Write as ordinary number 3 x 10^0


Answer 1




10⁰ = 1 because anything to the power of 0 is 1.

3×1= 3

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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.

  Tcos(41°) =Ucos(63°) . . . . . balance of horizontal components

  U = Tcos(41°)/cos(63°) . . . . write an expression for U


The vertical components must total 200 N, so we have ....

  Tsin(41°) +Usin(63°) = 200

  Tsin(41°) +Tcos(41°)sin(63°)/cos(63°) = 200

  T(sin(41°)cos(63°) +cos(41°)sin(63°))/cos(63°) = 200

  T = 200cos(63°)/sin(41° +63°) ≈ 93.6 . . . newtons

  U = 200cos(41°)/sin(41° +63°) ≈ 155.6 . . . newtons


The vertical cable must have sufficient tension to balance the weight of the traffic light, so its tension is 200 N.

Then the tensions in the 3 cables are ...

  41°: 93.6 N

  63°: 155.6 N

  90°: 200 N

The tension in each of the three cables are 94.29, 155.56 and 200 Newton respectively.

Given the following data:

  • Force = 200 Newton.
  • Angle 1 = 41°
  • Angle 2 = 63°

How to calculate the tension.

First of all, we would determine the third tension force based on the vertical component as follows:

\sum F_y = 0\n\nT_3 - F_g =0\n\nT_3 - F_g=200\;N

Next, we would apply Lami's theorem to resolve the forces acting on the traffic light at equilibrium:

For the horizontal component:

\sum F_x = -T_1cos41+T_2cos 63=0\n\n0.7547T_1=0.4540T_2\n\nT_1=(0.4540T_2)/(0.7547)\n\nT_1 = 0.6016T_2   ....equation 1.

For the vertical component:

\sum F_y = T_1sin41+T_2sin 63-T_3=0\n\n\sum F_y = T_1sin41+T_2sin 63-200=0\n\n0.6561T_1+0.8910T_2 =200   ...equation 2.

Substituting eqn. 1 into eqn. 2, we have:

0.6561 * (0.6016T_2)+0.8910T_2 =200\n\n0.3947T_2+0.8910T_2 =200\n\n1.2857T_2 =200\n\nT_2 = (200)/(1.2857) \n\nT_2 = 155.56\;Newton

For the first tension:

T_1 = 0.6061T_2\n\nT_1 = 0.6061 * 155.56\n\nT_1 = 94.29\;Newton

Read more on tension here: brainly.com/question/4080400

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  


Given that,

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=\frac{t}{\sqrt{1-(v^2)/(c^2)} }

t is observed time.

t=T* \sqrt{1-(v^2)/(c^2)} \n\nt=8.89* \sqrt{1-(14^2)/(18^2)} \n\nt=5.58\ s

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  g =76.2 m/s^2


From the question we are told that

     The mass of the weight is m =  1.30 kg

      The spring  constant  k =  1.73 g/m = 1.73 *10^(-3) \ kg/m

       The second harmonic frequency is f =  100 \ Hz

       The number of oscillation is N  =  200

        The time taken is  t =  315 \ s

Generally the frequency is  mathematically represented as

           f =  (v)/(\lambda)

At second harmonic frequency the length of the string vibrating is equal to  the wavelength of the wave generated

         l  =  \lambda

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  L

So      l =  (L)/(2)

The velocity of the vibrating string is mathematically represented as

             v  =  \sqrt{(T)/(\mu) }

Where T is the tension on the string which can be mathematically represented as

             T  = mg


           v =  \sqrt{(mg)/(k) }


          f =  (v)/((L)/(2) )

=>       v =  (fL )/(2)

=>      \sqrt{(mg)/(k) } =  (fL)/(2)

=>        g  =  (f^2 L^2 \mu)/(4m)

substituting values

             g =  ((100) * (1.73 *10^(-3) ))/((4 * 1.30))  L^2

              g =   3.326  m^(-1) s^(-2) L^2

Generally the period of oscillation is mathematically represented as

       T_p  =  2 \pi \sqrt{(L)/(g) }

=>   L  =  (T^2 g)/(4 \pi ^2)

   The period can be mathematically evaluated as

                T_p  =  (t)/(N)

 substituting values

             T_p  =  (315)/(200)

             T_p  = 1.575 \ s


          L = (1.575^2 * g )/(4 \pi ^2)

           L = 0.0628 ^2 g


      g =   3.326  m^(-1) s^(-2) L^2

substituting for L

        g =   3.326   ((0.0628) g)^2

=>    g = (1)/((3.326)* (0.0628)^2)

       g =76.2 m/s^2

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


Hi there,

To get started, recall the Center of Mass formula for two masses:

x_c_m = (m_1x_1+m_2x_2)/(m_1+m_2)  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:x_c_m=((-L/2)m+(L/2)2m)/(m+2m) =((L/2)(-m+2m))/(3m) \n x_c_m=(L(m))/(6m) =(L)/(6)

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.


Final answer:

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:



A 0.500-kg mass suspended from a spring oscillates with a period of 1.18 s. How much mass must be added to the object to change the period to 2.07 s?



The add mass = 5.465 kg


Note: Since the spring is the same, the length and Tension are constant.

f ∝ √(1/m)........................ Equation 1  (length and Tension are constant.)

Where f = frequency, m = mass of the spring.

But f = 1/T ..................... Equation 2

Substituting Equation 2 into equation 1.

1/T ∝ √(1/m)


T ∝ √(m)


T₁/√m₁ = k

where k = Constant of proportionality.

T₁/√m₁ = T₂/√m₂ ........................ Equation 3

making m₂ the subject of the equation

m₂ = T₂²(m₁)/T₁²........................... Equation 4

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:

\vec{F_B}=q\vec{v}\ X\ \vec{B}      (1)

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