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Pie
3 years ago
14

A body of volume 2000 cm3 has a mass 4 kg. Find the density of the body. Will the body sink or float in water, if the density of

water is 1 g/cm3.
Physics
1 answer:
Brums [2.3K]3 years ago
6 0

Answer:

2 g/cm^3, it will sink

Explanation:

The density of an object is given by

d=\frac{m}{V}

where

m is the mass of the object

V is its volume

For the body in the problem, we have

m = 4 kg = 4000 g

V=2000 cm^3

Therefore, its density is

d=\frac{4000}{2000}=2 g/cm^3

And the object will sink in water, because its density is larger than that of water, which is 1 g/cm^3. (an object sinks when its density is larger than that of water, otherwise it floats).

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An object is moving east with a constant speed of 30 m/s for 5 seconds. What is the object's acceleration
pentagon [3]

Answer:

<h3>The answer is 6 m/s²</h3>

Explanation:

The acceleration of an object given it's velocity and time taken can be found by using the formula

a =  \frac{v}{t}  \\

where

v is the velocity

t is the time

We have

a =  \frac{30}{5}  \\

We have the final answer as

<h3>6 m/s²</h3>

Hope this helps you

5 0
3 years ago
A 50 kg student climbs 3m to the top of a set of stairs. Calculate the change in the student’s gravitational potential energy fr
erastova [34]
Gpe = mgh
gpe = 50*10*3
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7 0
3 years ago
A well-thrown ball is caught in a well-padded mitt. If the deceleration of the ball is 2.10×104 m/s2 , and 1.85 ms (1 ms = 10−3
ankoles [38]

Answer:

u = - 38.85 m/s^-1

Explanation:

given data:

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time = 1.85*10^{-3} s

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from equation of motion we have following relation

v = u +at

0 =  u + 2.10*10^4 *1.85*10^{-3}

0 = u + (21 *1.85)

0 = u + 38.85

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negative sign indicate that the ball bounce in opposite directon

4 0
3 years ago
At higher speeds, how would you compensate for the decrease in field of vision
inysia [295]
When you ride a vehicle in a fast speed, then your peripheral vision will reduce that is why there is a need for you to follow the direction of the objects when you are travelling in order for you to compensate to the decrease in the field of vision.


5 0
4 years ago
Read 2 more answers
A real gas will behave most like an ideal gas under conditions of ________.
KengaRu [80]

Answer: high temperature and low pressure

Explanation:

The Ideal Gas equation is:  

P.V=n.R.T  

Where:  

P is the pressure of the gas  

V is the volume of the gas

n the number of moles of gas  

R=0.0821\frac{L.atm}{mol.K} is the gas constant  

T is the absolute temperature of the gas in Kelvin

According to this law, molecules in gaseous state do not exert any force among them (attraction or repulsion) and the volume of these molecules is small, therefore negligible in comparison with the volume of the container that contains them.  

Now, real gases can behave approximately to an ideal gas, under the conditions described above and taking into account the following:  

When <u>temperature is high</u> a real gas approximates to ideal gas, because the molecules move quickly, preventing the repulsion or attraction forces to take effect.  In addition, at <u>low pressures</u>, the volume of molecules is negligible.

4 0
3 years ago
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