Answer:
Double the current
Explanation:
The energy delivered by the heater is related to the current by the following relation:
E= 
let R * t = k ( ∴ R and t both are constant)
so E= k 
Now let:
E2= k I₂^2
E2= 4E
⇒ k I₂^2= 4* k
Cancel same terms on both sides.
I₂^2= 4*
taking square-root on both sides.
√I₂^2 = √4* I^2
⇒I₂= 2I
If we double the current the energy delivered each minute be 4E.
Answer:
Yes
Explanation:
There are two types of interference possible when two waves meet at the same point:
- Constructive interference: this occurs when the two waves meet in phase, i.e. the crest (or the compression, in case of a longitudinale wave) meets with the crest (compression) of the other wave. In such a case, the amplitude of the resultant wave is twice that of the original wave.
- Destructive interferece: this occurs when the two waves meet in anti-phase, i.e. the crest (or the compression, in case of a longitudinal wave) meets with the trough (rarefaction) of the other wave. In this case, the amplitude of the resultant wave is zero, since the amplitudes of the two waves cancel out.
In this problem, we have a situation where the compression of one wave meets with the compression of the second wave, so we have constructive interference.
Answer:
The acceleration of the ball is 4.18 [m/s^2]
Explanation:
By Newton's second law we can find the acceleration of the ball
![F = m*a\\where:\\F = force applied [N] or [kg*m/s^2]\\m = mass of the ball [kg]\\a = acceleration [m/s^s]](https://tex.z-dn.net/?f=F%20%3D%20m%2Aa%5C%5Cwhere%3A%5C%5CF%20%3D%20force%20applied%20%5BN%5D%20or%20%5Bkg%2Am%2Fs%5E2%5D%5C%5Cm%20%3D%20mass%20of%20the%20ball%20%5Bkg%5D%5C%5Ca%20%3D%20acceleration%20%5Bm%2Fs%5Es%5D)
Now we have:
![a = F/m\\a = \frac{1.8 [kg*m/s^s]}{0.43[kg]} \\a = 4.18 [kg]](https://tex.z-dn.net/?f=a%20%3D%20F%2Fm%5C%5Ca%20%3D%20%5Cfrac%7B1.8%20%5Bkg%2Am%2Fs%5Es%5D%7D%7B0.43%5Bkg%5D%7D%20%5C%5Ca%20%3D%204.18%20%5Bkg%5D)
Answer:
W = 1.06 MJ
Explanation:
- We will use differential calculus to solve this problem.
- Make a differential volume of water in the tank with thickness dx. We see as we traverse up or down the differential volume of water the side length is always constant, hence, its always 8.
- As for the width of the part w we see that it varies as we move up and down the differential element. We will draw a rectangle whose base axis is x and vertical axis is y. we will find the equation of the slant line that comes out to be y = 0.5*x. And the width spans towards both of the sides its going to be 2*y = x.
- Now develop and expression of Force required:
F = p*V*g
F = 1000*(2*0.5*x*8*dx)*g
F = 78480*x*dx
- Now, the work done is given by:
W = F.s
- Where, s is the distance from top of hose to the differential volume:
s = (5 - x)
- We have the work as follows:
dW = 78400*x*(5-x)dx
- Now integrate the following express from 0 to 3 till the tank is empty:
W = 78400*(2.5*x^2 - (1/3)*x^3)
W = 78400*(2.5*3^2 - (1/3)*3^3)
W = 78400*13.5 = 1058400 J
Answer:
242.85 Hz
Explanation:
For maximum intensity of sound, the path difference,ΔL = (n + 1/2)λ/2 where n = 0,1,2...
Since Abby is standing perpendicular to one speaker, the path length for the sound from the other speaker to him is L₁ = √(2.00² + 5.50²) = √(4.00 + 30.25) = √34.25 = 5.85 m.
The path difference to him is thus ΔL = 5.85 m - 5.50 m = 0.35 m.
Since ΔL = (n + 1/2)λ/2 and for lowest frequency n = 0,
ΔL = (n + 1/2)λ/2 = (0 + 1/2)λ/2 = λ/4
ΔL = λ = v/f and f = v/4ΔL where f = frequency of wave and v = velocity of sound wave = 340 m/s.
f = 340/(4 × 0.35) = 242.85 Hz
