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olganol [36]
2 years ago
14

Light and radio waves travel through a vacuum in a straight line at a speed of very nearly 3.00 ´ 108 m/s. How far is light year

(the distance light travels in a year)?
Physics
1 answer:
mezya [45]2 years ago
7 0
<span>We need to calculate the number of seconds in one year. time = (365 days) (24 hours/day) (3600 seconds/hour) time = 31536000 seconds We can calculate the distance that light travels in this many seconds. distance = speed x time distance = (3.00 x 10^8 m/s) (31536000 seconds) distance = 9.46 x 10^{15} meters A light year is 9.46 x 10^{15} meters</span>
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A 6.70 −μC particle moves through a region of space where an electric field of magnitude 1500 N/C points in the positive x direc
Alborosie

The given question is incomplete. The complete question is as follows.

A 6.70 −μC particle moves through a region of space where an electric field of magnitude 1500 N/C points in the positive x direction, and a magnetic field of magnitude 1.25 T points in the positive z direction.

A) If the net force acting on the particle is 6.21 \times 10^{-3} N in the positive x direction, find the components of the particle's velocity. Assume the particle's velocity is in the x-y plane.

Enter your answers numerically separated by commas.

Explanation:

The given data is as follows.

           Q = 6.50 \times 10^{-6} C

           E = 1300 N/C in the +x direction

           B = 1.02 T in the +z direction

and,    F_{net} = 6.25 \times 10^{-3} N in the +x direction

Also,       F_{net} = F_{E} - F_{b}

                         = qE - qvB

Now, we will calculate the value of v as follows.

             v = (\frac{1}{B}) \times (E - \frac{F_{net}}{q})

                 = (\frac{1}{1.02 T}) \times (1300 - \frac{6.25 \times 10^{-3}}{6.50 \times 10^{-6}})

                v = 458.507 m/s

Using the value for velocity, we need to know which direction it's going.

You know +x direction for E, +z direction for B and +x for F_{net}.

Using the right hand rule where:

your right thumb goes toward the F_{net}, then your index finger points to B (z direction) Then curl your middle, ring, and pink 90 angle. This shows where v is going which is -y direction.

Thus, we can conclude that v_{x}, v_{y}, v_{z} = 0, -(458.507), 0.

8 0
2 years ago
The actions of an employee are not attributable to the employer if the employer has not directly or indirectly encouraged the em
zepelin [54]

Answer:    the answer is d

Explanation: there are not more than 10 violations  within a twelve month period hope this helps

4 0
2 years ago
a crane lifts a 35000 N steel girder a distance of 25 m in 45 s. How much power did the crane require to lift the girder
Angelina_Jolie [31]
Hello.

The formula for Power is Work divided by Time; however, we do not have our value for Work - yet.
To find for the Work inputted, we need to use its formula: Force * Distance.

Let's multiply our Force by our Distance. Remember that our Force is always  measured in Newtons (N), and our Distance is measured by Meters (M).
35,000 * 25 = 875,000 J (Unit for Work is "J" or "Joules")

Now that we have the value for Work, let's apply it to our Power formula.
P = 875,000 / 45; 19,444.44~

The Power required to lift the girder is 1944.44~ W (Unit for Power is "W" or "Watts").

I hope this helps!
7 0
2 years ago
A battleship simultaneously fires two shells toward two identical enemy ships. One shell hits ship A, which is close by, and the
luda_lava [24]

Answer:

both cannonballs hit the ships with the same horizontal speed

Explanation:

Hello!

A parabolic motion is characterized in that its vertical component in Y is constantly changing, this is due to the constant downward acceleration of gravity.

When the movement starts the speed at Y is maximum, then when it reaches its maximum height point its speed is zero, and finally it begins to increase downwards until it touches the floor.

On the other hand, the horizontal speed remains constant AS THERE IS NO ACCELERATION IN HORIZONTAL DIRECTION.

therefore both cannonballs hit the ships with the same horizontal speed

regards!

8 0
2 years ago
A 0.900 kg ornament is hanging by a 1.50 m wire when the ornament is suddenly hit by a 0.400 kg missile traveling horizontally a
just olya [345]

Explanation:

The given data is as follows.

  Mass of the ornament (m_{1}) = 0.9 kg

  Length of the wire (l) = 1.5 m

 Mass of missile (m_{2}) = 0.4 kg

 Initial speed of missile (u_{2}) = 12 m/s

         r = 1.5 m

According to the law of conservation of momentum,

                   p_{i} = p_{f}  

     m_{1}u_{1} + m_{2}u_{2} = (m_{1} + m_{2})v

Putting the given values into the above formula as follows.

          m_{1}u_{1} + m_{2}u_{2} = (m_{1} + m_{2})v

         0.9 \times 0 + 0.4 \times 12 = (0.9 + 0.4)v

              0 + 4.8 = 1.3v

                  v = 3.69 m/s

Now, the centrifugal force produced is calculated as follows.

            F_{c} = (m_{1} + m_{2}) \times \frac{v^{2}}{r}

                       = (0.9 + 0.4) \times \frac{(3.69)^{2}}{1.5}

                       = 11.80 N

Hence, tension in the wire is calculated as follows.

              T = F_{c} + (m_{1} + m_{2})g

                 = 11.80 N + (0.9 + 0.4) \times 9.8

                 = 24.54 N

Thus, we can conclude that tension in the wire immediately after the collision is 24.54 N.

4 0
2 years ago
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