Answer:
The speed in the first point is: 4.98m/s
The acceleration is: 1.67m/s^2
The prior distance from the first point is: 7.42m
Explanation:
For part a and b:
We have a system with two equations and two variables.
We have these data:
X = distance = 60m
t = time = 6.0s
Sf = Final speed = 15m/s
And We need to find:
So = Inicial speed
a = aceleration
We are going to use these equation:
We are going to put our data:
With these equation, you can decide a method for solve. In this case, We are going to use an egualiazation method.
![[\sqrt{(15m/s)^2-(2*a*60m)}]^{2}=[15m/s-(a*6s)]^{2}](https://tex.z-dn.net/?f=%5B%5Csqrt%7B%2815m%2Fs%29%5E2-%282%2Aa%2A60m%29%7D%5D%5E%7B2%7D%3D%5B15m%2Fs-%28a%2A6s%29%5D%5E%7B2%7D)
If we analyze the situation, we need to have an aceleretarion greater than cero. We are going to choose a = 1.67m/s^2
After, we are going to determine the speed in the first point:
For part c:
We are going to use:
Answer:
the wave length becomes doubled or becomes two times the initial wavelength = 240 cm
Explanation:
From wave,
v = λf................ Equation 1
Where v = velocity of the wave, λ = wavelength of the wave, f = frequency of the wave.
Given: f = 1200 Hz, λ = 120 cm = 1.2 m
Substitute into equation 1
v = 1200(1.2)
v = 1440 m/s.
When the ship sent out a 600 Hz sound wave,
make λ the subject of formula in equation 1
λ = v/f............. Equation 2
Given: f = 600 Hz, v = 1440 m/s
Substitute into equation 2
λ = 1440/600
λ = 2.4 m or 240 cm.
When the ship sent out a 600 Hz sound wave instead, the wave length becomes doubled or becomes two times the initial wavelength = 240 cm
The magnitude of the force<span> a 1.5 x 10-3 C charge exerts on a 3.2 x 10-4 C charge located 1.5 m away is 1920 Newtons. The formula used to solve this problem is:
F = kq1q2/r^2
where:
F = Electric force, Newtons
k = Coulomb's constant, 9x10^9 Nm^2/C^2
q1 = point charge 1, C
q2 = point charge 2, C
r = distance between charges, meters
Using direct substitution, the force F is determined to be 1920 Newtons.</span>
Answer:
b ≈ 64 Kg/s
Explanation:
Given
Fd = −bv
m = 2.5 kg
y = 6.0 cm = 0.06 m
g = 9.81 m/s²
The object in the pan comes to rest in the minimum time without overshoot. this means that damping is critical (b² = 4*k*m).
m is given and we find k from the equilibrium extension of 6.0 cm (0.06 m):
∑Fy = 0 (↑)
k*y - W = 0 ⇒ k*y - m*g = 0 ⇒ k = m*g / y
⇒ k = (2.5 kg)*(9.81 m/s²) / (0.06 m)
⇒ k = 408.75 N/m
Hence, if
b² = 4*k*m ⇒ b = √(4*k*m) = 2*√(k*m)
⇒ b = 2*√(k*m) = 2*√(408.75 N/m*2.5 kg)
⇒ b = 63.9335 Kg/s ≈ 64 Kg/s
Answer:
X= 700 Joules
Explanation:
The question asked about the efficiency of the work done.
The formula for efficiency is: Efficiency = (Useful output / input work) * 100%
The useful output given in the question is 140J, the question asked for input work. Let X be the input work. It is also given that the efficiency is 20%.
Using the formula of efficiency,
20 = (140/X) * 100
So, we simply solve the above equation.
X= 140*100/20
X= 700 Joules
