Answer:
Explanation:
Take note that when the fuel of the rocket is consumed, the acceleration would be zero. However, at this phase the rocket would still be moving up until all the forces of gravity would dominate and change the direction of the rocket. Hence, there will be a need to calculate two distances, one from the ground until the point where the fuel is consumed and from that point to the point where the gravity would change the direction.
Given:
a = 76.8 m/s^2
t = 1.99 s
Solution:
d = vi (t) + 0.5 (a) (t^2)
d = (0) (1.99) + 0.5 (76.8) (1.99)^2
d = 0+38.4×3.9601
d = 152.068m
vf = vi + at
vf = 0 + 76.8 (1.99)
vf = 152.83 m/s (velocity when the fuel is consumed completely)
Then, we calculate the time it takes until it reaches the maximum height.
vf = vi + at
0 = 152.83+(-9.8) (t)
0 = 152.83 + (-9.8) (t)
-152.83 = -9.8t
t = 152.83/9.8 s
t = 15.59s
Then, the second distance
d= vi (t) + 0.5 (a) (t^2)
d = 152.83 (15.59) + 0.5 (-9.8) (15.59^2)
d = 2382.6197- 1190.93
d = 1191.68m
Then, we determine the maximum altitude:
d1 + d2 = 152.068 m + 1191.68m = 1343.748m
Answer:
49.3 N
Explanation:
Given that Pulling up on a rope, you lift a 4.25 kg bucket of water from a well with an acceleration of 1.80 m/s2 . What is the tension in the rope?
The weight of the bucket of water = mg.
Weight = 4.25 × 9.8
Weight = 41.65 N
The tension and the weight will be opposite in direction.
Total force = ma
T - mg = ma
Make tension T the subject of formula
T = ma + mg
T = m ( a + g )
Substitutes all the parameters into the formula
T = 4.25 ( 1.8 + 9.8 )
T = 4.25 ( 11.6 )
T = 49.3 N
Therefore, the tension in the rope is 49.3 N approximately.
Answer:
f = 6.57
×
10
¹⁵H
z
Explanation:
Given:
r
=
0.053
n
m = 5.3
×
10
⁻¹¹m is the electron orbit radius
In order to determine the frequency of the electron motion, we first equate the Coulomb force due to the electrostatic attraction between proton and electron to the centripetal force of the circular motion. We have:
mv²/r² = kq²/r²
Here we want to get v so we isolate it:
v
=
√
k
q
²/m
r
We then recall that the speed in uniform circular motion is:
v
=
2
π
r
f
So we replace everything and substitute:
2
π
r
f =
√
k
q
²/
m
r
We isolate the frequency here:
f
=
(
1
/2
π
r
) × (
√
k
q
²/m
r
)
We insert the term inside the square root to get a singular term of the form:
f
=
√
k
q
²/4
π
²m
r
³
We substitute and use the charge and mass of an electron here:
f
=
⎷
(
9
×
10
⁹
N
m
²/
C
²)
(
1.602
×
10
⁻¹⁹C
)²/4
π
²
(
9.109
×
10
⁻³¹k
g
)
(
5.3
×
10
⁻¹¹m
)
³
We will get:
f = 6.57
×
10
¹⁵H
z
Answer:
m₂ = 18 kg
Explanation:
Conceptual analysis
We apply Newton's second law on masses m₁ and m₂
∑F = m × a Formula (1)
∑F: algebraic sum of forces in Newton (N)
m: mass in kg
a: acceleration in m/s²
T: cord tension in Newton (N)
Known data
m₁ = 12 kg
F = 66 N
a₁ = a₂ = 2.2 m/s²
Problem development
We can see in the attached figure the free body diagram of the masses and we replace the data in formula (1), considering that positive forces are the ones that go in the same direction of acceleration::
Mass 1 (m₁) free body diagram:
∑Fₓ = m₁ × a₁
66 - T = 12 × 2.2
66 - T = 26.4
66 - 26.4 = T
T = 39.6 N
Mass 2 (m₂) free body diagram:
∑Fₓ = m₂ × a₂
T = m₂ × 2.2
39.6 = m₂ × 2.2
m₂ = 39.6 ÷ 2.2
m₂ = 18 kg
