The period of the pendulum is directly proportional to the square root of the length of the pendulum
Explanation:
The period of a simple pendulum is given by the equation
where
T is the period
L is the length of the pendulum
g is the acceleration of gravity
From the equation, we see that when the length of the pendulum increases, the period of the pendulum increases as the square root of L, 
. This means that
The period of the pendulum is directly proportional to the square root of the length of the pendulum
From the equation, we also notice that the period of a pendulum does not depend on its mass.
#LearnwithBrainly
Explanation:
It is given that, the metal with the highest melting temperature is tungsten which melts at around 3400 K, T = 3400 K
We need to find the wavelength of the peak of the black body distribution for this temperature. It can be calculated using Wein's displacement law as :
k is the constant, 
or
The wavelength of infrared is from 700 nm to 1 mm. So, the lies in infrared region of the spectrum. Hence, this is the required solution.
Answer:
Fx = 35.36 N
Fy = 35.36 N
Explanation:
From the question,
The X component of the force is
Fx = Fcos∅.................. Equation 1
Where Fx = X component of the force, F = Force, ∅ = Angle to the horizontal.
Give: F = 50 N, ∅ = 45°
Substitute into equation 1
Fx = 50(cos45°)
Fx = 50(0.7071)
Fx = 35.36 N
Similarly,
For Y component
Fy = Fsin∅
Where F y = Y component
Fy = 50(sin45°)
Fy = 50(0.7071)
Fy = 35.36 N
Answer:
The right response will be "450 volts".
Explanation:
The given values are:
R1 = 4.00 cm
R2 = 6.00 cm
q1 = +6.00 nC
q2 = −9.00 nC
As we know,
The potential difference between the two shell's difference will be:
⇒ ![\Delta V=K[(\frac{q1}{R1}+\frac{q2}{R2})-(\frac{q1}{R1} +(\frac{q2}{R2}))]](https://tex.z-dn.net/?f=%5CDelta%20V%3DK%5B%28%5Cfrac%7Bq1%7D%7BR1%7D%2B%5Cfrac%7Bq2%7D%7BR2%7D%29-%28%5Cfrac%7Bq1%7D%7BR1%7D%20%2B%28%5Cfrac%7Bq2%7D%7BR2%7D%29%29%5D)
![=K[\frac{q1}{R2}-\frac{q1}{R1} ]](https://tex.z-dn.net/?f=%3DK%5B%5Cfrac%7Bq1%7D%7BR2%7D-%5Cfrac%7Bq1%7D%7BR1%7D%20%5D)
On substituting the values, we get
![=(9\times 10^9)[\frac{6\times 10^{-9}}{0.04}-\frac{6\times 10^{-9}}{0.06}]](https://tex.z-dn.net/?f=%3D%289%5Ctimes%2010%5E9%29%5B%5Cfrac%7B6%5Ctimes%2010%5E%7B-9%7D%7D%7B0.04%7D-%5Cfrac%7B6%5Ctimes%2010%5E%7B-9%7D%7D%7B0.06%7D%5D)
Δ 
Answer: 272.82 drop/tile
Explanation:
Given that the Rain drops fall on a tile surface at a density of 4638 drops/ft2. There are 17 tiles/ft2. How many drops fall on each tile?
Tiles/ft^2 × drop/tiles = drop/ft^2
Tiles will cancel out. Leaving the answer to be drop/ ft^2
Substitutes all the magnitude of the above units.
17 × drop/tiles = 4638
Make drop/tiles the subject of formula
Drop/tiles = 4638/17
Drop/tiles = 272.82
Therefore, 272.82 drop/tile drops fall on each tile?
