Given :
A race car moves along a circular track of radius 100m at a velocity of 25m/s.
To Find :
(a) What is the time taken to complete one lap of the circular track.
(b) What is the time taken for 10 laps.
Solution :
Circumference of circular track,
[tex]C = 2\pi r\\\\C = 2\times \pi \times 100\ m\\\\C = 628.32 \ m[/tex]
a) Time taken to complete one lap is :
[tex]t= \dfrac{628.32}{25}\ s\\\\t =25.13 \ s[/tex]
b) Time taken to complete 10 laps is :
[tex]t_{10} = 10\times t\\\\t_{10} = 10\times 25.13\ s\\\\t_{10} = 251.3\ s[/tex]
Hence, this is the required solution.
How fast must a train accelerates from rest to cover 518 m in the first 7.48 s?
Heya!
For this problem, use the formula:
s = Vo * t + (at^2) / 2
Since the initial velocity is zero, the formula simplifies like this:
s = (at^2) / 2
Clear a:
2s = at^2
(2s) / t^2 = a
a = (2s) / t^2
Data:
s = Distance = 518 m
t = Time = 7,48 s
a = Aceleration = ¿?
Replace according formula:
a = (2*518 m) / (7,48 s)^2
Resolving:
a = 1036 m / 55,95 s^2
a = 23,34 m/s^2
The aceleration must be 23,34 meters per second squared
How will the motion of the arrow change after it leaves the bow?
The string moves to the right, as it restores its original position with the median plane of the bow. As a result, the string "pulls" on the arrow with a force F2. 2. The tip of the arrow T moves slightly to the left.
pls thank me and brainliest me
g Amateur radio operators in the United States can transmit on several bands. One of those bands consists of radio waves with a wavelength near . Calculate the frequency of these radio waves.
Answer:
7.5 × 10¹⁵ Hz
Explanation:
CHECK THE COMPLETE QUESTION BELOW;
Amateur radio operators in the United States can transmit on several bands. One of those bands consists of radio waves with a wavelength near 40nm.
Calculate the frequency of these radio waves.
The frequency of these radio waves can be calculated using the expression below
ν = c/λ
Where
v = Frequency of the radio waves
c= Speed of light= 3.00 × 10⁸ m/s
λ= Wavelength= 40 nm
= 40 × 10⁻⁹ m
ν = 3× 10⁸/40 × 10⁻⁹
= 7.5 × 10¹⁵ Hz
Hence, the frequency of these radio waves is 7.5 × 10¹⁵ Hz