Answer:
3,150,000N
Explanation:
According to Newton's second law;
F = mass * acceleration
Given
Mass = 45000kg
acceleration = 70m/s^2
Substitute
F = 45000 * 70
F = 3,150,000N
Hence the force required to be produced by the rocket engines is 3,150,000N
Answer:
Support at Cy = 1.3 x 10³ k-N
Support at Ay = 200 k-N
Explanation:
given:
fb = 300 k-N/m
fc = 100 k-N/m
D = 300 k-N
L ab = 6 m
L bc = 6 m
L cd = 6 m
To get the reaction A or C.
take summation of moment either A or C.
<em><u>Support Cy:</u></em>
∑ M at Ay = 0
(( x1 * F ) + ( D * Lab ) + ( D * L bc + D * L cd )
Cy = -------------------------------------------------------------------
( L ab + L bc )
Cy = 1.3 x 10³ k-N
<em><u>Support Ay:</u></em>
Since ∑ F = 0, A + C - F - D = 0
A = F + D - C
Ay = 200 k-N
Answer:
19.5°
Explanation:
The energy of the mass must be conserved. The energy is given by:
1) ![E=\frac{1}{2}mv^2+mgh](https://tex.z-dn.net/?f=E%3D%5Cfrac%7B1%7D%7B2%7Dmv%5E2%2Bmgh)
where m is the mass, v is the velocity and h is the hight of the mass.
Let the height at the lowest point of the be h=0, the energy of the mass will be:
2) ![E=\frac{1}{2}mv^2](https://tex.z-dn.net/?f=E%3D%5Cfrac%7B1%7D%7B2%7Dmv%5E2)
The energy when the mass comes to a stop will be:
3) ![E=mgh](https://tex.z-dn.net/?f=E%3Dmgh)
Setting equations 2 and 3 equal and solving for height h will give:
4) ![h=\frac{v^2}{2g}](https://tex.z-dn.net/?f=h%3D%5Cfrac%7Bv%5E2%7D%7B2g%7D)
The angle ∅ of the string with the vertical with the mass at the highest point will be given by:
5) ![cos\phi=\frac{l-h}{l}](https://tex.z-dn.net/?f=cos%5Cphi%3D%5Cfrac%7Bl-h%7D%7Bl%7D)
where l is the lenght of the string.
Combining equations 4 and 5 and solving for ∅:
6) ![\phi={cos}^{-1}(\frac{l-h}{l})={cos}^{-1}(1-\frac{h}{l})={cos}^{-1}(1-\frac{v^2}{2gl})](https://tex.z-dn.net/?f=%5Cphi%3D%7Bcos%7D%5E%7B-1%7D%28%5Cfrac%7Bl-h%7D%7Bl%7D%29%3D%7Bcos%7D%5E%7B-1%7D%281-%5Cfrac%7Bh%7D%7Bl%7D%29%3D%7Bcos%7D%5E%7B-1%7D%281-%5Cfrac%7Bv%5E2%7D%7B2gl%7D%29)
Answer:
It is sensible heat- the amount of heat absorbed by 1 kg of water when heated at a constant pressure from freezing point 0 degree Celsius to the temperature of formation of steam i.e. saturation temperature
So it is given as - mass× specific heat × rise in temperature
i.e. 4.2 × T
4.2 × (100–0)
So it is 420kj
If you ask how much quantity of heat is required to convert 1 kg of ice into vapour then you have to add latent heat of fusion that is 336 kj/kg and latent heat of vaporization 2257 kj/kg (these two process occur at constant temperature so need to add rise in tempeature)
So it will be
Q= 1×336 + 1× 4.18 ×100 + 1× 2257
Q = 3011 kj
Or 3.1 Mj
Hope you got this!!!!!!