Answer:
The kinetic energy of the mass at the instant it passes back through the equilibrium position is 0.06500 J.
Explanation:
Given that,
Mass = 2.15 kg
Distance = 0.0895 m
Amplitude = 0.0235 m
We need to calculate the spring constant
Using newton's second law
Where, f = restoring force
Put the value into the formula
We need to calculate the kinetic energy of the mass
Using formula of kinetic energy
Here, 
Here, 
Put the value into the formula
Hence, The kinetic energy of the mass at the instant it passes back through the equilibrium position is 0.06500 J.
False, you pass a light through a mixture If the light bounces off the particles, you will see the light shine through and you have a colloid mixture
The final speed of an airplane is v = 92.95 m/s
The rate of change of position of an object in any direction is known as speed i.e. in other word, Speed is measured as the ratio of distance to the time in which the distance was covered.
Solution-
Here given,
Acceleration a= 10.8 m/s2 .
Displacement (s)= 400m
Then to find final speed of airplane v=?
Therefore from equation of motion can be written as,
v²=u²+ 2as
where, u is initial speed, v is final speed ,a is acceleration and s is displacement of the airplane. Therefore by putting the value of a & s in above equation and (u =0) i.e. the initial speed of airplane is zero.
v²= 2×10.8 m/s²×400m
v²=8640m/s
v=92.95m/s
hence the final speed of airplane v =92.95m/s
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Answer:
t = 2.2 s
Explanation:
Given that,
A person observes a firework display for A safe distance of 0.750 km.
d = 750 m
The speed of sound in air, v = 340 m/s
We need to find the between the person see and hear a firework explosion. let it is t. So, using the formula of speed.
So, the required time is 2.2 seconds.
The one fact that needs to be mentioned but isn't given anywhere on or around the graph is: The distance, on the vertical axis, is the distance FROM home. So any point on the graph where the distance is zero ... the point is in the x-axis ... is a point AT home.
Segment D ...
Walking AWAY from home; distance increases as time increases.
Segment B ...
Not walking; distance doesn't change as time increases.
Segment C ...
Walking away from home, but slower than before; distance increases as time increases, but not as fast. Slope is less than segment-D.
Segment A ...
Going home; distance is DEcreasing as time increases. Walking pretty fast ... the slope of the line is steep.
