Answer:
8.85m/s
Explanation:
The potential energy the watermelon held before dropping is Ep=mgh=2*9.8*4=78.4J.
When it strikes the ground, all of its Ep will transfer into Ek, so 1/2*m*v^2=78.4.
We already knew that m=2, so insert that in, we will get the V^2=78.4 m/s, V=8.85 m/s
Answer:
Approximately
(assuming that the projectile was launched at angle of
above the horizon.)
Explanation:
Initial vertical component of velocity:
.
The question assumed that there is no drag on this projectile. Additionally, the altitude of this projectile just before landing
is the same as the altitude
at which this projectile was launched:
.
Hence, the initial vertical velocity of this projectile would be the exact opposite of the vertical velocity of this projectile right before landing. Since the initial vertical velocity is
(upwards,) the vertical velocity right before landing would be
(downwards.) The change in vertical velocity is:
.
Since there is no drag on this projectile, the vertical acceleration of this projectile would be
. In other words,
.
Hence, the time it takes to achieve a (vertical) velocity change of
would be:
.
Hence, this projectile would be in the air for approximately
.
Answer:
c. initial (x and y)
Explanation:
When a projectile is launched at a velocity with a launch angle, to solve it, we must first resolve the initial velocity into the x and y components. To do this will mean we have to treat it like a triangle due to the launch angle and the direction of the projectile.
Therefore, we will have to make use of trigonometric ratios which is also known by the mnemonic "SOH CAH TOA"
Thus, this method resolves the initial x and y velocities.
R = U : I. U is in Voltage and I is in Ampère. That gives you R = 36 : 8 = 4,5 Ohm