For the answer to the question above, each horse's force forms a right angle triangle with the barge and subtends an angle of 60/2 = 30°. The resultant in the direction of the barge's motion is:
Fx = Fcos(∅)
We can multiply this by 2 to find the resultant of both horses.
Fx = 2Fcos(∅)
Fx = 2 x 720cos(30)
Fx = 1247 N
Main Answer:
Given speed of spaceship v = 0.8c = 0.8 * 3 x 10^8 m/s
Speed of spaceship v = 2.4 x 10^8 m/s
Distance need to be travelled d = 4.3 light years
we know that 1 light year = 9.461 x 10^15 m
Distance need to be travelled d = 4.3 x 9.461 x 10^15
d = 40.6823 x 10^15 m
Time taken for the trip would elapse on a clock on board the spaceship
t = distance/ velocity
t = 40.6823 x 10^15 / 2.4 x 10^8
t = 16.95 x 10^7 sec
t = 4.71 x 10^4 hours
Explanation:
What is light year?
Light year is defined as the distance travelled by the light in one year. In a year, light travels through 300000 km per sec in the interstellar space.
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False, when wavelength waves increases the frequency will decreases.
<h2>The frequency of tuning fork B is 475 Hz</h2>
Explanation:
The number of beats produced is equal to the frequency difference between the two tuning forks .
The frequency of A = 480 Hz
Thus the frequency of B can be = 485 or 475 Hz
When B is filed , its frequency increases and it start producing 2 beats with A .
Thus it frequency must be 475 Hz , by filing , it increases to 478 Hz .
By which it gives 480 - 478 = 2 beats per second .
This cannot happen with frequency 485 , because by filing this difference increases in place of decreasing .
Answer: 16N
Explanation:
According to coulombs law which states that the force of attraction between two charges is directly proportional to the product of the charges and inversely proportional to the square of the distance between them. Mathematically,
F = kq1q2/r²
If two charges that are separated by one meter exert 1-N forces on each other, the equation will become
1 = kq1q2/1²
kq1q2 = 1 ... (1)
If the charges are pushed together so the separation is 25 centimeters, the force between them becomes,
F = kq1q2/0.25² (25cm converted to meters)
0.0625F = kq1q2 ... (2)
Dividing equation 1 by 2 to determine the force F on each charges, it becomes;
1/0.0625F = kq1q2/kq1q2
1/0.0625F = 1
0.0625F = 1
F = 1/0.0625
F= 16N