A force is a push or pull to an object
Answer:
- The temperature of 10°C will be experienced at an altitude of 2.52 km
- The temperature of 0°C will be experienced at an altitude of 4.15 km
Explanation:
Lapse Rate = -6.5°C/km of ascent.
Lapse Rate = Temperature difference/altitude difference
For the 10°C,
Temperature difference = 10 - 27 = -17°C
-6.5 = -17/(difference in altitude between the two points)
Difference in altitude = 17/6.5 = 2.52 km
For 0°C,
Temperature difference = 0 - 27 = - 27°C
-6.5 = -27/(difference in altitude between the two points)
Altitude difference = 27/6.5 = 4.15 km
Answer:
Boyle’s law and,Charles’s law
Explanation:
For a fixed mass of gas at constant pressure, the volume is directly proportional to the kelvin temperature. That means, for example, that if you double the kelvin temperature from, say to 300 K to 600 K, at constant pressure, the volume of a fixed mass of the gas will double as well.
Answer:
T_final = 279.4 [°C]
Explanation:
In order to solve this problem, we must use the following equation of thermal energy.
where:
Q = heat = 9457 [cal]
m = mass = 79 [g] = 0.079 [kg]
Cp = specific heat = 0.5 [cal/g*°C]
T_initial = initial temperature = 40 [°C]
T_final = final temperature [°C]
![9457 = 79*0.5*(T_{f}-40)\\239.41=T_{f}-40\\\\T_{f}=279.4[C]](https://tex.z-dn.net/?f=9457%20%3D%2079%2A0.5%2A%28T_%7Bf%7D-40%29%5C%5C239.41%3DT_%7Bf%7D-40%5C%5C%5C%5CT_%7Bf%7D%3D279.4%5BC%5D)
Answer:
Incomplete question. Complete question is: An electric drill starts from rest and rotates with a constant angular acceleration. After the drill has rotated through a certain angle, the magnitude of the centripetal acceleration of a point on the drill is twice the magnitude of the tangential acceleration. Determine the angle through which the drill rotates by this point.
The answer is : Δ θ = 1 rad
Explanation:
Ok, so the condition involves the centripetal acceleration and the tangential acceleration, so let’s start by writing expressions for each:
Ac= centripetal acceleration At= tangential acceleration
Ac = V² / r At = r α
Because we have to determine the angle ultimately, therefore we should convert the linear velocity into angular velocity in the expression for centripetal acceleration
V = r ω
Ac = (r ω)² / r = r² ω² / r
Ac = r ω²
now that we have expressions for the centripetal and tangential acceleration, we can write an equation that expresses the condition given: The magnitude of the centripetal acceleration is twice the magnitude of the tangential acceleration.
Ac = 2 At
That is,
r ω² = 2 r α
it is equivalent to;
ω² = 2 α
now we have the relation between angular speed and angular acceleration, but we also need to determine the angular displacement as well. Therefore choose a kinematics equation that doesn’t involve time because time is not mentioned in the question. Thus,
ω² – ω°² = 2 α Δ θ
such that ω° = 0
and ω² = 2 α
therefore;
2 α - 0 = 2 α Δ θ
2 α = 2 α Δ θ
So the angle will be : Δ θ = 1 rad
