Answer:
The velocity is 2661.5 m/s.
Explanation:
Radius, horizontal distance, d = 309 km
height, h = 25 km
acceleration due to gravity on moon, g =3.71 m/s^2
Let the time taken is t and the horizontal velocity is u.
horizontal distance = horizontal velocity x time
309 x 1000 = u t .... (1)
Use second equation of motion in vertical direction.
[tex]h = u_yt +0.5 gt^2\\\\25000 = 0 + 0.5\times 3.71\times t^2\\\\t =116.1 s[/tex]
So, put in (1)
309 x 1000 = u x 116.1
u = 2661.5 m/s
a concrete has a height of 5m and has unit area 3m² supports a mass of 30000kg.
Determine the stress, strain and change in height
Answer:
stress = 98000 N/m^2
strain = 3.92 x 10^-6
change in height = 0.0196 mm
Explanation:
Height, h = 5 m
Area, A = 3 m²
mass, m = 30000 kg
Stress is defined as the force per unit area.
[tex]stress = \frac{mg}{A}\\\\stress = \frac{30000\times 9.8}{3}\\\\stress = 98000 N/m^2[/tex]
Young's modulus of concrete is Y = 2.5 x 10^10 N/m^2
Young's modulus is defined as the ratio of stress to the strain.
[tex]Y = \frac{stress}{strain}\\\\2.5\times 10^{10}= \frac{98000}{strain}\\\\strain = 3.92\times 10^{-6}[/tex]
let the change in height is h'.
Strain is defined as the ratio of change in height to the original height.
[tex]3.92\times 10^{-6} = \frac{h'}{5}\\\\h' = 1.96\times 10^{-5}m = 0.0196 mm[/tex]
An object is suspended by a string from the ceiling of an elevator. If the tension in the string is equal to 25 N at an instant when the elevator is accelerating downward at a rate of 2.0 , what is the mass of the suspended object
By Newton's second law, the net force on the object is
∑ F = T - mg = - ma
where
• T = 25 N, the tension in the string
• m is the mass of the object
• g = 9.8 m/s², the acceleration due to gravity
• a = 2.0 m/s², the acceleration of the elevator-object system
Solve for m :
25 N - m (9.8 m/s²) = - m (2.0 m/s²)
==> m = (25 N) / (9.8 m/s² - 2.0 m/s²) ≈ 3.2 kg
A merry-go-round of radius R = 2.0 m has a moment of inertia I = 250 kg-m2
and is rotating at 10 rev/min. A 25-kilogram child at rest jumps onto the edge of the merry-go-round. What is the new angular speed of the merry-go-round?
Answer:
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Find the starting pressure of CCl4 at this temperature that produces a total pressure of 1.1 atm at equilibrium. Express the pressure in atmospheres to three significant figures.
The complete question is as follows: At 700 K, [tex]CCl_{4}[/tex] decomposes to carbon and chlorine. The Kp for the decomposition is 0.76.
Find the starting pressure of [tex]CCl_{4}[/tex] at this temperature that will produce a total pressure of 1.1 atm at equilibrium.
Answer: The starting pressure of [tex]CCl_{4}[/tex] is 0.79 atm.
Explanation:
The equation for decomposition of [tex]CCl_{4}[/tex] is as follows.
[tex]CCl_{4}(g) \rightleftharpoons C(s) + 2Cl_{2}(g)[/tex]
Let us assume that initial concentration of [tex]CCl_{4}[/tex] is 'a'. Hence, the initial and equilibrium concentrations will be as follows.
[tex]CCl_{4}(g) \rightleftharpoons C(s) + 2Cl_{2}(g)[/tex]
Initial: a 0 0
Equilibrium: (a - x) 0 2x
Total pressure = (a - x) + 2x = a + x
As it is given that the total pressure is 1.1 atm.
So, a + x = 1.1
a = 1.1 - x
Now, expression for equilibrium constant for this equation is as follows.
[tex]K_{p} = \frac{P^{2}_{Cl_{2}}}{P_{CCl_{4}}}\\0.76 = \frac{(2x)^{2}}{(a - x)}\\0.76 = \frac{4x^{2}}{1.1 - x - x}\\0.76 = \frac{4x^{2}}{1.1 - 2x}\\x = 0.31 atm[/tex]
Hence, the value of 'a' is calculated as follows.
a + x = 1.1 atm
a = 1.1 atm - x
= 1.1 atm - 0.31 atm
= 0.79 atm
Thus, we can conclude that starting pressure of [tex]CCl_{4}[/tex] is 0.79 atm.
A 300 kg block of dimensions 1.5 m × 1.0 m × 0.5 m lays on the table with its largest face.
Calculate:
Area of the largest face
Answer:
1.5
x 1.0
1.50
x 0.5
075.00
answer: 75.00m
Explanation:
I hope this help