Answer:
41.74 m/s
Explanation:
The energy used to draw the bowstring = the kinetic energy of the arrow.
Fd = 1/2mv²................................ Equation 1
Where F = force, d = distance move string, m = mass of the arrow, v = speed of the arrow.
make v the subject of the equation
v = √(2Fd/m)...................... Equation 2
Given: F = 201 N, m = 0.3 kg, d = 1.3 m.
Substitute into equation 2
v = √(2×201×1.3/0.3)
v = √(1742)
v = 41.74 m/s.
Hence the arrow leave the bow with a speed of 41.74 m/s
Answer:
number 3 The wind because the wind has strong pressure which causes to change peoples directions in where they are going
Explanation:
hope this helped by the way i think you answered b and 4 i don't know if this is a test and i sent this at the wrong time :(
Explanation:
It is given that,
Length of wire, l = 0.53 m
Current, I = 0.2 A
(1.) Approximate formula:
We need to find the magnitude of the magnetic field made by the current at a location 2.0 cm from the wire, r = 2 cm = 0.02 m
The formula for magnetic field at some distance from the wire is given by :


B = 0.000002 T

(2) Exact formula:


B = 0.00000199 T
or
B = 0.000002 T
Hence, this is the required solution.
To solve this problem it is necessary to apply the concepts related to the law of Malus which describe the intensity of light passing through a polarizer. Mathematically this law can be described as:

Where,
Indicates the intensity of the light before passing through the polarizer
I = Resulting intensity
= Indicates the angle between the axis of the analyzer and the polarization axis of the incident light
From the law of Malus when the light passes at a vertical angle through the first polarizer its intensity is reduced by half therefore

In the case of the second polarizer the angle is directly 60 degrees therefore



In the case of the third polarizer, the angle is reflected on the perpendicular, therefore, its angle of index would be

Then,



Then the intensity at the end of the polarized lenses will be equivalent to 0.09375 of the initial intensity.