Sound level at distance of 15 m is given as 20 dB
so intensity at this distance is given as
now if we move closer to some some distance the sound level is now 50 dB
now the intensity is given as
now we know that
so now the distance from friend must be 47 cm
Answer:
3g/cm³
Explanation:
<em>Use the formula:</em>
density = mass ÷ volume
<em>Substitute (plug in) the values:</em>
density = 3 ÷ 1 = 3g/cm³
Answer:
Wrong its B Use a different amount of mass in the cart for five different trials, roll the cart down a ramp with the same slope for each trial, and measure how long it takes the cart to roll one meter each time.
Explanation:
Answer:
a)
Y0 = 0 m
Vy0 = 15 m/s
ay = -9.81 m/s^2
b) 7.71 m
c) 3.06 s
Explanation:
The knowns are that the initial vertical speed (at t = 0 s) is 15 m/s upwards. Also at that time the dolphin is coming out of the water, so its initial position is 0 m. And since we can safely assume this happens in Earth, the acceleration is the acceleration of gravity, which is 9.81 m/s^2 pointing downwards
Y(0) = 0 m
Vy(0) = 15 m/s
ay = -9.81 m/s^2 (negative because it points down)
Since acceleration is constant we can use the equation for uniformly accelerated movement:
Y(t) = Y0 + Vy0 * t + 1/2 * a * t^2
To find the highest point we do the first time derivative (this is the speed:
V(t) = Vy0 + a * t
We equate this to zero
0 = Vy0 + a * t
0 = 15 - 9.81 * t
15 = 9.81 * t
t = 0.654 s
At this time it will have a height of:
Y(0.654) = 0 + 15 * 0.654 - 1/2 * 9.81 * 0.654^2 = 7.71 m
The doplhin jumps and falls back into the water, when it falls again it position will be 0 again. So we can equate the position to zero to find how long it was in the air knowing that it started the jump at t = 0s.
0 = Y0 + Vy0 * t + 1/2 * a * t^2
0 = 0 + 15 * t - 1/2 * 9.81 t^2
0 = 15 * t - 4.9 * t^2
0 = t * (15 - 4.9 * t)
t1 = 0 This is the moment it jumped into the air
0 = 15 - 4.9 * t2
15 = 4.9 * t2
t2 = 3.06 s This is the moment when it falls again.
3.06 - 0 = 3.06 s
Answer:
i. Cv =3R/2
ii. Cp = 5R/2
Explanation:
i. Cv = Molar heat capacity at constant volume
Since the internal energy of the ideal monoatomic gas is U = 3/2RT and Cv = dU/dT
Differentiating U with respect to T, we have
= d(3/2RT)/dT
= 3R/2
ii. Cp - Molar heat capacity at constant pressure
Cp = Cv + R
substituting Cv into the equation, we have
Cp = 3R/2 + R
taking L.C.M
Cp = (3R + 2R)/2
Cp = 5R/2
