Answers
Final answer:
The force of gravity that the space shuttle experiences is 9.8 x 10^5 Newtons.
Explanation:
To calculate the force of gravity that the space shuttle experiences, we can use the equation F = mg, where F represents the force of gravity, m is the mass of the object, and g is the acceleration due to gravity (approximately 9.8 m/s² on Earth). In this case, the mass of the space shuttle is given as 1.0 x 10^5 kg. However, we need to convert the altitude of the shuttle into meters, so 200.0 km becomes 200,000 meters.
Now we can calculate the force of gravity:
F = (1.0 x 10^5 kg)(9.8 m/s²)
F = 9.8 x 10^5 N
Therefore, the space shuttle experiences a force of gravity of 9.8 x 10^5 Newtons.
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Answers
-20t+ 80
Answers
Answer:
Fundamental frequency= 174.5 hz
Explanation:
We know
fundamental frequency=
velocity =
mass per unit length==0.00427
Now calculating velocity v=
=244.3
Distance between two nodes is 0.7 m.
Plugging these values into to calculate frequency
f = =174.5 hz
Answers
Explanation:
The relation between resistance and resistivity is given by :
is resistivity of material
l is length of wire
A is area of cross section of wire
Resistivity of a material is the hidden property. If one wire has 3 times the length of the other, then it doesn't affect its resistivity. Hence, the resistivity of two wires is
Answers
Answer:
Explanation:
Let m be the mass of the object and v be the maximum velocity . The tension will provide centripetal force for the circular motion .
T = mv² / R where R is radius of circular path . T is tension .
putting the values given in the equation above
26.9 = m v² / 1
m v² = 26.9
kinetic enrgy = 1/2 m v²
= 26.9 / 2
= 13.45 J
13 J .
Maximum kinetic energy = 13 J .
Answers
Answer:
3.44 rad
Explanation:
The rotational kinetic energy change of the disk is given by ΔK = 1/2I(ω² - ω₀²) where I = rotational inertia of solid sphere = MR²/2 where m = mass of solid disk = 4 kg and R = radius of solid disk = 4 m, ω₀ = initial angular speed of disk = 0 rad/s (since it starts from rest) and ω = final angular speed of disk
Since the kinetic energy is increasing at a rate of 21 J/s, the increase in kinetic energy in 3.3 s is ΔK = 21 J/s × 3.3 s = 69.3 J
So, ΔK = 1/2I(ω² - ω₀²)
Since ω₀ = 0 rad/s
ΔK = 1/2I(ω² - 0)
ΔK = 1/2Iω²
ΔK = 1/2(MR²/2)ω²
ΔK = MR²ω²/4
ω² = (4ΔK/MR²)
ω = √(4ΔK/MR²)
ω = 2√(ΔK/MR²)
Substituting the values of the variables into the equation, we have
ω = 2√(ΔK/MR²)
ω = 2√(69.3 J/( 4 kg × (4 m)²))
ω = 2√(69.3 J/[ 4 kg × 16 m²])
ω = 2√(69.3 J/64 kgm²)
ω = 2√(1.083 J/kgm²)
ω = 2 × 1.041 rad/s
ω = 2.082 rad/s
The angular displacement θ is gotten from
θ = ω₀t + 1/2αt² where ω₀ = initial angular speed = 0 rad/s (since it starts from rest), t = time of rotation = 3.3 s and α = angular acceleration = (ω - ω₀)/t = (2.082 rad/s - 0 rad/s)/3.3 s = 2.082 rad/s ÷ 3.3 s = 0.631 rad/s²
Substituting the values of the variables into the equation, we have
θ = ω₀t + 1/2αt²
θ = 0 rad/s × 3.3 s + 1/2 × 0.631 rad/s² (3.3 s)²
θ = 0 rad + 1/2 × 0.631 rad/s² × 10.89 s²
θ = 1/2 × 6.87159 rad
θ = 3.436 rad
θ ≅ 3.44 rad
Give your answer in standard form.
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Answer: