<h2>
a) Displacement of penny = 1300 i + 2400 j - 640 k</h2><h2>b) Magnitude of his displacement = 2729.47 m</h2>
Explanation:
a) He walks 1300 m east, 2400 m north, and then drops the penny from a cliff 640 m high.
1300 m east = 1300 i
2400 m north = 2400 j
Drops the penny from a cliff 640 m high = -640 k
Displacement of penny = 1300 i + 2400 j - 640 k
b) Displacement of man for return trip = -1300 i - 2400 j
Magnitude of his displacement = 2729.47 m
Answer:
a) D_ total = 18.54 m, b) v = 6.55 m / s
Explanation:
In this exercise we must find the displacement of the player.
a) Let's start with the initial displacement, d = 8 m at a 45º angle, use trigonometry to find the components
sin 45 = y₁ / d
cos 45 = x₁ / d
y₁ = d sin 45
x₁ = d sin 45
y₁ = 8 sin 45 = 5,657 m
x₁ = 8 cos 45 = 5,657 m
The second offset is d₂ = 12m at 90 of the 50 yard
y₂ = 12 m
x₂ = 0
total displacement
y_total = y₁ + y₂
y_total = 5,657 + 12
y_total = 17,657 m
x_total = x₁ + x₂
x_total = 5,657 + 0
x_total = 5,657 m
D_total = 17.657 i^+ 5.657 j^ m
D_total = Ra (17.657 2 + 5.657 2)
D_ total = 18.54 m
b) the average speed is requested, which is the offset carried out in the time used
v = Δx /Δt
the distance traveled using the pythagorean theorem is
r = √ (d1² + d2²)
r = √ (8² + 12²)
r = 14.42 m
The time used for this shredding is
t = t1 + t2
t = 1 + 1.2
t = 2.2 s
let's calculate the average speed
v = 14.42 / 2.2
v = 6.55 m / s
Answer:
i just need points rn sorry
Explanation:
Answer:
The time for final 15 cm of the jump equals 0.1423 seconds.
Explanation:
The initial velocity required by the basketball player to be able to jump 76 cm can be found using the third equation of kinematics as
where
'v' is the final velocity of the player
'u' is the initial velocity of the player
'a' is acceleration due to gravity
's' is the height the player jumps
Since the final velocity at the maximum height should be 0 thus applying the values in the above equation we get
Now the veocity of the palyer after he cover'sthe initial 61 cm of his journey can be similarly found as
Thus the time for the final 15 cm of the jump can be found by the first equation of kinematics as
where symbols have the usual meaning
Applying the given values we get
Answer:
(A) 10132.5Pa
(B)531kJ of energy
Explanation:
This is an isothermal process. Assuming ideal gas behaviour then the relation P1V1 = P2V2 holds.
Given
m = 10kg = 10000g, V1 = 0.1m³, V2 = 1.0m³
P1 = 101325Pa. M = 102.03g/mol
P2 = P1 × V1 /V2 = 101325 × 0.1 / 1 = 10132.5Pa
(B) Energy is transfered by the r134a in the form of thw work done in in expansion
W = nRTIn(V2/V1)
n = m / M = 10000/102.03 = 98.01mols
W = 98.01 × 8.314 × 283 ×ln(1.0/0.1)
= 531kJ.
