Conducting because it occurs when two objects touch and heat is transferred
Answer:
v=5.86 m/s
Explanation:
Given that,
Length of the string, l = 0.8 m
Maximum tension tolerated by the string, F = 15 N
Mass of the ball, m = 0.35 kg
We need to find the maximum speed the ball can have at the top of the circle. The ball is moving under the action of the centripetal force. The length of the string will be the radius of the circular path. The centripetal force is given by the relation as follows :
v is the maximum speed
Hence, the maximum speed of the ball is 5.86 m/s.
First, we have a change in the velocity from 85 to 164 m/s in 10 sec.
Then, we calculate the <u>acceleration </u>as:
Hence we need to calculate the velocity of the space vehicle at t = 2 sec using the first equation of motion:
Then, using the second equation of motion to calculate the distance:
Answer:
1.) Frequency F = 890.9 Hz
2.) Wavelength (λ) = 0.893 m
Explanation:
1.) Given that the wavelength = 0.385m
The speed of sound = 343 m / s
To predict the frequency, let us use the formula V = F λ
Where (λ) = wavelength = 0.385m
343 = F × 0.385
F = 343/0.385
F = 890.9 Hz
2.) Given that the frequency = 384Hz
Using the formula again
V = F λ
λ = V/F
Wavelength (λ) = 343/384
Wavelength (λ) = 0.893 m
The two questions can be solved with the use of formula
Answer:
Explanation:
Let the equilibrium position of third charge be x distance from q₁.
Force on third charge due to q₁
= 9 x 10⁹ x 5 x 10⁻⁹ x 15 x 10⁺⁹ / x²
Force on third charge due to q₂
= 9 x 10⁹ x 2 x 10⁻⁹ x 15 x 10⁺⁹ /( .40-x)²
Both the force will act in opposite direction and for balancing , they should be equal.
9 x 10⁹ x 5 x 10⁻⁹ x 15 x 10⁺⁹ / x² = 9 x 10⁹ x 2 x 10⁻⁹ x 15 x 10⁺⁹ /( .40-x)²
5 / x² = 2 / ( .4 - x )²
Taking square root on both sides
2.236 / x = 1.414 / .4 - x
2.236 ( .4 - x ) = 1.414 x
.8944 - 2.236 x = 1.414 x
.8944 = 3.65 x
x = .245 m
24.5 cm
So the third charge should be at a distance of 24.5 cm from q₁ .
