Answer:
- the effective lift area for the aircraft is 8.30 m²
- the required engine thrust is 1275 N
- required power is 79.7 kW
Explanation:
Given the data in the question;
Speed V = 225 km/hr = 62.5 m/s
The lift coefficient CL = 0.45
drag coefficient CD = 0.065
mass = 900 kg
g = 9.81 m/s²
a) the effective lift area for the aircraft
we know that for a steady level flight, weight = lift and thrust = drag
Using the equation for the lift force
F[tex]_L[/tex] = C[tex]_L[/tex][tex]\frac{1}{2}[/tex]ρV²A = W
we substitute
0.45 × [tex]\frac{1}{2}[/tex] × 1.21 × ( 62.5 )² × A = ( 900 × 9.81 )
1081.05 × A = 8829
A = 8829 / 1081.05
A = 8.30 m²
Therefore, the effective lift area for the aircraft is 8.30 m²
b) the required engine thrust and power to maintain level flight.
we use the expression for drag force
F[tex]_D[/tex] = T = C[tex]_D[/tex][tex]\frac{1}{2}[/tex]ρV²A
we substitute
= 0.065 × [tex]\frac{1}{2}[/tex] × 1.21 × ( 62.5 )² × 8.30
T = 1275 N
Since drag and thrust force are the same,
Therefore, the required engine thrust is 1275 N
Power required;
P = TV
p = 1275 × 62.5
p = 79687.5 W
p = ( 79687.5 / 1000 )kW
p = 79.7 kW
Therefore, required power is 79.7 kW
The strength of a magnetic field around a wire carrying a current of 20 A is 0.004 T. What is the strength of the
magnetic field if the current is changed to 40 A?
A 0.002 T
B 0.004 T
C 0.006 T
D 0.008 T
Answer:
D) 0.008
Explanation:
I just did it on Edg22 and I got it correct :D (Picture above)
Option D) 0.008T is the correct answer.
Hence, if the current is increased to the given value, the strength of the magnetic field around the wire carrying the current increases to 0.008T
Given the data in the question;
First scenario
Strength of magnetic field; [tex]B = 0.004T[/tex]Current; [tex]I = 20A[/tex]Second scenario
Strength of magnetic field; [tex]B' =\ ?[/tex]Current; [tex]I' = 40A[/tex]Magnetic FieldMagnetic field is a vector field or region around a magnet or electric charge upon which magnetic force is exerted.
To determine the strength of the magnetic field if the current is changed, we equate the two scenario.
[tex]\frac{B}{I} = \frac{B'}{I'}[/tex]
We substitute our given values into the expression
[tex]\frac{0.004T}{20A} = \frac{B'}{40A}\\ \\ B' * 20A = 0.004T * 40A\\\\B' = \frac{0.004T * 40A}{20A} \\\\B' = \frac{0.004T * 40}{20}\\ \\B' = \frac{0.16T}{20}\\ \\B' = 0.008T[/tex]
Option D) 0.008T is the correct answer.
Hence, if the current is increased to the given value, the strength of the magnetic field around the wire carrying the current increases to 0.008T
Learn more about magnetic field here: https://brainly.com/question/15695203
A satellite has a mass of 6463 kg and is in a circular orbit 4.82 × 105 m above the surface of a planet. The period of the orbit is 2.0 hours. The radius of the planet is 4.29 × 106 m. What would be the true weight of the satellite if it were at rest on the planet’s surface?
Answer:
The weight of the planet is 29083.5 N .
Explanation:
mass of satellite, m = 6463 kg
height of orbit, h = 4.82 x 10^5 m
period, T = 2 h
radius of planet, R = 4.29 x 10^6 m
Let the acceleration due to gravity at the planet is g.
[tex]T = 2\pi\sqrt\frac{(R+h)^3}{gR^2}\\\\2\times 3600 = 2\times3.14\sqrt\frac{(4.29+0.482)^3\times10^{18}}{g\times 4.29\times 4.29\times 10^{12} }\\\\24.2 g =108.67\\\\g = 4.5 m/s^2[/tex]
The weight of the satellite at the surface of the planet is
W = m g = 6463 x 4.5 = 29083.5 N
An ideal spring is fixed at one end. A variable force F pulls on the spring. When the magnitude of F reaches a value of 43.8 N, the spring is stretched by 15.5 cm from its equilibrium length. Calculate the additional work required by F to stretch the spring by an additional 10.4 cm from that position. (in J)
A force of 43.8 N is required to stretch the spring a distance of 15.5 cm = 0.155 m, so the spring constant k is
43.8 N = k (0.155 m) ==> k = (43.8 N) / (0.155 m) ≈ 283 N/m
The total work done on the spring to stretch it to 15.5 cm from equilibrium is
1/2 (283 N/m) (0.155 m)² ≈ 3.39 J
The total work needed to stretch the spring to 15.5 cm + 10.4 cm = 25.9 cm = 0.259 m from equilibrium would be
1/2 (283 N/m) (0.259 m)² ≈ 9.48 J
Then the additional work needed to stretch the spring 10.4 cm further is the difference, about 6.08 J.
An electric motor consumes 8.40 kJ of electrical energy in 1.00 min. Part A If one-third of this energy goes into heat and other forms of internal energy of the motor, with the rest going to the motor output, how much torque will this engine develop if you run it at 2900 rpmrpm
Answer:
The torque is 0.31 Nm.
Explanation:
Electrical energy, E = 8400 J
time, t = 1 min
Angular speed, w = 2900 rpm = 303.53 rad/s
efficiency = 2/3 of input power
The toque is given by
[tex]P =\tau w\\\\\frac{2}{3}\times \frac{E}{t}=\tau w\\\\\frac{2}{3}\times \frac{8400}{60}=\tau \times 303.53\\\\\tau =0.31 Nm[/tex]
A 0.200 m wire is moved parallel to a 0.500 T
magnetic field at a speed of 1.50 m/s. What emf is
induced across the ends of the wire?
Answer:
The required emf moved across the wire is zero
Explanation:
For a moving charge particle, the magnetic force can be determined by using the formula;
[tex]\varepsilon = Bvlsin \theta[/tex]
since the wire moves in parallel, the angle [tex]\theta[/tex] between magnetic field and velocity = 0°
B = 0.500 T
v = 1.50 m/s
l = 0.200 m
∴
[tex]\varepsilon = (0.500 \ T )(1.50 \ m/s) \times (0.200 \ m)\times sin (0)[/tex]
[tex]\varepsilon = 0.15\times sin (0)[/tex]
[tex]\varepsilon = 0[/tex]
A ball of mass=2kg is dropped from h=100m. What is the final velocity when it reaches the ground? *
A)45m/s
B)200m/s
C)32m/s
D)2000m/s
show your work please
Answer:
45m/s
Explanation:
(here I'm taking gravity due to acceleration as 10 m/s^2 and not 9.8 m/s^2 to match the options)
We know that when an object is dropped from a certain height it has initial velocity 0 m/s
so;
u= 0m/s
s= 100m
a= 10m/s^2
Using 4th law of motion;
v^2 - u^2 = 2× a× s
v^2 - 0^2 = 2 × 10 × 100
v^2 = 2000
v = √2000
v = 44.72 m/ s
Final velocity = 45 m/ s (rounding off to 45)
Please feel free to tell me if you have any confusion.
3. How can a generator that otherwise produces AC
current be modified to produce DC current?
Q2 A source of frequency 500 Hz emits waves of
wavelength 0.2m. How long does it take the waves to
travel 400m?
Answer:
4 secs
Explanation:
The first step is to calculate the velocity
V= frequency × wavelength
= 500× 0.2
= 100
Therefore the time can be calculated as follows
= distance/velocity
= 400/100
= 4 secs
An object of mass m 1 moving with speed v collides with another object of mass m 2 at rest and stick to it. Find the impulse to the second object.
Answer:
(m1+m2)vo=m1v+m2×0
⟹vo=m1+m2m1v
The impulse imparted to second object is equal to change is momentum is -
J=m2vo=m1+m2m2m1v
Consult Interactive Solution 7.10 for a review of problem-solving skills that are involved in this problem. A stream of water strikes a stationary turbine blade horizontally, as the drawing illustrates. The incident water stream has a velocity of 16.0 m/s, while the exiting water stream has a velocity of -16.0 m/s. The mass of water per second that strikes the blade is 36.0 kg/s. Find the magnitude of the average force exerted on the water by the blade.
Answer:
[tex]F=1152N[/tex]
Explanation:
From the question we are told that:
Mass/one sec [tex]m=36.0kg/s[/tex]
Incident water stream [tex]v=16.0m/s[/tex]
Exiting water stream [tex]v'=-16.0m/s[/tex]
Generally the equation for Force is mathematically given by
[tex]F=\frac{mv-mu}{t}[/tex]
[tex]F=\frac{36(-16-16)}{1}[/tex]
[tex]F=1152N[/tex]
What does this circle graph tell you about water on Earth? (2 points)
a pie graph with a big blue section covering seventy one percent and small gray section covering twenty nine percent with a key indicating that blue is water and gray is land
Fresh water covers 71 percent of Earth's surface.
Oceans covers 71 percent of Earth's surface.
Salt water covers 71 percent of Earth's surface.
Water covers 71 percent of Earth's surface.
Answer:
ocean covers 71 percent of the earth
Answer:
the ocean covers 71 percent of Earth's surface.
Explanation:
Ftension = 120 N
10 kg
Fg
What is the weight (not mass) of the box?
O 10 kg
0 19 N
98 N
O 98 kg
PLZ HELP
Answer:
98N
Explanation:
El peso se mide en kg y la fuerza no afecta
heat travel through vacuum by
a. conduction. b.convention
c. radiation. d. both a&b
Answer:
Option C
Explanation:
C. Radiation.....
Answer:
heat travel through vacuum by radiation
Does every magnet have a north and south pole?
yes
no
A truck on a straight road starts from rest, accelerating at 2.00m/s^2 until it reaches a speed of 20.0m/s. Then the truck travels for 20.0s at constant speed until the brakes are applied, stopping the truck in a uniform manner in an additional 5.00s.
(a) How long is the truck in motion?
(b) What is the average velocity of the truck for the motion described?
Answer:
Explanation:
a )
Time to reach the speed of 20 m/s with an acceleration of 2 m/s² can be calculated as follows .
v = u + a t
20 = 0 + 2 t
t = 20 /2 = 10 s .
Total time = 10 s + 20 s + 5 s = 35 s .
b) Average velocity = Total distance travelled / total time
Distance travelled in first 10 s
S₁ = ut + 1/2 a t²
= 0 + .5 x 2 x 10²
= 100 m
Distance travelled in next 20 s
S₂= 20s x 20 m/s = 400 m
Distance travelled in last 5 s .
deceleration in last 5 s
v = u + at
0 = 20 m/s + a x 5
a = - 4 m/s²
v² = u² - 2 a s
0 = (20 m/s)² - 2 x 4 m/s² x s
s = 50 m
S₃ = 50 m
Total distance = S₁ + S₂ + S₃
= 100 m + 400 m + 50 m
= 550 m .
Average velocity = 550 m / 35 s
= 15.71 m /s .
A magnetic field of magnitude 0.550 T is directed parallel to the plane of a circular loop of radius 43.0 cm. A current of 5.80 mA is maintained in the loop. What is the magnetic moment of the loop? (Enter the magnitude.)
Answer:
[tex]\mu = 3.36\times 10^{-3}\ A-m^2[/tex]
Explanation:
Given that,
The magnitude of magnetic field, B = 0.55 T
The radus of the loop, r = 43 cm = 0.43 m
The current in the loop, I = 5.8 mA = 0.0058 A
We need to find the magnetic moment of the loop. It is given by the relation as follows :
[tex]\mu = AI\\\\\mu=\pi r^2\times I[/tex]
Put all the values,
[tex]\mu=\pi \times (0.43)^2\times 0.0058\\\\=3.36\times 10^{-3}\ A-m^2[/tex]
So, the magnetic moment of the loop is equal to[tex]3.36\times 10^{-3}\ A-m^2[/tex].
A gazelle is running at a constant speed of 19.3 m/s toward a motionless hidden cheetah. At the instant the gazelle passes the cheetah, the cheetah accelerates at a rate of 7.1 m/s/s in pursuit of the gazelle. The gazelle maintains its constant speed. By the time the cheetah reaches a speed of 19.3 m/s to match the gazelle, how far apart are the two animals in units of m
Answer:
the animals are 26.2 meters apart.
Explanation:
Let's define t = 0s as the moment when the cheetah starts accelerating.
The gazelle moves with constant velocity, thus, it is not accelerating, then the acceleration of the gazelle is:
a₁(t) = 0m/s^2
where I will use the subscript "1" to refer to the gazelle and "2" to refer to the cheetah.
for the velocity of the gazelle we just integrate over time to get:
v₁(t) = V0
where V0 is the initial speed of the gazelle, which we know is 19.3 m/s
v₁(t) = 19.3 m/s
To get the position of the gazelle we integrate again:
p₁(t) = ( 19.3 m/s)*t + P0
where P0 is the position of the gazelle at t = 0s, let's define P0 = 0m
p₁(t) = ( 19.3 m/s)*t
The equations that describe the motion of the gazelle are:
a₁(t) = 0m/s^2
v₁(t) = 19.3 m/s
p₁(t) = ( 19.3 m/s)*t
Now let's do the same for the cheetah.
We know that its acceleration is 7.1 m/s^2
then:
a₂(t) = 7.1 m/s^2
for the velocity of the cheetah we integrate:
v₂(t) = (7.1 m/s^2)*t + V0
where v0 is the initial velocity of the cheetah, which we know its zero.
v₂(t) = (7.1 m/s^2)*t
Finally, for the position equation we integrate again, and remember that we have defined the initial position for the gazelle as zero, then the same happens for the cheetah.
p₂(t) = (1/2)*(7.1 m/s^2)*t^2
The equations for the cheetah are:
a₂(t) = 7.1 m/s^2
v₂(t) = (7.1 m/s^2)*t
p₂(t) = (1/2)*(7.1 m/s^2)*t^2
Now, we want to find the distance between both animals when the speed of the cheetah is 19.3 m/s, then first we need to solve:
v₂(t) = (7.1 m/s^2)*t = 19.3 m/s
t = (19.3 m/s)/(7.1 m/s^2) = 2.72s
Now, to find the distance between the two animals, we just compute the difference between the position equations for t = 2.72s
Distance = p₁(2.72s) - p₂(2.72s)
= ( 19.3 m/s)*2.72s - (1/2)*(7.1 m/s^2)*(2.72s)^2
= 26.2 m
So the animals are 26.2 meters apart.
If v = 5.00 meters/second and makes an angle of 60° with the negative direction of the y–axis, calculate all possible values of vx.
At what frequency would an inductor of inductance 0.8H have a reactance of 12000^?
Explanation:
The inductive reactance [tex]X_L[/tex] is given by
[tex]X_L = \omega L = 2 \pi fL[/tex]
Solving for f, we get
[tex]f = \dfrac{X_L}{2 \pi L} = \dfrac{12000\:\text{ ohms}}{2\pi (0.8\:H)}[/tex]
[tex]\:\:\:\:\:\:\:= 2387.3\:\text{Hz}[/tex]
write the full form of MBBS pliz now
Answer:
The full form of MBBS is Bachelor of Medicine, Bachelor of Surgery
pls mark me brainliest :))
The full form of MBBS in India is 'Bachelor of Medicine, Bachelor of Surgery'. However, MBBS is an abbreviation of Medicinae Baccalaureus Baccalaureus Chirurgiae, which is the term used for this course in Latin.
kinetic energy is associated with energy
True or false
Answer:
true ,the kinetic energy is associated with energy
During a long jump, an Olympic champion's center of mass rose about 1.2 m from the launch point to the top of the arc. 1) What minimum speed did he need at launch if he was traveling at 6.8 m/s at the top of the arc
Answer: Minimum speed needed by the Olympic champion at launch if he was traveling at 6.8 m/s at the top of the arc is 11.65 m/s.
Explanation:
Velocity is only in horizontal direction at the top most point which is similar to the velocity in the horizontal direction at the time of launch.
Now, according to the law of conservation of energy the formula used is as follows.
[tex]mgh = \frac{1}{2} mv^{2}_{y}\\v_{y} = \sqrt{2gh}\\= \sqrt{2 \times 9.8 m/s^{2} \times 1.2}\\= 4.85 m/s[/tex]
As speed at which the person is travelling was 6.8 m/s. Hence, the initial velocity will be calculated as follows.
[tex]v = \sqrt{v^{2}_{x} + v^{2}_{y}}\\= \sqrt{(6.8)^{2} + (4.85 m/s)^{2}}\\= 11.65 m/s[/tex]
Thus, we can conclude that minimum speed needed by the Olympic champion at launch if he was traveling at 6.8 m/s at the top of the arc is 11.65 m/s.
A tennis player swings at a ball at a constant speed, taking 0.50 s to rotate her arms and racquet from horizontal to vertical. What acceleration is felt by a small bug at the tip of her racquet if it is 1.3 m from her shoulder
Answer:
the acceleration of the small bug is 12.83 m/s²
Explanation:
Given;
time of motion, t = 0.5 s
radius of the circular path created by his arm, r = 1.3 m
if he rotates his arm from horizontal to vertical, the angular displacement = 90⁰
The centripetal acceleration of the ball is calculated as;
[tex]a_c = \omega^2 r\\\\a_c = (\frac{\theta}{t} )^2 r\\\\[/tex]
[tex]a_c = (\frac{90}{360} \times\frac{ 2\pi }{t} )^2r\\\\a_c = (\frac{\pi}{2t} )^2 r\\\\a_c = \frac{\pi^2r}{4t^2} = \frac{\pi^2 \times1.3 }{4\times 0.5^2} = 12.83 \ m/s^2[/tex]
Therefore, the acceleration of the small bug is 12.83 m/s²
The total power input to a pumped storage power station is 600 MW
The useful power output is 540 MW calculate the efficiency of this pumped storage power station.
Calculate how much power is wasted by the pumped storage power station.
Answer:
60MW wasted
Explanation:
600-540
=60MW
An electric generator has an 18-cmcm-diameter, 120-turn coil that rotates at 60 HzHz in a uniform magnetic field that is perpendicular to the rotation axis. Part A What magnetic field strength is needed to generate a peak voltage of 330 VV
Explanation:
OK ok nosepo [tex]2825.55[/tex]The magnetic field strength is needed to generate a peak voltage is 0.287 T.
The given parameters;
diameter of the generator, d = 18 cm = 0.18number of turn, N = 120 turnfrequency of the coil, f = 60 Hzmaximum voltage in the coil emf = 330 V;The maximum voltage in the coil is calculated as follows;
[tex]E_{max} = NAB\omega \\\\[/tex]
where;
N is the number of turnsA is the area of the coilB is the magnetic field strengthω is angular speedThe magnetic field strength is needed to generate a peak voltage is calculated as;
[tex]B = \frac{E_{max} }{NA \omega } \\\\B = \frac{E_{max} }{N (\frac{\pi d^2}{4} ) \times 2\pi f}\\\\ B = \frac{330}{120 \times (\frac{\pi \times 0.18^2}{4} ) \times 2\pi \times 60} \\\\B = 0.287 \ T[/tex]
Thus, the magnetic field strength is needed to generate a peak voltage is 0.287 T.
Learn more here:https://brainly.com/question/22784792
g to generate electricity, solar panels typically absorb visible light. How many photons of light with a frequency of 5045x10 14 hz does a solar panel absorb to create 360 kj
Answer:
the number of photons absorbed by the solar panel is 1.08 x 10²¹
Explanation:
Given;
frequency of each photon absorbed, f = 5045 x 10¹⁴ Hz
energy to be created by the solar panel, E = 360 kJ = 360,000 J
The energy of each photon absorbed is calculated as;
[tex]E_{photon} = hf\\\\where;\\\\h \ is \ Planck's \ constant = 6.626 \times 10^{-34} \ Js\\\\E_{photon} = (6.626 \times 10^{-34} )(5045 \times 10^{14})\\\\E_{photon} = 3.343 \times 10^{-16} \ J[/tex]
let the number of photons absorbed = n
[tex]n(E_{photon}) = 360,000 \ J\\\\n = \frac{360,000 \ J}{3.343 \times 10^{-16} \ J} \\\\n = 1.08 \times 10^{21} \ photons[/tex]
Therefore, the number of photons absorbed by the solar panel is 1.08 x 10²¹
Calculate how fast the ball would be moving at the instant it leaves the projectile launcher of the spring is compressed by 3.75 cm. Use a value of k = 500 N/m for the spring constant, 10 g for the mass of the ball, and 75 g for the effective mass of the ball holder. Show your work.
Answer:
V = 8.34m/s
Explanation:
Given that
1/2ke^2 = 1/2mv^2 ......1
Where e = 3.75cm = (3.75/100)m
e = 0.0375m
K = 500 N/m
m = 10g = 10/1000
= 0.01kg
Substitute the values into equation 1
0.5×500×(0.0375)^2 = 0.5×0.01×v^2
250×0.001395 = 0.005v^2
0.348 = 0.005v^2
v^2 = 0.348/0.005
v^2 = 69.6
V = √69.6
V = 8.34m/s
The ball launches at the speed of V = 8.34m/s
What is the vibrating source that creates sound waves?
Answer:
Sound is produced when an object vibrates, creating a pressure wave. This pressure wave causes particles in the surrounding medium (air, water, or solid) to have vibrational motion. As the particles vibrate, they move nearby particles, transmitting the sound further through the medium.
Three resistors 4ohms 6ohms8ohms are connected in series and are connected to cell of EMF 60volt and negilible resistance calculate the current in circuit , potential difference in resistors ,total terminal potential difference in circuit and lost volt
Explanation:
Given that,
Three resistors 4ohms 6ohms 8ohms are connected in series and are connected to cell of EMF 60volt.
The equivalent resistance in series combination is given by :
R = R₁ + R₂ +R₃
Put all the values,
R = 4 + 6 + 8
R = 18 ohms
Let I is the current in circuit. So,
[tex]I=\dfrac{V}{R}\\\\I=\dfrac{60}{18}\\\\I=3.33\ A[/tex]
Potential difference in 4 ohms,
[tex]V_1=IR_1\\\\V_1=3.33\times 4\\\\=13.32\ V[/tex]
Potential difference in 6 ohms,
[tex]V_2=IR_2\\\\V_2=3.33\times 6\\\\=19.98\ V[/tex]
Potential difference in 8 ohms,
[tex]V_3=IR_3\\\\V_3=3.33\times 8\\\\=26.64\ V[/tex]
Terminal potential difference in circuit is :
V = IR
Put all the values,
V = 3.33 × 18
V = 59.94 volts
Hence, this is the required solution.
Choose the smallest item from the list below.
1 glass of water
1 droplet of water
1 atom of oxygen
1 molecule of water
Answer
One molecule of water