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
Loudness of a sound wave is directly proportional to the intensity of the sound wave. In other words, when one increases, other also increases and vice-versa
Hope this helps!
Electron<span>. the central part of an atom containing </span>protons<span> and </span>neutrons<span> ... which of the following is necessary to calculate the atomic </span>mass<span> of an element? ... which of the </span>statements correctly compares<span>the relative size of an ion to its neutral atom?</span>
Use the equation I=V/R where I is current and V is the voltage plus R is the resistance so when voltage is the highest and resistance is lowest the current is the highest
Answer:
The mass of the beam is 0.074 kg
Explanation:
Given;
length of the uniform bar, = 1m = 100 cm
Set up this system with the given mass and support;
0-----------------33cm-----------------------------------100cm
↓ Δ ↓
0.15kg m
Where;
m is mass of the uniform bar
Apply the principle of moment to determine the value of "m"
sum of anticlockwise moment = sum of clockwise moment
0.15kg(33 - 0) = m(100 - 33)
0.15(33) = m(67)

Therefore, the mass of the beam is 0.074 kg