Answer:
velocity = 1527.52 ft/s
Acceleration = 80.13 ft/s²
Explanation:
We are given;
Radius of rotation; r = 32,700 ft
Radial acceleration; a_r = r¨ = 85 ft/s²
Angular velocity; ω = θ˙˙ = 0.019 rad/s
Also, angle θ reaches 66°
So, velocity of the rocket for the given position will be;
v = rθ˙˙/cos θ
so, v = 32700 × 0.019/ cos 66
v = 1527.52 ft/s
Acceleration is given by the formula ;
a = a_r/sinθ
For the given position,
a_r = r¨ - r(θ˙˙)²
Thus,
a = (r¨ - r(θ˙˙)²)/sinθ
Plugging in the relevant values, we obtain;
a = (85 - 32700(0.019)²)/sin66
a = (85 - 11.8047)/0.9135
a = 80.13 ft/s²
Answer: 16.22 m/s^2
Explanation: g= GM/r^2 G= (6.67x 10^-11) M= 1.66(6x 10^24) r=(6400x 10^3) so
((6.67x10^-11)(1.66x 6x 10^24))/ (6400x10^3)^2 = 16.22 m/s^2
Given the distance traveled and time elapsed, the average speed of the train is approximately 26.944m/s.
<h3>What is the average speed of the train?</h3>
Speed is simply referred to as distance traveled per unit time.
Mathematically, Speed = Distance ÷ time.
Given the data in the question;
- Distance traveled = 221miles
- Elapsed time = 3 hours and 40 minutes
First we convert miles to meters and Hours minutes to seconds.
221 miles = ( 221 × 1609.344 )m = 355665.024 meters
3 hours and 40 minutes = ( 3×60×60)s + ( 40×60)s
= 10800s + 2400s
= 13200s
Now, determine the average speed.
Speed = Distance ÷ time
Speed = 355665.024m / 13200s
Speed = 26.944m/s
Given the distance traveled and time elapsed, the average speed of the train is approximately 26.944m/s.
Learn more about speed here: brainly.com/question/7359669
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Answer:
v = 1.4 m /s
Explanation:
We shall apply law of conservation of mechanical energy
The kinetic energy of dart and block is converted into potential energy of both dart and block .
1 /2 (m+M) v² = ( m +M) gH
.5 x v² = 9.8 x .1
= v² = 1.96
v = 1.4
v = 1.4 m /s
Answer:
Explanation:
Let the magnetic field be B = B₁i + B₂j + B₃k
Force = I ( L x B ) , I is current , L is length and B is magnetic field .
In the first case
force = - 2.3 j N
L = 2.5 i
puting the values in the equation above
- 2.3 j = 8 [ 2.5 i x ( B₁i + B₂j + B₃k )]
= - 20 B₃ j + 20 B₂ k
comparing LHS and RHS ,
20B₃ = 2.3
B₃ = .115
B₂ = 0
In the second case
L = 2.5 j
Force = I ( L x B )
2.3i−5.6k = 8 ( 2.5 j x (B₁i + B₂j + B₃k )
= - 20 B₁ k + 20B₃ i
2.3i−5.6k = - 20 B₁ k + 20B₃ i
B₃ = .115
B₁ = .28
So magnetic field B = .28 i + .115 B₃
Part A
x component of B = .28 T
Part B
y component of B = 0
Part C
z component of B = .115 T .
