A traffic light weighing 200N hangs from a vertical cable tied to two other cables that are fastened to to a support ,as shown . The upper cables make angles 41° and 63° with the horizontal . Calculate the tension in of the three cables
93.6 N in the 41° cable
155.6 N in the 63° cable
200 N in the vertical cable
Let T and U represent the tensions in the 41° and 63° cables, respectively. In order for the system to be stationary, the horizontal components of these tensions must balance, and the vertical components of these tensions must total 200 N.
Assume that the speed of light in a vacuum has the hypothetical value of 18.0 m/s. A car is moving at a constant speed of 14.0 m/s along a straight road. A home owner sitting on his porch sees the car pass between two telephone poles in 8.89 s. How much time does the driver of the car measure for his trip between the poles?
Observed time, t = 5.58 s
Speed of light in a vacuum has the hypothetical value of, c = 18 m/s
Speed of car, v = 14 m/s along a straight road.
A home owner sitting on his porch sees the car pass between two telephone poles in 8.89 s.
We need to find the time the driver of the car measure for his trip between the poles. The relation between real and observed time is given by :
t is observed time.
So, the time observed by the driver of the car measure for his trip between the poles is 5.58 seconds.
As the captain of the scientific team sent to Planet Physics, one of your tasks is to measure g. You have a long, thin wire labeled 1.73 g/m and a 1.30 kg weight. You have your accurate space cadet chronometer but, unfortunately, you seem to have forgotten a meter stick. Undeterred, you first find the midpoint of the wire by folding it in half. You then attach one end of the wire to the wall of your laboratory, stretch it horizontally to pass over a pulley at the midpoint of the wire, then tie the 1.30 kg weight to the end hanging over the pulley. By vibrating the wire, and measuring time with your chronometer, you find that the wire's second harmonic frequency is 200 Hz . Next, with the 1.30 kg weight still tied to one end of the wire, you attach the other end to the ceiling to make a pendulum. You find that the pendulum requires 313 s to complete 200 oscillations. Pulling out your trusty calculator, you get to work.What value of g will you report back to headquarters?
The value of g is
From the question we are told that
The mass of the weight is
The spring constant
The second harmonic frequency is
The number of oscillation is
The time taken is
Generally the frequency is mathematically represented as
At second harmonic frequency the length of the string vibrating is equal to the wavelength of the wave generated
Noe from the question the vibrating string is just half of the length of the main string so
Let assume the length of the main string is
The velocity of the vibrating string is mathematically represented as
Where T is the tension on the string which can be mathematically represented as
Generally the period of oscillation is mathematically represented as
The period can be mathematically evaluated as
substituting for L
Three small objects are arranged along a uniform rod of mass m and length L. one of mass m at the left end, one of mass m at thecenter, and one of mass 2m at the right end. How far to the left or right of the rod's center should you place a support so that the rod with the attached objects will balance there?
See answer below
To get started, recall the Center of Mass formula for two masses:
where m is mass and x is displacement from the center of the shape.
Since masses at the center of a geometric shape have a displacement (x) value of 0, as the mass is already of the center, and does not affect Xcm. So, we can disregard the central mass, hence we use the above formula for two masses.
We can arbitrarily define left to be a negative (-) displacement, and vice versa for right direction. We proceed with the formula:
Since we defined left (-) and right (+), we notice the center of mass is (+) value. This makes sense, as there is slightly more mass on the right side. Hence, you should place a support 1/6 of the rod's length away from the rod's center.
Study well and persevere.
To balance the rod with the attached objects, place a support at a distance of L/3 from the left end of the rod.
To balance the rod with the attached objects, you need to place a support at a distance of L/3 from the left end of the rod. This is because the center of gravity of the system should be directly above the support.
The center of gravity is given by the equation xcg = (m*0 + m*L/2 + 2m*L)/ (m + m + 2m). Solving this equation, we get xcg = 2L/3. Therefore, the support should be placed at a distance of L/3 from the left end of the rod.
Learn more about Balancing a rod with objects here:
Where T₁ = initial, m₁ = initial mass, T₂ = final period, m₂ = final mass.
Given: T₁ = 1.18 s m₁ = 0.50 kg, T₂ = 2.07 s.
Substituting into equation 4
m₂ = (2.07)²(0.5)/(1.18)²
m₂ = 4.285(1.392)
m₂ = 5.965 kg.
Added mass = m₂ - m₁
Added mass = 5.965 - 0.5
Added mass = 5.465 kg.
Thus the add mass = 5.465 kg
) An electron moving along the x-axis enters a magnetic field. If the electron experiences a magnetic deflection in the -y direction, what is the direction of the magnetic field in this region
- z direction
To find the direction of the magnetic field, you take into account that the magnetic force over a charge, is given by the following cross product:
F_B: magnetic force
q: charge of the particle
v: velocity of the charge
B: magnetic field
In this case you have that the electron is moving along x-axis. You can consider this direction as the ^i direction. The electron experiences a magnetic deflection in the -y direction, that is, in the -^j direction.
By the cross products between unit vectors, you have that:
-^j = ^i X ^k
That is, the cross product between two vectors, one in the +x direction, and another one in the +z direction, generates a vector in the -y direction. However, it is necessary to take into account that the negative charge of the electron change the sign of the result of the cross product, which demands that the second vector is in the -z direction. That is:
-^i X -k^ = ^i X ^k = - ^j
Hence, the direction of the magnetic field is in the -z direction