Answer:
9.4 m
Explanation:
We can use a moving frame of reference with the same speed as the car. From this frame of reference the car doesn't move. The origin is at the back of the car, the positive X axis points back and the positive Y axis points up.
If the ballon is launched at 9.7 m/s at 39 degrees of elevation.
Vx0 = 9.7 * cos(39) = 7.5 m/s
Vy0 = 9.7 * sin(39) = 6.1 m/s
If we ignore air drag, the baloon will be subject only to the acceleration of gravity. We can use the equation of position under constant acceleration.
Y(t) = Y0 + Vy0 * t + 1/2 * a * t^2
Y0 = 0
a = -9.81 m/s^2
It will fall when Y(t) = 0
0 = 6.1 * t - 4.9 * t^2
0 = t * (6.1 - 4.9 * t)
t1 = 0 (this is when the balloon was launched)
0 = 6.1 - 4.9 * t2
4.9 * t2 = 6.1
t2 = 6.1 / 4.9 = 1.25 s
The distance from the car will be the horizonta distance it travelled in that time
X(t) = X0 + Vx0 * t
X(1.25) = 7.5 * 1.25 = 9.4 m
Answer:
The time taken by the rock to reach the ground is 0.569 seconds.
Explanation:
Given that,
A student throws a rock horizontally off a 5.0 m tall building, s = 5 m
The initial speed of the rock, u = 6 m/s
We need to find the time taken by the rock to reach the ground. Using second equation of motion to find it. We get :
So, the time taken by the rock to reach the ground is 0.569 seconds. Hence, this is the required solution.
Answer:
77.96dB
Explanation:
Recall that decibels are a unit of measuring intensity of sound, and depend on the logarithm of the intensity
the intensity, measured in decibels is given by:
I(db)=10log(I/I0)
I is the intensity in MKS units; I0 is the threshold intensity for human hearing (10^-12 W/m^2)
Thus, if the two sounds together have a dB of 81, we know:
81=10log(I/I0)
using the data above, we can find the intensity of the two sounds to be
0.000125 W/m^2
therefore, one firecracker has an intensity half of that, or 0.0000625W/m^2
now use this value to find the dB of one firecracker:
I(dB0=10log(0.0000625/10^-12)=77.96dB
Answer:
30.81°
Explanation:
θ₁ = angle of incidence = 50°
θ₂ = Angle of refraction
n₂ = Refractive index of glass = 1.5
n₁ = Refractive index of air = 1.0003
From Snell's Law
Using Snell's law as:
Angle of refraction= sin⁻¹ 0.5122 = 30.81°.
Answer:
Explanation:
For pendulum A: Length = L and gravity = g
The frequency of pendulum A is given by
Here, f is the frequency, L be the length
... (1)
For pendulum B: Length = 2L, gravity = g
The frequency of pendulum B is given by
.... (2)
Divide equation (1) by (2)
