Answer:
41.27m/s
Explanation:
According to law of conservation of momentum
m1u1+m2u2 = (m1+m2)v
m1 and m2 are the masses
u1 and u2 are the initial velocities
v is the velocity after impact
Given
m1 = 0.2kg
u1 = 43.7m/s
m2 = 45.9g = 0.0459kg
u2 = 30.7m/s
Required
Velocity after impact v
Substitute the given parameters into the formula
0.2(43.7)+0.0459(30.7) = (0.2+0.0459)v
8.74+1.409 = 0.2459v
10.149 = 0.2459v
v = 10.149/0.2459
v = 41.27m/s
Hence the speed of the golf ball immediately after impact is 41.27m/s
The type of waves used by bats are sound waves. Most of the species use their larynx to produce ultrasound waves in the frequency range of 20 to 200 kilohertz.
These sound waves are echoed, reflected, by surroundings, in this case food or prey. These reflections are received by the specialized receptor cells in the ears of bats. The reflections are analyzed by the brain to make an image.
Fun fact: The brain cells of bats are also specialized to better analyze the frequency of ultrasound used by the bat.
Answer:
λ = 2042 nm
Explanation:
given data
screen distance d = 11 m
spot s = 4.5 cm = 4.5 ×
m
separation L = 0.5 mm = 0.5 ×
m
to find out
what is λ
solution
we will find first angle between first max and central bright
that is tan θ = s/d
tan θ = 4.5 × / 11
θ = 0.234
and we know diffraction grating for max
L sinθ = mλ
here we know m = 1 so put all value and find λ
L sinθ = mλ
0.5 × sin(0.234) = 1 λ
λ = 2042.02 ×
m
λ = 2042 nm
Answer:
<h2>1.5 ohms</h2>
Explanation:
Power is expressed as P = V²/R
R = resistance
V = supplied voltage
Given P = 600W and V = 30V
R = V²/P
R = 30²/600
R = 900/600
R = 1.5ohms
magnitude of its resistance is 1.5ohms
