Answer:
Option B. 3660000 J
Explanation:
At the sea level, we'll assume that the height is 0 m. Hence, the potential energy at the sea level is zero.
Next, we shall determine the potential energy at a height of 366 m above the sea level. This can be obtained as follow:
Mass (m) = 1000 kg
Acceleration due to gravity (g) = 10 m/s²
Height (h) = 366 m
Potential energy (PE) =?
PE = mgh
PE = 1000 × 10 × 366
PE = 3660000 J
From the calculations made above, we can see clearly that the potential energy of the car at a height of 366 m above sea level is 3660000 J.
Hence, the potential energy of the car increases from 0 at the sea level to 3660000 J at 366 m above the sea level.
Answer:
![2.72\cdot 10^{-3} m/s^2](https://tex.z-dn.net/?f=2.72%5Ccdot%2010%5E%7B-3%7D%20m%2Fs%5E2)
Explanation:
The centripetal acceleration of an object in circular motion is the acceleration with which the object is attracted towards the center of the circular orbit. Mathematically, it is given by
![a=\frac{v^2}{r}](https://tex.z-dn.net/?f=a%3D%5Cfrac%7Bv%5E2%7D%7Br%7D)
where
v is the speed of the object
r is the radius of the orbit
The speed of the object is also given by the ratio between the circumference of the orbit and the orbital period, T:
![v=\frac{2\pi r}{T}](https://tex.z-dn.net/?f=v%3D%5Cfrac%7B2%5Cpi%20r%7D%7BT%7D)
Substituting into the previous equation, we find a new expression for the centripetal acceleration:
![a=\frac{4\pi^2 r}{T^2}](https://tex.z-dn.net/?f=a%3D%5Cfrac%7B4%5Cpi%5E2%20r%7D%7BT%5E2%7D)
In this problem:
- The radius of the orbit of the Moon is
![r = 384000000 m = 3.84\cdot 10^8 m](https://tex.z-dn.net/?f=r%20%3D%20384000000%20m%20%3D%203.84%5Ccdot%2010%5E8%20m)
- The period of the orbit is
![T=27.32 d \cdot 24\cdot 60\cdot 60 =2.36\cdot 10^6 s](https://tex.z-dn.net/?f=T%3D27.32%20d%20%5Ccdot%2024%5Ccdot%2060%5Ccdot%2060%20%3D2.36%5Ccdot%2010%5E6%20s)
Therefore, the centripetal acceleration is:
![a=\frac{4\pi^2 (3.84\cdot 10^8)}{(2.36\cdot 10^6)^2}=2.72\cdot 10^{-3} m/s^2](https://tex.z-dn.net/?f=a%3D%5Cfrac%7B4%5Cpi%5E2%20%283.84%5Ccdot%2010%5E8%29%7D%7B%282.36%5Ccdot%2010%5E6%29%5E2%7D%3D2.72%5Ccdot%2010%5E%7B-3%7D%20m%2Fs%5E2)
Answer:
The force needed will be greater to hold the object at rest.
Explanation:
For a better understanding let's take a look at the attached image.
In the image there is an example of this condition, we have an object of 3 kg-f mass, the first angle between the horizontal and the inclined plane is 35°. When we increase the angle to 45°. We can realize that we need to hold the object with a stronger force.
The calculation and the equations based on Newton's laws can be found in the attached image.
OPTION (C) IS CORRECT
ANSWER - UNIFORM MOTION
If there is no change in the velocity of the object then it is known to be in <u>UNIFORM MOTION</u>.
EXPLORE MORE:-
EXAMPLE - A CAR COVERS A DISTANCE OF 15 KM WE'D EVERY 2 HOURS
FOR AN UNIFORM MOTION, ACCELERATION IS ZERO , AS THERE IS NO CHANGE IN VELOCITY....
-THANKS.!!
Answer:
![f_o=592.36 Hz](https://tex.z-dn.net/?f=f_o%3D592.36%20Hz)
Explanation:
Given that
Train 1 (observer):
Speed = 109 km/h
Train 2 (source):
Speed = 99 km/h
Train 2 emitting frequency = 500 Hz
We know that observer and source are moving toward each other, then frequency heard by observer can be given as follows
![f_o=\left(\dfrac{C+V_o}{C-V_s}\right)f_s](https://tex.z-dn.net/?f=f_o%3D%5Cleft%28%5Cdfrac%7BC%2BV_o%7D%7BC-V_s%7D%5Cright%29f_s)
Where C is the velocity of sound (C=1225 Km/h)
Now by putting the values
![f_o=\left(\dfrac{C+V_o}{C-V_s}\right)f_s](https://tex.z-dn.net/?f=f_o%3D%5Cleft%28%5Cdfrac%7BC%2BV_o%7D%7BC-V_s%7D%5Cright%29f_s)
![f_o=\left(\dfrac{1225+109}{1225-99}\right)\times 500](https://tex.z-dn.net/?f=f_o%3D%5Cleft%28%5Cdfrac%7B1225%2B109%7D%7B1225-99%7D%5Cright%29%5Ctimes%20500)
![f_o=592.36 Hz](https://tex.z-dn.net/?f=f_o%3D592.36%20Hz)