The satellite dish follows the geometry of a paraboloid. Given the dimensions of a paraboloid that is 12 feet wide and 2 feet deep, the receiver for the satellite signals should be placed 0.5 feet from the base, along the axis of symmetry.
Explanation:This is a mathematical problem involving the geometry of paraboloids. A paraboloid must have a particular balance between its depth and its width that allows signals, such as those from satellites, to focus at a single point. This point is known as the focus of the paraboloid.
In this case, for a paraboloid that is 12 feet across and 2 feet deep, the focus will be at the position that is one-fourth the depth of the dish from its vertex. Therefore, 2 feet (the depth) divided by 4 equals 0.5 feet. Hence, the receiver should be placed 0.5 feet from the vertex of the paraboloid, along the axis of symmetry, in order for it to receive signals reflected off the dish.
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Carbon dioxide enters an adiabatic compressor at 100 kPa and 300K at a rate of 0.5 kg/s and leaves at 600 kPa and 450K. Neglecting kinetic energy changes, determine a) the volume flow rate of the carbon dioxide at the compressor inlet (Ans. around 0.3 m3/s) and b) the power input to the compressor (Ans. around 70 kW).
Explanation:
Below is an attachment containing the solution.
Answer:
The answers to the question are
a) The volume flow rate of the carbon dioxide at the compressor inlet is 0.2834 m³/s ≈ 0.3 m³/s
b) The power input to the compressor is 73.35 kW ≈ 70 kW
Explanation:
We note the following
Mass flow rate = 0.5 kg/s
Inlet pressure = 100 pKa
Outlet pressure = 600 kPa
Inlet temperature = 300 K
Outlet temperature = 450 K
Molar mass of CO₂ = 44.01 g/mol
R Universal Gas Constant = 8.314 4621. J K−1 mol−1
a) Number of moles = [tex]\frac{Mass}{Molar.Mass}[/tex] = [tex]\frac{500g}{44.01g}[/tex] = 11.361 moles
P·V= n·R·T ∴ V = [tex]\frac{n*R*T}{P}[/tex] = [tex]\frac{11.361*8.3145*300}{ 100 }[/tex] = 0.2834 m³
Therefore the volume flow rate = 0.2834 m³/s ≈ 0.3 m³/s
b) Cp at 300 K = 0.846 kJ/(kg K)
Cp at 600 K = 0.978 kJ/(kg K)
Cv = 0.657
K = 1.289
While the power input to the compressor can be calculated by
m'×Cp×(T₂-T₁)
Where m' = mass flow rate = 0.5 kg/s
Therefore power = 0.5 kg/s×0.978 kJ/(kg K)×(450 K - 300 K)
= 73.35 kJ/s = 73.35 kW ≈ 70 kW
A cyclist is coasting at 13 m/s when she starts down a 460 m long slope that is 30 m high. The cyclist and her bicycle have a combined mass of70 kg. A steady 12 N drag force due to air resistance acts on her as she coasts all the way to the bottom
What is her speed at the bottom of the slope?
Express your answer to two significant figures and include the appropriate units.
.
Answer:
Explanation:
Given that
Initial velocity u=13m/s
Length of slope
L=460m
Height of slope =30m
Mass of cyclist and bike =70kg
Drag force, fictional force=12 N
Final velocity?
Because the system is not isolated, there is some workdone by the drag force.
Therefore,
∆E=W
K.E(f) - K.E(i) + P.E(f) - P.E(i)=W
½mVf² - ½mVi² + mgy(f) - mgy(i)=W
Note, y(f) = 0, the cyclists is already on the floor
½mVf² -½mVi² - mgy(i) = -Fd × d
½×70×Vf² - ½×70×13²-70×9.81×30=-12×450
35Vf²- 5915 - 20601=-5400
35Vf²=-5400+5915+20601
35Vf²=21116
Vf²=21116/35
Vf²=603.314
Vf=√603.314
Vf=24.56m/s
The final velocity is 24.46m/s at the bottom of the track.
Jumping up before the elevator hits. After the cable snaps and the safety system fails, an elevator cab free-falls from a height of 30.0 m. During the collision at the bottom of the elevator shaft, a 86.0 kg passenger is stopped in 5.00 ms. (Assume that neither the passenger nor the cab rebounds.) What are the magnitudes of the (a) impulse and (b) average force on the passenger during the collision
Explanation:
Below is an attachment containing the solution.
A car of mass m = 1030 kg is traveling down a θ = 13-degree incline. When the car's speed is v0 = 14 m/s, a mechanical failure causes all four of its brakes to lock. The coefficient of kinetic friction between the tires and road is μk = 0.45.
a. Write an expression for the magnitude of the force of kinetic friction
b. Write an expression for the magnitude of the change in the car's height, h, along the y-direction, assuming it travels a distance L down the incline.
c. Calculate the distance the car travels down the hill 1 in meters until it comes to a stop at the end
Answer:
a. [tex]F_f = \mu mg cos\theta[/tex]
b. h = Lsinθ
c. 22.78 m
Explanation:
a. The kinetic friction is the product of kinetic coefficient and normal force N, which is the gravity force in the direction normal to the incline
[tex]F_f = \mu N = \mu mg cos\theta[/tex]
b. As the car travels a distance L down the incline of θ degrees, vertically speaking it would have traveled a distance of:
h = Lsinθ
As we can treat L and h in a right triangle where L is the hypotenuse and h is a side length in opposite of incline angle θ
c. Let g = 9.81 m/s2. the acceleration caused by kinetic friction according to Newton's 2nd law is
[tex]a = F_f/m = \mu g cos\theta = 0.45*9.81*cos13^o = 4.3 m/s^2[/tex]
We can use the following equation of motion to find out the distance traveled by the car:
[tex]v^2 - v_0^2 = 2a\Delta s[/tex]
where v = 0 m/s is the final velocity of the car when it stops, [tex]v_0[/tex] = 14m/s is the initial velocity of the car when it starts braking, a = -4.3 m/s2 is the deceleration of the car, and [tex]\Delta s[/tex] is the distance traveled, which we care looking for:
[tex]0^2 - 14^2 = 2(-4.3)\Delta s[/tex]
[tex]\Delta s = 14^2 / (2*4.3) = 22.78 m[/tex]
a. An expression for the magnitude of the force of kinetic friction is [tex]\(f_{\text{friction}} = \mu_k \cdot N\).[/tex]
b. An expression for the magnitude of the change in the car's height, h, along the y-direction, assuming it travels a distance L down the incline is: [tex]\(h = L \cdot \sin(\theta)\).[/tex]
c. The car will travel approximately 94.69 meters down the hill before coming to a stop due to the locked brakes.
The detailed explanation is as follows:
a. The magnitude of the force of kinetic friction can be calculated using the formula:
[tex]\(f_{\text{friction}} = \mu_k \cdot N\),[/tex]
Where:
[tex]\(f_{\text{friction}}\)[/tex] is the force of kinetic friction,
[tex]\(\mu_k\)[/tex] is the coefficient of kinetic friction (given as 0.45),
[tex]\(N\)[/tex] is the normal force.
The normal force can be calculated using the equation:
[tex]\(N = m \cdot g \cdot \cos(\theta)\),[/tex]
Where:
[tex]\(m\)[/tex] is the mass of the car (1030 kg),
[tex]\(g\)[/tex] is the acceleration due to gravity (approximately 9.81 m/s²),
[tex]\(\theta\)[/tex] is the angle of the incline (13 degrees converted to radians).
b. The magnitude of the change in the car's height [tex](\(h\))[/tex] along the y-direction can be found using trigonometry. When the car travels a distance [tex]\(L\)[/tex] down the incline, the vertical displacement [tex](\(h\))[/tex] can be calculated as:
[tex]\(h = L \cdot \sin(\theta)\).[/tex]
c. To calculate the distance the car travels down the hill until it comes to a stop, you can use the work-energy theorem. The work done by the force of kinetic friction will be equal to the initial kinetic energy of the car. The work-energy theorem is given as:
[tex]\(W = \Delta KE\),[/tex]
Where:
[tex]\(W\)[/tex] is the work done by friction (negative, as it opposes motion),
[tex]\(\Delta KE\)[/tex] is the change in kinetic energy.
The initial kinetic energy is:
[tex]\(KE_0 = \frac{1}{2} m v_0^2\).[/tex]
The final kinetic energy is zero because the car comes to a stop.
So, the work done by friction is:
[tex]\(W = -\frac{1}{2} m v_0^2\).[/tex]
Now, you can use the work-energy theorem to find the distance \(L\) down the incline:
[tex]\(W = -\frac{1}{2} m v_0^2 = \Delta KE = KE_f - KE_0\),[/tex]
Where [tex]\(KE_f = 0\)[/tex] (final kinetic energy).
Solve for [tex]\(L\):[/tex]
[tex]\(-\frac{1}{2} m v_0^2 = -\mu_k m g L \cos(\theta)\).[/tex]
Now, solve for [tex]\(L\):[/tex]
[tex]\[L = \frac{v_0^2}{2 \mu_k g \cos(\theta)}.\][/tex]
Substitute the known values:
[tex]\[L = \frac{(14 m/s)^2}{2 \cdot 0.45 \cdot 9.81 m/s^2 \cdot \cos(13^\circ)} \approx 94.69 \, \text{meters}.\][/tex]
So, the car will travel approximately 94.69 meters down the hill before coming to a stop due to the locked brakes.
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A bird sits on a high-voltage power line with its feet 5.000 cm apart. The wire is made from aluminum, is 2.00 cm in diameter, and carries a current of 170.0 A. What is the potential difference between the bird's feet?
The potential difference between the bird's feet is 0.71 mV
Explanation:
Given-
Distance between the feet, x = 5 cm = 0.05 m
Diameter of the wire = 2 cm
Radius, r = 2/2 = 1 cm = 0.01 m
Current, I = 170 A
Potential difference, ΔV = ?
We know,
ΔV = IR
Where,
ΔV is the potential difference
I is the current
R is the resistance of the wire
And also
R = ρ L/ A
where,
ρ is the resistivity of aluminium wire (ρ = 2.65 X 10⁻⁸ Ωm)
L is the length
A is the area
Equating both the equations,
ΔV = I * ρL/A (The value of R is replaced in the first equation)
ΔV =
[tex]170 * \frac{(2.65 X 10^-^8) * 0.05}{\pi (0.01)^2} \\\\\frac{22.525 X 10^-^8}{0.000314} \\\\\frac{22.525 X 10^6}{314 X 10^8} \\\\0.0717 X 10^-^2\\\\7.17 X 10^-^4[/tex]
Therefore, the potential difference between the bird's feet is 0.71 mV
The electron current in a horizontal metal wire is 7.4 × 1018 electrons/s, and the electrons are moving to the left. What are the magnitude and direction of the conventional current?
Answer:
The magnitude and direction of the conventional current is then 1.186 Amps moving to the left direction
Explanation:
To answer the question, it should be noted that the direction of conventional current is in the opposite direction of the flow of electrons. Therefore, the direction of flow of conventional current will be to the right
The magnitude of the electric current is equal to the rate of flow of the electrons or the time it takes for the electrons to flow past a section of the wire. Therefore the magnitude is that of the 7.4 × 1018 electrons/s
However the unit of the electricity = ampere which is = coulombs/seconds
The 7.4 × 1018 electrons carry
7.4 × 10¹⁸×1.60217662 × 10⁻¹⁹ coulombs = 1.1856106988 coulombs
Therefore the magnitude of electric current = 1.186 coulombs/Seconds
= 1.186 Amps
Explanation:
Below is an attachment containing the solution.
Which vessels have a tunica media with relatively more smooth muscle than elastic tissue, and an elastic membrane on each face of the tunica media?
Answer:
Muscular Arteries
Explanation:
Muscular arteries continue from elastic arteries and control the distribution of blood throughout the body.
Muscular arteries are the vessels that have more smooth muscle than elastic tissue in their tunica media, along with an elastic membrane on each face of the tunica media. They are found farther away from the heart and have a significant role in vasoconstriction due to their increased amount of smooth muscle. They have an internal and external elastic membrane.
Explanation:The vessels that have a tunica media with more smooth muscle than elastic tissue and an elastic membrane on each face of the tunica media are known as muscular arteries. These arteries exist farther from the heart and due to their increased amount of smooth muscle, they play a significant role in vasoconstriction.
In these arteries, the percentage of elastic fibers decreases, while the presence of smooth muscle increases. This results to the artery having a thick tunica media. It's important to note that the diameter of muscular arteries can range from 0.1mm to 10mm.
Additionally, muscular arteries possess an internal elastic membrane (also known as the internal elastic lamina) at the boundary with the tunica media, as well as an external elastic membrane in larger vessels. This gives these arteries increased structure while allowing them the ability to stretch. Due to the decreased blood pressure, muscles arteries can accommodate, elasticity is less crucial in these types of vessels.
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Most of the visible light we see coming from the sun originates from the
Answer:
The answer for this is photosphere.
Explanation:
Most of the visible light we see coming from the sun originates from photosphere.The Photosphere is 300km dense and the temperature at the bottom of the Photosphere is 6400K and the top of the Photosphere is 4600K respectively.
Following are the feature of photosphere that is given below.
Limb Darkening: The edges are darker than the centre part of the sun.Sunspots: The size of the sunspots is similar to the size of the Earth.
Two small balls, A and B, attract each other gravitationally with a force of magnitude F. If we now double both masses and the separation of the balls, what will now be the magnitude of the attractive force on each one?A) 16F
B) 8F
C) 4F
D) F
E) F/4
Answer:
D) F
Explanation:
Let m and M be the mass of the balls A and B respectively and r be the distance between the two balls. The magnitude of attractive gravitational force experienced by the balls due to each other is given by the relation :
[tex]F=\frac{GMm}{r^{2} }[/tex] ......(1)
Now, if the masses of both the balls gets doubled as well as there separation distance also gets doubled, then let F₁ be the new gravitational force acting on them.
Since, New mass of ball A = 2M
New mass of ball b = 2m
Distance between the two balls = 2r
Substitute these values in equation (1).
[tex]F_{1} =\frac{G(2M)(2m)}{(2r)^{2} }[/tex]
[tex]F_{1} =\frac{4GMm}{4r^{2} }=\frac{GMm}{r^{2} }[/tex]
Using equation (1) in the above equation.
F₁ = F
The area of each plate of a parallel plate capacitor is 0.021 m2. The plates are 2.75 mm apart with a dielectric material (κ = 3.0) between them. The maximum possible electric field between the plates is 3.25 ✕ 105 V/m.What is the maximum energy that can be stored in the capacitor?
Explanation:
The given data is as follows.
Dielectric constant, K = 3.0
Area of the plates (A) = 0.021 [tex]m^{2}[/tex]
Distance between plates (d) = [tex]2.75 \times 10^{-3} m[/tex]
Maximum electric field (E) = [tex]3.25 \times 10^{5} V/m[/tex]
Now, we will calculate the capacitance as follows.
C = [tex]\frac{k \epsilon_{o} \times A}{d}[/tex]
= [tex]\frac{3.0 \times 8.85 \times 10^{-12} \times 0.021}{2.75 \times 10^{-3}}[/tex]
= [tex]\frac{0.55755 \times 10^{-12}}{2.75 \times 10^{-3}}[/tex]
= [tex]0.203 \times 10^{-9}[/tex] F
Formula to calculate electric charge is as follows.
E = [tex]\frac{\sigma}{k \epsilon_{o}}[/tex]
or, Q = [tex]E \times k \times \epsilon_{o}A[/tex] (as [tex]\frac{\sigma}{\epsilon_{o}} = \frac{Q}{A}[/tex])
= [tex]3.25 \times 10^{5} \times 3.0 \times 8.85 \times 10^{-12} \times 0.021[/tex]
= [tex]181.2 \times 10^{-9} C[/tex]
Formula to calculate the energy is as follows.
U = [tex]\frac{1 \times Q^{2}}{2 \times C}[/tex]
= [tex]\frac{(181.2 \times 10^{-9} C)^{2}}{2 \times 1.6691 \times 10^{-9}}[/tex]
= [tex]\frac{32833.44 \times 10^{-18}}{3.3382 \times 10^{-9}}[/tex]
= [tex]9835.67 \times 10^{-9}[/tex]
or, = [tex]98.35 \times 10^{-7} J[/tex]
Thus, we can conclude that the maximum energy that can be stored in the capacitor is [tex]98.35 \times 10^{-7} J[/tex].
You are driving home from school steadily at 95 km/h for 180km. It then begins to rain and you slow to 65 km/h. You arrive home after driving 4.5 h. (a) How far is your hometown from school? (b) What was your average speed?
Answer:
the total trajectory length is 349.39 km
Explanation:
for the first trajectory
Time taken in first trajectory = First trajectory length /velocity = 180 km/95 km/h = 1.894 hours
therefore since the total time is 4.5 hours
Time taken in second trajectory = Second trajectory length /velocity
4.5 hours- 1.894 hours = Second trajectory length / 65 km/h
Second trajectory length = 169.39 km
therefore the total trajectory length is 180 km + 169.39 km = 349.39 km
Explanation:
Below is an attachment containing the solution.
Using a good pair of binoculars, you observe a section of the sky where there are stars of many different apparent brightnesses. You find one star that appears especially dim. This star looks dim because it is:_______.
Answer:
Farther,
Because the stars are far from one another
Explanation:
The star look dim because a star's brightness also depends on its proximity to us. The more distant an object is, the dimmer it appears.
The sun appears very bright to us because it is closer to us, the sun distance from the earth is one light year which is around 92,955,807 miles. Now the closest star to the earth is 4.22 light-years, which is four times that of the sun and so it slowly spread out over time.
Therefore, if two objects have the same level of brightness, but one is farther away, the closer star will appear brighter than the more distant star - even though they are equally bright!
The same applies to star.
Show that the kinetic energy of a particle of mass m is related to the magnitude of the momentum p of that particle by KE 5 p2/2m. (Note: This expression is invalid for particles traveling at speeds near that of light.)
Answer:
Kinetic energy: [tex]E=\frac{1}{2}mv^{2}[/tex]
Momentum: p = mv
Kinetic energy in terms of momentum: [tex]E=\frac{1}{2}\frac{(mv)^{2}}{m}=\frac{p^{2}}{2m}[/tex]
Explanation:
The kinetic energy is given by this equation:
[tex]E=\frac{1}{2}mv^{2}[/tex] (1)
Now, we know that the momentum of a particle is p = m*v. This equation is true only with a classical particle, it meas particles with a speed less than the speed of light. If we had a particle traveling at speeds near that of light, the momentum would be p = γm₀v, where γ is the Lorentz factor.
So, if we see, we can rewrite the equation (1) to get this expression in terms of p.
Let's multiply and divide by mass (m) in the equation (1).
[tex]E=\frac{1}{2}\frac{m^{2}v^{2}}{m}[/tex]
[tex]E=\frac{1}{2}\frac{(mv)^{2}}{m}[/tex]
Using the p = mv here:
[tex]E=\frac{1}{2}\frac{p^{2}}{m}[/tex]
[tex]E=\frac{p^{2}}{2m}[/tex]
Therefore the kinetic energy can express in terms of momentum.
Let's see that it could not be possible using the the relativistic momentum, because it has a relativistic factor.
I hope it helps you!
The free throw line in basketball is 4.570 m (15 ft) from the basket, which is 3.050 m (10 ft) above the floor. A player standing on the free throw line throws the ball with an initial speed of 7.157 m/s, releasing it at a height of 2.440 m above the floor. At what angle above the horizontal must the ball be thrown to exactly hit the basket?
Answer:
[tex]\theta = 86.491^{\textdegree}[/tex]
Explanation:
The equations for the horizontal and vertical position of the ball are, respectivelly:
[tex]4.570\,m = [(7.157\,\frac{m}{s})\cdot\cos \theta]\cdot t\\3.050\,m = 2.440\,m +[(7.157\,\frac{m}{s})\cdot \sin \theta]\cdot t - \frac{1}{2}\cdot (9.807\,\frac{m}{s^{2}} )\cdot t^{2}[/tex]
By isolating each trigonometric component and summing each equation:
[tex]20.885\,m^{2} = [51.223\,\frac{m^{2}}{s^{2}}\cdot \cos^{2} \theta]\cdot t^{2}[/tex]
[tex][0.61\,m + \frac{1}{2}\cdot (9.807\,\frac{m}{s^{2}} )\cdot t^{2}]^{2} = [51.223\,\frac{m^{2}}{s^{2}}\cdot \sin^{2} \theta]\cdot t^{2}[/tex]
[tex]21.257\,m^{2} + (5.982\,\frac{m^{2}}{s^{2}})\cdot t^{2}+(24.044\,\frac{m^{2}}{s^{4}} )\cdot t^{4} = (2623.796\,\frac{m^{2}}{s^{2}})\cdot t^{2}[/tex]
[tex]21.257\,m^{2} - (2617.814\,\frac{m^{2}}{s^{2}})\cdot t^{2}+(24.044\,\frac{m^{2}}{s^{4}} )\cdot t^{4} = 0[/tex]
The positive real roots are:
[tex]t_{1} = 10.434\,s,t_{2} = 0.09\,s[/tex]
The needed angle is:
[tex]\theta = \cos^{-1} [\frac{4.570\,m}{(7.157\,\frac{m}{s} )\cdot t} ]\\\theta_{1} = 86.491^{\textdegree}\\\theta_{2} = NaN[/tex]
10. A satellites is in a circular orbit around the earth at a height of 360 km above the earth’s surface. What is its time period? What is its orbital speed?
Answer:
Orbital speed=8102.39m/s
Time period=2935.98seconds
Explanation:
For the satellite to be in a stable orbit at a height, h, its centripetal acceleration V2R+h must equal the acceleration due to gravity at that distance from the center of the earth g(R2(R+h)2)
V2R+h=g(R2(R+h)2)
V=√g(R2R+h)
V= sqrt(9.8 × (6371000)^2/(6371000+360000)
V= sqrt(9.8× (4.059×10^13/6731000)
V=sqrt(65648789.18)
V= 8102.39m/s
Time period ,T= sqrt(4× pi×R^3)/(G× Mcentral)
T= sqrt(4×3.142×(6.47×10^6)^3/(6.673×10^-11)×(5.98×10^24)
T=sqrt(3.40×10^21)/ (3.99×10^14)
T= sqrt(0.862×10^7)
T= 2935.98seconds
You have been hired to help improve the material movement system at a manufacturing plant. Boxes containing 16 kg of tomato sauce in glass jars must slide from rest down a frictionless roller ramp to the loading dock, but they must not accelerate at a rate that exceeds 2.6 m/s2 because of safety concerns.a. What is the maximum angle of inclination of the ramp?b. If the vertical distance the ramp must span is 1.4 m, with what speed will the boxes exit the bottom of the ramp?c. What is the normal force on a box as it moves down the ramp?
a) [tex]15.4^{\circ}[/tex]
b) 5.2 m/s
c) 151.2 N
Explanation:
a)
When the box is on the frictionless ramp, there is only one force acting in the direction along the ramp: the component of the forc of gravity parallel to the ramp, which is given by
[tex]mg sin \theta[/tex]
where
m =16 kg is the mass of the box
[tex]g=9.8 m/s^2[/tex] is the acceleration due to gravity
[tex]\theta[/tex] is the angle of the ramp
According to Newton's second law of motion, the net force on the box is equal to the product of mass and acceleration, so:
[tex]F=ma\\mgsin \theta = ma[/tex]
where a is the acceleration.
From the equation above we get
[tex]a=g sin \theta[/tex]
And we are told that the acceleration must not exceed
[tex]a=2.6 m/s^2[/tex]
Substituting this value and solving for [tex]\theta[/tex], we find the maximum angle of the ramp:
[tex]\theta=sin^{-1}(\frac{a}{g})=sin^{-1}(\frac{2.6}{9.8})=15.4^{\circ}[/tex]
b)
Here we are told that the vertical distance of the ramp is
[tex]h=1.4 m[/tex]
Since there are no frictional forces acting on the box, the total mechanical energy of the box is conserved: this means that the initial gravitational potential energy of the box at the top must be equal to the kinetic energy of the box at the bottom of the ramp.
So we have:
[tex]GPE=KE\\mgh=\frac{1}{2}mv^2[/tex]
where:
m = 16 kg is the mass of the box
[tex]g=9.8 m/s^2[/tex]
h = 1.4 m height of the ramp
v = final speed of the box at the bottom of the ramp
Solving for v,
[tex]v=\sqrt{2gh}=\sqrt{2(9.8)(1.4)}=5.2 m/s[/tex]
c)
There are two forces acting on the box in the direction perpendicular to the ramp:
- The normal force, N, upward
- The component of the weight perpendicular to the ramp, downward, of magnitude
[tex]mg cos \theta[/tex]
Since the box is in equilibrium along the perpendicular direction, the net force is zero, so we can write:
[tex]N-mg cos \theta[/tex]
and by substituting:
m = 16 kg
[tex]g=9.8 m/s^2[/tex]
[tex]\theta=15.4^{\circ}[/tex]
We can find the normal force:
[tex]N=mg cos \theta=(16)(9.8)cos(15.4^{\circ})=151.2 N[/tex]
The maximum angle of inclination of the ramp and the final velocity of the boxes are calculated based on the given acceleration and distance. The normal force on a box moving down the ramp remains unaffected by the angle of inclination in this case as the ramp is devoid of friction. It's an application of physics concepts in real-world situations.
Explanation:This problem is a practical application of the concepts of physics, specifically mechanics. Let's break it down.
a. The maximum angle of inclination of the ramp can be found by utilizing the relationship between the acceleration, the gravitational constant, and the angle of inclination. The formula is as follow:
sin(θ) = acceleration / g
Substitute the given acceleration (2.6 m/s²) and the gravitational constant g (9.8 m/s²) and solve for θ.
b. To find the final velocity of the boxes, we can apply the equations of motion. Using the formula v² = u² + 2gs (where u is the initial velocity, g is the gravitational constant, s is the distance) and substituting the given values (u=0, g=2.6 m/s², s=1.4m), we find the final velocity.
c. The normal force on a box moving down the ramp can be found from the formula: Normal force = mg cos (θ) (where m is the mass, g is gravitational constant and θ is the angle of inclination). Here, θ does not affect the normal force because the box is moving down a frictionless ramp.
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What is the angular velocity of the second hand on a clock? (Hint: It takes the second hand 6060 seconds to rotate 2π2π radians. Divide the number of radians by the number of seconds so that your answer has units of radians/second.)
Explanation:
Below is an attachment containing the solution.
A cosmic ray (an electron or nucleus moving ar speeds close to the speed of light) travels across the Milky Way at a speed of 0.98 times the spee of light. If the Milky way is 30,000 pc across, how long does this journey take in our (essentially stationary) frame of reference and the cosmic ray's frame of reference, in years?
Answer:
Cosmic ray's frame of reference: 99,875 years
Stationary frame of reference: 501,891 years
Explanation:
First of all, we convert the distance from parsec into metres:
[tex]d=30,000 pc =9.26\cdot 10^{20} m[/tex]
The speed of the cosmic ray is
[tex]v=0.98 c[/tex]
where
[tex]c=3.0 \cdot 10^8 m/s[/tex] is the speed of light. Substituting,
[tex]v=(0.98)(3.0\cdot 10^8)=2.94\cdot 10^8 m/s[/tex]
And so, the time taken to complete the journey in the cosmic's ray frame of reference (called proper time) is:
[tex]T_0 = \frac{d}{v}=\frac{9.26\cdot 10^{20}}{2.94\cdot 10^8}=3.15\cdot 10^{12} s[/tex]
Converting into years,
[tex]T_0 = \frac{3.15\cdot 10^{12}}{(365\cdot 24\cdot 60 \cdot 60}=99,875 years[/tex]
Instead, the time elapsed in the stationary frame of reference is given by Lorentz transformation:
[tex]T=\frac{T_0}{\sqrt{1-(\frac{v}{c^2})^2}}[/tex]
And substituting v = 0.98c, we find:
[tex]T=\frac{99,875}{\sqrt{1-(\frac{0.98c}{c})^2}}=501,891 years[/tex]
On your first trip to Planet X you happen to take along a 290g mass, a 40-cm-long spring, a meter stick, and a stopwatch. You're curious about the free-fall acceleration on Planet X, where ordinary tasks seem easier than on earth, but you can't find this information in your Visitor's Guide. One night you suspend the spring from the ceiling in your room and hang the mass from it. You find that the mass stretches the spring by 21.1cm . You then pull the mass down 11.2cm and release it. With the stopwatch you find that 11 oscillations take 18.2sCan you now satisfy your curiosity?what is the new g?
Answer:
Explanation:
11 oscillations in 18.2 s
Time period is defined as the time taken to complete one oscillation.
T = 18.2 / 11 = 1.655 s
mass, m = 290 g = 0.29 kg
Δx = 21.1 cm = 0.211 m
ω = 2π / T = (2 x 3.14) / 1.655 = 3.796 rad/s
[tex]\omega =\sqrt{\frac{K}{m}}[/tex]
Where, K is the spring constant
K = ω² m = 3.796 x 3.796 x 0.29 = 4.18 N/m
Now, mg = K Δx
0.29 x g = 4.18 x 0.211
g = 3.04 m/s²
Two cars having different weights are traveling on a level surface at different constant velocities. Within the same time interval, greater force will always be required to stop the car that has greater
Answer:
The answer to the question is
Momentum
Explanation:
Momentum in physics refer to the attribute that a body has by virtue of its mass and velocity. It is found by multiplying the mass of the moving object and the velocity, hence
Momentum = Mass × Velocity = m·v
Newton first law of motion states that the force acting on an object is proportional to the rate of change of momentum produced
Therefore when we find the momentum of the two cars, the one that has the greater momentum will require the most force to stop it.
Momentum is a physics term; it refers to the quantity of motion that an object has. A sports team that is on the move has the momentum. If an object is in motion (on the move) then it has momentum
Momentum
A generator uses a coil that has 100 turns and a 0.50-T magnetic field. The frequency of this generator is 60.0 Hz, and its emf has an rms value of 120 V. Assuming that each turn of the coil is a square (an approximation), determine the length of the wire from which the coil is made
Answer:
38 m
Explanation:
Number of turns=N=100
Magnetic field=B=0.50 T
Frequency of the generator=f=60 Hz
Rms value of emf=[tex]E_{rms}=120 V[/tex]
We have to find the length of the wire from which the coil is made.
Peak value of emf=[tex]E_0=E_{rms}\sqrt 2=120\times \sqrt 2=169.7 V[/tex]
Length of wire=[tex]4\sqrt{\frac{NE_0}{2\pi fB}}[/tex]
Substitute the values
Length of wire=[tex]4\times \sqrt{\frac{169.7\times 100}{0.50\times 2\pi\times 60}}[/tex]
Length of wire=38 m
Hence, the length of wire from which the coil is made=38 m
To determine the length of wire in a generator coil with 100 turns in a 0.50-T field and 120 V rms at 60 Hz, one can use the formula for rms emf of a generator and solve for the area of one turn to find the length per turn and multiply by the number of turns.
Explanation:The question is asking to determine the length of wire used to make a coil in a generator. The generator has 100 turns of wire, operates with a 0.50-T magnetic field, and has an rms value of the emf of 120 V with a 60.0 Hz frequency. Assuming the turns are squares, we can use the formula for the rms value of the emf (Erms) for a generator, which is Erms = NABωrms, where N is the number of turns, A is the area of the turn, B is the magnetic field, and ωrms is the rms angular velocity. The rms angular velocity ωrms is related to the frequency (f) by the equation ωrms = 2πf/√2.
To find the side length (L) of the square turns, we rearrange the formula to solve for A and then take the square root. Once we have L, we multiply by 4 to get the perimeter of one turn and then by 100 to find the total length of wire needed for all turns.
(a) Use the de Broglie relation λ = h/p to nd the wavelength of a raindrop with mass 1 mg and speed 1 cm/s. Is there a way to set up a raindrop-diraction experiment and detect the wave-like properties of rain?
Question:
a). Use the de Broglie relation λ=h/p to find the wavelength of a raindrop with mass m=1 mg and speed 1cm/s.
ii). Does it seem likely that the wave properties of a raindrop could be easily detected?
b). Find the wavelength of electrons with KE = 500 eV.
c). If a neutron has the same wavelength as blue light (λ=450 nm) what is it's KE?
ii). What if it's an electron?
Answer:
The answers to the question are
a). The wavelength of the raindrop is 6.626*10⁻²⁶ m
The properties of the rain drop will be hardly detected
b). The wavelength of the electrons is 5.491×10⁻¹¹ m
c). The KE of the neutron is 5.242510⁻²⁸ J
ii). For an electron it will increase to be KE (electron) = 9.6392639×10⁻²⁵ J
Explanation:
Using de Broglie relations, we have
p = h/λ and E = h·f also E = 1/2·m·v²
a). λ= h/p, E= p²/2·m, p = √(2·m·E), λ = h/√(2·m·E)
Where
λ=wavelength
E = energy
p = momentum
m = mass
The kinetic energy of the rain drop is [tex]\frac{1}{2}[/tex]×m×v² = 0.5×(1×10⁻⁶)(0.01)2
= 5× 10⁻¹¹ J
λ = h/√(2·m·E) = 6.626*10-34 Js/√(2×1×10⁻⁶×5× 10⁻¹¹)
= 6.626*10⁻²⁶ m
The properties of the rain drop will not be easily detected
b). The electron energy is equivalent to 500 eV ⇒500 eV × 1.6×10⁻¹⁹ J/eV
= 8×10⁻¹⁷ J
λ = h/√(2·m·E) = 6.626×10⁻³⁴ Js/√(2*×9.1×10⁻³¹×8×10⁻¹⁷)
= 5.491×10⁻¹¹ m
c). λ = h/√(2·m·E) then √(2·m·E) = h/ λ or E = (h/λ)²/(2·m)
= (6.626×10⁻³⁴/5.0×10⁻⁷)²/(2×1.674927471×10⁻²⁷)
E = 5.242510⁻²⁸ J
ii). For an electron, we have m = 9.10938356 × 10⁻³¹ kg
λ = (6.626×10⁻³⁴/5.0×10⁻⁷)²/(2×9.10938356×10⁻³¹) = 9.6392639×10⁻²⁵ J
A boulder is raised above the ground, so that its potential energy relative to the ground is 200 J. Then it is dropped. Estimate what its kinetic energy will be just before hitting the ground.
Answer:
200 J
Explanation:
In this problem, I assume there is no air resistance, so the system is isolated (=no external forces).
For an isolated system, the total mechanical energy is constant, and it is given by:
[tex]E=KE+PE[/tex]
where
KE is the kinetic energy
PE is the potential energy
The kinetic energy is the energy due to the motion of the object, while the potential energy is the energy due to the position of the object relative to the ground.
At the beginning, when the boulder is raised above the ground, its height above the ground is maximum, while its speed is zero; it means that all its mechanical energy is just potential energy, and it is:
[tex]E=PE_{max}=200 J[/tex]
As the boulder falls down, its altitude decreases, so its potential energy decreases, while the speed increases, and the kinetic energy increases. Therefore, potential energy is converted into kinetic energy.
Eventually, just before the boulder hits the ground, the height of the object is zero, and the speed is maximum; this means that all the energy has now converted into kinetic energy, and we have
[tex]E=KE_{max}=200 J[/tex]
Therefore, the kinetic energy just before hitting the ground is 200 J.
A proton (????p = +????, mp = 1.0 u; where u = unified mass unit ≃ 1.66 × 10−27kg), a deuteron (???????? = +????, m???? = 2.0 u) and an alpha particle (???????? = +2????, m???? = 4.0 u) are accelerated from rest through the same potential difference ????, and then enter the same region of uniform magnetic field ????⃗⃗ , moving perpendicularly to the direction of the magnetic field.
a. What is the ratio of the proton’s kinetic energy Kp to the alpha particle’s kinetic energy K?????
b. What is the ratio of the deuteron’s kinetic energy K???? to the alpha particle’s kinetic energy K?????
c. If the radius of the proton’s circular orbit ????p = 10 cm, what is the radius of the deuteron’s orbit ?????????
d. What is the radius of the alpha particle’s orbit ?????????
Answer:
a. 1/2 b. 1/2 c, 20 cm d. 40 cm
Explanation:
Here is the complete question
A proton ( = +, = 1.0 u; where u = unified mass unit ≃ 1.66 × 10−27kg), a deuteron ( = +, = 2.0 u) and an alpha particle ( = +2, = 4.0 u) are accelerated from rest through the same potential difference , and then enter the same region of uniform magnetic field ⃗⃗ , moving perpendicularly to the direction of the magnetic field.
A) What is the ratio of the proton’s kinetic energy to the alpha particle’s kinetic energy ?
B) What is the ratio of the deuteron’s kinetic energy to the alpha particle’s kinetic energy ?
C) If the radius of the proton’s circular orbit = 10 cm, what is the radius of the deuteron’s orbit ?
D) What is the radius of the alpha particle’s orbit ?
Solution
a. For both particles, kinetic energy = electric potential energy
For proton K.E= K₁ = 1/2m₁v₁² = +eV , for alpha particle K.E = K₂ = 1/2m₂v₂²= +2eV
where m₁, m₂ and v₁, v₂ are the respective masses and velocities of the proton and alpha particle. So, the ratio of their kinetic energies is
1/2m₁v₁²/1/2m₂v₂² = +eV/+2eV
m₁v₁²/m₂v₂² = 1/2.
So the ratio K₁/K₂ = 1/2
b. For both particles, kinetic energy = electric potential energy
For deuteron K₁ = 1/2m₁v₁² = +eV , for alpha particle K₂ = 1/2m₂v₂²= +2eV
where m₁, m₂ and v₁, v₂ are the respective masses and velocities of the deuteron and alpha particle. So, the ratio of their kinetic energies is
1/2m₁v₁²/1/2m₂v₂² = +eV/+2eV
m₁v₁²/m₂v₂² = 1/2.
So the ratio K₁/K₂ = 1/2
c. The radius of the proton's circular is gotten from the centripetal force which equal the magnetic force. So,
mv²/r = Bev
r₁ = mv/Be
Since mass of deuteron m₂ equals twice mass of proton m₁, m₂ = 2m₁
So, radius of deuteron's circular orbit equals
r₂ = m₂v/Be = 2m₁v/Be = 2r₁ = 2 × 10 cm = 20 cm
d. The radius of the alpha particle is given by r₃ = m₃v/Be. Since mass of alpha particle equal four times mass of proton, m₃ = 4m₁.
So, radius of alpha particle's circular orbit equals
r₃ = m₃v/Be = 4m₁v/Be = 4r₁ = 4 × 10 cm = 40 cm
Why is magnesium the limiting reactant in this experiment
Answer:
Explanation:
Magnesium is being oxidized by the oxygen in the air to magnesium oxide. This is a highly exothermic combustion reaction, giving off intense heat and light. The reaction of the combustion of magnesium in oxygen is given below: 2Mg(s) + O2(g) → 2MgO(s) The stoichiometric factor is 2 moles of magnesium are burned for every 1 mole of oxygen (2mol Mg/1mol O2). If the magnesium strip weighs 1 gram, then there is 0.04 mol of magnesium (1 gram divided by 24.3 grams/mol Mg) available in the reaction. The amount of oxygen required to completely react with the magnesium strip is:0.04 mol Mg x (1 mol O2 / 2 mol Mg) = 0.02 mol O2 x 16 g/mol O2 = 0.32 gram O2.The magnesium will burn until consumed entirely. There is much more oxygen available in the atmosphere than needed to consume the magnesium. Thus the magnesium is the limiting reactant because it determines the amount of product formed.If the raindrops hit at 8.3 m/s , estimate the magnitude of the force on the bottom of a 1.0-m2 pan due to the impacting rain which we assume does not rebound.
Answer:
F = 0.1153 N
Explanation:
Given:
- Rain fall rate = 5 cm/hr
- Velocity before impact vi = 8.3 m/s
- The Area of the pan A = 1.0 m^2
- The density of water ρ = 1000 kg/m^3
Find:
Estimate the magnitude of the force
Solution:
- Consider rain drops impacting a surface looses all its momentum. So the change in its momentum is just the momentum with which it impacted.
- We do not know the size of each drop. But the rain fall rate allows us to calculate the rate of change of momentum.
- Total Force: The total force experience by the surface due to the momentum transfer from the impacting rain drops is:
F = m(Δv) / t = ρ*v*A*Δh/Δt
Where
Δh/Δt = rain-fall rate.
F = 1000*8.3*1*0.05 / 3600
F = 0.1153 N
The magnitude of the force is 0.1153N
Given-
The velocity of the raindrop=8.3m/sec
Let when the raindrop hits the surface it loses its momentum
Now the magnitude of the force on the bottom of the given area can be calculated as
[tex]F=\dfrac{m\bigtriangleup v}{t}[/tex]
[tex]F= \rho vA \frac{\bigtriangleup h}{\bigtriangleup t}[/tex]
here rainfall rate
=[tex]\dfrac{\bigtriangleup h}{\bigtriangleup t}[/tex]
therefore,
[tex]F=1000\times 8.3\times 1 \times \dfrac{0.05}{3600}[/tex]
[tex]F=0.1153N[/tex]
Hence the magnitude of the force is 0.1153N
For more detail about the momentum, follo9w the link
https://brainly.com/question/4956182
A large truck collides head-on with a small car. The car is severely damaged as a result of the collision. According to Newton's third law, how do the forces acting between the truck and car compare during the collision
Answer: Force on the truck is equal to force on the car.
Explanation: According to the Newton's third law of motion which states that; For every action, there is an equal and opposite reaction. These pair of forces are regarded as action - reaction forces. These size or magnitude of the forces on the colliding objects are equal or the same, while the direction of the colliding objects are opposite.
In the scenario above, both truck and carry have the same mass, however, the damage suffered by the car is based on its smaller mass which makes it unable to withstand the acceleration resulting from the collision.
ma = m(-a)
m= mass, a= acceleration
an object of unknown mass oscillates on the end of a spring wit period 8 s. A 10kg object is attached to the first object, changing te period to 12 s. What is the mass of the first object
Answer:
Explanation:
Let the first object have a mass of M
And a period of T1=8sec
The second object has a mass 10kg and a period of T2=12 sec
It is know that,
The period of a spring-mass system is proportional to the square root of the mass and inversely proportional to the square root of the spring constant.
T=2π√(m/k)
Then the constant in this equation is the spring constant (k) and 2π, which does not change for the same material.
Then, make k subject of formulas
T²=4π²(m/k)
T²k=4π²m
Then, k/4π²=m/T²
So the k is directly proportional to m and inversely proportional to T²
M1/T1²=M2/T2²
Since, M1 is unknown, M2=10kg, T1=8sec and T2=12
Then,
M1/T1²=M2/T2²
M1/8²=10/12²
M1/64=0.06944
M1=0.06944×64
M1=4.444kg
The mass of the first object is 4.44kg
Calculate the work done (in J) by a 90.0 kg man who pushes a crate 4.25 m up along a ramp that makes an angle of 20.0° with the horizontal (see below). He exerts a force of 535 N on the crate parallel to the ramp and moves at a constant speed. Be certain to include the work he does on the crate and on his body to get up the ramp.
Answer:
W = 3.4x0³ J.
Explanation:
The work done by the man is given by the following equation:
[tex] W = F_{t}\cdot d [/tex] (1)
where W: is the work, Ft is the total force and d: is the displacement = 4.25 m.
We need to find first the total force Ft, which is:
[tex] Ft = Fm + W [/tex]
where Fm: is the force exerted by the man = 535 N, W: is the weight = m*g*sin(θ), m: is the mass of the man, g: is the gravitational acceleration = 9.81 m/s², and θ: is the angle = 20.0°.
[tex] F_{t} = Fm + W = 500 N + 90.0 kg*9.81 m/s^{2} * sin(20.0) = 802.0 N [/tex]
Hence, the work is:
[tex] W = 802.0 N \cdot 4.25 m = 3.4 \cdot 10 ^{3} J [/tex]
Therefore, the work done by the man is 3.4x10³ J.
I hope it helps you!
A car accelerates uniformly from rest and reaches a speed of 21.5 m/s in 11.4 s. The diameter of a tire is 66.5 cm. Find the number of revolutions the tire makes during this motion, assuming no slipping. Answer in units of rev.
Answer:
57.39 rev
Explanation:
From circular motion,
s = rθ................... Equation 1
Where s = distance, r = radius, θ = angular distance.
make θ the subject of the equation
θ = s/r............... Equation 2
Where can look for s using any of the equation of motion
s = (v+u)t/2............ Equation 3
Where v and u = Final and initial velocity respectively, t= time.
Given: v = 21.5 m/s, u = 0 m/s (at rest), t = 11.4 s
Substitute into equation 3
s = (21.5+0)11.4/2
s = 122.55 m.
given: r = 66.5/2 = 33.25 cm = 0.3325 m
Substitute into equation 2
θ = 122.55/0.3325
θ = 368.57 rad
θ = (360.57×0.159155) rev
θ = 57.39 rev
Answer:
58.6886 revolutions
Explanation:
First we need to know the total distance travelled by the car, and we can do that using Torricelli formula:
V2= Vo2 + 2aDS
V = 21.5
Vo = 0
a = 21.5/11.4 = 1.886
(21.5)^2 = 2*1.886*DS
DS = 462.25/3.772 = 122.5477 m
For each revolution of the tire, the car moves the circunference of the tire, which is pi*d = 3.14*66.5 = 208.81 cm = 2.0881 m
So, to know the number of revolutions, we divide the total travel distance by the circunference of the tire:
122.5477/2.0881 = 58.6886