Hi there!
A.
Since the can was launched from ground level, we know that its trajectory forms a symmetrical, parabolic shape. In other words, the time taken for the can to reach the top is the same as the time it takes to fall down.
Thus, the time to its highest point:
[tex]T_h = \frac{T}{2}[/tex]
Now, we can determine the velocity at which the can was launched at using the following equation:
[tex]v_f = v_i + at[/tex]
In this instance, we are going to look at the VERTICAL component of the velocity, since at the top of the trajectory, the vertical velocity = 0 m/s.
Therefore:
[tex]0 = v_y + at\\\\0 = vsin\theta - g\frac{T}{2}[/tex]
***vsinθ is the vertical component of the velocity.
Solve for 'v':
[tex]vsin(\alpha_0) = g\frac{T}{2}\\\\v = \frac{gT}{2sin(\alpha_0)}[/tex]
Now, recall that:
[tex]W = \Delta KE = \frac{1}{2}m(\Delta v)^2[/tex]
Plug in the expression for velocity:
[tex]W = \frac{1}{2}M (\frac{gT}{2sin(\alpha_0)})^2\\\\\boxed{W = \frac{Mg^2T^2}{8sin^2(\alpha _0)}}[/tex]
B.
We can use the same process as above, where T' = 2T and Th = T.
[tex]v = \frac{gT}{sin(\alpha _0)} }\\\\W = \frac{1}{2}M(\frac{gT}{sin(\alpha _0)})^2\\\\\boxed{W = \frac{Mg^2T^2}{2sin^2(\alpha _0)}}[/tex]
C.
The work done in part B is 4 times greater than the work done in part A.
[tex]\boxed{\frac{W_B}{W_A} = \frac{4}{1} = 4.0}[/tex]
A. To find the work done on the can by the launching device, we need to calculate the change in kinetic energy of the can. When the can is launched, it has an initial kinetic energy of 0 (since it starts from rest).
1/2 * M * [(initial velocity)^2 * cos^2(α0) + (g * T)^2]
WA = 1/2 * M * [(initial velocity)^2 * cos^2(α0) + (g * T)^2]
B. If the can stays in the air twice as long, the time of flight becomes 2T. We can repeat the above steps with the new value of T to find the initial velocity: Initial vertical velocity = g * (2T)
WB [tex]= 1/2 * M * [(initial velocity)^2 * cos^2(α0) + (2gT)^2][/tex]
C. To compare the results, we can calculate the ratio of WB to WA:
WB/WA [tex]= [1/2 * M * [(initial velocity)^2 * cos^2(α0) + (2gT)^2]] / [1/2 * M * [(initial velocity)^2 * cos^2(α0) + (g * T)^2]][/tex]
Notice that the mass M and the angle α0 cancel out:
WB/WA [tex]= [((initial velocity)^2 * cos^2(α0) + (2gT)^2)] / [((initial velocity)^2 * cos^2(α0) + (g * T)^2)][/tex]
A. At the highest point of its trajectory, the kinetic energy is again 0 (the vertical velocity component is 0). So the work done by the launching device equals the initial kinetic energy. The initial kinetic energy can be expressed as: Initial kinetic energy = [tex]1/2 * M * (initial velocity)^2[/tex]
The initial velocity can be calculated using the launch angle and the time of flight T. At the highest point, the vertical component of the velocity is 0, so we only consider the horizontal component of the velocity. Horizontal component of velocity = initial velocity * cos(α0). The time T is the time taken to reach the highest point, so we can write:
T = time taken to reach the highest point = (initial vertical velocity) / g
Initial vertical velocity = g * T
Now, the initial velocity can be written as:
initial velocity = + (Initial vertical [tex])√[(Horizontal component of velocity)^2[/tex]
=[tex]√[(initial velocity * cos(α0))^2 + (g * T)^2][/tex]
=[tex]√[(initial velocity)^2 * cos^2(α0) + (g * T)^2][/tex]
Since the initial velocity is equal to the change in kinetic energy, we have: Initial kinetic energy :[tex]1/2 * M * [(initial velocity)^2 * cos^2(α0) + (g * T)^2] \\ WA = 1/2 * M * [(initial velocity)^2 * cos^2(α0) + (g * T)^2][/tex]
B. Initial velocity:
[tex]√[(initial velocity * cos(α0))^2 + (g * (2T))^2]\\ = √[(initial velocity)^2 * cos^2(α0) + (2gT)^2][/tex]
Now, the initial kinetic energy for this case is: Initial kinetic energy (new) [tex]= 1/2 * M * [(initial velocity)^2 * cos^2(α0) + (2gT)^2][/tex]
WB [tex]= 1/2 * M * [(initial velocity)^2 * cos^2(α0) + (2gT)^2][/tex]
C. This shows that the mass and angle do not affect the work done on the can; only the time of flight and the acceleration due to gravity influence it. Since we know that T is doubled in part B (2T), we can write:
WB/WA [tex]= [((initial velocity)^2 * cos^2(α0) + (2g * 2T)^2)] / [((initial velocity)^2 * cos^2(α0) + (g * T)^2)][/tex]
WB/WA [tex]= [((initial velocity)^2 * cos^2(α0) + (4gT)^2)] / [((initial velocity)^2 * cos^2(α0) + (gT)^2)][/tex]
Now, we can see that WB is larger than WA by a factor of [tex][((initial velocity)^2 * cos^2(α0) + (4gT)^2)] / [((initial velocity)^2 * cos^2(α0) + (gT)^2)].[/tex]
The value of this factor will depend on the specific values of the initial velocity, launch angle, and time of flight, but this ratio is greater than 1, indicating that more work is done on the can when it stays in the air for twice the time.
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When trying to simplify and find the equivalent resistance you
should first simplify resistors in_before
simplifying those in_
Answer:When trying to simplify and find the equivalent resistance you
should first simplify resistors in PARALLEL before
simplifying those in SERIES
Explanation: hope this helps:)
How do I go about this?
Hi there!
(a)
Recall that:
[tex]W = F \cdot d = Fdcos\theta[/tex]
W = Work (J)
F = Force (N)
d = Displacement (m)
Since this is a dot product, we only use the component of force that is IN the direction of the displacement. We can use the horizontal component of the given force to solve for the work.
[tex]W =248(56)cos(30) = 12027.36 J[/tex]
To the nearest multiple of ten:
[tex]W_A = \boxed{12030 J}[/tex]
(b)
The object is not being displaced vertically. Since the displacement (horizontal) is perpendicular to the force of gravity (vertical), cos(90°) = 0, and there is NO work done by gravity.
Thus:
[tex]\boxed{W_g = 0 J}[/tex]
(c)
Similarly, the normal force is perpendicular to the displacement, so:
[tex]\boxed{W_N = 0 J}[/tex]
(d)
[tex]F_{f} =\mu_k N[/tex]
In this instance, the normal force is equivalent to the downward force of gravity and the vertical component of the applied force.
[tex]N = F_g + F_A sin(30)\\\\N = mg + F_A sin(30)\\\\N = 56(9.8) + 248sin(30) = 672.8 N[/tex]
Since the force of friction resists the applied force (assigned the positive direction), the work due to friction is NEGATIVE because energy is being LOST. Thus:
[tex]W_f = -\mu_k Nd\\W_f = - (0.1)(672.8)(56) = -3767.68J[/tex]
In multiples of ten:
[tex]\boxed{W_f = -3770 J}[/tex]
(e)
Simply add up the above values of work to find the net work.
[tex]W_{net} = W_A + W_f \\\\W_{net} = 12027.36 + (-3767.68) = 8259.68 J[/tex]
Nearest multiple of ten:
[tex]\boxed{W_{net} = 8260 J}}[/tex]
(f)
Similarly, we can use a summation of forces in the HORIZONTAL direction. (cosine of the applied force)
[tex]F_{net} = F_{Ax} - F_f[/tex]
[tex]W = F_{net} \cdot d = (F_{Ax} - F_f)[/tex]
[tex]W = (F_Acos(30) - \mu_k N)d\\W = (248cos(30) - 0.1(672.8)) * 56 \\\\W = 8259.68 J[/tex]
Nearest multiple of ten:
[tex]\boxed{W_{net} = 8260 J}[/tex]
How does microwave technology cook food?
Question 4 options:
By heating all the water molecules which changes the thermal energy of the food
By heating surface water or fat molecules only, and the remaining food is cooked by conduction.
Microwaves cannot cook food.
By heating the internal water molecules, the food cooks from the inside out.
Answer:
By heating all the water molecules which changes the thermal energy of the food
Explanation:
How to calculate Density of tetragonal unit cell by formula?
Answer:
[tex]\huge\underline{\red{A}\blue{n}\pink{s}\purple{w}\orange{e}\green{r} -}[/tex]
The formula of Density for a random unit cell is as follows -
[tex]Density = \frac{ Z _{effective} \times M }{N _{a} \times a {}^{3} } \\ [/tex]
where ,
★ [tex]Z _{effective}[/tex] = number of atoms in a unit cell
★ M = molar mass
★ [tex] N _{a} = Avogadro's \: Number = 6 × 10^{23}[/tex]
★ a = edge length
now ,
In a tetragonal unit cell ,
the atoms are present over the corners as well as the at the body centre.
Therefore ,
[tex]Z _{effective} = contribution \: by \: the \: 8 \: corners \: + contribution \: by \: the \: body \: centre[/tex]
[tex]\implies \: (\frac{1}{8} \times 8) \: + \: 1 \\ \\ \implies \: 1 + 1 \\ \\ \implies2[/tex]
And ,
[tex]N _{a} = 6 \times 10 {}^{23} [/tex]
Substituting the values of the following in the formula of Density , we get
[tex]\bold\purple{Density = \frac{2 \times M}{6 \times 10 {}^{23} \times a {}^{3} }} \\[/tex]
hope helpful :D
A trailer is pulling a car with a force of 900N on a level road. The car is moving at 70 km/h how much work is done by the trailer on the car in 15mm?
Answer:
W=F⋅l=F⋅vt=900(N)⋅19.44(m/s)⋅900(s)=15.75⋅10
6 (J)
Explanation:
Please help me, Thank you
Answer:
D. 2.50
Explanation:
Frequency by definition is the number of waves that pass a fixed point in the given amount of time.
A 3m plank AB of mass 20kg has a center of gravity 2m from end A and rests on supports R1 and R2 both 0,5m from either end. It also carries a 10kg block at end A. Calculate the reactions at R1 and R2
Answer:
130.7 N
Explanation:
Answer ASAP and only if you know its correct
this is science btw
Answer: It is brighter
Explanation: We are getting closer to the light because the earth is expanding.
the answer is it is brighter cause the light from the object is known as blueshift
An ideal gas initially is allowed to expand isothermally until its volume of 1.6 L and pressure is 5 kPa, undergoes isothermal expansion until its volume is 8 L and its pressure is 1 kPa.
1. Calculate the work done by the gas. Answer in units of kJ.
2. Find the heat added to the gas during this process. Answer in units of kJ.
(a) The work done by the gas during the isothermal expansion is -25.6 J.
(b) The heat added to the gas during this process is 25.6 J.
Net work done by the ideal gas against the external pressureThe net work done by the ideal gas in the isothermal expansion is calculated as follows;
W(net) = ΔP x ΔV
W(net) = ( 1 kPa - 5 kPa) x (8L - 1.6 L)
W(net) = -25.6 kPa.L
W(net) = -25.6 J
Head added to the gasThe heat added to the gas is calculated as follows;
W = -Q
-25.6 J = -Q
Q = 25.6 J
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what is tofu name one tofu.
Answer:
Tofu is also known as Bean Curd .Tofu is a food prepared by coagulating soya milk.
what is acceleration due to gravity?
the speed at which the gravity pulls a freely falling object is known as it's acceleration due to gravity, it is represented by the letter (g)
the acceleration due to gravity on Earth is 9.8m/s²
By 2005, the world's tallest building will be the International Finance Center in Taipei,
Republic of China. Suppose a 1.00 C charge is placed at both the base and the top of the
International Finance Center. If the magnitude of the electric force stretching the building
is 4.48 x 104N, how tall is the International Finance Center?
Electric field refers to the region in space where the influence of a charge is felt, the height of the building is 447 m.
What is electric field?
The term electric field refers to the region in space where the influence of a charge is felt.
Since we know that;
E = F/q = 4.48 x 104N/1C = 4.48 x 10^4 N/C
E = Kq/r^2
4.48 x 10^4 = 9 * 10^9 * 1/r^2
r = √9 * 10^9/4.48 x 10^4
r = √2 * 10^5
r = 447 m
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Which standing wave has exactly three antinodes?
A.
B.
C.
D.
Answer:
Explanation:
C looks good, b/c antinodes are where the waves is at a max displament.
The wave has exactly three antinodes is option C.
What number of antinodes are in 3 status waves?
As in all status wave styles, each node is separated through an antinode. This sample with 3 nodes and two antinodes is known as the second one harmonic and is depicted within the animation.
Where are the antinodes?The other of a node is an anti-node, a factor where the amplitude of the status wave is at most. those occur midway among the nodes. In a full wavelength of a status wave, there are two loops. So, there should be nodes midway of every of the two loops.
What's a wave in technology?Wave, propagation of disturbances from location to vicinity in a ordinary and prepared manner. maximum acquainted are floor waves that journey on water, however sound, mild, and the movement of subatomic particles all exhibit wavelike residences.
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When scientists look at very distant galaxies through powerful telescopes, they see the galaxies as they were millions or billions of year ago. Why is this?
Answer:
speed of light
Explanation:
the sun's light is 8 mins behind imaging that a billion fold
Changing the thickness of the myelin sheath surrounding an axon changes its capacitance and thus the conduction speed. A myelinated nerve fiber has a conduction speed of 53 m/s . If the spacing between nodes is 1.0 mm and the resistance of segments between nodes is 26 MΩ , what is the capacitance of each segment?
For a myelinated nerve fiber has a conduction speed of 53 m/s, the capacitance of each segment is mathematically given as
C=8.2*10^{-13}F
What is the capacitance of each segment?Generally, the equation for the Constant is mathematically given as
t=RC
Therefore
t=V/d
Hence
C=V/Rd
C=1.2*1)^-3/26*10^6*56
C=8.2*10^{-13}
In conclusion, the capacitance
C=8.2*10^{-13}F
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what is dollar effect
Answer:
Above is correct
A p.d. of 6 V is applied across a 2 resistor. What current flows?
Answer: current is 3 A.
Explanation:
V = 6V
R = 2Ω
I = ?
∴ V = IR
I = V/R
I = 6V/2Ω
I = 6/2
I = 3 A ...answer...
hope that helps...
Answer:
3 amps
Explanation:
V = IR
V/R = I
6/2 = 3 amps
what are the equivalent celsius and kelvin temperatures of -128 f?
Answer:
128 Kelvin = 128 - 273.15 = -145.15 Celsius. Temperature conversion chart Sample temperature conversions 103.55 Kelvin to degrees Fahrenheit 39.82 degrees Fahrenheit to Kelvin
Explanation:
hope this helps have a good day
Temperate marine climates are normally:
a.hot
c. cold
b. cool to warm
d. warm to hot
Answer:
Temperate marine climates are humid and have mild winters. There are three kinds of temperate marine climates: marine west coast, humid subtropical, and mediterranean. Marine west coast climates are the coolest temperate marine climates. Humid ocean air brings cool, rainy summers and mild, rainy winters.
6.
What happens at the condensation point?
Answer:
Condensation point is when the temperature at which a vapor condenses into a liquid without a change in the temperature of the substance.
An electricAn electric field E = 100,000i N/C causes the 5.0 g point charge to hang at a 20°
angle. What is the charge on the ball?
E= 100,000i N/C
to find:the charge on the ball
solution:Using force balance on mass in vertical direction
T= tension in rope
in x-direction
Fx= T*sin A= q*E
in y-direction
Fy = T*cos A = m*g
divide both equation
tan A = q*E/(m*g)
q = (m*g*tan A)/E
Using given values:
q (5.0*10^-3*9.81*tan 20 deg)/100000
q=1.78*10^-7
therefore, the charge on the ball is 1.78*10^-7
Which of the following was NOT one of Ghana's chief trading exports?
A. Gold
B. Ivory
C. Horses
Think of the human body composition. Why do humans sink much slower than a boulder weighing the same mass? What makes humans less dense?
Answer:
because of the oxygen in our bodies.
Find the total resistance ....
The Equivalent resistance is :
[tex]\qquad \tt \dashrightarrow \: \dfrac{14}{9} \: \: ohms[/tex]
The solution is in attachment for solution ~
If two runners are running in a 500 meter dash and runner #1 finishes in 25 seconds and runner #2 finishes in 27 seconds. What are the speeds of both runners? Which one is faster
For two runners running in a 500 meter dash and runner, the speeds of both runners is mathematically given as
v1=25m/s
v2=20m/s
and the first runner is faster
What are the speeds of both runners?Generally, the equation for the speed is mathematically given as
v=d/t
Therefore, for first runner
v=500/20
v1=25m/s
For 2nd runner
v=500/25
v2=20m/s
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The radius of the aorta is about 1.4cm , and the blood passing through it has a speed of about 40cm/s .
Calculate the average speed of blood flow in the major arteries of the body, which have a total cross-sectional area of about 2.1cm2 .
Express your answer to two significant figures and include the appropriate units.
For the average speed of blood flow in the major arteries of the body is mathematically given as
v2 = 117.29m/s
What is the average speed of blood flow in the major arteries of the body?Generally, the equation for the average speed is mathematically given as
A1 v1 = A2 v2
(pi r1^2) v1 = A2 v2
(3.14x(1.4)^2 )x 40 = (2.1) xV2
v2 = 117.29m/s
In conclusion, the average speed of blood flow
v2 = 117.29m/s
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Age, height, and weight are examples of numerical data.
Miguel is holding a 5 kg box.
How much force is the box exerting on him? In what direction?
How much force is he exerting on it? In what direction?
hope this answer helps...!!!
Answer ASAP and only if you know its correct
this is science btw also, brainliest to the correct answer
Answer:
scientists see emission spectra 'shifted' towards the red end of the electromagnetic spectrum—
Explanation:
hoped this helped
the doppler effect shows stars with red shifted light
cart a having a mass of 150 kg initially moving to the right at a speed of 8 m/s collides with cart b with a mass of 150 kg, initially moving to the right at 6 m/s after a collision cart a continues to move to the right but with a speed of 5 m/s
a. what is the speed of cart b after collision
b. what is the total momentum of the system before and after collision
The speed of cart b is 6m/s while the total momentum of the systmen is 4200 kg m/s
Conservation of Linear MomentumGiven Data
Mass of cart one M1 = 150kgInitial Velocity U1 = 8m/sFinal VelocityV1 = 5 m/sMass of cart two M2 = 150kg
Velocity U2 = 6m/s
Applying the principle of conservation of linear momentum we have
M1U1+M2U2 = M1V1+ M2V2
a. what is the speed of cart b after collision
substituting our given data we have
150*8+ 150*6 = 150*5+150*V2
1200 + 900 = 1200+ 150V2
2100 - 1200 = 150V2
900 = 150V2
Divide both sides by 150
V2 = 900/150
V2 = 6m/s
b. what is the total momentum of the system before and after collision
Total Momentum in the system is
Total momentum = Momentum before Impact+ Momentum after Impact
Total momentum = M1U1+M2U2 + M1V1+ M2V2
Total momentum = 1200 + 900 + 1200+ 900
Total momentum = 4200 kg m/s
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