Answer:
it will option option A hope it helps
A jet aircraft is traveling at 233 m/s in horizontal flight. The engine takes in air at a rate of 122 kg/s and burns fuel at a rate of 3.36 kg/s. The exhaust gases are ejected at 487 m/s relative to the aircraft. Find the thrust of the jet engine. Answer in units of N.
Answer:
33624.32 N
Explanation:
From second law of motion, thrust of engine can be computed as
F= Δp/ Δt
= m1 * Δv + m2 v(o)
F= m1 *[v(o) - v] + m2 *v(o).............eqn(1)
Where
v(o) = initial velocity of the gas when ejected= 487 m/s
v= final velocity=233 m/s
m1= initial rate The engine takes in air at =122 kg/s
m2= rate of burning fuel= 3.36 kg/s.
If we input the given values into eqn (1) we have
F= 122×[ 487-233] + [3.36×487]
F=30988+1636.33
=33624.32 N
Hence, the thrust of the jet engine is 33624.32 N
A circus tightrope walker weighing 800 N is standing in the middle of a 15 meter long cable stretched between two posts. The cable was originally horizontal. The lowest point of the cable is now at his feet and is 30 cm below the horizontal. Assume the cable is massless. What is the tension in the cable
Answer:
T = 10010 N
Explanation:
To solve this problem we must use the translational equilibrium relation, let's set a reference frame
X axis
Fₓ-Fₓ = 0
Fₓ = Fₓ
whereby the horizontal components of the tension in the cable cancel
Y Axis
[tex]F_{y} + F_{y} - W =0[/tex]
2[tex]F_{y}[/tex] = W
let's use trigonometry to find the angles
tan θ = y / x
θ = tan⁻¹ (0.30 / 0.50 L)
θ = tan⁻¹ (0.30 / 0.50 15)
θ = 2.29º
the components of stress are
F_{y} = T sin θ
we substitute
2 T sin θ = W
T = W / 2sin θ
T = [tex]\frac{ 800}{ 2sin 2.29}[/tex]
T = 10010 N
why should we not use nuclear energy
Answer:
Barriers to and risks associated with an increasing use of nuclear energy include operational risks and the associated safety concerns, uranium mining risks, financial and regulatory risks, unresolved waste management issues, nuclear weapons proliferation concerns, and adverse public opinion.
Explanation:
Lenz’s Law allows us to find _______.
the direction of the induced current.
the magnitude of the induced emf.
the direction of the induced emf.
the magnitude of the induced current.
Answer:
Explanation:
a
A magnifying glass uses a convex lens of focal length 6.25 cm. When it is held 5.20 cm in front of an object, what is the image distance?
(Mind your minus signs)
(Unit=cm)
Answer:
The answer is -30.95.
Explanation:
Use the lens equation: 1/focal length = 1/object distance + 1/image distance. The answer comes out to -30.95. This is correct on Acellus.
When it is held 5.20 cm in front of an object, the image distance will be "-30.95 cm". To understand the calcultaion, check below.
Convex lensAccording to the question,
Object distance, u = -5.20 cm
Focal length, f = 6.25 cm
By using the Lens formula, we get
→ [tex]\frac{1}{f} = \frac{1}{v} - \frac{1}{u}[/tex]
or,
→ [tex]\frac{1}{v} = \frac{1}{f} + \frac{1}{u}[/tex]
By substituting the values, we get
[tex]= \frac{1}{6.25} - \frac{1}{5.20}[/tex]
[tex]\frac{1}{v} = -\frac{21}{650}[/tex]
By applying cross-multiplication,
v = -30.95 cm
Thus the above answer is correct.
Find out more information about Convex lens here:
https://brainly.com/question/13102253
Under what conditions will a moving 0.030 kg marble and a moving 2.43 kg rock have the same kinetic energy
Answer:
To have the same kinetic energy the speed of the marble must be 9 times the speed of rock.
Explanation:
The general formula of kinetic energy is given as follows:
[tex]K.E = \frac{1}{2}mv^{2}[/tex]
where,
K.E = Kinetic Energy
m = mass of the object
v = speed of the object
So, for the marble and rock to have same kinetic energy, we can write:
[tex]K.E_{marble} = K.E_{rock}\\\\\frac{1}{2}m_{marble}v_{marble}^{2} = \frac{1}{2}m_{rock}v_{rock}^{2}\\\\(0.03\ kg)v_{marble}^{2} = (2.43\ kg)v_{rock}^{2}\\\\taking\ square\ root\ on\ both\ sides:\\v_{marble} = \sqrt{\frac{2.43\ kg}{0.03\ kg}}v_{rock}\\\\v_{marble} = 9\ v_{rock}[/tex]
Hence, to have the same kinetic energy the speed of the marble must be 9 times the speed of rock.
Does changing the height of point C affect the speed of the coaster car at point D?
Without friction, NO.
The speed at D depends only on the difference in height between A and D. Whatever happens between them doesn't matter.
The speed of the coaster car at point D will be affected if the height of point C is changed.
Potencial Energy:
It is the enrgy in a body due to the position of differnt part of the object or system.
As we increase the the hight of the car the potetial enrgy increase, the gravitational acceleration on car will be more due to the high of the point C.
Therefore, the speed of the coaster car at point D will be affected if the height of point C is changed.
To know more about speed of the coaster car,
https://brainly.com/question/9178285
A fisherman notices that his boat is moving up and down periodically without any horizontal motion, owing to waves on the surface of the water. It takes a time of 2.30 s for the boat to travel from its highest point to its lowest, a total distance of 0.660 m . The fisherman sees that the wave crests are spaced a horizontal distance of 5.50 m apart. How fast are the waves traveling
Answer:
v = 1.2 m/s
Explanation:
The wavelength of the waves is given as the horizontal distance between the crests:
λ = wavelength = 5.5 m
Now, the time period is given as the time taken by boat to move from the highest point again to the highest point. So it will be equal to twice the time taken by the boat to travel from highest to the lowest point:
T = Time Period = 2(2.3 s) = 4.6 s
Now, the speed of the wave is given as:
[tex]v = f\lambda[/tex]
where,
v= speed of wave = ?
f = frequency of wave = [tex]\frac{1}{T} = \frac{1}{4.6\ s} = 0.217\ Hz[/tex]
Therefore,
[tex]v = (0.217\ Hz)(5.5\ m)\\[/tex]
v = 1.2 m/s
Traumatic brain injury such as concussion results when the head undergoes a very large acceleration. Generally, an acceleration less than 800 m/s2 lasting for any length of time will not cause injury, whereas an acceleration greater than 1,000 m/s2 lasting for at least 1 ms will cause injury. Suppose a small child rolls off a bed that is 0.55 m above the floor. If the floor is hardwood, the child's head is brought to rest in approximately 1.7 mm. If the floor is carpeted, this stopping distance is increased to about 1.4 cm. Calculate the magnitude and duration of the deceleration in both cases, to determine the risk of injury. Assume the child remains horizontal during the fall to the floor. Note that a more complicated fall could result in a head velocity greater or less than the speed you calculate.
Answer:
a) wooden floor
a = 3170.6 m / s², t = 1.03 10⁻³ s
the child's traumatic injury to the brain
b) the floor is carpeted
a = 385 m / s²
no injuries are created in the child
Explanation:
To solve this exercise we can use the energy conservation relations, let's start by looking for the speed of the child when he is just reaching the ground
starting point. When you get out of bed
Em₀ = U = m g h
final point. Just when it hits the floor
[tex]Em_{f}[/tex] = K = ½ m v²
as there is no friction, energy is conserved
Em₀ = Em_{f}
mgh = ½ m v²
v² = 2 gh
let's calculate
v² = 2 9.8 0.55
v² = 10.78
v = 3.28 m / s
Now let's use the concepts of kinematics to find the deceleration. The case of the wooden floor, where the distance for the deceleration is
x = 1.7 mm = 0.0017 m
v² = v₀² - 2 a y
as the child stops the final velocity is zero
0 = v₀² - 2a y
a = v₀² / 2y
let's calculate
a = [tex]\frac{10.78}{2 \ 0.0017}[/tex]
a = 3170.6 m / s²
Let's find the time that braking lasts
v = v₀ - a t
0 = v₀ - at
t = v₀ / a
t = 3.28 / 3170.6
t = 1.03 10⁻³ s
hence the child's traumatic injury to the brain
second case the floor is carpeted, in this case the stopping distance is
x = 1.4 cm = 0.014 m
we look for acceleration
a = v₀² / 2y
a = [tex]\frac{10.78}{2 \ 0.014}[/tex]
a = 385 m / s²
therefore no injuries are created in the child
In conclusion we see that with the wooden floor there is silence and with the carpeted floor there is no
Leticia leaves the grocery store And walks 150 M’s to parking lot then she turns 90° to the right and walks an additional 70 M’s to her car what is the magnitude of displacement of her car from the grocery store at exit
Answer:
Explanation:
Its 165.5m
A level test track has a coefficient of road adhesion of 0.80, and a car being tested has a coefficient of rolling friction that is approximated as 0.018 for all speeds. The vehicle is tested unloaded and achieves the theoretical minimum stop in 180 ft (from brake application). The initial speed was 60 mi/h. Ignoring aerodynamic resistance, what is the unloaded braking efficiency
Answer:
the unloaded braking efficiency is 84.6 %
Explanation:
Given the data in the question;
by Ignoring aerodynamic resistance; we can find the theoretical stopping distance using the following formula
S = (Y[tex]_{b}[/tex]( V₁² - V₂²)) / ( 2g( ηbμ + [tex]f_{rl}[/tex] ± sin∅[tex]_{g}[/tex]))
now given that the tracked is levelled, ∅[tex]_{g}[/tex] = 0, also Y[tex]_{b}[/tex] = 1.04 for level or flat road
Speed V₁ = 60mil/hr = (60×5280)/(1×60×60) = 316800ft/3600s = 88ft/s
now, we substitute in our values to get the braking efficiency;
180ft = (1.04( (88ft/s)² - 0²)) / ( 2(32.2( (ηb/100)(0.80) + (0.018) ± sin(0°)))
180ft = 8053.76 / ( 64.4)(0.008ηb + 0.018)
180ft = 8053.76 / ( 0.5152ηb + 1.1592)
180( 0.5152ηb + 1.1592) = 8053.76
( 0.5152ηb + 1.1592) = 8053.76 /180
0.5152ηb + 1.1592 = 44.7431
0.5152ηb = 44.7431 - 1.1592
0.5152ηb = 43.5839
ηb = 43.5839 / 0.5152
ηb = 84.596 ≈ 84.6 %
Therefore, the unloaded braking efficiency is 84.6 %
Curtis, a student in our class, makes the following statement: The puck reached a slightly higher location on the ramp than I predicted. This is because I used the wrong mass for the puck when I did all my calculations. I accidentally used the mass of the smaller puck rather than the mass of the larger puck in my video." Is this a plausible explanation? Would the using the wrong mass for the puck during the calculations mean the puck would reach a greater height? Explain your reasoning.
Answer and Explanation: No, the explanation is not plausible. The puck sliding on the ice is an example of the Principle of Conservation of Energy, which can be enunciated as "total energy of a system is constant. It can be changed or transferred but the total is always the same".
When a player hit the pluck, it starts to move, gaining kinetic energy (K). As it goes up a ramp, kinetic energy decreases and potential energy (P) increases until it reaches its maximum. When potential energy is maximum, kinetic energy is zero and vice-versa.
So, at the beginning of the movement the puck only has kinetic energy. At the end, it gains potential energy until its maximum.
The representation is as followed:
[tex]K_{i}+P_{i}=K_{f}+P_{f}[/tex]
[tex]K_{i}+0=0+P_{f}[/tex]
[tex]\frac{1}{2}mv^{2} = mgh[/tex]
As we noticed, mass of the object can be cancelled from the equation, making height be:
[tex]h=\frac{v^{2}}{2g}[/tex]
So, the height the puck reaches depends on velocity and acceleration due to gravity, not mass of the puck.
A 56-kg skater initially at rest throws a 5.0-kg medicine ball horizontally to the left. Suppose the ball is accelerated through a distance of 1.0 m before leaving the skater's hand at a speed of 7.0 m/s. Assume the skater and the ball to be point-like and the surface to be frictionless and ignore air resistance. Use a vertical y-axis with the positive direction pointing up and a horizontal x-axis with the positive direction pointing to the right. What will happen to the skater and the ball after the ball is thrown
Answer:
[tex]a_2=24.5\ \text{m/s}^2[/tex] towards right
[tex]a_1=2.1875\ \text{m/s}^2[/tex] towards left
Explanation:
[tex]m_1[/tex] = Mass of skater = 56 kg
[tex]m_2[/tex] = Mass of ball = 5 kg
v = Final velocity of ball = 7 m/s
u = Initial velocity of ball = 0
s = Distance the ball moved in the hand of the skater = 1 m
Moving left is considered and moving right is considered positive.
From kinematic equations of motion we have
[tex]v^2-u^2=2as\\\Rightarrow a=\dfrac{v^2-u^2}{2s}\\\Rightarrow a=\dfrac{7^2-0^2}{2\times 1}\\\Rightarrow a=24.5\ \text{m/s}^2[/tex]
So, the ball will move towards right with a magnitude of acceleration [tex]a_2=24.5\ \text{m/s}^2[/tex].
The force on the ball will be
[tex]F_2=m_2a_2\\\Rightarrow F_2=5\times 24.5\\\Rightarrow F_2=122.5\ \text{N}[/tex]
The force on the ball is [tex]122.5\ \text{N}[/tex]
The reaction force on the skater will be equal to the force on the ball but will have opposite direction.
[tex]-F_1=F_2\\\Rightarrow -m_1a_1=F_2\\\Rightarrow a_1=-\dfrac{122.5}{56}\\\Rightarrow a_1=-2.1875\ \text{m/s}^2[/tex]
So, the skater will move towards left with a magnitude of acceleration [tex]2.1875\ \text{m/s}^2[/tex]
After the ball is thrown, the skater will accelerate to the left at 2.19 m/s² while the ball will accelerate to the right at 24.5 m/s².
The given parameters
Mass of the skater, m1 = 56 kgMass of the ball, m2 = 5 kgDistance traveled by the ball, d = 1 mSpeed of the ball, v = 7 m/sThe acceleration of the ball as it leaves the skaters hand is calculated as follows;
[tex]v^2 = u^2 + 2as\\\\ v^2 = 0 + 2as\\\\ a = \frac{v^2}{2s} \\\\ a = \frac{7^2}{2(1)} \\\\ a = 24.5 \ m/s^2[/tex]
The force experienced by the ball is calculated as follows;
[tex]F = ma\\\\ F_b = 5 \times 24.5\\\\ F_b = 122.5 \ N[/tex]
The force experienced by the skater is equal in magnitude to that of the ball but opposite in direction.
[tex]F_s = -F_b\\\\ F_s = -122.5 \ N[/tex]
The acceleration of the skater after the ball was thrown is calculated as follows;
[tex]F = ma \\\\ a = \frac{F}{m} \\\\ a = \frac{-122.5 }{56} \\\\ a =-2.19 \ m/s^2[/tex]
Thus, after the ball is thrown, the skater will accelerate to the left at 2.19 m/s² while the ball will accelerate to the right at 24.5 m/s².
Learn more about Newton's third law of motion here: https://brainly.com/question/25998091
When researchers replicate a study, they are seeking to __________.
A.
prove that the hypothesis upon which the study was founded is untestable
B.
develop a new hypothesis
C.
change the study to provide new results
D.
support or reject the hypothesis upon which the study was founded
Please select the best answer from the choices provided
A
B
C
D
Answer:
D
Explanation:
right edge 2022
Cathode ray tubes (CRTs) used in old-style televisions have been replaced by modern LCD and LED screens. Part of the CRT included a set of accelerating plates separated by a distance of about 1.54 cm. If the potential difference across the plates was 26.5 kV, find the magnitude of the electric field (in V/m) in the region between the plates.
Answer:
E = 1,720,779.221 or 1.720779221 * 10^ 6V/m
Explanation:
The electric field between the parallel conducting plates is given by
E =V / d
where V is the potential difference and d is the distance between the plates.
E = 26.5 kV/ 1.54 cm
Now we have to convert into proper units
26.5 kv= 26.5 * 1000 v= 26500 volts
1 kv= 1000 volts
1.54 cm = 1.54/ 100 m= 0.0154m
1m = 100cm
Now putting the values
E= 26500/0.0154 = 1,720,779.221 V/m
The Electric field is equal to E= 1,720,799.221 or 1.7220799221 * 10 ^6 Volts per meter.
In scientific notation this can be written as 1.7220799221 *10^6 V/m
A truck travelling down the street suddenly brakes, applying a 14 N force over 3.5 seconds. What was the impulse over the given time.
Answer:
49 Ns
Explanation:
Given data
Force= 14N
time = 3.5seconds
Applying the expression for impulse
P= Ft
substitute
P=14*3.5
P=49 Ns
Hence the impulse is 49 Ns
Which landform is produced at location E where the Mississippi River enters the Gulf of
Mexico?
a delta a drumlin an out wash an escarpment
Answer:
a delta
Explanation:
The landform produced at the location E where the Mississippi River enters the Gulf of Mexico is a delta.
A delta is a depositional landform where a smaller body of water enters into a larger one.
The Gulf of Mexico contains a larger body of water and as the Mississippi river enters into it, it splits up into many distributaries.
So, this feature is a delta.
In the graph, which two regions show the particle undergoing zero acceleration and negative acceleration respectively?
A.
BC shows zero acceleration, and AB shows negative acceleration.
OB.
AB shows zero acceleration, and CD shows negative acceleration.
O C.
BC shows zero acceleration, and CD shows negative acceleration.
D.
AB shows zero acceleration, and BC shows negative acceleration.
Answer:
c
Explanation:
Answer:
c
Explanation:
"45 meters north" is an example of
Answer:
Displacement
Explanation:
The quantity 45m north is a typical example of displacement.
Displacement is the distance traveled by a body in a specific direction. Displacement is a vector quantity with both magnitude and direction.
When we are specifying the displacement of a body, the direction must be indicated accurately. Therefore, the quantity given is displacementIf a person weighs 140 lb'on Earth, their mass in kilograms is
Answer:
70 kg
Explanation:
divide it by 2
Hope this helped!
Answer:
63.502932 Kilograms
Explanation:
Please help will mark as brainiest.
12. An object with a mass of 1250 g has a speed of 17 m/s. What is the object's momentum? Round
answer to the nearest hundredth.
13. How fast is Team USA's Oracle sailing during the America's Cup? The boat has a mass of 1320 kg and
travels with a momentum of 32023.2 kg x m/s Round answer to the nearest hundredth.
Answer:
Explanation:
Momentum is mass multiplied by velocity, p = mv
Velocity is just speed but with a direction. So they can essentially be considered the same thing as long as you acknowledge this fact and understand where each is applied.
12)
First convert the grams to kilograms, those are the standard units us physicists use.
p = (1.25 kg)(17 m/s)
=21.25 kg*m/s
13)
p = mv
32,023.2 kg*m/s = (1,320 kg)(x m/s)
x = 24.26 m/s
If an ice cube with the mass of 5.0 grams melts in a closed system such as a closed glass jar what is the mass of the liquid water after the ice cube completely melts
Two trains, each having a speed of 22 km/h, are headed at each other on the same straight track. A bird that can fly 60 km/h flies off the front of one train when they are 51 km apart and heads directly for the other train. On reaching the other train it flies directly back to the first train, and so forth. (We have no idea why a bird would behave in this way.) What is the total distance the bird travels before the trains collide
Answer:
the total distance the bird travels before the trains collide is 69.54 km
Explanation:
Given the data in the question,
dA = 22km/h × t
dB = 51km - 22km/h × t
now we find the time, when the two trains collide
dA = Db
22km/h × t = 51km - 22km/h × t
44km/h × t = 51km
t = 51km / 44km/h
t = 1.159 hrs
so the bird can fly back and forth for 1.159 hrs before the train collide.
hence, distance travelled by the bird in total will be;
d = v × t
we substitute
d = 60 km/h × 1.159 h
d = 69.54 km
Therefore, the total distance the bird travels before the trains collide is 69.54 km
In this system, ______kinetic_______ energy from the moving water of the faucet turns the turbine. The Mechanical energy of the spinning turbine generates Electical energy which is transformed into Thermal energy that causes the temperature of the water to increase. The water then becomes steam and gives off more Thermal energy into the atmosphere.
Answer:
The MECHANICAL energy of the water is transformed into KINETIC ENERGY
Explanation:
In this exercise you are asked to complete the sentence so that it makes sense.
When the tap is opened, the water has a mechanical energy given by the potential energy due to the height and the energy accumulated by the pressure of the system inside the pipe.
Therefore the sentence must be completed with
The MECHANICAL energy of the water is transformed into KINETIC ENERGY from the moving tap that turns the turbine ...
Answer:
dawg this helped me
Explanation:
i had to do the exact same question thanks homie
A box has a mass of 150 kg. If a net force of 3,000 N acts
on the box, what is the box’s acceleration?
If you carry a box across the floor, you don't do work. Why not?
The force exerted by the person is an upward force equal to the weight of the box, and that force is perpendicular to the motion. If there is no motion in the direction of the force, then no work is done by that force. Yet you certainly feel like you are doing work if you carry a heavy box.
In order to accomplish work on an object there must be a force exerted on the object and it must move in the direction of the force. Energy is required to do work and the basic SI unit of energy is the joule, the amount of energy required to exert a force of 1 Newton through a distance of 1 meter (1 joule = 1 newton meter).*picture 1*For the special case of a constant force, the work may be calculated by multiplying the distance times the component of force which acts in the direction of motion.*picture 2*
In order to accomplish work on an object there must be a force exerted on the object and it must move in the direction of the force.*picture 3*For a constant force F which moves an object in a straight line from x1 to x2 , the work done by the force can be visualized as the area enclosed under the force line below*picture 4*For the more general case of a variable force F(x) which is a function of x, the work is still the area under the force curve, and the work expression becomes an integral. *picture 5*credits-goo.gle,hyper physics
Match each letter to the description
A woman walks in a
straight line with the sun
to her right at six o'clock in
the morning. Toward
which pole does she walk?
A line of latitude.
A man walks in a straight
line with the sun to his
right at six o'clock in the
evening. Toward which
pole does he walk?
A line of longitude.
A woman walks in a straight line with the sun to her right at six o'clock in the morning.
The sun rises East of her, so the woman is walking toward the North pole.
A man walks in a straight line with the sun to his right at six o'clock in the evening.
The sun sets West of him, so the man is walking toward the South pole.
The woman and the man are both walking along lines of constant longitude.
A beaker is filled with water to the rim. Gently you place a toy duck in the beaker and some of the water spills out. The duck floats. The weight of the beaker with the duck floating in it is Group of answer choices less than its weight before adding the duck. the same as its weight before adding the duck. greater or less than its weight before adding the duck, depending on the weight of the duck. greater than its weight before adding the duck.
Answer:
Explanation:
The same as its weight before adding the duck
A boat is cruising in a straight line at a constant speed of 2.5 m/s when it is shifted into neutral. After coasting 14 mm the engine is engaged again, and the boat resumes cruising at the reduced constant speed of 1.5 m/s. Assuming constant acceleration while coasting,
a. How much time did it take for the boat to coast the 12 m?
b. What was the boat’s acceleration while it was coasting?
c. What was the speed of the boat when it had coasted for 6.0 m? Explain.
Answer:
Explanation:
a ) distance s = 12 m .
constant speed = 1.5 m/s
time taken to coast 12 m
= distance / speed
= 12 / 1.5 = 8 s
b ) initial velocity u = 2.5 m /s
final velocity v = 1.5 m /s
displacement s = 14 m
acceleration a = ?
v² = u² + 2 as
1.5² = 2.5² + 2 a x 14
a = - .1428 m /s²
= - 14.28 cm / s²
c )
v = ?
u = 2.5 m /s
s = 6 m
a = - .1428 m /s²
v² = u² - 2 as
= 2.5² - 2 x 6 x .1428
= 6.25 - 1.71
= 4.54
v = 2.13 m /s .
A iron block with a mass of 4.8 kg initially slides over a rough horizontal surface with a speed of 1.2 m/s. Friction slows the block to rest. While slowing to rest, 85.0% of the kinetic energy of the block is absorbed by the block itself as internal energy. What is the temperature increase of the block
Answer:
Explanation:
Kinetic energy of block will be converted into heat energy by friction .
Heat energy produced = 1/2 m v²
= .5 x 4.8 x 1.2²
= 3.456 J
85% of energy is converted into heat energy , so heat energy produced
= .85 x 3.456 = 2.9376 J .
If Q heat is given to m mass of object having s as specific heat and Δt is increase in temperature
Q = msΔt
specific heat of iron s = 462 J / kg C
Putting the values ,
2.9376 = 4.8 x 462 x Δt
Δt = 13.24 x 10⁻⁴ ⁰C.