Answer:
1.08 s
Explanation:
From the question given above, the following data were obtained:
Height (h) reached = 1.45 m
Time of flight (T) =?
Next, we shall determine the time taken for the kangaroo to return from the height of 1.45 m. This can be obtained as follow:
Height (h) = 1.45 m
Acceleration due to gravity (g) = 9.8 m/s²
Time (t) =?
h = ½gt²
1.45 = ½ × 9.8 × t²
1.45 = 4.9 × t²
Divide both side by 4.9
t² = 1.45/4.9
Take the square root of both side
t = √(1.45/4.9)
t = 0.54 s
Note: the time taken to fall from the height(1.45m) is the same as the time taken for the kangaroo to get to the height(1.45 m).
Finally, we shall determine the total time spent by the kangaroo before returning to the earth. This can be obtained as follow:
Time (t) taken to reach the height = 0.54 s
Time of flight (T) =?
T = 2t
T = 2 × 0.54
T = 1.08 s
Therefore, it will take the kangaroo 1.08 s to return to the earth.
In other words a infinitesimal segment dV caries the charge
<span>dQ = ρ dV </span>
<span>Let dV be a spherical shell between between r and (r + dr): </span>
<span>dV = (4π/3)·( (r + dr)² - r³ ) </span>
<span>= (4π/3)·( r³ + 3·r²·dr + 3·r·(dr)² + /dr)³ - r³ ) </span>
<span>= (4π/3)·( 3·r²·dr + 3·r·(dr)² + /dr)³ ) </span>
<span>drop higher order terms </span>
<span>= 4·π·r²·dr </span>
<span>To get total charge integrate over the whole volume of your object, i.e. </span>
<span>from ri to ra: </span>
<span>Q = ∫ dQ = ∫ ρ dV </span>
<span>= ∫ri→ra { (b/r)·4·π·r² } dr </span>
<span>= ∫ri→ra { 4·π·b·r } dr </span>
<span>= 2·π·b·( ra² - ri² ) </span>
<span>With given parameters: </span>
<span>Q = 2·π · 3µC/m²·( (6cm)² - (4cm)² ) </span>
<span>= 2·π · 3×10⁻⁶C/m²·( (6×10⁻²m)² - (4×10⁻²m)² ) </span>
<span>= 3.77×10⁻⁸C </span>
<span>= 37.7nC</span>
Answer:
Explanation:
Light rays coming from moon is blocked by the pencil
so as per figure we know that angle subtended by pencil and angle subtended by moon must be same
so we have
so we have
so we have
Answer:
a) From definition a transverse wave is which one where the elements moves perpendicular to the direction of the wave. For example is a wave is moving from the left to the right the elements would be wibrating or moving upward or downward.
We have a lot examples for a transverse wave. For example water waves, strings on the musical instruments , light and radio waves.
b) We can identify a transverse wave if the particles are displaced perpendicular to the direction of the wave. Usually these types of wave occur in elastic solids. And we can identify it when we see a pattern perpendicular between the wave direction and the particles motion. In simple words we need to see that the wave is moving down and up.
Explanation:
Part a
From definition a transverse wave is which one where the elements moves perpendicular to the direction of the wave. For example is a wave is moving from the left to the right the elements would be wibrating or moving upward or downward.
We have a lot examples for a transverse wave. For example water waves, strings on the musical instruments , light and radio waves.
Part b
We can identify a transverse wave if the particles are displaced perpendicular to the direction of the wave. Usually these types of wave occur in elastic solids. And we can identify it when we see a pattern perpendicular between the wave direction and the particles motion. In simple words we need to see that the wave is moving down and up.
