Answer:
The inductance of the solenoid is 71.98 mH
Explanation:
Given;
number of loops per length of the solenoid, n = 100 loops/cm
radius of the solenoid, r = 4.53 cm
length of the solenoid, L = 8.88 cm
current carried by the solenoid, I = 500 mA
The inductance of the solenoid is calculated as;
[tex]L = \frac{N^2\times \mu_0 \times A}{l}[/tex]
where;
A is the area of the coil
[tex]A =\pi r^2 = \pi (0.0453)2 = 0.00645 \ m^2[/tex]
N is the number of turns, = 100 x 8.88 = 888 loops
l is the length = 0.0888 m
[tex]L = \frac{N^2\times \mu_0 \times A}{l}\\\\L = \frac{(888)^2\times \(4\pi \times 10^{-7}) \times (0.00645)}{0.0888}\\\\L = 0.07198 \ H\\\\L = 71.98 \ mH[/tex]
Therefore, the inductance of the solenoid is 71.98 mH
Pete is investigating the solubility of salt (NaCl) in water. He begins to add 50 grams of salt to 100 grams of
room temperature tap water in a beaker. After adding all of the salt and stirring for several minutes, Pete
notices a solid substance in the bottom of the beaker. Which statement best explains why there is a solid
substance in the bottom of the beaker?
A. The salt he is using is not soluble in water.
B. The salt is changing into a new substance that is not soluble in water,
C. The dissolving salt is causing impurities in the water to precipitate to the bottom
D. The water is saturated and the remaining salt precipitates to the bottom
Answer:
would the answer be c
Explanation: that what i think in my opian
Answer:
A
Explanation:
A child on a tricycle is moving at a speed of 1.40 m/s at the start of a 2.25 m high and 12.4 m long incline. The total mass is 48.0 kg, air resistance and rolling resistance can be modeled as a constant friction force of 41.0 N, and the speed at the lower end of the incline is 6.50 m/s. Determine the work done (in J) by the child as the tricycle travels down the incline.
Answer:
The work done by the child as the tricycle travels down the incline is 416.96 J
Explanation:
Given;
initial velocity of the child, [tex]v_i[/tex] = 1.4 m/s
final velocity of the child, [tex]v_f[/tex] = 6.5 m/s
initial height of the inclined plane, h = 2.25 m
length of the inclined plane, L = 12.4 m
total mass, m = 48 kg
frictional force, [tex]f_k[/tex] = 41 N
The work done by the child is calculated as;
[tex]\Delta E_{mech} = W - f_{k} \Delta L\\\\W = \Delta E_{mech} + f_{k} \Delta L\\\\W = (K.E_f - K.E_i) + (P.E_f - P.E_i) + f_{k} \Delta L\\\\W = \frac{1}{2} m(v_f^2 - v_i^2) + mg(h_f - h_i) + f_{k} \Delta L\\\\W = \frac{1}{2} \times 48(6.5^2 - 1.4^2) + 48\times 9.8(0-2.25) + (41\times 12.4)\\\\W = 966.96 \ - \ 1058.4 \ + \ 508.4\\\\W = 416.96 \ J[/tex]
Therefore, the work done by the child as the tricycle travels down the incline is 416.96 J
Which of the following describes the relationship between the weight of fluid
displaced by an object and the buoyant force exerted on the object?
A. Archimedes' principle
B. Flow rate equation
C. Pascal's principle
D. Bernoulli's principle
¿cual es la presión que se aplica sobre un líquido encerrado en un tanque, por medio de un pistón que tiene un aria de 0.02 metros cuadrados ya aplica una fuerza con una magnitud de 100 newtons?
Answer:
Podemos decir que la presión que se aplica sobre un liquido encerrado en un tanque es de 5000 Pa.
Explanation:
Answer:
Podemos decir que la presión que se aplica sobre un liquido encerrado en un tanque es de 5000 Pa.
Explanation:
How many times will the temperature of oxygen with a mass of 1 kg increase if its volume is increased by 4 times, and the pressure is decreased by 2 times?
Round off the answer to the nearest whole number.
Answer:
9.2 Relating Pressure, Volume,
Figure 1. In 1783, the first (a) hydrogen-filled balloon flight, (b) manned hot air balloon flight, and (c) manned hydrogen-filled balloon flight occurred. When the hydrogen-filled balloon depicted in (a) landed, the frightened villagers of Gonesse reportedly destroyed it with pitchforks and knives. The launch of the latter was reportedly viewed by 400,000 people in Paris.
Explanation:
hope its help :)
nicsfrom #philippines
Define Kinetic Energy:
Begin by defining kinetic energy in your own words within one concise eight word sentence
Answer:
kinetic energy is the energy an object has beause of it's motion
Answer:
See explanation below
Explanation:
Kinetic energy is energy stored in a moving object.
A car is traveling at a constant velocity of 20 m/s for 45 seconds. How far does the car go?
Answer: My bad bro. Just tryna get me some points g
Explanation: tuff
Answer:
Explanation:
D=s*t
D=(20)(45)
D=900m.
hope it helps
c) You wish to put a 1000-kg satellite into a circular orbit 300 km above the earth's surface. (a)
What speed, period, and radial acceleration will it have? (b) How much work must be done to the
satellite to put it in orbit? (c) How much additional work would have to be done to make the
Answer:
Scalar
Explanation:
No direction
If you drive first at 40 km/h west and later at 60 km/h west, your average velocity is 50 km/h west.
and what else? is that all?
please help me and i will mark u as brainlist
Answer:
a) 8 secs I think
b)2m/s^2
When two bodies at different temperatures are placed in thermal contact with each other, heat flows from the body at higher temperature to the body at lower temperature until them both acquire the same temperature. Assuming that there is no loss of heat to the surroundings, the heatSingle choice.
(1 Point)
(a) gained by the hotter body will be equal to the heat lost by the colder body
(b) the heat gained by the hotter body will be less than the heat lost by the colder body
(c) the heat gained by the hotter body will be greater than the heat lost by the colder body
(d) the heat lost by the hotter body will be equal to the heat gained by the colder body.
Answer:
Part d is correct.
An amusement park ride whisks you vertically upward. You travel at a constant speed of 15 m/s during the entire ascent. You drop your phone 4.0 s after you (and your phone) begin your ascent from ground level.
a. How high above the ground is your phone when you drop it?
b. Find the maximum height above the ground reached by your phone.
Answer:
a. 60 m
b. 71.48 m
Explanation:
Below are the calculations:
a. The phone's height above the ground = Speed x Time
The phone's height above the ground = 15 x 4 = 60 m
b. Speed when phone drops, u = 15 m/s
At maximum height, v = 0
Use below formula:
v² = u² -2gh
0 = 15² + 2 × 9.8 × h
h = 11.48 m
Total height = 60 + 11.48 = 71.48 m
A cement block accidentally falls from rest from the ledge of a 53.4-m-high building. When the block is 19.4 m above the ground, a man, 2.00 m tall, looks up and notices that the block is directly above him. How much time, at most, does the man have to get out of the way
Answer:
The time required by the man to get out of the way is 0.6 s.
Explanation:
height of building, H = 53.4 m
height of block, h = 19.4 m
height of man, h' = 2 m
Let the velocity of the block at 19.4 m is v.
use third equation of motion
[tex]v^2 = u^2 + 2 gh\\\\v^2 = 0 + 2 \times 9.8 \times (53.4 - 19.4)\\\\v = 25.8 m/s[/tex]
Now let the time is t.
Use second equation of motion
[tex]h = u t + 0.5 gt^2\\\\19.4 - 2 = 25.8 t + 4.9 t^2\\\\4.9 t^2 + 25.8 t - 17.4= 0 \\\\t = \frac{-25.8\pm\sqrt{665.64 + 341.04}}{9.8}\\\\t = \frac{-25.8\pm31.7}{9.8}\\\\t = 0.6 s, - 5.9 s[/tex]
Time cannot be negative so time t = 0.6 s.
Equilibrium of forces
Answer:
If the size and direction of the forces acting on an object are exactly balanced, then there is no net force acting on the object and the object is said to be in equilibrium. Because the net force is equal to zero, the forces in Example 1 are acting in equilibrium.
Equilibrium of forces means that the net force is 0. It can either be when there is no force acting on the object or when the force acting on the object are balanced.
Ramesh announced in class: ''Yesterday I had fever and my body temperature was 100 degrees.'' Ravi said: ''We learnt in the last period that water boils at a temperature of 100 degrees'' Sonal said: "So Ramesh’s temperature yesterday was close to the boiling point of water.'' What can we say about that conversation?Single choice.
(1 Point)
(a) All are correct: human body temperature during fever is close to the boiling point of water.
(b) Ramesh is making some mistake - he is not remembering his temperature correctly.
(c) Ravi is incorrectly recalling the boiling point of water he learnt about in class.
(d) Ramesh and Ravi are correct, but they are using different measurement scales.
Answer:
D. Ramesh and Ravi are correct, but they are using different measurement scales.
\Huge{\underline{\textrm{Explanation}}}Explanation
Here, Ravi says that his body temperature is 100 degrees, but does not mention that whether it is 100 degrees Celsius or 100 degrees Fahrenheit. When the temperature of a human body is more than 100.4 degree Fahrenheit (38°C), or near to it, the person is considered to have fever.
The boiling point of water is 100 degrees Celsius and not 100 degrees Fahrenheit.
Thus, they both are using different measurement scales.
You wish to make a simple amusement park ride in which a steel-wheeled roller-coaster car travels down one long slope, where rolling friction is negligible, and later slows to a stop through kinetic friction between the roller coaster's locked wheels sliding along a horizontal plastic (polystyrene) track. Assume the roller-coaster car (filled with passengers) has a mass of 756.5 kg and starts 88.2 m above the ground. (a) Calculate how fast the car is going when it reaches the bottom of the hill.
Answer:
The speed of roller coaster at the ground is 41.6 m/s.
Explanation:
mass of roller coaster, m = 756.5 kg
height, h = 88.2 m
(a) Let the speed of car at the ground is v.
Use conservation of energy
Potential energy at height = kinetic energy at bottom
[tex]m gh = \frac{1}{2}mv^2\\\\9.8\times 88.2= 0.5\times v^2\\\\v = 41.6 m/s[/tex]
The wave functions for states of the hydrogen atom with orbital quantum number l=0 are much simpler than for most other states, because the angular part of the wave.
a. True
b. False
Transfer of thermal energy between air molecules in closed room is an example of
conduction
convection
radiation
Answer and I will give you brainiliest
Answer: Conduction
Explanation: Conduction is the process by which heat energy is transmitted through collisions between neighboring atoms or molecules. Conduction occurs more readily in solids and liquids, where the particles are closer to together, than in gases, where particles are further apart.
5. Tests performed on a 16.0 cm strip of the donated aorta reveal that it stretches 3.37 cm when a 1.80 N pull is exerted on it. (a) What is the force constant of this strip of aortal material
Answer:
53.41 N/m
Explanation:
From Hooke's law,
Applying,
F = ke............. Equation 1
Where F = Force, e = extension, k = force constant of the aortal material
Make k the subject of the equation
k = F/e............. Equation 2
From the question,
Given: F = 1.8 N, e = 3.37 cm = 0.0337 m
Substitute these values into equation 2
k = 1.8/(0.0337)
k = 53.41 N/m
Hence the force constant of the aortal material is 53.41 N/m
Its volume is 20 cm3, and its mass is 100 grams. What is the sample’s density?
help asap PLEASE I will give u max everything all that
steps if possible
Explanation:
2. [tex]R_T = R_1 + R_2 + R_3 = 625\:Ω + 330\:Ω + 1500\:Ω[/tex]
[tex]\:\:\:\:\:\:\:= 2455\:Ω = 2.455\:kΩ[/tex]
3. Resistors in series only need to be added together so
[tex]R_T = 8(140\:Ω) = 1120\:Ω = 1.12\:kΩ[/tex]
b) When the muscles connected to the crystalline lens contract fully, its focal length is 16.5000 cm. With this focal length, how far away must an object be to form sharply focused images on the retina? (Note: this distance is called the far point of vision.)
c) When the muscles connected to the crystalline lens relax, the focal length is 9.0000 cm. With this focal length, how close must an object be to form sharply focused images on the retina? (Note: this distance is called the near point of vision.)
d) As people age, the crystalline lens hardens (a condition called presbyopia or “old-age” eyes) and can only vary in focal length from 12 to 15.60 cm. Calculate range of vision (the new near point and far point) for this older eye.
e) Based on part d) why might an older person hold the newspaper at arm’s length to read it?
Answer:
I have to go to work and figure it out
A box of bananas weighing 40 N rests on a horizontal surface. The coefficient of static friction between the box and the surface is 0.4, and the coefficient of kinetic friction is 0.2. If no horizontal force is applied to the box and the box is at rest, how large is the friction force exerted on the box?
8 N
16 N
6 N
0 N
Answer:
[tex]0.4x40 \div 0.2 = 80[/tex]
8N
Suppose the coefficient of static friction between a quarter and the back wall of a rocket car is 0.383. At what minimum rate would the car have to accelerate so that a quarter placed on the back wall would remain in place?
Answer:
25.59 m/s²
Explanation:
Using the formula for the force of static friction:
[tex]f_s = \mu_s N[/tex] --- (1)
where;
[tex]f_s =[/tex] static friction force
[tex]\mu_s =[/tex] coefficient of static friction
N = normal force
Also, recall that:
F = mass × acceleration
Similarly, N = mg
here, due to min. acceleration of the car;
[tex]N = ma_{min}[/tex]
From equation (1)
[tex]f_s = \mu_s ma_{min}[/tex]
However, there is a need to balance the frictional force by using the force due to the car's acceleration between the quarter and the wall of the rocket.
Thus,
[tex]F = f_s[/tex]
[tex]mg = \mu_s ma_{min}[/tex]
[tex]a_{min} = \dfrac{mg }{ \mu_s m}[/tex]
[tex]a_{min} = \dfrac{g }{ \mu_s }[/tex]
where;
[tex]\mu_s = 0.383[/tex] and g = 9.8 m/s²
[tex]a_{min} = \dfrac{9.8 \ m/s^2 }{0.383 }[/tex]
[tex]\mathbf{a_{min}= 25.59 \ m/s^2}[/tex]
In a circuit, a source of voltage...
A
converts electrical energy to heat.
B
impedes the flow of electricity through the circuit.
C
leaves the circuit open so that current does not flow.
D
provides the power to move charge through the circuit.
Answer:
yes provides the power to move through the circuit
Explanation:
A factory worker pushes a 32.0 kgkg crate a distance of 4.0 mm along a level floor at constant velocity by pushing horizontally on it. The coefficient of kinetic friction between the crate and floor is 0.24.
(a) What magnitude of force must the worker apply?
(b) How much work is done on the crate by this force?
(c) How much work is done on the crate by friction?
(d) How much work is done on the crate by the normal force? By gravity?
(e) What is the total work done on the crate?
Answer:
a) The force that the worker must apply has a magnitude of 75.317 newtons.
b) The external force does a work of 0.301 joules.
c) The friction force does a work of -0.301 joules.
d) Both normal force and gravity have done a work of 0 joules.
e) The total work done on the crate is 0 joules.
Explanation:
a) As the crate is moving at constant velocity, we know that magnitude of the force done on the crate must be equal to the friction force. Hence, we must use the following formula:
[tex]F = \mu\cdot m\cdot g[/tex] (1)
Where:
[tex]F[/tex] - External force, in newtons.
[tex]\mu[/tex] - Coefficient of kinetic friction, no unit.
[tex]m[/tex] - Mass, in kilograms.
[tex]g[/tex] - Gravity acceleration, in meters per square second.
If we know that [tex]\mu = 0.24[/tex], [tex]m= 32\,kg[/tex] and [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], then the external force is:
[tex]F = (0.24)\cdot (32\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)[/tex]
[tex]F = 75.317\,N[/tex]
The force that the worker must apply has a magnitude of 75.317 newtons.
b) The direction of the force is parallel to the direction of motion. The work done by this force ([tex]W_{F}[/tex]), in joules, is determined by this formula:
[tex]W_{F} = F\cdot \Delta s[/tex] (2)
Where [tex]\Delta s[/tex] is the distance travelled by the crate, in meters.
If we know that [tex]F = 75.317\,N[/tex] and [tex]\Delta s = 0.004\,m[/tex], then the work done by the force is:
[tex]W_{F} = (75.317\,N)\cdot (0.004\,m)[/tex]
[tex]W_{F} = 0.301\,J[/tex]
The external force does a work of 0.301 joules.
c) The direction of the friction force is antiparallel to the direction of motion. The work done by this force ([tex]W_{f}[/tex]), in joules, is determined by this formula:
[tex]W_{f} = -W_{F}[/tex] (3)
[tex]W_{f} = -0.301\,J[/tex]
The friction force does a work of -0.301 joules.
d) The direction of the normal force is perpendicular to the direction of motion. Therefore, no work is done due to normal force.
[tex]W_{N} = 0\,J[/tex]
Likewise, no work is done by gravity.
[tex]W_{g} = 0\,J[/tex]
Both normal force and gravity have done a work of 0 joules.
e) The total work is the sum of the works done by the external force and the friction force:
[tex]W = W_{F}+W_{f}[/tex]
[tex]W = 0.301\,J - 0.301\,J[/tex]
[tex]W = 0\,J[/tex]
The total work done on the crate is 0 joules.
One way families influence healthy technology use is when siblings explain the use of media to each other. Which of these outfits would you expect if this guideline was followed?
Answer:
The answer would be C.
Explanation:
This is what I would expect when you show someone else how to do something then is also known as teaching.
Please Mark as Brainliest
Hope this Helps
A flywheel with radius of 0.400 mm starts from rest and accelerates with a constant angular acceleration of 0.600 rad/s2rad/s2. For related problem-solving tips and strategies, you may want to view a Video Tutor Solution of Throwing a discus. Part A For a point on the rim of the flywheel, what is the magnitude of the tangential acceleration after 2.00 ss of acceleration
Answer: [tex]0.00024\ m/s^2[/tex]
Explanation:
Given
Radius of flywheel is [tex]r=0.4\ mm[/tex]
Angular acceleration [tex]\alpha=0.6\ rad/s^2[/tex]
For no change in radius, tangential acceleration is given as
[tex]\Rightarrow a_t=a\lpha \times r[/tex]
Insert the values
[tex]\Rightarrow a_t=0.6\times 0.4\times 10^{-3}\ m/s^2\\\Rightarrow a_t=2.4\times 10^{-4}\ m/s^2\ \text{or}\ 0.00024\ m/s^2[/tex]
By what amount does the 52-cmcm-long femur of an 85 kgkg runner compress at this moment? The cross-section area of the bone of the femur can be taken as 5.2×10−4m25.2×10−4m2 and its Young's modulus is 1.6×1010N/m2.1.6×1010N/m2.
Answer:
0.156 mm
Explanation:
Here is the complete question
The normal force of the ground on the floor can reach three times a runner's body weight when the foot strikes the pavement. By what amount does the 52-cm-long femur of an 85 kg runner compress at this moment? The cross-section area of the bone of the femur can be taken as 5.2 × 10⁻⁴ m² and its Young's modulus is 1.6 × 10¹⁰ N/m²
The Young's modulus of the bone Y = stress/strain = σ/ε = F/A ÷ ΔL/L = FL/AΔL where F = force on bone = 3mg(since it is 3 times his weight) where m = mass of runner = 85 kg and g = acceleration due to gravity = 9.8 m/s². L = length of femur = 52 cm = 0.52 m, A = cross-sectional area of femur = 5.2 × 10⁻⁴ m² and ΔL = compression of femur.
Making ΔL subject of the formula,
ΔL = FL/AY
ΔL = 3mgL/AY
Substituting the values of the variables into the equation, we have
ΔL = 3mgL/AY
ΔL = 3 × 85 kg × 9.8 m/s² × 0.52 m/(5.2 × 10⁻⁴ m² × 1.6 × 10¹⁰ N/m²)
ΔL = 1299.48 kgm²/s² ÷ 8.32 × 10⁻⁶ N
ΔL = 156.1875 × 10⁻⁶ m
ΔL = 0.1561875 × 10⁻³ m
ΔL = 0.1561875 mm
ΔL ≅ 0.156 mm
The amount does the 52-cm long femur of 85 kg is 0.156 mm.
Calculation of the amount:Since
The Young's modulus of the bone Y should be
= stress/strain
= σ/ε
So,
= F/A ÷ ΔL/L
here F = force on bone = 3mg
m = mass of runner = 85 kg
and g = acceleration due to gravity = 9.8 m/s²
L = length of femur = 52 cm = 0.52 m,
A = cross-sectional area of femur = 5.2 × 10⁻⁴ m²
and ΔL = compression of femur.
Now
ΔL = FL/AY
ΔL = 3mgL/AY
Now
ΔL = 3mgL/AY
= 3 × 85 kg × 9.8 m/s² × 0.52 m/(5.2 × 10⁻⁴ m² × 1.6 × 10¹⁰ N/m²)
= 1299.48 kgm²/s² ÷ 8.32 × 10⁻⁶ N
= 156.1875 × 10⁻⁶ m
= 0.1561875 × 10⁻³ m
= 0.1561875 mm
= 0.156 mm
Learn more about moment here: https://brainly.com/question/24717686
How is the kinetic energy of the particles that make up a substance different
after it changes state?
Answer:
the kinetic energy is less after a substance changes from liquid to gas
Explanation:
the kinetic energy is greater after a substance changes from a solid to liquid
Answer:
The kinetic energy is less after a substance changes from a gas to a liquid.
Explanation: