations:

6.3\frac{m}{s^2}=a

Therefore, the acceleration is 6.3 meters per square second.

8 0
1 year ago
A satellite is to be launched into an orbit of radius,r Show that v = 2gr, where V is the
Snezhnost [94]

Explanation:

We start by using the conservation law of energy:

\Delta{K} + \Delta{U} = 0

or

\dfrac{1}{2}mv^2 - G\dfrac{mM}{r} = 0

Simplifying the above equation, we get

v^2 = 2G\dfrac{M}{r}

We can rewrite this as

v^2 = 2\left(G\dfrac{M}{r^2}\right)r

Note that the expression inside the parenthesis is simply the acceleration due to gravity g so we can write

v^2 = 2gr

where v is the launch velocity.

6 0
3 years ago
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