Answer:
the diameter of the bright circular spot formed is 0.787 m
Explanation:
Given that;
Radius of the flat circular mirror = 0.100 m
height of small ight source = 0.920 m
high ceiling = 2.70 m
now;
Diameter(mirror) = 2×r = 2 × 0.100 = 0.2 m
D(spot) = [Diameter(mirror) × ( 2.70m + 0.920 m)] / 0.920 m
so
D(spot) = 0.2m × 3.62m / 0.920 m
D(spot) = 0.724 m / 0.920 m
D(spot) = 0.787 m
Therefore, the diameter of the bright circular spot formed is 0.787 m
Sandy is riding a bicycle with tires that have a diameter of 650 mm. A small twig, caught in the spokes, causes the tire to click once each revolution. Of Sandy hears 8 such clicks every 3 seconds then how fast is she cycling (to the nearest km/hr)
Answer:
Explanation:
Sandy hears 8 such clicks every 3 seconds and a small twig, caught in the spokes, causes the tire to click once each revolution that means the wheel of the cycle is rotating at 8 rotations every 3 seconds or 8/3 rotation per second . In each rotation , it moves distance equal to its circumference .
circumference = 2π r = 2 x 3.14 x .65 / 2 m
= 2.041 m
In 8/3 rotation , distance covered = 8/3 x 2.041 = 5.44 m
So speed of cycle is 5.44 m per second
5.44 x 60 x 60 m per hour
19584 m per hour
= 19.584 km per hour .
= 20 km per hour approx.
which of the following is used to answer scientific questions?
A. Experiments
B. Intuition
C. Opinion polls
D. Imagination
A three-phase line, which has an impedance of (2 + j4) ohm per phase, feeds two balanced three-phase loads that are connected in parallel. One of the loads is Y-connected with an impedance of (30 + j40) ohm per phase, and the other is connected with an impedance of (60 - j45) ohm per phase. The line is energized at the sending end from a 60-Hz, three-phase, balanced voltage source of 120 √3V (rms, line-to-line).
Determine:
a. the current, real power and reactive power delivered by the sending-end source
b. the line-to-line voltage at the load
c. the current per phase in each load
d. the total three-phase real and reactive powers absorbed by each load and by the
Answer:
hello your question has a missing information
The other is Δ-connected with an impedance of (60 - j45) ohm per phase.
answer : A) 5A ∠0° ,
p( real power ) = 1800 and Q ( reactive power ) = 0 VAR
B) 193.64 v
C) current at load 1 = 2.236 A , current at load 2 = 4.472 A
D) Load 1 : 450 watts(real power ) , 600 VAR ( reactive power )
Load 2 : 1200 watts ( real power ), -900 VAR ( reactive power )
Explanation:
First convert the Δ-connection to Y- connection attached below is the conversion and pre-solution
A) determine the current, real power and reactive power delivered by the sending-end source
current power delivered (Is) = 5A ∠0°
complex power delivered ( s ) = 3vs Is
= 3 * 120∠0° * 5∠0° = 1800 + j0 ---- ( 1 )
also s = p + jQ ------ ( 2 )
comparing equation 1 and 2
p( real power ) = 1800 and Q ( reactive power ) = 0 VAR
B) determine Line-to-line voltage at the load
Vload = √3 * 111.8
= 193.64 v
c) Determine current per phase in each load
[tex]I_{l1} = Vl1 / Zl1[/tex]
= [tex]\frac{111.8<-10.3}{50<53.13}[/tex] = 2.236∠ 63.43° A hence current at load 1 = 2.236 A
[tex]I_{l2} = V_{l2}/Z_{l2}[/tex]
= [tex]\frac{111.8<-10.3}{25<-36.87}[/tex] = 4.472 ∠ 26.57° A hence current at load 2 = 4.472 A
D) Determine the Total three-phase real and reactive powers absorbed by each load
For load 1
3-phase real power = [tex]3I_{l1} ^{2} R_{l1}[/tex] = 3 * 2.236^2 * 30 = 450 watts
3-phase reactive power = [tex]3I_{l1} ^{2} X_{l1}[/tex] = 3 * 2.236^2 * 40 = 600 VAR
for load 2
3-phase real power = [tex]3I_{l1} ^{2} R_{l2}[/tex] = 1200 watts
3-phase reactive power = [tex]3I_{l1} ^{2} X_{l2}[/tex] = -900 VAR
The sum of load powers and line losses, 1800 W+ j0 VAR and The line voltage magnitude at the load terminal is 193.64 V.
(a) The impedance per phase of the equivalent Y,
[tex]\bar{Z}_{2}=\frac{60-j 45}{3}=(20-j 15) \Omega[/tex]
The phase voltage,
[tex]\bold { V_{1}=\frac{120 \sqrt{3}}{\sqrt{3}}=120 VV }[/tex]
Total impedance from the input terminals,
[tex]\bold {\begin{aligned}&\bar{Z}=2+j 4+\frac{(30+j 40)(20-j 15)}{(30+j 40)+(20-j 15)}=2+j 4+22-j 4=24 \Omega \\&\bar{I}=\frac{\bar{V}_{1}}{\bar{Z}}=\frac{120 \angle 0^{\circ}}{24}=5 \angle 0^{\circ} A\end{aligned} }[/tex]
The three-phase complex power supplied [tex]\bold {=\bar{S}=3 \bar{V}_{1} \bar{I}^{*}=1800 W}[/tex]
P =1800 W and Q = 0 VAR delivered by the sending-end source.
(b) Phase voltage at load terminals will be,
[tex]\bold {\begin{aligned}\bar{V}_{2} &=120 \angle 0^{\circ}-(2+j 4)\left(5 \angle 0^{\circ}\right) \\&=110-j 20=111.8 \angle-10.3^{\circ} V\end{aligned} }[/tex]
The line voltage magnitude at the load terminal,
[tex]\bold{\left(V_{ LOAD }\right)_{L-L}=\sqrt{3} 111.8=193.64 V(V }[/tex]
(c) The current per phase in the Y-connected load,
[tex]\bold {\begin{aligned}&\bar{I}_{1}=\frac{\bar{V}_{2}}{\bar{Z}_{1}}=1-j 2=2.236 \angle-63.4^{\circ} A \\&\bar{I}_{2}=\frac{\bar{V}_{2}}{\bar{Z}_{2}}=4+j 2=4.472 \angle 26.56^{\circ} A\end{aligned} }[/tex]
The phase current magnitude,
[tex]\bold {\left(I_{p h}\right)_{\Delta}=\frac{I_{2}}{\sqrt{3}}=\frac{4.472}{\sqrt{3}}=2.582 }[/tex]
(d) The three-phase complex power absorbed by each load,
[tex]\bold {\begin{aligned}&\bar{S}_{1}=3 \bar{V}_{2} \bar{I}_{1}^{*}=430 W +j 600 VAR \\&\bar{S}_{2}=3 \bar{V}_{2} \bar{I}_{2}^{*}=1200 W -j 900 VAR\end{aligned}}[/tex]
The three-phase complex power absorbed by the line is
[tex]\bold{\bar{S}_{L}=3\left(R_{L}+j X_{L}\right) I^{2}=3(2+j 4)(5)^{2}=150 W +j 300 VAR }[/tex]
Since, the sum of load powers and line losses,
[tex]\bold {\begin{aligned}\bar{S}_{1}+\bar{S}_{2}+\bar{S}_{L} &=(450+j 600)+(1200-j 900)+(150+j 300) \\&=1800 W +j 0 VAR\end{aligned} }[/tex]
To know more about voltage,
https://brainly.com/question/2364325
If the speed of an object does NOT change, the object is traveling at a
constant speed
increasing speed
decreasing speed
Answer:
If the speed does not change at all, the object would be moving at a constant speed.
Energy from the Sun is transferred from the Earth’s surface to the atmosphere, resulting in
atmospheric convection currents that produce winds. How do physical properties of the air
contribute to convection currents?
a -The warmer air sinks because it is more dense than cooler air.
b -The warmer air rises because it is more dense than cooler air.
c- The warmer air sinks because it is less dense than cooler air.
d -The warmer air rises because it is less dense than cooler air.
A 2028 kg Oldsmobile traveling south on Abbott Road at 14.5 m/s is unable to stop on the ice covered intersection for a red light at Saginaw Street. The car collides with a 4146 kg truck hauling animal feed east on Saginaw at 9.7 m/s. The two vehicles remain locked together after the impact. Calculate the velocity of the wreckage immediately after the impact. Give the speed for your first answer and the compass heading for your second answer. (remember, the CAPA abbreviation for degrees is deg) -1.75
Answer:
v = 8.1 m/s
θ = -36.4º (36.4º South of East).
Explanation:
Assuming no external forces acting during the collision (due to the infinitesimal collision time) total momentum must be conserved.Since momentum is a vector, if we project it along two axes perpendicular each other, like the N-S axis (y-axis, positive aiming to the north) and W-E axis (x-axis, positive aiming to the east), momentum must be conserved for these components also.Since the collision is inelastic, we can write these two equations for the momentum conservation, for the x- and the y-axes:We can go with the x-axis first:[tex]p_{ox} = p_{fx} (1)[/tex]
⇒ [tex]m_{tr} * v_{tr}= (m_{olds} + m_{tr}) * v_{fx} (2)[/tex]
Replacing by the givens, we can find vfx as follows:[tex]v_{fx} = \frac{m_{tr}*v_{tr} }{(m_{tr} + m_{olds)} } = \frac{4146kg*9.7m/s}{2028kg+4146 kg} = 6.5 m/s (3)[/tex]
We can repeat the process for the y-axis:[tex]p_{oy} = p_{fy} (4)[/tex]
⇒[tex]m_{olds} * v_{olds}= (m_{olds} + m_{tr}) * v_{fy} (5)[/tex]
Replacing by the givens, we can find vfy as follows:[tex]v_{fy} = \frac{m_{olds}*v_{olds} }{(m_{tr} + m_{olds)} } = \frac{2028kg*(-14.5)m/s}{2028kg+4146 kg} = -4.8 m/s (6)[/tex]
The magnitude of the velocity vector of the wreckage immediately after the impact, can be found applying the Pythagorean Theorem to vfx and vfy, as follows:[tex]v_{f} = \sqrt{v_{fx} ^{2} +v_{fy} ^{2} }} = \sqrt{(6.5m/s)^{2} +(-4.8m/s)^{2}} = 8.1 m/s (7)[/tex]
In order to get the compass heading, we can apply the definition of tangent, as follows:[tex]\frac{v_{fy} }{v_{fx} } = tg \theta (8)[/tex]
⇒ tg θ = vfy/vfx = (-4.8m/s) / (6.5m/s) = -0.738 (9)
⇒ θ = tg⁻¹ (-0.738) = -36.4º
Since it's negative, it's counted clockwise from the positive x-axis, so this means that it's 36.4º South of East.What energy store is in the human
BEFORE he/she lifts the hammer?
I believe the answer would be protentional because they have the potential energy in them to lift the hammer.
F=9 N, a=3 m/s², m=?
Answer:
3kg
Explanation:
Given parameters:
Force = 9N
Acceleration = 3m/s²
Unknown:
Mass = ?
Solution:
From Newton's second law of motion:
Force = mass x acceleration
So;
9 = mass x 3
mass = 3kg
g Incandescent bulbs generate visible light by heating up a thin metal filament to a very high temperature so that the thermal radiation from the filament becomes visible. One bulb filament has a surface area of 30 mm2 and emits 60 W when operating. If the bulb filament has an emissivity of 0.8, what is the operating temperature of the filament
Answer:
2577 K
Explanation:
Power radiated , P = σεAT⁴ where σ = Stefan-Boltzmann constant = 5.6704 × 10⁻⁸ W/m²K⁴, ε = emissivity of bulb filament = 0.8, A = surface area of bulb = 30 mm² = 30 × 10⁻⁶ m² and T = operating temperature of filament.
So, T = ⁴√(P/σεA)
Since P = 60 W, we substitute the vales of the variables into T. So,
T = ⁴√(P/σεA)
= ⁴√(60 W/(5.6704 × 10⁻⁸ W/m²K⁴ × 0.8 × 30 × 10⁻⁶ m²)
= ⁴√(60 W/(136.0896 × 10⁻¹⁴ W/K⁴)
= ⁴√(60 W/(13608.96 × 10⁻¹⁶ W/K⁴)
= ⁴√(0.00441 × 10¹⁶K⁴)
= 0.2577 × 10⁴ K
= 2577 K
A car is traveling on a straight road at a constant 35 m/sm/s, which is faster than the speed limit. Just as the car passes a police motorcycle that is stopped at the side of the road, the motorcycle accelerates forward in pursuit. The motorcycle passes the car 13.5 ss after starting from rest. What is the acceleration of the motorcycle (assumed to be constant)
Answer:
2.59m/s
Explanation:
Using the equation of motion
v = u+at
v is the final velocity = 35ms
u is the initially velocity = 9m/s
t is the time = 13.5s
a is the acceleration
Substitute into the formula
35 = 0+13.5a
a = 35/13.5
a = 2.59m/s²
Hence the acceleration of the motorcycle is 2.59m/s
Un autobús en una autopista lleva una magnitud de la velocidad de 95 km/h, el conductor observa que debido a un derrumbe la carretera está cerrada, en ese instante acciona los frenos, deteniendo el autobús después de recorrer 60 m. a) ¿Cuál es el valor de la aceleración en el autobús?
Answer:
La aceleración del autobús es -5.80 m/s².
Explanation:
Podemos encontrar la aceleración del autobús usando la siguiente ecuación:
[tex] v_{f}^{2} = v_{0}^{2} + 2ad [/tex]
Where:
[tex]v_{f}[/tex]: es la velocidad final = 0 (se detiene al final)
[tex]v_{0}[/tex]: es la velocidad inicial = 95 km/h
d: es la distancia recorrida = 60 m
Por lo tanto, la aceleración es:
[tex] a = \frac{v_{f}^{2} - v_{0}^{2}}{2d} = \frac{0 - (95 \frac{km}{h}*\frac{1000 m}{1 km}*\frac{1 h}{3600 s})^{2}}{2*60 m} = -5.80 m/s^{2} [/tex]
El signo negativo se debe a que el autobús está desacelerando (hasta que se detiene).
Entonces, la aceleración del autobús es -5.80 m/s².
Espero que te sea de utilidad!
Which of the following is a mixture?
a air
biron
Chydrogen
d nickel
Answer:
it will option option A hope it helps
Determine the magnitude of the electric field at the point P. Express your answer in terms of Q, x, a, and k. Express your answer in terms of the variables Q, x, a, k, and appropriate constants.
Complete Question
The question image is in the first uploaded image
Answer:
[tex]E=\frac{KQ*4xa}{(x^2-a^2)^2}[/tex]
Explanation:
From the question we are told that
Distance b/w Q mid point and P is given as x
Generally the equation for magnitude of the electric field at the point P is given as
[tex]E=\frac{kQ}{d^2}[/tex]
where
[tex]k=\frac{1}{4\pi e_0}[/tex]
[tex]d=x^2-a^2[/tex]
Therefore
[tex]E= \frac{1}{4\pi e_0} \frac{Q}{(x^2-a^2)^2}- \frac{1}{4\pi e_0} \frac{Q}{(x^2+a^2)^2}[/tex]
[tex]E= \frac{Q}{4\pi e_0} (\frac{1}{(x^2-a^2)^2}- \frac{1}{(x^2+a^2)^2})[/tex]
Therefore equation for magnitude of the electric field at the point P is
[tex]E=\frac{KQ*4xa}{(x^2-a^2)^2}[/tex]
A group of 25 particles have the following speeds: two have speed 11 m/s, seven have 16 m/s , four have 19 m/s, three have 26 m/s, six have 31 m/s, one has 37 m/s, and two have 45 m/s.
Requiredd:
a. Determine the average speed.
b. Determine the rms speed.
c. Determine the most probable speed.
Answer:
a) Average speed is 24.04 m/s
b) the rms speed is 25.84 m/s
c) the most probable speed is 16 m/s
Explanation:
Given the data in the question;
a) Determine the average speed.
To determine the average speed, we simply divide total some of speed by number of particles;
Average speed = [(2×11 m/s)+(7×16 m/s)+(4×19 m/s)+(3×26 m/s)+(6×31 m/s)+(1×37 m/s)+(2×45 m/s)] / 25
= 601 / 25
= 24.04 m/s
Therefore, Average speed is 24.04 m/s
b) Determine the rms speed
we know that (rms speed)² = sum of square speed / total number of particles
so
(rms speed)² = [(2×11²)+(7×16²)+(4×19²)+(3×26²)+(6×31²)+(1×37²)+(2×45²)] / 25
(rms speed)² = 16691 / 25
(rms speed)² = 667.64
(rms speed) = √ 667.64
(rms speed) = 25.84 m/s
Therefore, the rms speed is 25.84 m/s
c) Determine the most probable speed.
Most particles (7) have velocity 16 m/s
i.e 7 is the maximum number of particle for a particular speed ,
Therefore, the most probable speed is 16 m/s
According to Newton's law of universal gravitation, which statements are true?
As we move to higher altitudes, the force of gravity on us decreases.
O As we move to higher altitudes, the force of gravity on us increases,
O As we gain mass, the force of gravity on us decreases.
O Aswe gain mass, the force of gravity on us increases.
DAs we move faster, the force of gravity on us increases.
Four cylindrical wires of different sizes are made of the same material. Which of the following combinations of length and cross-sectional area of one of the wires will result in the smallest resistance?
a. Length Area
3L 3a
b. Length Area
3L 6a
c. Length Area
6L 3a
d. Length Area
6L 6a
Answer:
Explanation:
For resistance of a wire , the formula is as follows .
R = ρ L/S
where ρ is specific resistance , L is length and S is cross sectional area of wire .
for first wire resistance
R₁ = ρ 3L/3a = ρ L/a
for second wire , resistance
R₂ = ρ 3L/6a
= .5 ρ L/a
For 3 rd wire resistance
R₃ = ρ 6L/3a
= 2ρ L/a
For fourth wire , resistance
R₄ = ρ 6L/6a
= ρ L/a
So the smallest resistance is of second wire .
Its resistance is .5 ρ L/a
How much kinetic energy does a 0.104 kg hamster have if it is moving at 24.0 m/s?
Answer:
30J
Explanation:
Given parameters:
Mass of hamster = 0.104kg
Velocity = 24m/s
Unknown:
Kinetic energy = ?
Solution:
Kinetic energy is the energy due to the motion of a body. It is mathematically derived by;
Kinetic energy = [tex]\frac{1}{2}[/tex] m v²
m is the mass
v is the velocity
Kinetic energy = [tex]\frac{1}{2}[/tex] x 0.104 x 24² = 30J
When a drag strip vehicle reaches a velocity of 60 m/s, it begins a negative acceleration by releasing a drag chute and applying its brakes. While reducing its velocity back to zero, its acceleration along a straight line path is a constant -7.5 m/s2 . What displacement does it undergo during this deceleration period
Answer:
240 meters
Explanation:
The distance traveled by the vehicle can be calculated using the following equation:
[tex] v_{f}^{2} = v_{0}^{2} + 2ax [/tex] (1)
Where:
x: is the displacement
[tex]v_{f}[/tex]: is the final speed = 0 (reduces its velocity back to zero)
[tex]v_{0}[/tex]: is the initial speed = 60 m/s
a: is the acceleration = -7.5 m/s²
By solving equation (1) for x we have:
[tex] x = \frac{v_{f}^{2} - v_{0}^{2}}{2a} = \frac{0 - (60 m/s)^{2}}{2*(-7.5 m/s^{2})} = 240 m [/tex]
Therefore, the vehicle undergoes 240 meters of displacement during the deceleration period.
I hope it helps you!
6 A test of a driver's perception/reaction time is being conducted on a special testing track with level, wet pavement and a driving speed of 50 mi/h. When the driver is sober, a stop can be made just in time to avoid hitting an object that is first visible 385 ft ahead. After a few drinks under exactly the same conditions, the driver fails to stop in time and strikes the object at a speed of 30 mi/h. Determine the driver's perception/reaction time before and after drinking. (Assume practical stopping distance.)
Answer:
a. 10.5 s b. 6.6 s
Explanation:
a. The driver's perception/reaction time before drinking.
To find the driver's perception time before drinking, we first find his deceleration from
v² = u² + 2as where u = initial speed of driver = 50 mi/h = 50 × 1609 m/3600 s = 22.35 m/s, v = final speed of driver = 0 m/s (since he stops), a = deceleration of driver and s = distance moved by driver = 385 ft = 385 × 0.3048 m = 117.35 m
So, a = v² - u²/2s
substituting the values of the variables into the equation, we have
a = v² - u²/2s
a = (0 m/s)² - (22.35 m/s)²/2(117.35 m)
a = - 499.52 m²/s²/234.7 m
a = -2.13 m/s²
Using a = (v - u)/t where u = initial speed of driver = 50 mi/h = 50 × 1609 m/3600 s = 22.35 m/s, v = final speed of driver = 0 m/s (since he stops), a = deceleration of driver = -2.13 m/s² and t = reaction time
So, t = (v - u)/a
Substituting the values of the variables into the equation, we have
t = (0 m/s - 22.35 m/s)/-2.13 m/s²
t = - 22.35 m/s/-2.13 m/s²
t = 10.5 s
b. The driver's perception/reaction time after drinking.
To find the driver's perception time after drinking, we first find his deceleration from
v² = u² + 2as where u = initial speed of driver = 50 mi/h = 50 × 1609 m/3600 s = 22.35 m/s, v = final speed of driver = 30 mi/h = 30 × 1609 m/3600 s = 13.41 m/s, a = deceleration of driver and s = distance moved by driver = 385 ft = 385 × 0.3048 m = 117.35 m
So, a = v² - u²/2s
substituting the values of the variables into the equation, we have
a = v² - u²/2s
a = (13.41 m/s)² - (22.35 m/s)²/2(117.35 m)
a = 179.83 m²/s² - 499.52 m²/s²/234.7 m
a = -319.69 m²/s² ÷ 234.7 m
a = -1.36 m/s²
Using a = (v - u)/t where u = initial speed of driver = 50 mi/h = 50 × 1609 m/3600 s = 22.35 m/s, v = final speed of driver = 30 mi/h = 30 × 1609 m/3600 s = 13.41 m/s, a = deceleration of driver = -1.36 m/s² and t = reaction time
So, t = (v - u)/a
Substituting the values of the variables into the equation, we have
t = (13.41 m/s - 22.35 m/s)/-1.36 m/s²
t = - 8.94 m/s/-1.36 m/s²
t = 6.6 s
A 5-kg object is moving with a speed of 4 m/s at a height of 2 m. The potential energy of the object is approximately
J.
Answer:
P.E = 98 Joules
Explanation:
Given the following data;
Mass = 5kg
Speed = 4m/s
Height = 2m
We know that acceleration due to gravity is equal to 9.8m/s²
To find the potential energy;
Potential energy can be defined as an energy possessed by an object or body due to its position.
Mathematically, potential energy is given by the formula;
[tex] P.E = mgh[/tex]
Where, P.E represents potential energy measured in Joules.
m represents the mass of an object.
g represents acceleration due to gravity measured in meters per seconds square.
h represents the height measured in meters.
Substituting into the equation, we have;
[tex] P.E = 5*9.8*2[/tex]
P.E = 98 Joules
A 35 kg box initially sliding at 10 m/s on a rough surface is brought to rest by 25 N
of friction. What distance does the box slide?
Answer:
the distance moved by the box is 70.03 m.
Explanation:
Given;
mass of the box, m = 35 kg
initial velocity of the box, u = 10 m/s
frictional force, F = 25 N
Apply Newton's second law of motion to determine the deceleration of the box;
-F = ma
a = -F / m
a = (-25 ) / 35
a = -0.714 m/s²
The distance moved by the box is calculated as follows;
v² = u² + 2ad
where;
v is the final velocity of the box when it comes to rest = 0
0 = 10² + (2 x - 0.714)d
0 = 100 - 1.428d
1.428d = 100
d = 100 / 1.428
d = 70.03 m
Therefore, the distance moved by the box is 70.03 m.
What is the period of an objects motion?
A compact car has a mass of 1310 kg . Assume that the car has one spring on each wheel, that the springs are identical, and that the mass is equally distributed over the four springs.
Required:
a. What is the spring constant of each spring if the empty car bounces up and down 2.0 times each second?
b. What will be the car’s oscillation frequency while carrying four 70 kg passengers?
Answer:
a) k= 3232.30 N / m, b) f = 4,410 Hz
Explanation:
In this exercise, the car + spring system is oscillating in the form of a simple harmonic motion, as the four springs are in parallel, the force is the sum of the 4 Hocke forces.
The expression for the angular velocity is
w = √k/m
the angular velocity is related to the period
w = 2π / T
we substitute
T = 2[tex]\pi[/tex] √m/ k
a) empty car
k = 4π² m / T²
k = 4 π² 1310/2 2
k = 12929.18 N / m
This is the equivalent constant of the short springs
F1 + F2 + F3 + F4 = k_eq x
k x + kx + kx + kx = k_eq x
k_eq = 4 k
k = k_eq / 4
k = 12 929.18 / 4
k= 3232.30 N / m
b) the frequency of oscillation when carrying four passengers.
In this case the plus is the mass of the vehicle plus the masses of the passengers
m_total = 1360 + 4 70
m_total = 1640 kg
angular velocity and frequency are related
w = 2pi f
we substitute
2 pi f = Ra K / m
in this case the spring constant changes us
k_eq = 12929.18 N / m
f = 1 / 2π √ 12929.18 / 1640
f = π / 2 2.80778
f = 4,410 Hz
The radius of the Sun is 6.96 x 108 m and the distance between the Sun and the Earth is roughtly 1.50 x 1011 m. You may assume that the Sun is a perfect sphere and that the irradiance arriving on the Earth is the value for AMO, 1,350 W/m2. Calculate the temperature at the surface of the Sun.
Answer:
5766.7 K
Explanation:
We are given that
Radius of Sun , R=[tex]6.96\times 10^{8} m[/tex]
Distance between the Sun and the Earth, D=[tex]1.50\times 10^{11}m[/tex]
Irradiance arriving on the Earth is the value for AMO=[tex]1350W/m^2[/tex]
We have to find the temperature at the surface of the Sun.
We know that
Temperature ,T=[tex](\frac{K_{sc}D^2}{\sigma R^2})^{\frac{1}{4}}[/tex]
Where [tex]K_{sc}=1350 W/m^2[/tex]
[tex]\sigma=5.67\times 10^{-8}watt/m^2k^4[/tex]
Using the formula
[tex]T=(\frac{1350\times (1.5\times 10^{11})^2}{5.67\times 10^{-8}\times (6.96\times 10^{8})^2})^{\frac{1}{4}}[/tex]
T=5766.7 K
Hence, the temperature at the surface of the sun=5766.7 K
what is momentum of a train that is 60,000 kg that is moving at velocity of 17m/s?
explain your answer
Anyone can help me out with this question ? Just number 2,
Answer:
- 21⁰C .
Explanation:
Speed of jet = 2.05 x 10³ km /h
= 2050 x 1000 / (60 x 60 ) m /s
= 569.44 m / s
Mach no represents times of speed of sound , the speed of jet
1.79 x speed of sound = 569.44
speed of sound = 318.12 m /s
speed of sound at 20⁰C = 343 m /s
Difference = 343 - 318.12 = 24.88⁰C
We know that 1 ⁰C change in temperature changes speed of sound
by .61 m /s
So a change in speed of 24.88 will be produced by a change in temperature of
24.88 / .61
= 41⁰C
temperature = 20 - 41 = - 21⁰C .
A particle with charge Q and mass M has instantaneous speed uy when it is at a position where the electric potential is V. At a later time, the particle has moved a distance R away to a position where the electric potential is V2 ) Which of the following equations can be used to find the speed uz of the particle at the new position?
a. 1/2M(μ2^2-μ1^2)=Q (v1-v2)
b. 1/2M(μ2^2-μ1^2)^2=Q(v1-v2)
c. 1/2Mμ2^2=Qv1
d. 1/2Mμ2^2=1/4πx0 (Q^2/R)
Answer:
A
Explanation:
Ke = 1/2 MV^2
How does speed and mass effect kinetic energy ?
Answer:
in fact, kinetic energy is directly proportional to mass: if you double the mass, then you double the kinetic energy. Second, the faster something is moving, the greater the force it is capable of exerting and the greater energy it possesses. ... Thus a modest increase in speed can cause a large increase in kinetic energy.
Explanation:
Answer: The more mass of an object has, the more Kinetic energy it has.
Explanation:
Kinetic energy is comparable to mass. If you double the mass then you double the kinetic energy. The faster the object is moving the greater the energy possesses. A large increase in speed can have a large increase in kinetic energy.
Statement A: 2.567 km, to two significant figures. Statement B: 2.567 km, to three significant figures. Determine the correct relationship between the statements. View Available Hint(s) Determine the correct relationship between the statements. Statement A is greater than Statement B. Statement A is less than Statement B. Statement A is equal to Statement B.
Answer:
Statement A is greater than Statement B.
Explanation:
Statement A: 2.567 km, to two significant figures..
To 2 sig figures means only 2 whole numbers should be left after approximation. Thus, 2.567 to 2 significant figures is 2.6 km
Statement B: 2.567 km, to three significant figures. To 3 sig figures means only 3 whole numbers should be left after approximation. Thus, 2.567 to 3 significant figures is 2.57 km
Comparing both values, statement A is obviously greater than Statement B
A remote controlled airplane moves 7.2 m in 2.5seconds what is the plane’s velocity
Answer:
2.88m/s
Explanation:
Given parameters:
Displacement = 7.2m
Time taken = 2.5s
Unknown:
Velocity of the plane = ?
Solution:
Velocity is the displacement divided by the time taken.
Velocity = [tex]\frac{displacement}{time taken}[/tex]
So;
Velocity = [tex]\frac{7.2}{2.5}[/tex] = 2.88m/s