Answer:
I do I make a brinliest can you please can me
To solve this problem we will apply the concepts of equilibrium and Newton's second law.
According to the description given, it is under constant ascending acceleration, and the balance of the forces corresponding to the tension of the rope and the weight of the elevator must be equal to said acceleration. So
Here,
T = Tension
m = Mass
g = Gravitational Acceleration
a = Acceleration (upward)
Rearranging to find T,
Therefore the tension force in the cable is 10290.15N
Answer:
The dart with the small mass will travel the farthest distance.
Explanation:
Acceleration is proportional to force times mass, and inertia is proportional to mass. Inertia is the reluctance of a moving body to stop, and a stationary body to start moving (inertia increses with mass). Assuming they both have the same aerodynamic design, and that they are both launched with the same force applied for the same time duration, the dart with less small mass will accelerate faster than the big mass dart. From this we can see that the small dart will have covered a longer distance before the effect of the force stops, when compared to the more massive dart.
Answer:
In the table, 1=46.7 °C, 1=165 J, 2=819 J, 3=1510 J, and 4=2830 J.
Other experiments determine that the material has a temperature of fusion of
fusion =235 °C and a temperature of vaporization of vapor=481 °C.
If the sample of material has a mass of =8.60 g, calculate the specific heat when this material is a solid, and when it is liquid, l
Solution:
With reference to Fig. 1
Let 'x' be the distance from the wall
Then for 
DAC:
⇒ 
Now for the BAC:
⇒ 
Now, differentiating w.r.t x:
![\frac{d\theta }{dx} = \frac{d}{dx}[tan^{-1} \frac{d + h}{x} - tan^{-1} \frac{d}{x}]](https://tex.z-dn.net/?f=%5Cfrac%7Bd%5Ctheta%20%7D%7Bdx%7D%20%3D%20%5Cfrac%7Bd%7D%7Bdx%7D%5Btan%5E%7B-1%7D%20%5Cfrac%7Bd%20%2B%20h%7D%7Bx%7D%20-%20%20tan%5E%7B-1%7D%20%5Cfrac%7Bd%7D%7Bx%7D%5D)
For maximum angle, 
= 0
Now,
0 = [/tex]\frac{d}{dx}[tan^{-1} \frac{d + h}{x} - tan^{-1} \frac{d}{x}][/tex]
0 = 
After solving the above eqn, we get
x = 
The observer should stand at a distance equal to x =
