Answer:
The value is 
Explanation:
From the question we are told that
The frequency of each sound is 
The speed of the sounds is 
The distance of the first source from the point considered is 
The distance of the second source from the point considered is 
Generally the phase angle made by the first sound wave at the considered point is mathematically represented as
![\phi_a = 2 \pi [\frac{a}{\lambda} + ft]](https://tex.z-dn.net/?f=%5Cphi_a%20%3D%20%202%20%5Cpi%20%5B%5Cfrac%7Ba%7D%7B%5Clambda%7D%20%20%2B%20ft%5D)
Generally the phase angle made by the first sound wave at the considered point is mathematically represented as
![\phi_b = 2 \pi [\frac{b}{\lambda} + ft]](https://tex.z-dn.net/?f=%5Cphi_b%20%3D%20%202%20%5Cpi%20%5B%5Cfrac%7Bb%7D%7B%5Clambda%7D%20%20%2B%20ft%5D)
Here b is the distance o f the first wave from the considered point
Gnerally the phase diffencence is mathematically represented as
![\Delta \phi= \phi_a - \phi_b = 2 \pi [\frac{ a}{\lambda} + ft ] - 2 \pi [\frac{b}{\lambda} + ft ]](https://tex.z-dn.net/?f=%5CDelta%20%5Cphi%3D%20%5Cphi_a%20-%20%5Cphi_b%20%20%3D%20%202%20%5Cpi%20%5B%5Cfrac%7B%20a%7D%7B%5Clambda%7D%20%20%2B%20ft%20%5D%20-%202%20%5Cpi%20%5B%5Cfrac%7Bb%7D%7B%5Clambda%7D%20%20%2B%20ft%20%5D)
=> ![\Delta \phi = \frac{2\pi [ a - b]}{ \lambda }](https://tex.z-dn.net/?f=%5CDelta%20%20%5Cphi%20%20%20%3D%20%20%20%5Cfrac%7B2%5Cpi%20%5B%20a%20-%20b%5D%7D%7B%20%5Clambda%20%7D)
Gnerally the wavelength is mathematically represented as
=> 
=> 
=> ![\Delta \phi = \frac{2* 3.142 [ 4.40 - 4.0 ]}{ 0.611 }](https://tex.z-dn.net/?f=%5CDelta%20%20%5Cphi%20%20%20%3D%20%20%20%5Cfrac%7B2%2A%203.142%20%5B%204.40%20-%204.0%20%5D%7D%7B%20%200.611%20%20%7D)
=>
Answer:
-6.0 m/s, 10.4 m/s
Explanation:
To find the x- and y- components, we have to apply the formulas:
where
v = 12.0 m/s is the magnitude of the vector
is the angle between the direction of the vector and the positive x-axis
Here, the angle given is the angle above the negative x-axis; this means that the angle with respect to the positive x-axis is
So, the two components are:
Answer: 339.148N
Explanation:
Data
Time (t) = 47s
U = 0m/s
V = 9.5m/s
Mass of B = 540kg
Frictional force on B = 230N
Both boats are connected so if A moves, B moves too.
Acceleration of boat A =?
Using equation of motion,
V = u + at
9.5 = 0 + a*47
a = 9.5 / 47
a = 0.2021 m/s²
The force required to accelerate boat B since it's the same force moving both boats =?
F = Mass * acceleration
F = 540 * 0.2021 = 109.14N
A frictional force of 230N exists on boat B
Total force (Tension) = frictional force + normal force = (109.15 + 230)N = 339.148N
The force needed to the stop the car is -3.79 N.
Explanation:
The force required to stop the car should have equal magnitude as the force required to move the car but in opposite direction. This is in accordance with the Newton's third law of motion. Since, in the present problem, we know the kinetic energy and velocity of the moving car, we can determine the mass of the car from these two parameters.
So, here v = 30 m/s and k.E. = 3.6 × 10⁵ J, then mass will be
Now, we know that the work done by the brake to stop the car will be equal to the product of force to stop the car with the distance travelled by the car on applying the brake.Here it is said that the car travels 95 m after the brake has been applied. So with the help of work energy theorem,
Work done = Final kinetic energy - Initial kinetic energy
Work done = Force × Displacement
So, Force × Displacement = Final kinetic energy - Initial Kinetic energy.
Thus, the force needed to the stop the car is -3.79 N.
