the car travels 34 mi in one hour.
then, in 6 hours car travels
34 x 6 mi
= 204 mi
The motorbike reaches 100 km/h in 3.5 seconds
Explanation:
The motion of the motorbike is a uniformly accelerated motion (= constant acceleration), therefore we can use the following suvat equation:
where
v is the final velocity
u is the initial velocity
a is the acceleration
t is the time
For the motorbike in this problem,
u = 0 (it starts from rest)
is the final velocity
is the acceleration
Solving for t, we find the time it takes for the bike to reach that velocity:
Learn more about accelerated motion:
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Answer:
Explanation:
Given
Temperature of the gas is increased from 100 to 200
Also we know that average kinetic energy of the molecules is
Where
R=Gas constant
=Avogadro's number
T=Temperature in kelvin
So kinetic energy increases by
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:
Please find the answer in the explanation
Explanation:
When you calculate the SLOPE of a line segment, what does the SLOPE represent? (Choose all that apply) the Distance traveled the Displacement the Velocity the Acceleration None of the above
The slope of any time graph can not give you distance or displacement except for position - time graph.
When you plot either distance or displacement against time, that is, distance time graph or displacement time graph, you can get speed or velocity as the slope of the line segment.
You can only acceleration as a slope in a line of best fit if velocity is plotted against time. That is, in a velocity time graph.
