Answer:
polar
Explanation:
they said it was extreme so atleast it would be polar
Answer:
Approximately , assuming that .
Explanation:
Under the assumptions, the package would start with an initial upward velocity of and accelerate downward at a constant (negative because acceleration points downward.)
Right before landing, the package would be below where it was released. Hence, the displacement of the package at that moment would be (negative since this position is below the initial position.)
The duration of the motion can be found in the following steps:
Rearrange the SUVAT equation to find , the velocity of the package right before reaching the ground. Notice that because the package would be travelling downward, the value of should be negative.
.
Subtract the initial velocity from the new value to find the change in velocity. Divide this change in velocity by acceleration (rate of change in velocity) to find the duration of the motion:
.
Answer:
4.281 kgm/s upward
Explanation:
Impulse:This can be defined product of force and time. The S.I unit of impulse is Ns.
From Newton's second law of motion,
Impulse = Change in momentum.
I = mΔv....................... Equation 1.
Where m = mass of the ball, Δv = change in velocity of the ball
and Δv = v -u
Where u = velocity of the ball before it hit the floor, v = velocity of the ball after if hit the floor
I = m(v-u) -------------- Equation 2
But
the initial kinetic energy of the ball = potential energy at the initial height (1.2 m above)
1/2mu² = mgh₁
Where h₁ = initial height. or height of the ball before collision
making u the subject of the equation,
u = √(2gh₁)........................ Equation 3
Where h₁ = 1.2 m g = 9.81 m/s²
Substitute into equation 3
u = √(2×1.2×9.81)
u =√(23.544)
u = -4.852 m/s.
Note: u is negative because the ball was moving downward at the first instance.
Similarly,
v = √(2gh₂)............................. Equation 3
h₂ = height of the ball after collision
Given: h₂ = 0.7 m, g = 9.81 m/s²
Substitute into equation
v = √(2×9.81×0.7)
v = √13.734
v = 3.71 m/s.
Also given: m = 0.5 kg,
Substituting into equation 2
I = 0.5(3.71-(4.852)
I = 0.5(8.562)
I = 4.281 kgm/s. Upward.
Thus the impulse = 4.281 kgm/s upward
Answer:
C. work function
Explanation:
In the photoelectric effect, the energy of the incident photon is used in part to extract the electron from the metal (and this energy is called work function) and the rest is converted into kinetic energy of the electron. In formula:
where
hf is the energy of the incident photon, which is the product between h (the Planck constant) and f (the photon's frequency)
is the work function
K is the kinetic energy of the photoelectron as it leaves the material
tied together
Answer:
a=6m/sec²
Explanation:
f=5 N
a1=8m/sec²
a2=24m/sec
m1=f/a1=5/8kg
m2=f/a2=5/24kg
New, f=(m1+m2)a
a=5/5/8+5/24=5/15+5/24=24*5/20
a=6m/sec²