Answer:
Vertical Height = 0.784 meter, Speed back at starting point = 10 m/s
Explanation:
Given Data:
V is the overall velocity vector, 
and 
are its initial vertical and horizontal components
To find:
Max Height 
achieved
Calculation:
1) Using the 
equation of motion, we know
2) In terms of gravity 
height and the vertical component of Velocity 
.
3) As 
as at maximum height the vertical component of velocity is zero maximum height achieved
putting values
4) 
5) As for the speed when it reaches back its starting point, it will have a speed similar to its launching speed, the reason being the absence of air friction (Air drag)
a) Density at 100 degrees: 
Explanation:
The density of mercury at 0 degrees is 
Let's take 1 kg of mercury. Its volume at 0 degrees is
The formula to calculate the volumetric expansion of the mercury is:
where
is the cubic expansivity of mercury
V is the initial volume
is the increase in temperature
In this part of the problem, 
So, the expansion is
So, the new density is
b) Density at 22 degrees: 
We can apply the same formula we used before, the only difference here is that the increase in temperature is
And the volumetric expansion is
So, the new density is
Heat transfer in a closed system is the addition of changes in internal energy and the total amount of work done by it. The final energy of the system is 35.5kJ.
<h3>What is heat transfer? </h3>
Heat transfer is the transfer of heat energy due to temperature differences.
The paddle-wheel paintings are quantities of workdone, 500 N.m or 0.5kJ.
The preliminary (initial) power of the device is 10 kJ.
Total warmness transferred in the course of the method is 30 kJ
Total warmness misplaced in the course of the method to the encompassing air is 5 kJ.
The energy of the system is given as:
The energy of the system = Energy in - Energy out
The energy of the system = Initial energy + Energy transferred + Work done - Energy lost
Energy of the system = 10 + 30 + 0.5 - 5 kJ
Energy of the system = 35.5 kJ
Read more about energy:
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Answer:
The apparent depth of (a) the fish is 5.3 cm and (b) the image of the fish is 24.8 cm.
Explanation:
According to the following equation:
where <em>nw</em> and <em>na</em> is the refractive indices of water (1.33) and air (1.00) respectively; <em>s</em> is the depth of the fish below the surface of the water; s' is the apparent depth of the fish from normal incidence and Rc is the radius of curvature of the mirror at the bottom of the tank.
Note that the bottom of the tank is assumed to be a flat mirror, therefore the radius of curvature is very large (R⇒∞).
Therefore, the above equation can be expressed as:
Now we can solve for the apparent depth of the fish.
(a) 
(Make s' subject of the formula from the above equation)
∴ 
(b) The motionless fish floats 13 cm above the mirror, therefore the image of the fish will be situated at 13 + 20 =33 cm away from the real fish.
Therefore, s = 33 cm
NB: Here, it is assumed that the water is pure, as impurities may alter the refractive index of water.
Answer:
p= 400.29N
........the horizontal force
Explanation:
Given data
mass=25 kg
small cube mass mass=4 kg
Us (The Coefficient of static b/w two cubes) = 0.71
to find
The horizontal force to keep the small cube from sliding downward
Solution
F=ma.........................from Newton Second law
Where F=force
a=acceleration
m=mass
we can write equation in form of acceleration
a=F/m
The acceleration on small box is same as that on the large box.
Let P be force to find.
then:
a=p/(25kg+4kg)
a=p/(29kg)m/s²
The force acting on small box:
F=ma
f=4*(p/29)N........................normal force
friction force= Us*(normal force).........where Us is coefficient of static friction.
friction force= 0.71*(4*p/29)
Now to find weight
weight= mg
weight= 4*9.8
for the object(small box) not to slide down the friction force b/w the two objects have to be exactly the same as the weight of the object.
0.71*(4*p/29)=4*9.8
solving for p(force)
p= 400.29N
........the horizontal force
