Answer:
The hot temperature is 157.5 K
The cold temperature is 48.8 K
Explanation:
Step 1: Data given
The working substance of a certain Carnot engine is 1.50 mol of an ideal monatomic gas.
The volume increases by a factor of 5.7
The work output of the engine is 940 J in each cycle.
During the isothermal expansion portion of this engine's cycle, the volume of the gas doubles. This means V2 = 2*V1 (and V4 = 2*V3)
Step 2:For a carnot engine:
V2/V1 = V4/V3
Work = nR((T1)ln(V2/V1) - (T2)ln(V4/V3))
⇒with Work = the work done in the cycle = 940J
⇒with n = the number of moles = 1.50 moles
⇒with R = the gas constant = 8.314 J/mol*K
⇒with T1 = the hot temperature
⇒With T2⇒ the cold temperature
where R = 8.31 J/mol K Gas Constant
940J = 1.5moles * 8.314 J/mol*K * (T1*ln(2) - T2*ln(2)))
940 = 1.5 * 8.314 ln(2) * (T1-T2)
(T1-T2) = 940 / (1.5*8.314*ln(2))
(T1-T2) = 108.7K
For the reversible adiabatic expansion: T2 = T1*(V1/V2)^(R/Cv). Where V2/V1 = 5.7 (Because during the adiabatic expansion the volume increases by a factor of 5.7)
For a monatomic ideal gas, Cv = 3/2R
When we combine both, we'll have:
T2 = T1*(1/5.7)^(R/3/2R)
T2 = T1*(1/5.7)^(2/3)
T2= T1 * 0.31
Since we know that (T1-T2) = 108.7K
we have:
T1 - 0.31T1= 108.7K
0.69T1 = 108.7K
T1 = 157.5K
T2 = 157.5*0.31 = 48.8K
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
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
Explain the following defects of a simple electric cell:
a.Polarization,
ß. Local action.
Answer:
Explanation:
The two major defects of simple electric cells causes current supplied to be for short time. These defects are: polarization and local action.
a. Polarization: This is a defect caused by an accumulation of hydrogen bubbles at the positive electrode of the cell. It can be prevented by the use of vent, using a hydrogen absorbing material or the use of a depolarizer.
b. Local Action: This is the gradual wearing away of the electrode due to impurities in the zinc plate. It can be controlled by the amalgamation of the zinc plate before it is used.
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
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.
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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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
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]
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]
can some one help me :< its music
Need ur help,,, :-[ :-{
...... ............ .. ..
Answer:
Graph B express the magnetic relationship of magnetic flux and electronic flow
A uniformly dense solid disk with a mass of 4 kg and a radius of 4 m is free to rotate around an axis that passes through the center of the disk and perpendicular to the plane of the disk. The rotational kinetic energy of the disk is increasing at 21 J/s. If the disk starts from rest through what angular displacement (in rad) will it have rotated after 3.3 s?
Answer:
3.44 rad
Explanation:
The rotational kinetic energy change of the disk is given by ΔK = 1/2I(ω² - ω₀²) where I = rotational inertia of solid sphere = MR²/2 where m = mass of solid disk = 4 kg and R = radius of solid disk = 4 m, ω₀ = initial angular speed of disk = 0 rad/s (since it starts from rest) and ω = final angular speed of disk
Since the kinetic energy is increasing at a rate of 21 J/s, the increase in kinetic energy in 3.3 s is ΔK = 21 J/s × 3.3 s = 69.3 J
So, ΔK = 1/2I(ω² - ω₀²)
Since ω₀ = 0 rad/s
ΔK = 1/2I(ω² - 0)
ΔK = 1/2Iω²
ΔK = 1/2(MR²/2)ω²
ΔK = MR²ω²/4
ω² = (4ΔK/MR²)
ω = √(4ΔK/MR²)
ω = 2√(ΔK/MR²)
Substituting the values of the variables into the equation, we have
ω = 2√(ΔK/MR²)
ω = 2√(69.3 J/( 4 kg × (4 m)²))
ω = 2√(69.3 J/[ 4 kg × 16 m²])
ω = 2√(69.3 J/64 kgm²)
ω = 2√(1.083 J/kgm²)
ω = 2 × 1.041 rad/s
ω = 2.082 rad/s
The angular displacement θ is gotten from
θ = ω₀t + 1/2αt² where ω₀ = initial angular speed = 0 rad/s (since it starts from rest), t = time of rotation = 3.3 s and α = angular acceleration = (ω - ω₀)/t = (2.082 rad/s - 0 rad/s)/3.3 s = 2.082 rad/s ÷ 3.3 s = 0.631 rad/s²
Substituting the values of the variables into the equation, we have
θ = ω₀t + 1/2αt²
θ = 0 rad/s × 3.3 s + 1/2 × 0.631 rad/s² (3.3 s)²
θ = 0 rad + 1/2 × 0.631 rad/s² × 10.89 s²
θ = 1/2 × 6.87159 rad
θ = 3.436 rad
θ ≅ 3.44 rad
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
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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The working substance of a certain Carnot engine is 1.90 of an ideal
monatomic gas. During the isothermal expansion portion of this engine's
cycle, the volume of the gas doubles, while during the adiabatic expansion
the volume increases by a factor of 5.7. The work output of the engine is
930 in each cycle.
Compute the temperatures of the two reservoirs between which this engine
operates.
Answer:
Explanation:
The energy for an isothermal expansion can be computed as:
[tex]\mathsf{Q_H =nRTIn (\dfrac{V_b}{V_a})}[/tex] --- (1)
However, we are being told that the volume of the gas is twice itself when undergoing adiabatic expansion. This implies that:
[tex]V_b = 2V_a[/tex]
Equation (1) can be written as:
[tex]\mathtt{Q_H = nRT_H In (2)}[/tex]
Also, in a Carnot engine, the efficiency can be computed as:
[tex]\mathtt{e = 1 - \dfrac{T_L}{T_H}}[/tex]
[tex]e = \dfrac{T_H-T_L}{T_H}[/tex]
In addition to that, for any heat engine, the efficiency e =[tex]\dfrac{W}{Q_H}[/tex]
relating the above two equations together, we have:
[tex]\dfrac{T_H-T_L}{T_H} = \dfrac{W}{Q_H}[/tex]
Making the work done (W) the subject:
[tex]W = Q_H \Big(\dfrac{T_H-T_L}{T_H} \Big)[/tex]
From equation (1):
[tex]\mathsf{W = nRT_HIn(2) \Big(\dfrac{T_H-T_L}{T_H} \Big)}[/tex]
[tex]\mathsf{W = nRIn(2) \Big(T_H-T_L} \Big)}[/tex]
If we consider the adiabatic expansion as well:
[tex]PV^y[/tex] = constant
i.e.
[tex]P_bV_b^y = P_cV_c^y[/tex]
From ideal gas PV = nRT
we can have:
[tex]\dfrac{nRT_H}{V_b}(V_b^y)= \dfrac{nRT_L}{V_c}(V_c^y)[/tex]
[tex]T_H = T_L \Big(\dfrac{V_c}{V_b}\Big)^{y-1}[/tex]
From the question, let us recall aw we are being informed that:
If the volumes changes by a factor = 5.7
Then, it implies that:
[tex]\Big(\dfrac{V_c}{V_b}\Big) = 5.7[/tex]
∴
[tex]T_H = T_L (5.7)^{y-1}[/tex]
In an ideal monoatomic gas [tex]\gamma = 1.6[/tex]
As such:
[tex]T_H = T_L (5.7)^{1.6-1}[/tex]
[tex]T_H = T_L (5.7)^{0.67}[/tex]
Replacing the value of [tex]T_H = T_L (5.7)^{0.67}[/tex] into equation [tex]\mathsf{W = nRIn(2) \Big(T_H-T_L} \Big)}[/tex]
[tex]\mathsf{W = nRT_L In(2) (5.7 ^{0.67 }-1}})[/tex]
From in the question:
W = 930 J and the moles = 1.90
using 8.314 as constant
Then:
[tex]\mathsf{930 = (1.90)(8.314)T_L In(2) (5.7 ^{0.67 }-1}})[/tex]
[tex]\mathsf{930 = 15.7966\times 1.5315 (T_L )})[/tex]
[tex]\mathsf{T_L= \dfrac{930 }{15.7966\times 1.5315}}[/tex]
[tex]\mathbf{T_L \simeq = 39 \ K}[/tex]
From [tex]T_H = T_L (5.7)^{0.67}[/tex]
[tex]\mathsf{T_H = 39 (5.7)^{0.67}}[/tex]
[tex]\mathbf{T_H \simeq 125K}[/tex]
The earth's radius is about 4000 miles. Kampala, the capital of Uganda, and Singapore are both nearly on the equator. The distance between them is 5000 miles as measured along the earth's surface.
a. Through what angle do you turn, relative to the earth, if you fly from Kampala to Singapore? Give your answer in both radians and degrees.
b. The flight from Kampala to Singapore take 9 hours. What is the plane's angular speed relative to the earth?
Answer:
a) the required angle in both radian and degree is 1.25 rad and 71.6°
b) the plane's angular speed relative to the earth is 3.86 × 10⁻⁵ rad/sec
Explanation:
Given the data in the question;
a)
we know that The expression for the angle subtended by an arc of circle at the center of the circle is,
θ = Length / radius
given that Length is 5000 miles and radius is 4000 miles
we substitute
θ = 5000 miles / 4000 miles
θ = 1.25 rad
Radian to Degree
θ = 1.25 rad × ( 180° / π rad )
θ = 71.6°
Therefore, required angle in both radian and degree is 1.25 rad and 71.6°
b)
The flight from Kampala to Singapore take 9 hours.
the plane's angular speed relative to the earth = ?
we know that, the relation between angular velocity and angular displacement is;
ω = θ / t
given that θ is 1.25 rads and time t is 9 hours or ( 9 × 3600 sec ) = 32400 sec
we substitute
ω = 1.25 rad / 32400 sec
ω = 3.86 × 10⁻⁵ rad/sec
Therefore, the plane's angular speed relative to the earth is 3.86 × 10⁻⁵ rad/sec
A grade 12 Physics student shoots a basketball
from the ground at a hoop which is 2.0 m above
her release. The shot was at a velocity of 10 m/s
and at an angle of 80° to the ground.
a. Determine the vertical velocity of the ball
when it is at the level of the net. You
should get two answers.
Please show ALL steps
Answer:
7.84 m/s
Explanation:
Height, h = 2 m
Initial velocity, u = 10 m/s
Angle, A = 80°
(a) Let the time taken to go to the net is t.
Use second equation of motion
[tex]h = u t + 0.5 at^2\\\\- 2 = - 10 sin 80 t - 4.9 t^2\\\\4.9 t^2 + 9.8 t - 2 = 0 \\\\t= \frac{- 9.8\pm\sqrt{9.8^2 + 4\times 4.9\times 2}}{9.8}\\\\t = \frac{- 9.8 \pm 11.6}{9.8}\\\\t = - 2.2 s , 0.2 s[/tex]
Time cannot be negative.
So, t = 0.2 s
The vertical velocity at t = 0.2 s is
v = u + at
v = 10 sin 80 - 9.8 x0.2
v = 9.8 - 1.96 = 7.84 m/s
A bullet with mass 5.35 g is fired horizontally into a 2.174-kg block attached to a horizontal spring. The spring has a constant 6.17 102 N/m and reaches a maximum compression of 6.34 cm.
(a) Find the initial speed of the bullet-block system.
(b) Find the speed of the bullet.
Answer:
a)[tex]V=1.067\: m/s[/tex]
b)[tex]v=434.65\: m/s [/tex]
Explanation:
a)
Using the conservation of energy between the moment when the bullet hit the block and the maximum compression of the spring.
[tex]\frac{1}{2}MV^{2}=\frac{1}{2}k\Delta x^{2}[/tex]
Where:
M is the bullet-block mass (0.00535 kg + 2.174 kg = 2.17935 kg)V is the speed of the systemk is the spring constant (6.17*10² N/m)Δx is the compression of the spring (0.0634 m)Then, let's find the initial speed of the bullet-block system.
[tex]V^{2}=\frac{k\Delta x^{2}}{M}[/tex]
[tex]V=\sqrt{\frac{6.17*10^{2}*0.0634^{2}}{2.17935}}[/tex]
[tex]V=1.067\: m/s[/tex]
b)
Using the conservation of momentum we can find the velocity of the bullet.
[tex]mv=MV[/tex]
[tex]v=\frac{MV}{m}[/tex]
[tex]v=\frac{2.17935*1.067}{0.00535}[/tex]
[tex]v=434.65\: m/s [/tex]
I hope it helps you!
SCALCET8 3.9.018.MI. A spotlight on the ground shines on a wall 12 m away. If a man 2 m tall walks from the spotlight toward the building at a speed of 1.7 m/s, how fast is the length of his shadow on the building decreasing when he is 4 m from the building
Answer:
The length of his shadow is decreasing at a rate of 1.13 m/s
Explanation:
The ray of light hitting the ground forms a right angled triangle of height H, which is the height of the building and width, D which is the distance of the tip of the shadow from the building.
Also, the height of the man, h which is parallel to H forms a right-angled triangle of width, L which is the length of the shadow.
By similar triangles,
H/D = h/L
L = hD/H
Also, when the man is 4 m from the building, the length of his shadow is L = D - 4
So, D - 4 = hD/H
H(D - 4) = hD
H = hD/(D - 4)
Since h = 2 m and D = 12 m,
H = 2 m × 12 m/(12 m - 4 m)
H = 24 m²/8 m
H = 3 m
Since L = hD/H
and h and H are constant, differentiating L with respect to time, we have
dL/dt = d(hD/H)/dt
dL/dt = h(dD/dt)/H
Now dD/dt = velocity(speed) of man = -1.7 m/s ( negative since he is moving towards the building in the negative x - direction)
Since h = 2 m and H = 3 m,
dL/dt = h(dD/dt)/H
dL/dt = 2 m(-1.7 m/s)/3 m
dL/dt = -3.4/3 m/s
dL/dt = -1.13 m/s
So, the length of his shadow is decreasing at a rate of 1.13 m/s
A star has a declination of approximately -90°. in what direction is the Star located from the celestial equator?
East
North
South
West
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.
See more about dipole at brainly.com/question/12757739
A tank is full of water. Find the work (in J) required to pump the water out of the spout. (Use 9.8 m/s2 for g. Use 1,000 kg/m3 as the density of water. Round your answer to the nearest whole number.)
A rescue plane spots a survivor 132 m directly below and releases an emergency kit with a parachute. If the package descends at a constant vertical acceleration of 6.89 m/s2 the initial plane horizontal speed was 86.9 m/s, how far away from the survivor will it hit the waves
Answer: 19.15 meters on the waves away from the survivor.
Explanation:
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
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 the volume of cuboid of side 4cm. Convert it in SI form
Answer:
0.000064 cubic meters.
Explanation:
Given the following data;
Length of side = 4 centimeters
Conversion:
100 centimeters = 1 meters
4 cm = 4/100 = 0.04 meters
To find the volume of cuboid;
Mathematically, the volume of a cuboid is given by the formula;
Volume of cuboid = length * width * height
However, when all the sides are equal the formula is;
Volume of cuboid = L³
Volume of cuboid = 0.04³
Volume of cuboid = 0.000064 cubic meters.
a circuit shown below is Wheastone Bridge used to determine the valve of unknown resistor X by comparison with three resistors M,N,P whose resistances can be varied. For each setting, the resistances of each resistor is precisely known. With switches k1and k2 closed, these resistors are varied until the current in the galvanometer G is zero; the bridge is then said to be balanced. (a) if the galvanometer G shows zero deflection when M=850.0, N=15.00 and P=33.48, what is the unknown resistance X?
Answer:
X = 0.6
Explanation:
The resistance of the unknown resistor can be found by using the formula of the Wheatstone bridge:
[tex]\frac{M}{N}=\frac{P}{X}\\\\\frac{850}{15} = \frac{33.48}{X}\\\\X = \frac{(33.48)(15)}{850}[/tex]
X = 0.6
Hence, the unknown value of resistance is found to be 0.6 units.
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
what is the force of a body which have mass of 7 kg
Answer:
Force acting on a body of mass 7 kg which produces an accceleration of 10 m/s2 is 70 N
Answer:
10 m/s2 or 70 newtons.
Explanation:
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