Answer:
Stretch can be obtained using the Elastic potential energy formula.
The expression to find the stretch (x) is 
Explanation:
Given:
Elastic potential energy (EPE) of the spring mass system and the spring constant (k) are given.
To find: Elongation in the spring (x).
We can find the elongation or stretch of the spring using the formula for Elastic Potential Energy (EPE).
The formula to find EPE is given as:
Rewriting the above expression in terms of 'x', we get:
Example:
If EPE = 100 J and spring constant, k = 2 N/m.
Elongation or stretch is given as:
Therefore, the stretch in the spring is 10 m.
So, stretch in the spring can be calculated using the formula for Elastic Potential Energy.
Answer:
Answer:u=66.67 m/s
Explanation:
Given
mass of meteor m=2.5 gm\approx 2.5\times 10^{-3} kg
velocity of meteor v=40km/s \approx 40000 m/s
Kinetic Energy of Meteor
K.E.=\frac{mv^2}{2}
K.E.=\frac{2.5\times 10^{-3}\times (4000)^2}{2}
K.E.=2\times 10^6 J
Kinetic Energy of Car
=\frac{1}{2}\times Mu^2
=\frac{1}{2}\times 900\times u^2
\frac{1}{2}\times 900\times u^2=2\times 10^6
900\times u^2=4\times 10^6
u^2=\frac{4}{9}\times 10^4
u=\frac{2}{3}\times 10^2
u=66.67 m/s
I already answered this quesiton. The fact is that there are only two kind of poles and since the two taped poles of the magnets labeled A and B attracts one to each other, we know that the two taped poles of the first two magnets are oppsosite.
Then, the taped pole of the third magnet has to be equal to one of the first two taped poles and opposite to the other of the first two taped poles.
That drives you to conclude (predict) that when she brings the taped end of the third magnet (magnet C) near each of the first two magntes, in one case they will attract each other and in the other case they will repele mutually.
Answer:
0.777m
Explanation:
The sound wave has a wavelength of 0.773m.
Explanation:
To solve this problem we have to use the wave equation that is given below:
We know the frequency and the velocity, both of which have good units. All we have to do is rearrange the equation and solve for
λ
:
λ
=
v
f
Let's plug in our given values and see what we get!
λ
=
340
m
s
440
s
−
1
λ
=
0.773
m
Hope this helps, Mark as brainliest if u want
(a) 3.56 m/s
(b) 11 - 3.72a
(c) t = 5.9 s
(d) -11 m/s
For most of these problems, you're being asked the velocity of the rock as a function of t, while you've been given the position as a function of t. So first calculate the first derivative of the position function using the power rule.
y = 11t - 1.86t^2
y' = 11 - 3.72t
Now that you have the first derivative, it will give you the velocity as a function of t.
(a) Velocity after 2 seconds.
y' = 11 - 3.72t
y' = 11 - 3.72*2 = 11 - 7.44 = 3.56
So the velocity is 3.56 m/s
(b) Velocity after a seconds.
y' = 11 - 3.72t
y' = 11 - 3.72a
So the answer is 11 - 3.72a
(c) Use the quadratic formula to find the zeros for the position function y = 11t-1.86t^2. Roots are t = 0 and t = 5.913978495. The t = 0 is for the moment the rock was thrown, so the answer is t = 5.9 seconds.
(d) Plug in the value of t calculated for (c) into the velocity function, so:
y' = 11 - 3.72a
y' = 11 - 3.72*5.913978495
y' = 11 - 22
y' = -11
So the velocity is -11 m/s which makes sense since the total energy of the rock will remain constant, so it's coming down at the same speed as it was going up.
