Answer:
a = 2.5 [m/s²]
Explanation:
To solve this problem we must use the following equation of kinematics.
![v_{f} =v_{o} +a*t](https://tex.z-dn.net/?f=v_%7Bf%7D%20%3Dv_%7Bo%7D%20%2Ba%2At)
where:
Vf = final velocity = 25 [m/s]
Vo = initial velocity = 0 (star from the rest)
a = acceleration [m/s²]
t = time = 10 [s]
25 = 0 + (a*10)
a = 25/10
a = 2.5 [m/s²]
Answer:
W = 600 N
Explanation:
The mechanical advantage of any machine is the ratio of the load or weight to the effort applied. Hence, the general formula for mechanical advantage is as follows:
M.A = W/P
where,
M.A =Mechanical Advantage of Hammer = 10
W = Maximum force that could be generated = ?
P = Force created by me = 60 N
Therefore,
10 = W/60 N
W = 10*60 N
<u>W = 600 N</u>
The transfer of thermal energy as heat requires a difference in temperature between the two points of transfer. Heat may be transferred by means of conduction, convection, or radiation. Conduction is the transfer of thermal energy (heat in transfer) due to collisions between the molecules in the object.
The correct answer to the question is - A). Light waves can travel in a vacuum and travel at a constant speed even if the light source is moving.
EXPLANATION:
Before going to answer this question, first we have to understand the fundamental postulate of quantum mechanics.
As per the fundamental postulate of quantum mechanics, the speed of light is always constant irrespective of the nature of frame of reference i.e speed of light is same both in inertial and non-inertial frame of reference. It is independent of the velocity of source also.
Again we know that light is an electromagnetic wave. Hence, it won't require any medium for its propagation. It can travel in space or vacuum also.
Hence, the correct option out of four options is that light waves can travel in a vacuum and travel at a constant speed even if the light source is moving.
Answer:
The pressure altitude is ![A = 10\ km](https://tex.z-dn.net/?f=A%20%3D%20%2010%5C%20km)
The density altitude is ![\rho = 0.429 kg/m^3](https://tex.z-dn.net/?f=%5Crho%20%3D%200.429%20kg%2Fm%5E3)
Explanation:
From the question we are told that
The outside pressure is ![P_o = 2.65*10^{4} N/m^2](https://tex.z-dn.net/?f=P_o%20%20%3D%20%202.65%2A10%5E%7B4%7D%20N%2Fm%5E2)
The outside temperature is ![T_o = 215 \ K](https://tex.z-dn.net/?f=T_o%20%3D%20%20215%20%5C%20%20K)
Generally the density altitude is mathematically represented as
![\rho = \frac{P_o }{RT}](https://tex.z-dn.net/?f=%5Crho%20%3D%20%20%5Cfrac%7BP_o%20%7D%7BRT%7D)
Here R is the gas constant with value ![R = 287\ J/kg/K](https://tex.z-dn.net/?f=R%20%3D%20%20287%5C%20%20J%2Fkg%2FK)
So
![\rho = \frac{2.65 (*10^{4}}{287 * 215}](https://tex.z-dn.net/?f=%5Crho%20%3D%20%20%5Cfrac%7B2.65%20%28%2A10%5E%7B4%7D%7D%7B287%20%2A%20215%7D)
=> ![\rho = 0.429 kg/m^3](https://tex.z-dn.net/?f=%5Crho%20%3D%200.429%20kg%2Fm%5E3)
Generally the altitude at which the pressure in the question is experience is
![A = 10\ km](https://tex.z-dn.net/?f=A%20%3D%20%2010%5C%20km)
Thus the pressure altitude is ![A = 10\ km](https://tex.z-dn.net/?f=A%20%3D%20%2010%5C%20km)