Why is it important not to present a biased argument as a public speaker?
A.
Because it is unconvincing
B.
Because it is immoral
C.
Because it is unfair
D.
Because it is pointless
Answer:
Explanation:
a) because it's immoral as you're trying to convince people of your views when you should be giving both sides of the story so people are able to come up with their own opinions on what you're talking about
A pendulum is 0.760 m long, and the bob has a mass of 1,00 kg. At the bottom of its swing, the bob's speed is 1.60 m/s. The tension is greater than the weight of the bob because Multiple Choice 0 its horizontal component is increasing from zero. O the bob has a downward acceleration, so the net Fy must be downward O O the bob has zero acceleration. O ( the bob has an upward acceleration, so the net Fy must be upward Required information A pendulum is 0.760 m long, and the bob has a mass of 1.00 kg. At the bottom of its swing, the bob's speed is 1.60 m/s. What is the tension in the string at the bottom of the swing?
Answer:
T = 13.17 N
the correct one is there is a digested centripetal acceleration towards upward
Explanation:
For this exercise we can use Newton's second law at the bottom of the pendulum's path
we will assume that the upward direction is positive
T - W = m a
in this case it is describing a circle the acceleration is central
a = v² / R
where the radius of the trajectories equals the length of the pendulum
R = L
we substitute
T = mg + m v² / L
T = m (g + v² / L)
When reviewing the different statements, the correct one is there is a digested centripetal acceleration towards upward
let's calculate the tension of the rope
T = 1 (9.8 + 1.6 2 / 0.760)
T = 13.17 N
A marble rolls 269cm across the floor with a constant speed of in 44.1cm/s.
Answer:
what's the question ,so I can answer it right ?
When a piano tuner strikes both the A above middle C on the piano and a 440 Hz tuning fork, he hears 4 beats each second. The frequency of the piano's:____________.
A) 444 Hz
B) 880 Hz
C) 436 Hz
D) either 436 Hz or 444 Hz
Answer:
D) either 436 Hz or 444 Hz
Explanation:
frequency of the tuning fork, F₁ = 440 Hz
frequency of the piano, F₂ = ?
Beat frequency, F = 4 Hz
Beat frequency is given as the difference between the frequency of the two instruments and it is given by;
F = F₂ - F₁ or F = F₁ - F₂
F₂ = F + F₁ or F - F₁ = - F₂
F₂ = 4 Hz + 440 Hz or 4 - 440 = - F₂
F₂ = 444 Hz or - 436 = - F₂
F₂ = 444 Hz or F₂ = 436 Hz
Therefore, the frequency of the piano is 444 Hz or 436 Hz
Part A
Your GPS shows that your friend’s house is 10.0 km away (Figure 2). But there is a big hill between your houses and you don’t want to bike there directly. You know your friend’s street is 6.0 km north of your street. How far do you have to ride before turning north to get to your friend’s house?
8 km
Part B
Referring to the diagram in Part A, what is the sine of the angle
θ at the location of the friend's house?
Answer:
Part A
You have to ride 8.0 km before turning north to get to your friend’s house.
Part B
The sine of the angle θ at the location of the friend's house is 0.8
Explanation:
The remaining part of the question which is an image is attached below
Explanation:
Part A
To determine how far you will ride ride before turning north,
From the diagram, that is the distance of your street.
Let the distance of your street be [tex]A[/tex]
and the distance of your friend's street be [tex]B[/tex]
and let the displacement between your friends house and your house be [tex]C[/tex]
The relation in the diagram shows a right angle triangle.
The sides of the right angle triangle are represented as [tex]A,B[/tex] and [tex]C[/tex].
To find [tex]A[/tex], which is the distance of your street,
From Pythagorean theorem, 'The square of hypotenuse is the sum of squares of the other two sides'
That is,
[tex]/Hypoyenuse/^{2} = /Adjacent/^{2} + /Opposite/^{2}[/tex]
[tex]C[/tex] is the hypotenuse, which is the displacement between your friends house and your house,
Hence, [tex]C = 10.0 km[/tex]
[tex]B[/tex] is adjacent, which is the distance of your friends street
then, [tex]B = 6.0 km[/tex]
and [tex]A[/tex] is the opposite, which is the distance of your house
From Pythagoras theorem, we can then write that,
[tex]C^{2} = B^{2} + A^{2}[/tex]
Then, [tex]10.0^{2} = 6.0^{2} + A^{2}[/tex]
[tex]A^{2} = 100.0 - 36.0\\A^{2} = 64.0\\A = \sqrt{64.0}[/tex]
[tex]A = 8.0km[/tex]
Hence, you have to ride 8.0 km before turning north to get to your friend’s house.
Part B
To find the sine of the angle θ at the location of the friend's house,
In the diagram, the sine of the angle θ is given by
[tex]Sin\theta = \frac{Opposite}{Hypotenuse}[/tex]
Hence, [tex]Sin\theta = \frac{A}{C}[/tex]
Then,
[tex]Sin\theta = \frac{8.0}{10}[/tex]
[tex]Sin\theta = 0.8[/tex]
Hence, the sine of the angle θ at the location of the friend's house is 0.8
A. The amount of distance you have to ride before turning North to get to your friend’s house is 8 kilometers.
B. The sine of the angle (θ) at the location of your friend's house is 0.8.
Let your friend's house be a.Let your friend's street be b.Let the distance between your house and your friend be c.Given the following data:
Distance c = 10 kmDistance a = 6 kmA. To determine the amount of distance you have to ride before turning North to get to your friend’s house, we would apply Pythagorean's theorem:
Mathematically, Pythagorean's theorem is given by the formula:
[tex]c^2 = a^2 + b^2\\\\10^2 =6^2+b^2\\\\100=36+b^2\\\\b^2 =100-36\\\\b^2 =64\\\\b=\sqrt{64}[/tex]
b = 8 kilometers
B. To find the sine of the angle (θ) at the location of the friend's house:
Mathematically, the sine of an angle is given by the formula:
[tex]Sin\theta = \frac{opposite}{hypotenuse}[/tex]
Substituting the given parameters into the formula, we have;
[tex]Sin\theta = \frac{8}{10} \\\\Sin\theta = 0.8[/tex]
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Using the equation for Impact, can you explain the following:
Why are car steering rods designed to collapse?
Why are highway guard rails designed to crumple up on impact?
Why are traffic saftey barrels filled with water or sand?
Explanation:
Equation for Impact
FΔt = ΔP,
F = force
Δt = Impact of time
ΔP = Change in momentum
Car steering is engineered to fail in order to maximize the time of contact and hence reduce the initial impact and mitigate the damage incurred.
Road guard railing crumple on contact to maximize impact time and hence reduce impact intensity and mitigate damage.
Road safety containers are loaded with liquid or sand as they improve the period of impact.
Oppositely charged parallel plates are separated by 4.49 mm. A potential difference of 600 V exists between the plates. (a) What is the magnitude of the electric field between the plates? N/C (b) What is the magnitude of the force on an electron between the plates? N (c) How much work must be done on the electron to move it to the negative plate if it is initially positioned 3.14 mm from the positive plate?
Answer:
A. Using
E=V/d
= 600/4.49*10^-3
= 1.336 x10^5 N/C
b) F = E*q = 1.33610^5 x 1.6*10^-19
= 2.17 x 10^-14 N
c) Work = Fs distance = 2.17 x 10^-14 N (4.49-3.14)*.001= 1.35 x 10^-17 J
Newtons Second Law of motion states that for acceleration to stay the same...
Answer:
newton force is the state of gravity force they abstract when they came closer
r = (2+2+1) i - (t+1)] + t3 k
what is the direction of initial velocity
Answer:
In the - j direction, that is negative of the y-axis
Explanation:
As typed in the question, the position of the object is given by the expression in three component ( i, j, k) form:
r (t) = 5 i - (t + 1 ) j + t^3 k
and since the velocity is the derivative of position with respect to time, by doing the derivative of this expression we get:
v(t) = 0 i - 1 j +3 t^2 k
which for the initial velocity requested (that is at time zero) we have:
v(t) = 0 i - 1 j +3 (0)^2 k = = 1 j
Then the direction of the initial velocity is entirely in the direction of the j versor, that is pointing to the negative of the y-axis.
ou are walking down a straight path in a park and notice there is another person walking some distance ahead of you. The distance between the two of you remains the same, so you deduce that you are walking at the same speed of 1.05 m/s. Suddenly, you notice a wallet on the ground. You pick it up and realize it belongs to the person in front of you. To catch up, you start running at a speed of 2.75 m/s. It takes you 18.5 s to catch up and deliver the lost wallet. How far ahead of you was this person when you started running
Answer:
The value is [tex]d = 31.45 \ m [/tex]
Explanation:
Generally the relative speed at which you are moving with respect to the person ahead of you is mathematically represented as
[tex]v_r = v_s - v_c[/tex]
substituting 1.05 m/s for [tex] v_c [/tex] and 2.75 m/s for [tex]v_s[/tex]
So
[tex]v_r = 2.75 - 1.05[/tex]
=> [tex]v_r = 1.7 \ m/s [/tex]
Generally the distance by which the person is ahead of you is mathematically represented as
[tex]d = v_r * t[/tex]
substituting 18.5 s for [tex] t [/tex]
[tex]d = 1.7 * 18.5[/tex]
=> [tex]d = 31.45 \ m [/tex]
The chilled water system for a 27-story building has a pump located at ground level. The lost head in a vertical riser from the pump to an equipment room on the twenty-seventhfloor is 40ftof water, and the pump produces 270ft of head. What is the pressure on the suction side of the pump for a pressure of 8 psig to exist in the riser on the twenty-fifth floor
This question is incomplete, the complete question is;
The chilled water system for a 27-story building has a pump located at ground level. The lost head in a vertical riser from the pump to an equipment room on the twenty-seventh floor is 40ft of water, and the pump produces 270ft of head. What is the pressure on the suction side of the pump for a pressure of 8 psig to exist in the riser on the twenty-fifth floor
Assume 12ft of elevation per floor
Answer: 48.68 psig
Explanation:
First we calculate the elevation of the building
hb = 27 story * 12ft per floor/story
hb = 324 ft
given that the head lost in the vertical riser hL = 40 ft
now the delivery head required in the riser on he 27th floor;
hd = 8 psig * (2.31 ft / 1 psig)
hd = 18.46 ft
Now calculate the suction head required by balancing the energy per unit weight of water, considering pump as the control volume
hp = (hb + hL + hd) - hs
hs = hb + hL + hd - hp
where hp is the head developed by the pump (270 ft)
hb is the elevation of the 27th floor of the building ( 324 ft)
hL is the head lost in the vertical riser ( 40 ft)
hd is the head required to exist in the riser on the 27th floor (18.46 ft)
so we substitute
hs = 324 ft + 40 ft + 18.46 ft - 270 ft
hs = 112.46
so 112.46ft * (1 psig / 2.31 ft)
= 48.68 psig
A sports car starts from rest at an intersection and accelerates toward the east on a straight road at 9.0 m/s2. Just as the sports car starts to move, a bus traveling east at a constant 16 m/s on the same straight road passes the sports car. When the sports car catches up with and passes the bus, how much time has elapsed
Answer:
3.6s
Explanation:
Have In mind that the sports car will catch up with bus when their positions are equal
So
S = ut + 1/2at²
where u is initial velocity
For sports car we have
x = 0 + 1/2 x 9 x t²
For Bus
S= 16ts + 0
Equating the two
1/2 x 9 x t² = 16t
4.5 x t = 16
t = 3.6s
When is the kinetic energy of the ball being transformed into gravitational potential energy?
Answer:
The second option
Explanation:
I am not sure sry
The hot reservoir for a Carnot engine has a temperature of 889 K, while the cold reservoir has a temperature of 657 K. The heat input for this engine is 4710 J. The 657-K reservoir also serves as the hot reservoir for a second Carnot engine. This second engine uses the rejected heat of the first engine as input and extracts additional work from it. The rejected heat from the second engine goes into a reservoir that has a temperature of 406 K. Find the total work delivered by the two engines.
Answer:
Explanation:
Efficiency of first engine
= T₁ - T₂ / T₁ where T₁ is temperature of hot reservoir , T₂ is temperature of cold reservoir
= (889 - 657 ) / 889
= ,261 or 26.1 %
output of work = .261 x 4710 = 1229.15 J .
Efficiency of second engine
= (657 - 406 ) / 657
= .382
Heat rejected in engine one is heat input of second engine
heat input of second engine = 4710 - 1229.15 = 3480.85
output of work of second engine
= .382 x 3480.85 = 1329.68 J
Total work delivered by two engine
= 1229.15 + 1329.68 J
= 2558.83 J
what would happen to life on earth if all the plants died? plz help
Answer:
we would all die
Explanation:
If all the plants on earth died, so would the people. ... When green plants make food, they give off oxygen. This is a gas that all animals must breathe in order to stay alive. Without plants, animals would have no oxygen to breathe and would die.
(7%) Problem 14: A robot cheetah can jump over obstacles. Suppose the launch speed is vo = 4.74 m/s, and the launch angle is 0 = 25.5
degrees above horizontal.
What is the maximum height in meters?
An object, initially at rest, moves 250 m in 17 s. What is its acceleration?
Answer:
1.73 m/s²
Explanation:
Given:
Δx = 250 m
v₀ = 0 m/s
t = 17 s
Find: a
Δx = v₀ t + ½ at²
250 m = (0 m/s) (17 s) + ½ a (17 s)²
a = 1.73 m/s²
The acceleration of this object is 1.730 meter per seconds square.
Given the following data:
Initial velocity = 2.5 m/s (since the object is starting from rest).Time = 17 seconds.To find the acceleration of this object, we would use the second equation of motion.
Mathematically, the second equation of motion is given by the formula;
[tex]S = ut + \frac{1}{2} at^2[/tex]
Where:
S is the displacement or distance covered.u is the initial velocity.a is the acceleration.t is the time measured in seconds.Substituting the given values into the formula, we have;
[tex]250 = 0(17) + \frac{1}{2} (a)(17^2)\\\\250 = \frac{1}{2} (289)a\\\\250 = 144.5a\\\\a = \frac{250}{144.5}[/tex]
Acceleration, a = 1.730 [tex]m/s^2[/tex]
Therefore, the acceleration of this object is 1.730 meter per seconds square.
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Drive-reduction theory states that motivation comes from a combination of both reinforcement and drive.
ОА.
True
OB. False
A car starts at 80 m/s but sees a cop and hits the brakes slowing down to 50 m/s in 2 sec.
Answer:
-15m/s/s
Explanation:
Acceleration = change in speed/ change in time
The change in speed is calculated by subtracting the initial speed from the final speed, so the change in speed is: 50 - 80 = -30m/s. The change in time is 2 - 0 = 2.
So the car acceleration is -30/2 = -15m/s/s
It is negative because it is decelerating.
Hope this helped!
A hockey puck initially travelling to the right at 34 m/s. It moves for 7 before
coming to a stop. How far did it move in 7 seconds?
You can use kinematic equations
Answer:
[tex]x=119m[/tex]
Explanation:
Hello,
In this case, since the hockey puck was moving at 34 m/s and suddenly stopped (final velocity is zero) in 7 seconds, we can first compute the acceleration via:
[tex]a=\frac{v_f-v_o}{t}=\frac{0m/s-34m/s}{7s}\\ \\a=-4.86m/s^2[/tex]
In such a way, we can compute the displacement via:
[tex]x=\frac{v_f^2-v_o^2}{2a}\\ \\x=\frac{0^2-(34m/s)^2}{2*-4.86m/s^2}\\ \\x=119m[/tex]
Best regards.
Whenever an object is moving at a constant rate, the value(s) that equal zero is (are):
a) Speed
b) Acceleration
c) Velocity
d) All of the above
Whenever an object is moving at a constant rate . . .
a). its speed is that constant rate.
b). Its acceleration MAY BE zero, if it's also moving in a straight line.
c). Its velocity is that constant rate if it's moving in a straight line. Otherwise, its velocity could have many different values, depending on the path it's following.
A bird flies 3.7 meters in 46 seconds, what is its speed?
Answer:
Speed is 0.08 m/s.
Explanation:
Given the distance that the bird flies = 3.7 meters
The time is taken by the bird to fly the 3.7 meters = 46 seconds
We have given distance and time. Now we have to find the speed at which the bird flies. So, to calculate the speed of the bird we have to divide the distance by the time.
Below is the formula to find the speed.
Speed = Distance / Time
Now insert the given value in the formula.
Speed = 3.7 / 46 = 0.08 m/s
.A hard rubber ball, released at chest height, falls to the pavement and bounces back to nearly the same height. When it is in contact with the pavement, the lower side of the ball is temporarily flattened. Suppose the maxi-mum depth of the dent is on the order of 1 cm. Find the order of magnitude of the maximum acceleration of the ball while it is in contact with the pavement. State your assumptions, the quantities you estimate, and the values you estimate for them.
Answer:
a = 1.1 10⁵ m / s²
Explanation:
This is a momentum exercise, where we use the relationship between momentum and momentum
I = ∫ F dt = Δp
= p_f - p₀
as they indicate that the ball bounces at the same height, we can assume that the moment when it reaches the ground is equal to the moment when it bounces, but in the opposite direction
F t = 2 (m v)
therefore the average force is
F = 2 m v / t
where in general the mass of the ball unknown, the velocity of the ball can be calculated using the conservation of energy
starting point. Done the ball is released with zero initial velocity
Em₀ = U = mgh
final point. Upon reaching the ground, just before the deformation begins
Em_f = K = ½ m v²
energy is conserved in this system
Em₀ = Em_f
m g h = ½ m v²
v = √ (2gh)
This is the velocity of the body when it reaches the ground, so the force remains
F = 2m √(2gh) /t
where the height of the person's chest is known and the time that the impact with the floor lasts must be estimated in general is of the order of milli seconds
knowing this force let's use Newton's second law
F = m a
a = F / m
a = 2 √(2gh) / t
We can estimate the order of magnitude of this acceleration, assuming the person's chest height of h = 1.5 m and a collision time of t = 1 10⁻³ s
a = 2 √ (2 9.8 1.5) / 10⁻³
a = 1.1 10⁵ m / s²
Air that initially occupies 0.22 m3 at a gauge pressure of 86 kPa is expanded isothermally to a pressure of 101.3 kPa and then cooled at constant pressure until it reaches its initial volume. Compute the work done by the air. (Gauge pressure is the difference between the actual pressure and atmospheric pressure.)
Answer:
total work done = -5960.8 J
Explanation:
given data
initial volume v1 = 0.22 m³
initial pressure p1 = 86 kPa
final pressue p2 = 101.3 kPa
solution
we apply here isothermal expansion that is express as
p1 × v1 = p2 × v2 ......................1
put here value
86 × 0.22 = 101.3 × v2
v2 = 0.1867 m³
and
work done will be here
w1 = p1 × v1 × ln([tex]\frac{p1}{p2}[/tex]) ....................2
w1 = 86 × 10³ × 0.22 × [tex]ln(\frac{86}{101.3})[/tex]
w1 = -3.097 × 10³ J
and
it is cooled to initial volume at constant pressure so here work done will be
w2 = p(v2 - v1) .................3
w2 = 86 × 10³ × ( 0.1867 - 0.22 )
w2 = -2863.8 J
so
total work done is
total work done = w1 + w2
total work done = -3097 + -2863.8
total work done = -5960.8 J
Let's start by calculating what acceleration the rocket must produce to launch into earth orbit. In order to attain orbit around earth, the ATLAS V rocket must accelerate up to a speed of about 7700 meters per second in about 4.2 minutes. What average acceleration is required to accomplish this
Answer:
30.56 m/s^2
Explanation:
Given that In order to attain orbit around earth, the ATLAS V rocket must accelerate up to a speed of about 7700 meters per second in about 4.2 minutes.
The average acceleration that is required to accomplish this will be
Average acceleration = change in velocity / time
Average acceleration = 7700/ 4.2 × 60
Average acceleration = 7700/252
Average acceleration = 30.56 m/s^2
You are trying to get to class on time using the UCF Shuttle. You are later than usual getting to the stop and see the shuttle pulling away from the stop while you are still 3.9 m behind the bus stop. In 40.9 m you will reach a barrier and you must catch the shuttle before that point. The shuttle has a constant acceleration of 4.5 m/s2. What is the minimum velocity you have to run at to catch the bus before it reaches the barrier
Answer:
20.1 m/s
Explanation:
Since You are later than usual getting to the stop and see the shuttle pulling away from the stop while you are still 3.9 m behind the bus stop. And In 40.9 m you will reach a barrier and you must catch the shuttle before that point.
Given that the shuttle has a constant acceleration of 4.5 m/s2.
The total distance to cover is:
Total distance = 40.9 + 3.9 = 44.8 m
Assuming you are starting from rest. Then initial velocity U = 0
Using the 3rd equation of motion to calculate the minimum velocity.
V^2 = U^2 + 2as
V^2 = 0 + 2 × 4.5 × 44.8
V^2 = 403.2
V = sqrt (403.2)
V = 20.1 m/s
Therefore, the minimum velocity you have to run at to catch the bus before it reaches the barrier is 20.1 m/s
a) How long (in ns) does it take light to travel 1.0 m in a vacuum?
b) What distance does light travel in water, glass, cubic zirconia during the time it travels in 1.0 m vacuum?
Answer:
a
[tex]t = 3.33 \ ns[/tex]
b
i [tex]D_w =0.75 \ m [/tex]
ii [tex]D_g =0.67 \ m [/tex]
iii [tex]D_c =0.46 \ m [/tex]
Explanation:
Generally the time taken to travel 1 m in a vacuum is mathematically represented as
[tex]t = \frac{1}{c}[/tex]
Here c is the speed of light with value [tex]c = 3.0*10^{8} \ m/s[/tex]
So
[tex]t = \frac{1}{3.0*10^{8}}[/tex]
[tex]t = 3.33*10^{-9} \ s[/tex]
[tex]t = 3.33 \ ns[/tex]
The distance light travels in water is mathematically represented as
[tex]D_w = \frac{c}{n_w} * t[/tex]
Here n_w is the refractive index of water with value 1.333
So
[tex]D_w = \frac{3.0*10^{8}}{1.333} * 3.33*10^{-9}[/tex]
[tex]D_w =0.75 \ m [/tex]
The distance light travels in glass is mathematically represented as
[tex]D_g = \frac{c}{n_g} * t[/tex]
Here n_g is the refractive index of glass with value 1.5
So
[tex]D_g = \frac{3.0*10^{8}}{1.5} * 3.33*10^{-9}[/tex]
[tex]D_g =0.67 \ m [/tex]
The distance light travels in cubic zirconia is mathematically represented as
[tex]D_c = \frac{c}{n_g} * t[/tex]
Here n_c is the refractive index of cubic zirconia with value 2.15
So
[tex]D_c = \frac{3.0*10^{8}}{2.15} * 3.33*10^{-9}[/tex]
[tex]D_c =0.46 \ m [/tex]
An airplane with a hot-wire anemometer mounted on its wing tip is to fly through the turbulent boundary layer of the atmosphere at a speed of 50 m/sec. The velocity fluctuations in the atmosphere are of order 0.5 m/sec, the length scale of the large eddies is about 100 m. The hot-wire anemometer is to be designed so that it will register the motion of the smallest eddies.What is the highest frequency the anemometer will encounter
Answer:
0.55 hz
Explanation:
Given that the plane fly through the turbulent boundary layer of the atmosphere at a speed of 50 m/sec. And the velocity fluctuations in the atmosphere are of order 0.5 m/sec, the length scale of the large eddies is about 100 m.
The maximum speed attained will be
Maximum speed = 50 + 0.5 = 5.5 m/s
The Length = 100m
Speed = FL
Where F = frequency
Substitute speed and distance length into the formula
55 = 100F
F = 55/100
F = 0.55 Hz
Therefore, the highest frequency the anemometer will encounter will be 0.55 Hz
Answer:
40079 Hz
Explanation:
1. The first step is to calculate energy dissipation ∈=u^3/l and here u is fluctuating velocity
∈=(0.5^3)/100 = 0.00125 m^2/s^3
2. Find out the length scale of the small eddies
η=(viscosity/∈)^1/4
η=(1.470e-5/0.00125)^1/4 = 0.00126 m
3. The frequency associated with these small-scale eddies will be the greatest frequency the anemometer will encounter, thus:
u_max=f_max * η
u_max = u + u' = 50+0.5=50.5 m/s
f_max = u_max/η = 50.5/0.00126 = 40079 Hz
This is the heighest frequency the anemometer will encounter.
You drop a ball from a window located on an upper floor of a building. It strikes the ground with speed v. You now repeat the drop, but you ask a friend down on the ground to throw another ball upward at speed v. Your friend throws the ball upward at the same moment that you drop yours from the window. At some location, the balls pass each other. Is this location.
Answer:
y = y₀ (1 - ½ g y₀ / v²)
Explanation:
This is a free fall problem. Let's start with the ball that is released from the window, with initial velocity vo = 0 and a height of the window i
y = y₀ + v₀ t - ½ g t²
y = y₀ - ½ g t²
for the ball thrown from the ground with initial velocity v₀₂ = v
y₂ = y₀₂ + v₀₂ t - ½ g t²
in this case y₀ = 0
y₂2 = v t - ½ g t²
at the point where the two balls meet, they have the same height
y = y₂
y₀ - ½ g t² = vt - ½ g t²
y₀i = v t
t = y₀ / v
since we have the time it takes to reach the point, we can substitute in either of the two equations to find the height
y = y₀ - ½ g t²
y = y₀ - ½ g (y₀ / v)²
y = y₀ - ½ g y₀² / v²
y = y₀ (1 - ½ g y₀ / v²)
with this expression we can find the meeting point of the two balls
If two tug boats are towing a ship with force of 5 tons each and the angle between the two ropes is 60 degrees, what is the resultant force on the ship? Explain how to use a force table to verify answer.
Answer:
8.6602 tons
Explanation:
We first draw the known vector forces.
2fcos30⁰
We have f to be equal to 5tons
Inserting into formula
Σfx = 2(5)cos30⁰
= 8.6602 tons
Σfy is equal to 0, this is because in the y direction, the forces cancel themselves out.
Therefore the resultant force on the ship is equal to 8.6602 tons
I hope this helps!
Please check attachment for diagram.