Answer:
Xcm = (6 * 0 + 4 & 3 + 2 * 0) / 12 = 1
Ycm = (6 * 0 + 4 * 0 + 2 * 3) / 12 = 1/2
(Xcm , Ycm) = (1 , 1/2)
Using definition of center of mass
A caris initially at rest starts moving with a constant acceleration of 0.5 m/s2 and travels a distance of 5 m. Find
(i) Final velocity
(ii)The time taken
Answer:
(I)
[tex] { \bf{ {v}^{2} = {u}^{2} - 2as }} \\ {v}^{2} = {0}^{2} - (2 \times 0.5 \times 5) \\ {v}^{2} = 5 \\ { \tt{final \: velocity = 2.24 \: {ms}^{ - 1} }}[/tex]
(ii)
[tex]{ \bf{v = u + at}} \\ 2.24 = 0 + (0.5t) \\ { \tt{time = 4.48 \: seconds}}[/tex]
A magnetohydrodynamic (MHD) drive works by applying a magnetic field to a fluid which is carrying an electric current.
a. True
b. False
Answer:
True
Explanation:
A magnetohydrodynamic drive or MHD accelerator is a method which is used for propelling the vehicles using only by applying the electric and magnetic fields. It has no moving parts. It accelerates an electrically conductive propellant (liquid or gas) with magnetohydrodynamics.
Its working principle is same as an electric motor except that in an MHD drive, the moving rotor is replaced by the fluid acting directly as the propellant.
An MHD accelerator is reversible.
So, the statement is true.
4. Paper is solid in packets labelled 80 g/m2. This means that a sheet of paper of area
10 000cm? has a mass of 80 g. The thickness of each sheet is 0.11mm. What is the
density of the paper?
A 0.073 g/cm?
B 0.088 g/cm
C 0.73 g/cm3
D 0.88 g/cm
B
с
Answer:
Option C. 0.73 g/cm³
Explanation:
From the question given above, the following data were obtained:
Mass = 80 g
Area (A) = 10000 cm²
Thickness = 0.11 mm
Density =?
Next, we shall convert 0.11 mm to cm. This can be obtained as follow:
10 mm = 1 cm
Therefore,
0.11 mm = 0.11 mm × 1 cm / 10 mm
0.11 mm = 0.011 cm
Thus, 0.11 mm is equivalent to 0.011 cm.
Next, we shall determine the volume of the paper. This can be obtained as follow:
Area (A) = 10000 cm²
Thickness = 0.011 cm
Volume =?
Volume = Area × Thickness
Volume = 10000 × 0.011
Volume = 110 cm³
Finally, we shall determine the density of the paper. This can be obtained as follow:
Mass = 80 g
Volume = 110 cm³
Density =?
Density = mass / volume
Density = 80 / 110
Density = 0.73 g/cm³
Therefore the density of the paper is 0.73 g/cm³
A beam of light has a wavelength of 549nm in a material of refractive index 1.50. In a different material of refractive index 1.07, its wavelength will be:_________.
Explanation:
someone to check if the answer is correct
Betelgeuse (in Orion) has a parallax of 0.00451 + 0.00080 arcsec,as measured by the Hipparcos satellite. What is the distance to Betelgeuse, and what is the uncertainty in that measurement?
We have that the distance to Betelgeuse, and the uncertainty in that measurement is
[tex]d=(221.7\pm39.33)pc[/tex]Uncertainty U = 0.00080
From the Question we are told that
Betelgeuse (in Orion) has a parallax of 0.00451 + 0.00080
Generally
[tex]Distance\ in\ parsecs =\frac{ 1}{(parallax\ measured\ in\ arcseconds}[/tex]
Where
Parallax [tex]P =0.00451[/tex]
Uncertainty [tex]U = 0.00080[/tex]
Generally the equation for the distance is mathematically given as
[tex]d=(\frac{1}{P}pc\pm(\frac{U}{P}*100\%))[/tex]
Therefore
[tex]d=(\frac{1}{0.00451}pc\pm(\frac{0.00080}{0.00451}*100\%))[/tex]
[tex]d=(221.7\pm39.33)pc[/tex]
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Which indicates the first law of thermodynamics
Answer:
(d)
Explanation:
because dU = Q -W so ,that the option d(D) is correct
Please show steps as to how to solve this problem
Thank you!
Explanation:
Let x = distance of [tex]F_1[/tex] from the fulcrum and let's assume that the counterclockwise direction is positive. In order to attain equilibrium, the net torque [tex]\tau_{net}[/tex] about the fulcrum is zero:
[tex]\tau_{net} = -F_1x + F_2d_2 = 0[/tex]
[tex] -m_1gx + m_2gd_2 = 0[/tex]
[tex]m_1x = m_2d_2[/tex]
Solving for x,
[tex]x = \dfrac{m_2}{m_1}d_2[/tex]
[tex]\:\:\:\:=\left(\dfrac{105.7\:\text{g}}{65.7\:\text{g}} \right)(13.8\:\text{cm}) = 22.2\:\text{cm}[/tex]
There are two possible alignments of a dipole in an external electric field where the dipole is in equilibrium: when the dipole moment is parallel to the electric field and when the dipole moment is oriented opposite the electric field.
Part A
Are both alignments stable? (Consider what would happen in each case if you gave the dipole a slight twist.)
a) Yes
b) No
Part B
Based on your answer to the previous part and your experience in mechanics, in which orientation does the dipole have less potential energy?
a) The arrangement with the dipole moment parallel to the electric field has less potential energy.
b) The arrangement with the dipole moment opposite the electric field has less potential energy.
c) Both arrangements have the same potential energy.
Answer:
A. (b)
B. (a)
Explanation:
The electric dipole moment is the product of charge and the length of the dipole.
The torque on the dipole placed in the external electric field is given by
torque = p E sin A
where, p is the electric dipole moment, E is the electric field, A is the angle between the field and dipole moment.
When the dipole moment is parallel to the electric field, the net torque is zero and it is said to be in stable equilibrium.
When the dipole moment is anti parallel to the electric field, the net torque is zero but the dipole is in unstable equilibrium.
So, the option (b) is correct.
Teh energy is given by
U = - p E cos A
When the angle A is zero , the potential energy is negative and it is minimum.
In this exercise we have to use the knowledge about dipole to be able to mark the correct alternative for each question, in this way we find that:
A) Letter b
B) Letter a
So knowing that the electric dipole moment is the product of charge and the length of the dipole and the torque on the dipole placed in the external electric field is given by:
[tex]torque = p E sin (A)[/tex]
where:
p: the electric dipole momentE: the electric fieldA: the angle between the field and dipole momentWhen the dipole moment is parallel to the electric field, the net torque is zero and it is said to be in stable equilibrium. When the dipole moment is anti parallel to the electric field, the net torque is zero but the dipole is in unstable equilibrium.
Now the energy is given by:
[tex]U = - p E cos (A)[/tex]
We can say that when the angle A is zero , the potential energy is negative and it is minimum.
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A uniform horizontal bar of mass m1 and length L is supported by two identical massless strings. String A Both strings are vertical. String A is attached at a distance d
Answer:
a) T_A = [tex]\frac{g}{d}\ ( m_2 x + m_1 \ \frac{L}{2} )[/tex] , b) T_B = g [m₂ ( [tex]\frac{x}{d} -1[/tex]) + m₁ ( [tex]\frac{L}{ 2d} -1[/tex]) ]
c) x = [tex]d - \frac{m_1}{m_2} \ \frac{L}{2d}[/tex], d) m₂ = m₁ ( [tex]\frac{ L}{2d} -1[/tex])
Explanation:
After carefully reading your long sentence, I understand your exercise. In the attachment is a diagram of the assembly described. This is a balancing act
a) The tension of string A is requested
The expression for the rotational equilibrium taking the ends of the bar as the turning point, the counterclockwise rotations are positive
∑ τ = 0
T_A d - W₂ x -W₁ L/2 = 0
T_A = [tex]\frac{g}{d}\ ( m_2 x + m_1 \ \frac{L}{2} )[/tex]
b) the tension in string B
we write the expression of the translational equilibrium
∑ F = 0
T_A - W₂ - W₁ - T_B = 0
T_B = T_A -W₂ - W₁
T_ B = [tex]\frac{g}{d}\ ( m_2 x + m_1 \ \frac{L}{2} )[/tex] - g m₂ - g m₁
T_B = g [m₂ ( [tex]\frac{x}{d} -1[/tex]) + m₁ ( [tex]\frac{L}{ 2d} -1[/tex]) ]
c) The minimum value of x for the system to remain stable, we use the expression for the endowment equilibrium, for this case the axis of rotation is the support point of the chord A, for which we will write the equation for this system
T_A 0 + W₂ (d-x) - W₁ (L / 2-d) - T_B d = 0
at the point that begins to rotate T_B = 0
g m₂ (d -x) - g m₁ (0.5 L -d) + 0 = 0
m₂ (d-x) = m₁ (0.5 L- d)
m₂ x = m₂ d - m₁ (0.5 L- d)
x = [tex]d - \frac{m_1}{m_2} \ \frac{L}{2d}[/tex]
d) The mass of the block for which it is always in equilibrium
this is the mass for which x = 0
0 = d - \frac{m_1}{m_2} \ \frac{L}{2d}
[tex]\frac{m_1}{m_2} \ (0.5L -d) = d[/tex]
[tex]\frac{m_1}{m_2} = \frac{ d}{0.5L-d}[/tex]
m₂ = m₁ [tex]\frac{0.5 L -d}{d}[/tex]
m₂ = m₁ ( [tex]\frac{ L}{2d} -1[/tex])
Click Stop Using the slider set the following: coeff of restitution to 1.00 A velocity (m/s) to 6.0 A mass (kg) to 6.0 B velocity (m/s) to 0.0 Calculate what range can the mass of B be to cause mass A to bounce off after the collision. Calculate what range can the mass of B be to cause mass A to continue forward after the collision. Check your calculations with the simulation. What are the ranges of B mass (kg)
Answer:
[tex]M_b=6kg[/tex]
Explanation:
From the question we are told that:
Coefficient of restitution [tex]\mu=1.00[/tex]
Mass A [tex]M_a=6kg[/tex]
Initial Velocity of A [tex]U_a=6m/s[/tex]
Initial Velocity of B [tex]U_b=0m/s[/tex]
Generally the equation for Coefficient of restitution is mathematically given by
[tex]\mu=\frac{V_b-V_a}{U_a-U_b}[/tex]
[tex]1=\frac{v_B}{6}[/tex]
[tex]V_b=6*1[/tex]
[tex]V_b=6m/s[/tex]
Generally the equation for conservation of linear momentum is mathematically given by
[tex]M_aU_a+M_bU_b=M_aV_a+M_bV_b[/tex]
[tex]6*6+=M_b*6[/tex]
[tex]M_b=6kg[/tex]
A horizontal force of P=100 N is just sufficient to hold the crate from sliding down the plane, and a horizontal force of P=350 N is required to just push the crate up the plane. Determine the coefficient of static friction between the plane and the crate, and find the mass of the crate.
"down/up the plane" suggests an inclined plane, but no angle is given so I'll call it θ for the time being.
The free body diagram for the crate in either scenario is the same, except for the direction in which static friction is exerted on the crate. With the P = 100 N force holding up the crate, static friction points up the incline and keeps the crate from sliding downward. When P = 350 N, the crate is pushed upward, so static friction points down. (see attached FBDs)
Using Newton's second law, we set up the following equations.
• p = 100 N
∑ F (parallel) = f + p cos(θ) - mg sin(θ) = 0
∑ F (perpendicular) = n - p sin(θ) - mg cos(θ) = 0
• P = 350 N
∑ F (parallel) = P cos(θ) - F - mg sin(θ) = 0
∑ F (perpendicular) = N - P sin(θ) - mg cos(θ) = 0
(where n and N are the magnitudes of the normal force in the respective scenarios; ditto for f and F which denote static friction, so that f = µn and F = µN, with µ = coefficient of static friction)
Solve for n and N :
n = p sin(θ) + mg cos(θ)
N = P sin(θ) - mg cos(θ)
Substitute these into the corresponding equations containing µ, and solve for µ :
µ = (mg sin(θ) - p cos(θ)) / (mg cos(θ) + p sin(θ))
µ = (P cos(θ) - mg sin(θ)) / (P sin(θ) + mg cos(θ))
Next, you would set these equal and solve for m :
(mg sin(θ) - p cos(θ)) / (mg cos(θ) + p sin(θ)) = (P cos(θ) - mg sin(θ)) / (P sin(θ) + mg cos(θ))
...
Once you find m, you back-substitute and solve for µ, but as you might expect the result will be pretty complicated. If you take a simple angle like θ = 30°, you would end up with
m ≈ 36.5 kg
µ ≈ 0.256
The coefficient of static friction between the plane and the crate is μ = 0.256 and the mass of the crate is m=36.4 kg.
From the given,
The force that opposes the crate by sliding is P = 100N
In X-axis, the sum of forces is zero.
ΣF = 0
Pcosθ - mgsinθ-Ff = 0
Ff = Pcosθ - mgsinθ
In Y-axis
Psinθ - mgcosθ - N = 0
N = Psinθ-mgcosθ
Frictional force, Ff = μN, μ is the coefficient of friction
Ff = μN
Pcos30- mgsin30 + μ( Psin30+mgcos30) = 0
μ = mgsin30-Pcos30/Psin30+mgcos30 ------1
The block is sliding with the horizontal force, F = 350N
X-axis
P₂cosθ - mgsinθ-Ff = 0
Y-axis
P₂sinθ - mgcosθ - N = 0
N = P₂sinθ-mgcosθ
μ = P₂cos30-mgsin30/P₂sin30-mgcos30 -----2
Equate equations 1 and 2
mgsin30-Pcos30/Psin30+mgcos30 =P₂cos30-mgsin30/P₂sin30-mgcos30
4.905m-86.6/50+8.49 = 303.1-4.905m/175+8.49
41.7m² + 123m - 1.516×10⁴ = 0
-41.7m² +2330m -1.516×10⁴(4.905-86.6)(175+8.49) =(303.1-4.905)(50+8.49)
83.4m² - 2207m -3.03×10⁴ = 0
m= 36.4 kg
Hence, the mass of the crate is 36.4 Kg.
Substitute the value of m in equation 1,
μ = 4.905(36.4) - 86.6 / 50 + 8.49
μ = 0.256
Thus, the coefficient of static friction is 0.256.
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A tennis player receives a shot with the ball (0.0600 kg) traveling horizontally at 59.4 m/s and returns the shot with the ball traveling horizontally at 37.2 m/s in the opposite direction. (Take the direction of the ball's final velocity (toward the net) to be the +x-direction).
(a) What is the impulse delivered to the ball by the racket?
(b) What work does the racket do on the ball?
5 9 . 4
- 3 7 . 2
2 2 . 2
Explanation:
Use the algorithm method.
5 9 . 4
- 3 7 . 2
2 2 . 2
2 Therefore, 59.4-37.2=22.259.4−37.2=22.2.
22.2
22.2
What is the maximum distance the train can travel if it accelerates from rest until it reaches its cruising speed and then runs at that speed for 15 minutes
The question is incomplete. The complete question is :
A high-speed bullet train accelerates and decelerates at the rate of 4 ft/s^2. Its maximum cruising speed is 90 mi/h. What is the maximum distance the train can travel if it accelerates from rest until it reaches its cruising speed and then runs at that speed for 15 minutes?
Solution :
Given :
Speed of the bullet train, v = 90 mi/h
= [tex]$90 \times \frac{5280}{3600}$[/tex]
= 132 ft/s
Time = 15 minutes
= 15 x 60
= 900 s
Acceleration from rest,
[tex]$a(t) = 4 \ ft/s^2$[/tex]
[tex]$v(t) = 4t + C$[/tex]
Since, v(0) = 0, then C = 0, so velocity is
v(t) = 4t ft/s
Then find the position function,
[tex]$s(t) = \frac{4}{2}t^2 + C$[/tex]
[tex]$=2t^2+C$[/tex]
It is at position 0 when t = 0, so C = 0, and the final position function for only the time it is accelerating is :
[tex]$s(t) = 2t^2$[/tex]
Time to get maximum cruising speed is :
4t = 132
t = 33 s
Distance travelled (at cruising speed) by speed to get the remaining distance travelled.
[tex]$900 \ s \times 132 \ \frac{ft}{s} = 118800 \ ft$[/tex]
Total distance travelled, converting back to miles,
[tex]$2178 + 118800 = 120978\ ft . \ \frac{mi}{5280 \ ft}$[/tex]
= 22.9125 mi
Therefore, the distance travelled is 22.9125 miles
ASK YOUR TEACHER A 2.0-kg mass swings at the end of a light string with the length of 3.0 m. Its speed at the lowest point on its circular path is 6.0 m/s. What is its kinetic energy at an instant when the string makes an angle of 50 degree with the vertical
Answer:
K_b = 78 J
Explanation:
For this exercise we can use the conservation of energy relations
starting point. Lowest of the trajectory
Em₀ = K = ½ mv²
final point. When it is at tea = 50º
Em_f = K + U
Em_f = ½ m v_b² + m g h
where h is the height from the lowest point
h = L - L cos 50
Em_f = ½ m v_b² + mg L (1 - cos50)
energy be conserve
Em₀ = Em_f
½ mv² = ½ m v_b² + mg L (1 - cos50)
K_b = ½ m v_b² + mg L (1 - cos50)
let's calculate
K_b = ½ 2.0 6.0² + 2.0 9.8 6.0 (1 - cos50)
K_b = 36 +42.0
K_b = 78 J
PLEASE HELP ME WITH THIS ONE QUESTION
A photon has 2.90 eV of energy. What is the photon’s wavelength? (h = 6.626 x 10^-19, 1 eV = 1.6 x 10^-19 J)
Explanation:
First, we convert the energy from eV to Joules:
[tex]2.90\:\text{eV}×\left(\dfrac{1.6×10{-19}\:J}{1\:\text{eV}} \right)[/tex]
[tex]= 4.64×10^{-19}\:\text{eV}[/tex]
We know from definition that
[tex]E=h\nu = \dfrac{hc}{\lambda}[/tex]
so the wavelength of the photon is
[tex]\lambda = \dfrac{hc}{E} = 4.28×10^8\:\text{m}[/tex]
You have two identical beakers A and B. Each beaker is filled with water to the same height. Beaker B has a rock floating at the surface (like a pumice stone). Which beaker, with all its contents, weighs more. Or are they equal?
Answer:
a) if we assume that the water does not spill, Beaker B weighs more than beaker S, or which in this case Beaker A weighs more
b) If it is spilled in water the weight of the two beakers is the same
Explanation:
The beaker weight is
beaker A
W_total = W_ empty + W_water
Beaker B
W_total = W_ empty + W_water + W_roca
a) if we assume that the water does not spill, Beaker B weighs more than beaker S, or which in this case Beaker A weighs more
b) If it is spilled in water, the weight of the two beakers is the same because the amount of liquid spilled and equal to the weight of the stone, therefore the two beakers weigh the same
Two cylindrical resistors are made from copper. The first one is of length L and of radius r . The 2nd resistor is of length 6L and of radius 2r. The ratio of these two resistances R1/R2 is:
Answer:
[tex]R1/R2=\frac{2}{3}[/tex]
Explanation:
From the question we are told that:
1st's Length [tex]l=L[/tex]
1st's radius [tex]r=r[/tex]
2nd's Length [tex]l=6L[/tex]
2nd's radius [tex]r=2r[/tex]
Generally the equation for Resistance R is mathematically given by
[tex]R=\frac{\rho L}{\pi r^2}[/tex]
Therefore
[tex]R_1=\frac{\rho L}{\pi r^2}[/tex]
And
[tex]R_2=\frac{\rho 6L}{\pi (2r)^2}[/tex]
Therefore
[tex]R1/R2=\frac{\frac{\rho L}{\pi r^2}}{\frac{\rho 6L}{\pi (2r)^2}}[/tex]
[tex]R1/R2=\frac{2}{3}[/tex]
if 145kl of energy is added to water, what mass of water can be heated from 35C to 100C then vaporized at 100C
Answer:
m = 0.057 kg = 57 g
Explanation:
Energy Added to Water = Heat added to raise the temperature of water + Heat used to vaporize water
[tex]E = mC\Delta T + mH\\E = m(C\Delta T + H)[/tex]
where,
E = Energy added to water = 145 KJ
m = mass of water = ?
C = specific heat capacity of water = 4.2 KJ/kg.°C
ΔT = change in temperature = 100°C - 35°C = 65°C
H = Latent heat of vaporization of water = 2260 KJ/kg
Therefore,
[tex]145\ KJ = m[(4.2\ KJ/kg.^oC)(65^oC)+2260\ KJ/kg]\\\\145\ KJ = m(2533\ KJ/kg)\\\\m = \frac{145\ KJ}{2533\ KJ/kg}[/tex]
m = 0.057 kg = 57 g
The mass of water that can be heated is equal to 0.527 kilograms.
Given the following data:
Quantity of energy = 145 kJ = 145,000 Joules.Initial temperature = 35.0°CFinal temperature = 100.0°CScientific data:
Specific heat capacity of water = 4200 J/kg°CLatent heat of vaporization of water = 2260 KJ/kgTo calculate the mass of water that can be heated:
The quantity of energy and heat.Note: The quantity of energy added to water is equal to the quantity of heat used to vaporize water and the quantity of heat that is added to raise the temperature of water.
Mathematically, this is given by this expression:
[tex]E=mc\theta + mH\\\\E= m(c\theta + H)[/tex]
Making m the subject of formula, we have:
[tex]m=\frac{E}{c\theta + H}[/tex]
Substituting the parameters into the formula, we have;
[tex]m=\frac{145000}{[42000\times (100-35)] + 2260}\\\\m=\frac{145000}{(4200\times 65) + 2260}\\\\m=\frac{145000}{273000 + 2260}\\\\m=\frac{145000}{275260}[/tex]
Mass, m = 0.527 kilograms.
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Find the refractive index of a medium
having a velocity of 1.5 x 10^8*
Explanation:
someone to check if the answer is correct
A star has a declination of approximately -90°. in what direction is the Star located from the celestial equator?
East
North
South
West
Match each term with the best description.
a. Tightly woven fabric used to smother and extinguish a fire.
b. Consists of absorbent material that can be ringed around a chemical spill until the spill can be neutralized.
c. Device used to control small fires in an emergency situation
d. Provides chemical. physical. Health, and safety information regarding chemical reagents and supplies
1. Spill containment kit
2. Safety Data sheet
3. Fume hood
4. Fire extinguisher
5. Fire blanket
Answer:
A - 5
B - 1
C - 4
D -2
Explanation:
I don't have one i just know...
The fire blanket is a tightly woven fabric. The spill containment kit consists of absorbent material. Fire extinguishers control small fires and the safety data sheet provides chemical, health, and safety information.
(a) The fire blanket is a blanket, which may be quickly thrown over a fire to snuff out the flames, and comprises fire-resistant materials.
Hence, option (a) matches with option (5)
(b) In order to contain a chemical spill, absorbent items like pads, socks, or booms are frequently included in spill containment kits.
Hence, option (b) correctly matches with option (1).
(c) A fire extinguisher is a tool used to put out small fires during emergencies.
Hence, option (c) correctly matches with option (4).
(d) A Safety Data Sheet (SDS) gives in-depth details regarding a specific chemical or chemical mixture. It provides information about the physical characteristics of the chemical, any potential risks, safe handling and storage practices, emergency response strategies, and more.
Hence, option (d) correctly matches option (2).
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can some one help me :< its music
Need ur help,,, :-[ :-{
...... ............ .. ..
Answer:
Graph B express the magnetic relationship of magnetic flux and electronic flow
A horse gallops a distance of 10 kilometers in a time of 30 minutes its average speed is?
Answer:
20 km/hr
Explanation:
Distance = 10km
Time = 30 minutes = 1/2 hour
Average Speed = Total distance / Total Time Taken
= 10 ÷ 1/2
= 10 x 2
= 20 km/hr
Average speed = (distance covered) / (time to cover the distance)
Average speed = (10 km) / (30 minutes)
Average speed = 1/3 km/min
Most people would probably want to see it in a more convenient, more familiar unit, such as km/hour or m/second.
(10 km / 30 min) x (60 min / hour) = (10 x 60 / 30) (km-min / min-hour)
Average speed = 20 km/hour
AvgSpd = (10 km / 30 min) x (1,000 m / km) x (min / 60 sec)
AvgSpd = (10x1,000 / 30x60) (km-m-min / min-km-sec)
Averge Speed = 5.56 m/s
Which of the following is a noncontact force?
O A. Friction between your hands
O B. A man pushing on a wall
O C. Air resistance on a car
D. Gravity between you and the Sun
Answer:
Gravity between you and the sun
NASA is giving serious consideration to the concept of solar sailing. A solar sailcraft uses a large, low mass sail and the energy and momentum of sunlight for propulsion. (a) Should the sail be absorbing or reflective
Answer:
Reflective
Explanation:
The radiation pressure of the wave that totally absorbed is given by;
[tex]P_{abs}= \frac{I}{C}[/tex]
and While the radiation pressure of the wave totally reflected is given by;
[tex]P_{ref}= \frac{2I}{C}[/tex]
Now compare the two-equation you can clearly see that the pressure due to reflection is larger than absorption therefore the sail should be reflective.
find out the odd one and give reason (length, volume, time, mass
Answer:
Time
Explanation:
The answer to the question is actually time. Time is not needed when you calculate the mass or volume of an object, a square, sphere, rectangle, or any other 3D shape. You must also calculate the length to know what numbers you will be multiplying by. The answer to the question is time.
Two cars are moving. The first car has twice the mass of the second car but only half as much kinetic energy. When both cars increase their speed by 2.76 m/s, they then have the same kinetic energy. Calculate the original speeds of the two cars.
Let m be the mass of the second car, so the first car's mass is 2m.
Let K be the kinetic energy of the second car, so the first car's kinetic energy would be K/2.
Let u and v be the speeds of the first car and the second car, respectively. At the start,
• the first car has kinetic energy
K/2 = 1/2 (2m) u ² = mu ² ==> K = 2mu ²
• the second car starts with kinetic energy
K = 1/2 mv ²
It follows that
2mu ² = 1/2 mv ²
==> 4u ² = v ²
When their speeds are both increased by 2.76 m/s,
• the first car now has kinetic energy
1/2 (2m) (u + 2.76 m/s)² = m (u + 2.76 m/s)²
• the second car now has kinetic energy
1/2 m (v + 2.76 m/s)²
These two kinetic energies are equal, so
m (u + 2.76 m/s)² = 1/2 m (v + 2.76 m/s)²
==> 2 (u + 2.76 m/s)² = (v + 2.76 m/s)²
Solving the equations in bold gives u ≈ 1.95 m/s and v ≈ 3.90 m/s.
A bag of cement of Weight 1000N hangs from ropes. Two of the ropes make angles of 1=60 and 2=30 with the horizontal.if the system is in equilibrium,find the tension T1,T2andT3 in the ropes
Answer:
T1 = 499.9N, T2 = 865.8N, T3 = 1000N
Explanation:
To find the tensions we need to find the vertical and horizontal components of T1 and T2
T1x = T1 cos60⁰, T1y = T1 sin60⁰
Also, T2x = T2 cos30⁰, T2y = T2 sin30⁰
For the forces to be in equilibrium,
the sum of vertical forces must be zero and the sum of horizontal forces must also be zero
Sum of Fx = 0
That is, T1x - T2x=0
NB: T2x is being subtracted because T1x and T2x are in opposite directions
T1 cos60⁰ - T2 cos30⁰ = 0
0.866T1 - 0.5T2 = 0 ............ (1)
Sum of Fy = 0
T1y + T2y - 1000 = 0
T1 sin60⁰ + T2 sin30⁰ - 1000 = 0
NB: The weight of the bag of cement is also being subtracted because it's in an opposite direction.
0.5T1 - 0.866T2 - 1000 = 0 ........(2)
From (1)
make T1 the subject
T1 = 0.5T2/0.866
Substitute T1 into (2)
0.5 (0.5T2/0.866) - 0.866T2 = 1000
(0.25/0.866)T2 - 0.866T2 = 1000
0.289T2 - 0.866T2 = 1000
1.155T2 = 1000
T2 = 865.8N
Then T1 = 0.5 x 865.8 / 0.866
T1 = 499.9N
T3 = 1000N
NB: The weight of the bag is the Tension above the rope, which is T3
A person with a near point of 85 cm, but excellent distant vision, normally wears corrective glasses. But he loses them while traveling. Fortunately, he has his old pair as a spare.
(a) If the lenses of the old pair have a power of +2.25 diopters, what is his near point (measured from his eye) when he is wearing the old glasses if they rest 2.0 cm in front of his eye?
(b) What would his near point be if his old glasses were contact lenses instead?
Answer:
a) p = 95.66 cm, b) p = 93.13 cm
Explanation:
For this problem we use the constructor equation
[tex]\frac{1}{f} = \frac{1}{p} + \frac{1}{q}[/tex]
where f is the focal length, p and q are the distances to the object and the image, respectively
the power of the lens is
P = 1 / f
f = 1 / P
f = 1 / 2.25
f = 0.4444 m
the distance to the object is
[tex]\frac{1}{p} = \frac{1}{f} -\frac{1}{q}[/tex]
the distance to the image is
q = 85 -2
q = 83 cm
we must have all the magnitudes in the same units
f = 0.4444 m = 44.44 cm
we calculate
[tex]\frac{1}{p} = \frac{1}{44.44} - \frac{1}{83}[/tex]
1 / p = 0.010454
p = 95.66 cm
b) if they were contact lenses
q = 85 cm
[tex]\frac{1}{p} = \frac{1}{44.44} - \frac{1}{85}[/tex]
1 / p = 0.107375
p = 93.13 cm