We really can't tell from the given information.
We don't know HOW MUCH Marv enlarged his cannonballs,
or HOW MUCH faster Seymour's balls became.
If we assume that they both, let's say, DOUBLED something,
then Seymour accomplished more, and the destructive capability
of his balls has increased more.
I say that because the destructive capability of a cannonball is
pretty much just its kinetic energy when it arrives and hits the target.
Now, we all know the equation for kinetic energy.
K.E. = (1/2) (mass) (speed-SQUARED) .
We can see right away that if Marv started shooting balls with
double the mass but at the same speed, then they have double
the kinetic energy of the old ones.
But if Seymour started shooting the same balls with double the SPEED,
then they have (2-SQUARED) as much kinetic energy as they used to.
That's 4 times as much destructive capability as before.
So we can say that when it comes to cannons and their balls and
smashing things to bits and terrorizing your opponents, if making
a bigger mess is better, then more mass is better, but more speed
is better-squared.
Answer:
r = 0.114 m
Explanation:
To find the speed of the proton, from conservation of energy, we know that
KE = PE
Thus, we have;
(1/2)mv² = qV
Where;
V is potential difference = 1kv = 1000V
q is charge on proton which has a value of 1.6 x 10^(-19) C
m is mass of proton with a constant value of 1.67 x 10^(-27) kg
Let's make the velocity v the subject;
v =√(2qV/m)
v = √(2•1.6 x 10^(-19)•1000)/(1.67 x 10^(-27))
v = 4.377 x 10^(5) m/s
Now to calculate the radius of the circular motion of charge we know that;
F = mv²/r = qvB
Thus, mv²/r = qvB
Divide both sides by v;
mv/r = qB
Thus, r = mv/qB
Value of B from question is 0.04T
Thus,
r = (1.67 x 10^(-27) x 4.377 x 10^(5))/(1.6 x 10^(-19) x 0.04)
r = 0.114 m
r = 8.76 m
Answer:
The marble was moving in a projectile and the speed of the engine was 2.716 m/s
Explanation:
The vertical component of the marble's flight path relative to the train
is given by the equation y(t) = v*t - (4.9)*t^2,
where, v is the initial upward velocity of the marble relative to the train.
So with y(1) = v - 4.9 = 0 we have
v = 4.9 m/s.
The marble will reach maximum height after 0.5 seconds, at which the
height will be y(0.5) = (4.9)*(0.5) - (4.9)*(0.5)^2 = (4.9)*(0.25) = 1.225 m.
Now, the marble has a vertical velocity component of 4.9 m/s and a horizontal velocity component
of V m/s such that tan(61) = 4.9 / V
V = 4.9 / tan(61) = 2.716 m/s
This horizontal velocity component of the marble is the same as the
speed of the train i.e. 2.716 m/s.
Answer:
Explanation:
The free body diagram of the block on the slide is shown in the below figure
Since the block is in equilibrium we apply equations of statics to compute the necessary unknown forces
N is the reaction force between the block and the slide
For equilibrium along x-axis we have
Using value of N from equation β in α we get value of force as
Applying values we get
