What does increasing the impulse on an object due to its momentum?
1 answer:
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Answer:
123 J transfer into the gas
Explanation:
Here we know that 123 J work is done by the gas on its surrounding
So here gas is doing work against external forces
Now for cyclic process we know that
so from 1st law of thermodynamics we have
so work done is same as the heat supplied to the system
So correct answer is
123 J transfer into the gas
Answer:
∅ = 89.44°
Explanation:
In situations like this air resistance are usually been neglected thereby making g= 9.81 m/
Bring out the given parameters from the question:
Initial Velocity () = 1000 m/s
Target distance (d) = 2000 m
Target height (h) = 800 m
Projection angle ∅ = ?
Horizontal distance = tcos ∅ .......................... Equation 1
where = velocity in the X - direction
t = Time taken
Vertical Distance = y = t -
g
................... Equation 2
Where = Velocity in the Y- direction
t = Time taken
=
sin∅
Making time (t) subject of the formula in Equation 1
t = d/(cos ∅)
t = =
=
...................Equation 3
substituting equation 3 into equation 2
Vertical Distance = d =
-
g
Vertical Distance = h = sin∅ -
g
Vertical Distance = h = dtan∅ - g
Applying geometry
=
+ 1
Vertical Distance = h = d tan∅ - 2 g ( + 1)
substituting the given parameters
800 = 2000 tan ∅ - 2 (9.81)( + 1)
800 = 2000 tan ∅ - 19.6( + 1) Equation 4
Replacing tan ∅ = Q .....................Equation 5
In order to get a quadratic equation that can be easily solve.
800 = 2000 Q - 19.6 + 19.6
Rearranging 19.6 - 2000 Q + 780.4 = 0
= 101.6291
= 0.411
Inserting the value of Q Into Equation 5
tan ∅ = 101.63 or tan ∅ = 0.4114
Taking the Tan inverse of each value of Q
∅ = 89.44° ∅ = 22.37°
Corrected and Formatted Question:
A wave on a string is described by y(x, t) = (3.0 cm) × cos[2π(x/(2.4 m) + t/(0.20 s))], where x is in m and t in s.
(a) In what direction is this wave traveling?
(b) What are the wave speed, frequency, and wavelength?
(c) At t = 0.50 , what is the displacement of the string at x = 0.20 m?
Answer:
The wave is travelling in the negative x direction
The wave speed = 12.0m/s
The frequency = 5Hz
The wavelength = 2.4m
The displacement at t = 0.50s and x = 0.20m is -0.029m
Explanation:
The general wave equation is given by;
y(x, t) = y cos (2(x/λ) - 2
ft) --------------------------------(i)
Where;
y(x, t) is the displacement of the wave at position x and a given time t
y = amplitude of the wave
f = frequency of the wave
λ = wavelength of the wave
Given;
y(x, t) = (3.0 cm) × cos[2π(x/(2.4 m) + t/(0.20 s))] ------------------(ii)
Which can be re-written as;
y(x, t) = (3.0 cm) × cos[2π(x/(2.4 m)) + 2π(t/(0.20 s))] -------------(iii)
Comparing equations (i) and (iii) we have that;
=> 2π(x/(2.4 m) = 2π(x/λ)
=> λ = 2.4m
Therefore the wavelength of the wave is 2.4m
Also, still comparing the two equations;
=> 2π(t/(0.20 s) = 2πft
=> f = 1 / 0.20
=> f = 5Hz
Therefore the frequency of the wave is 5Hz
To get the wave speed (v), it is given by;
v = f x λ
Where f = 5Hz and λ = 2.4m
=> v = 5 x 2.4
=> v = 12.0m/s
Therefore, the speed of the wave is 12.0m/s
At t = 0.50s and x = 0.20m;
The displacement, y(x,t) of the string wave is given by
y(x, t) = (3.0 cm) × cos[2π(x/(2.4 m) + t/(0.20 s))]
<em>Convert the amplitude of 3.0cm to m</em>
=> 3.0cm = 0.03m
<em>Substitute this back into the equation</em>
=> y(x, t) = (0.03m) × cos[2π(x/(2.4 m) + t/(0.20 s))]
<em>Substitute the values of t and x into the equation above;</em>
=> y(x, t) = (0.03m) × cos[2π((0.20)/(2.4 m) + 0.50/(0.20 s))]
<em>Carefully solve the equation</em>
=> y(x, t) = (0.03m) × cos[2π((0.20)/(2.4 m)) + 2π(0.50/(0.20 s))]
=> y(x, t) = (0.03m) × cos[0.08π + 5π]
=> y(x, t) = (0.03m) × cos[5.08π]
=> y(x, t) = (0.03m) × cos[15.96]
=> y(x, t) = (0.03m) × cos[15.96]
=> y(x, t) = (0.03m) × -0.9684
=> y(x, t) = 0.029m
Therefore the displacement at those points is -0.029m
Also, the sign of the displacement shows that the direction of the wave is in the negative x direction.