Answer:
C. It decreases
Explanation:
The pressure law states that for a constant volume of gas in a sealed container the temperature of the gas is directly proportional to its pressure. This can be easily understood by visualizing the particles of gas in the container moving with a greater energy when the temperature is increased.
A common example is cooking gas when refilled, there is a perceptible change in the temperature of the cylinder.
In 50 words or more, describe the scientific process that led to the development of the big bang theory. Include the key ideas and observations of the various scientists you've learned about.
Answer:
Explanation:
The Big Bang theory is a cosmological model of the observable universe from the earliest known periods through its subsequent large-scale evolution.
These theories were based on the hypothesis that all the matter in the universe was created in one big bang at a particular time in the remote past.
What led to big bang theory was that free electrons would have caused light (photons) to scatter the way sunlight scatters from the water droplets in clouds," NASA stated. Over time, however, the free electrons met up with nuclei and created neutral atoms. This allowed light to shine through about 380,000 years after the Big Bang.
The best-supported theory of our universe's origin centers on an event known as the big bang. This theory was born of the observation that other galaxies are moving away from our own at great speed in all directions, as if they had all been propelled by an ancient explosive force.
Origins of the universe, explained
The most popular theory of our universe's origin centers on a cosmic cataclysm unmatched in all of history—the big bang.
The best-supported theory of our universe's origin centers on an event known as the big bang. This theory was born of the observation that other galaxies are moving away from our own at great speed in all directions, as if they had all been propelled by an ancient explosive force.
A Belgian priest named Georges Lemaître first suggested the big bang theory in the 1920s, when he theorized that the universe began from a single primordial atom. The idea received major boosts from Edwin Hubble's observations that galaxies are speeding away from us in all directions, as well as from the 1960s discovery of cosmic microwave radiation—interpreted as echoes of the big bang—by Arno Penzias and Robert Wilson.
Further work has helped clarify the big bang's tempo. Here’s the theory: In the first 10^-43 seconds of its existence, the universe was very compact, less than a million billion billionth the size of a single atom.
It's thought that at such an incomprehensibly dense, energetic state, the four fundamental forces—gravity, electromagnetism, and the strong and weak nuclear forces—were forged into a single force, but our current theories haven't yet figured out how a single, unified force would work. To pull this off, we'd need to know how gravity works on the subatomic scale, but we currently don't.
Two machines, A and B, each working at a constant rate, can complete a certain task working together in 6 days. In how many days, working alone, can machine A complete the task?
your question is missing the given conditions, here is the complete question;
Two machines, A and B, each working at a constant rate, can complete a certain task working together in 6 days. In how many days, working alone, can machine A complete the task?
(1) The average time A and B can complete the task working alone is 12.5 days.
(2) It would take machine A 5 more days to complete the task alone than it would take machine B to complete the task
Answer:
Machine A, working alone, can complete the task in 15 days.
Explanation:
As the average is given as 12.5 so,
A+B=2*12.5
A+ B=25
second condition says machine A takes 5 more days to complete the task alone than B so
A=B+5........Eq1
Now from the question statement we can get that
1/A+1/B=1/6, where A is time needed for A to do the task alone and B is the time needed for B to complete the task alone
thus, simplyfing above equation we get as
AB/A+B=6; \\ putting A+B=25 we get
AB/25=6
AB=6*25
AB=150; \\ putting Eq1 we get
(B+5)B=150
B^2+5B-150=0
simplifying the quadratic equation above we get B=10
putting B=10 in Eq1, we get
A+10=25
A=15
Thus, Machine A, working alone, can complete the task in 15 days.
A man stands on the roof of a building of height 14.9 m and throws a rock with a velocity of magnitude 31.9 m/s at an angle of 25.2 ∘ above the horizontal. You can ignore air resistance.Assume that the rock is thrown from the level of the roof.
Calculate the maximum height above the roof reached by the rock.
Calculate the magnitude of the velocity of the rock just before it strikes the ground.
Calculate the horizontal distance from the base of the building to the point where the rock strikes the ground.
Answer:
(a) 9.402 meters above the roof
(b) Velocity is 31.193 m/s just before hitting the ground
(c) Horizontal distance traveled is 104.2 meters
Explanation:
Let's first find the horizontal and vertical velocity components of the rock when it is thrown. These are:
[tex]V_x=31.9 * Cos(25.2)[/tex]
[tex]V_y=31.9*Sin(25.2)[/tex]
So we have,
Horizontal velocity = 28.864 m/s
Vertical velocity = 13.582 m/s
Now let's solve the three parts with this information:
(a) To find the maximum height, we need to use the fact that vertical velocity at this point will be zero, as the rock is just about to start falling downward. So we have:
[tex](V_f)^2 - (V_i)^2 = 2*a*s[/tex]
where [tex]V_f[/tex] is the final velocity = 0 m/s
[tex]V_i[/tex] is the initial velocity = 13.582 m/s
and a is the acceleration = -9.81 m/s^2
Solving, we get:
[tex]0^2-13.582^2=2*(-9.81)*s[/tex]
s = 9.402 m (distance above roof)
(b) Using the maximum height from (a), we can solve the following equation:
[tex](V_f)^2 - (V_i)^2 = 2*a*s[/tex]
where [tex]V_f[/tex] is the final velocity we need to find,
[tex]V_i[/tex] is the initial velocity = 0 m/s (from maximum height)
a is the acceleration = 9.81 m/s^2
and s = 14.9 + 9.402 = 24.302 m
Solving, we get:
[tex](V_f)^2 - (0)^2 = 2*(9.81)*(24.302)[/tex]
[tex]V_f = 21.836[/tex] m/s
As this is just the vertical velocity, to find the total velocity we have:
V = [tex]\sqrt{(V_x)^2+(V_y)^2}[/tex]
V = [tex]\sqrt{28.864^2 + 21.836^2}[/tex]
V = 36.193 m/s (Total velocity just before it hits the ground)
(c) To solve for this, we need to know the total time for this projectile motion, we can calculate this as follows:
[tex]s=u*t+\frac{1}{2} (a*t^2)[/tex]
here, s = -14.9 m
u = 13.582 m/s (initial vertical velocity)
a = - 9.81 m/s^2 (acceleration due to gravity)
Solving, we get:
[tex]-14.9 = 13.582t+0.5(-9.81t^2)[/tex]
and get the answers:
t1 = 3.61 s
t2 = -0.84 s
Since t2 isn't possible, our total time is t1 = 3.61 seconds.
Using this and our horizontal velocity, we can find the total distance traveled:
Distance = 28.864 * 3.61
Distance = 104.2 m (horizontal)
The blade of a lawn mower is rotating at an angular speed of 128 rad/s. The tangential speed of the outer edge of the blade is 32 m/s. What is the radius of the blade?
Answer:
0.25m
Explanation:
Using the expression that relates the angular velocity(w) and the linear velocity (v).
v= wr where;
w is the angular speed= 128rad/s
v is the linear speed = 32m/s
r is the radius
r = v/w
r = 32/128
r = 0.25m
The radius of the blade is 0.25m
(a) (i) Find the gradient of f. (ii) Determine the direction in which f decreases most rapidly at the point (1, −1). At what rate is f decreasing? (b) (i) Find the gradient of F. (ii) Find the directional derivative of F at the point (1, 1, −5) in the direction of the vector a = 2 i + 3 j − √ 3 k.
Question:
Problem 14. Let f(x, y) = (x^2)y*(e^(x−1)) + 2xy^2 and F(x, y, z) = x^2 + 3yz + 4xy.
(a) (i) Find the gradient of f.
(ii) Determine the direction in which f decreases most rapidly at the point (1, −1). At what rate is f decreasing?
(b) (i) Find the gradient of F.
(ii) Find the directional derivative of F at the point (1, 1, −5) in the direction of the vector a = 2 i + 3 j − √ 3 k.
Answer:
The answers to the question are
(a) (i) the gradient of f = ((y·x² + 2·y·x)·eˣ⁻¹ + 2·y² )i + (x²·eˣ⁻¹+4·y·x) j
(ii) The direction in which f decreases most rapidly at the point (1, −1), ∇f(x, y) = -1·i -3·j is the y direction.
The rate is f decreasing is -3 .
(b) (i) The gradient of F is (2·x+4·y)i + (3·z+4·x)j + 3·y·k
(ii) The directional derivative of F at the point (1, 1, −5) in the direction of the vector a = 2 i + 3 j − √ 3 k is ñ∙∇F = 4·x +⅟4 (8-3√3)y+ 9/4·z at (1, 1, −5)
4 +⅟4 (8-3√3)+ 9/4·(-5) = -6.549 .
Explanation:
f(x, y) = x²·y·eˣ⁻¹+2·x·y²
The gradient of f = grad f(x, y) = ∇f(x, y) = ∂f/∂x i+ ∂f/∂y j = = (∂x²·y·eˣ⁻¹+2·x·y²)/∂x i+ (∂x²·y·eˣ⁻¹+2·x·y²)/∂y j
= ((y·x² + 2·y·x)·eˣ⁻¹ + 2·y² )i + (x²·eˣ⁻¹+4·y·x) j
(ii) at the point (1, -1) we have
∇f(x, y) = -1·i -3·j that is the direction in which f decreases most rapidly at the point (1, −1) is the y direction.
The rate is f decreasing is -3
(b) F(x, y, z) = x² + 3·y·z + 4·x·y.
The gradient of F is given by grad F(x, y, z) = ∇F(x, y, z) = = ∂f/∂x i+ ∂f/∂y j+∂f/∂z k = (2·x+4·y)i + (3·z+4·x)j + 3·y·k
(ii) The directional derivative of F at the point (1, 1, −5) in the direction of the vector a = 2·i + 3·j −√3·k
The magnitude of the vector 2·i +3·j -√3·k is √(2²+3²+(-√3)² ) = 4, the unit vector is therefore
ñ = ⅟4(2·i +3·j -√3·k)
The directional derivative is given by ñ∙∇F = ⅟4(2·i +3·j -√3·k)∙( (2·x+4·y)i + (3·z+4·x)j + 3·y·k)
= ⅟4 (2((2·x+4·y))+3(3·z+4·x)- √3∙3·y) = 4·x +⅟4 (8-3√3)y+ 9/4·z at point (1, 1, −5) = -6.549
Although solid matter is mostly empty space, we don't fall through the floor because atoms are constantly vibrating, even at absolute zero. of electrical forces. of gravitational forces. of nuclear forces. none of the above
We don't fall through the floor because of the electrical forces, specifically the electrostatic repulsion between negatively charged electrons surrounding atoms in both our bodies and the floor.
Although solid matter is mostly empty space, we don't fall through the floor because of electrical forces. Atoms consist mostly of empty space, with the volume of protons, neutrons, and electrons comprising less than 1% of an atom's total volume. The reason we perceive matter as solid and do not pass through the floor is that the electrons surrounding atoms are negatively charged, and the same charges repel each other. This repulsive electrostatic force is stronger at the scale we experience than gravitational force, which is comparatively weak.
A soccer ball kicked on a level field has an initial vertical velocity component of 15.0 mps. assuming the ball lands at the same height from which it was kicked was the total time of the ball in the air?
Answer:
3seconds
Explanation:
Time of flight of an object is the time taken by the object to spend in the air before landing. Mathematically it is represented as;
T = 2u/g
Where g is the acceleration due to gravity = 10m/s²
If the vertical velocity component (u) is given as 15m/s, the time of flight will be;
T = 2(15)/10
T = 30/10
T = 3seconds.
Therefore the total time by the ball in the air is 3seconds
Explanation:
Below is an attachment containing the solution.
An electrical cable consists of 200 strands of fine wire, each having 2.26 µΩ resistance. The same potential difference is applied between the ends of all the strands and results in a total current of 0.876 A. (a) What is the current in each strand? (b) What is the applied potential difference? (c) What is the resistance of the cable?
Answer:
4.38 × 10⁻³ A
0.395mΩ
452 µΩ
Explanation:
(a)the current in each strand is = the total current/ number of strand
0.876/ 200 = 4.38 × 10⁻³ A
(b) potential difference = IR
the total resistance = 2.26 µΩ × 0.876
= 0.395mΩ
c) resistance of the cable = 2.26 µΩ × 200
=452 µΩ
Consider a current carrying a wire coming out of your computer screen towards you. Which statement below correctly describes the magnetic field created by the current in the wire?
1. The magnetic field encircles the wire in a counterclockwise direction
2. The magnetic field encircles the wire in a clockwise direction
Answer:
The magnetic field encircles the wire in a reversal or counterclockwise direction
Explanation:
Lenz's law states that the direction of the current induced in a conductor by a changing magnetic field is such that the magnetic field created by the induced current opposes the initial changing magnetic field which produced it. According to Lenz law, the magnetic field encircles the wire in a reversal or counterclockwise direction.
Therefore, the first option is the correct answer.
The prominent semicircular space above a doorway in a Romanesque church portal is referred to as a ___________ and was often covered in elaborate carvings
Answer: Tympanum
Explanation: A Tympanum could be described as the semicircular space above the doorway or window or entrance of a building. The space is usually designed using elaborate sculptural designs, Rocky costumes and ornaments. The tympanum could also be triangular-shaped and attributed to classical Greece and Roman architecture. It is usually located between the portals archivolt and the lintel. The tympanum is commonly observed in the churches and temples of ancient Rome and Greece.
The semicircular space above a doorway in a Romanesque church portal is called a tympanum and it was often decorated with intricate carvings relating to religious themes.
Explanation:The prominent semicircular space above a doorway in a Romanesque church portal is referred to as a tympanum. These were often covered in elaborate carvings that represented religious imagery or important biblical stories. The Romanesque style of architecture, prevalent in Europe from the 9th to 12th centuries, was known for its massive quality, thick walls, and robust pillars. This style also romanticized elaborate decorative details, among them the beautifully carved tympanum.
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fessor drives off with his car (mass 840 kg), but forgot to take his coffee mug (mass 0.41 kg) off the roof. The coefficient of static friction between the mug and the roof is 1.1, and the coefficient of kinetic friction is 0.4. What is the maximum acceleration of the car, so the mug does not slide off
Answer:
Acceleration =4.42m/s^2
Explanation:
As the mug is not sliding on the car roof ,there must be static friction between the car and the mug. The maximum limit of static friction that can act is given by:
Kmg
Where I is coefficient of static friction between car roof and mug
m = mass of mug
g= acceleration due to gravity
Acceleration of the car= Kmg
Acceleration = 1.1 × 0.41 × 9.8 =4.42m/s
Answer:
10.791 m/s^2
Explanation:
Solution:
- As the mug is not sliding on the car roof , there will be static friction between car and mug.
- The maximum limit of static friction that can act is given by = k*m*g
- Where, k = coefficient of static friction between the mug and the roof
m = mass of mug
g = acc. due to gravity
- So the maximum acc. of car ,
(m*a = k*m*g)
a = k*g
- Therefore max. a = 1.1*g = 1.1*9.81 = 10.791 m/s^2
What are the similarities between the operation of a radio telescope and that of an optical reflecting telescope?
Answer:
They both intercept,focus and determine the intensity of an incoming radiation but at different degree.
Explanation:
Radio telescope is much larger than optical telescope because radio wavelengths are much longer than optical wavelengths. The longer the wavelengths implies that the radio waves has a lower energy than optical light waves. So, in order to collect enough radio photons to detect a signal, the radio dishes must be very large.
A good baseball pitcher can throw a baseball toward home plate at 93 mi/h with a spin of 1510 rev/min. How many revolutions does the baseball make on its way to home plate? For simplicity, assume that the 60 ft path is a straight line.
Answer:
[tex]\Delta \theta =11.073\ rev[/tex]
Explanation:
Given,
Speed of baseball = 93 mi/h
spin = 1510 rev/min
[tex]\Delta x = 60\ ft[/tex]
[tex]v =93\times \dfrac{88}{60} = 136.4 ft/s[/tex]
[tex]\omega = 1510\times \dfrac{1}{60}= 25.167\ rad/s[/tex]
[tex]\Delta t = \dfrac{\Delta x}{v}[/tex]
[tex]\Delta t = \dfrac{60}{136.4 }[/tex]
[tex]\Delta t = 0.44\ s[/tex]
Revolutions of ball
[tex]\Delta \theta = \omega t[/tex]
[tex]\Delta \theta = 25.167\times 0.44[/tex]
[tex]\Delta \theta =11.073\ rev[/tex]
Revolution of the ball is equal to [tex]\Delta \theta =11.073\ rev[/tex].
MIT’s robot cheetah can jump over obstacles 46. cm high and has speed of 12.0 km/h. a) If the robot launches itself at an angle of 60.° at this speed, what is its maximum height? b) What would the launch angle have to be to reach a height of 46. cm?
Answer:
(a) [tex]y_{max}=0.423m[/tex]
(b) [tex]\alpha =64.3^{o}[/tex]
Explanation:
Given data
[tex]v_{i}=12km/h=3.33m/s\\\alpha =60^{o}\\g=9.8m/s^{2}\\Required\\(a)y_{max}\\(b)Angle[/tex]
Solution
For Part (a)
As the velocity component in direction of y is given by:
[tex]v_{yi}=v_{i}Sin\alpha \\v_{yi}=3.33Sin60\\v_{yi}=2.88m/s[/tex]
The maximum displacement is given by:
[tex]v_{yf}^{2}=v_{yi}^{2}-2gy_{max}\\ y_{max}=\frac{(2.88)^{2}}{2(9.8)}\\ y_{max}=0.423m[/tex]
For Part (b)
To reach y=46cm =0.46m apply:
[tex]0=v_{yi}^{2}-2(9.8)(0.46)\\v_{yi}=3m/s\\As\\Sin\alpha =\frac{v_{yi}}{v_{i}}\\\alpha =Sin^{-1}(\frac{v_{yi}}{v_{i}})\\\alpha =Sin^{-1}(\frac{3}{3.33} )\\\alpha =64.3^{o}[/tex]
a) When launched at a 60° angle with a speed of 3.33 m/s, the robot cheetah can reach a maximum height of about 0.715 meters.
b) To reach a height of 0.46 meters, the launch angle needs to be approximately 31.74 degrees.
a) To find the maximum height (h_max) when the robot cheetah launches at a 60-degree angle with a speed of 3.33 m/s, we can use the following formula:
h_max = (u^2 * sin^2(θ)) / (2 * g)
Where:
u is the initial speed (3.33 m/s)
θ is the launch angle (60 degrees)
g is the acceleration due to gravity (9.8 m/s²)
First, calculate sin(60 degrees):
sin(60 degrees) = √3 / 2 ≈ 0.866
Now, plug in the values and calculate h_max:
h_max = (3.33^2 * (0.866)^2) / (2 * 9.8) ≈ 0.715 meters
b) To determine the launch angle (θ) required to reach a height of 0.46 meters (46 cm), we'll rearrange the same formula:
sin^2(θ) = (2 * g * h_desired) / u^2
Plug in the values:
sin^2(θ) = (2 * 9.8 * 0.46) / (3.33^2)
Now, find sin(θ):
sin(θ) ≈ √(0.2794) ≈ 0.5289
Finally, calculate θ:
θ ≈ arcsin(0.5289) ≈ 31.74 degrees
So, the launch angle required to reach a height of 0.46 meters is approximately 31.74 degrees.
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A strong base ______ in solution.
A.) dissociates completely
B.) dissociates partially
C.) does not dissociate
D.) always solidifies
Answer:
a) dissociates completely
Explanation: strong base is a base that is completely dissociated in an aqueous solution.
A uniform cylindrical grindstone has a mass of 15.0 kg and a radius of 18.6 cm. The grindstone reaches a rotational velocity of 1.83 rev/min before its motor turns off. After the grindstone motor is turned off, a knife blade is pressed against the outer edge of the grindstone with a perpendicular force of 8.82 N. The coefficient of kinetic friction between the grindstone and the blade is 0.80. Use the work energy theorem to determine how many turns the grindstone makes before it stops.
Answer:
[tex]n \approx 5.778\times 10^{-4}\,rev[/tex]
Explanation:
The Work-Energy Theorem is applied herein:
[tex]-\frac{1}{4}\cdot m_{cyl}\cdot R^{2}\cdot \omega^{2} = - \mu_{k}\cdot N\cdot r \cdot 2\pi \cdot n[/tex]
The number of turns needed to stop the grindstone is:
[tex]n = \frac{m_{cyl}\cdot R^{2}\cdot \omega^{2}}{4\cdot \mu_{k}\cdot N \cdot r\cdot 2\pi}[/tex]
[tex]n = \frac{(15\,kg)\cdot (0.186\,m)^{2}\cdot [(1.83\,\frac{rev}{min} )\cdot (\frac{2\pi\,rad}{1\,rev} )\cdot (\frac{1\,min}{60\,s} )]^{2}}{4\cdot (0.80)\cdot (8.82\,N)\cdot(0.186\,m)\cdot 2\pi}[/tex]
[tex]n \approx 5.778\times 10^{-4}\,rev[/tex]
In a circus performance, a monkey is strapped to a sled and both are given an initial speed of 2.0 m/s up a 20.0° inclined track. The combined mass of monkey and sled is 24 kg, and the coefficient of kinetic friction between sled and incline is 0.20. How far up the incline do the monkey and sled move? m
Answer:
2.65 m
Explanation:
From work-kinetic energy principle,
workdone by friction + workdone by gravity on sled = kinetic energy change of sled
Let h be the vertical height moved and d the distance moved along the incline. h = dsinθ where θ is the angle of the incline = 20°.
The workdone by gravity on the sled is mghcos180 = -mgh = -mgdsinθ
The frictional force = -μmgcosθ where μ = 0.20 and the work done by friction = -μmgcosθd
The kinetic energy change = 1/2m(v₂² - v₁²) where v₁ = initial speed = 2.0 m/s and v₂ = final speed = 0 m/s (since the sled stops)
So, -mgdsinθ - μmgcosθd = 1/2m(v₂² - v₁²)
-gd(sinθ - μcosθ) = 1/2(v₂² - v₁²)
d = (v₂² - v₁²)/-2g(sinθ - μcosθ)
substituting the values for the variables,
d = (0 - 2²) /[- 2 × 9.8(sin20 - 0.2cos20)]
d = -4 /[- 2 × 9.8(sin20 - 0.2cos20)]
d = -4/ - 1.51 = 2.65 m
It has moved up the incline a distance of 2.65 m.
Some Canadian troops are sent (as part of a U.N. peacekeeping force) to a country located on the Earth's equator. At night, when homesickness makes them gaze sleeplessly at the stars, which of the following will be familiar to them (the same at the equator as in Canada):
a. the celestial poles are on the north and south points of the horizon
b. the celestial equator is overhead and passes through the zenith
c. all stars rise and set (none remain in the sky all night long)
d. all-stars are above the horizon exactly half a day
e. none of the above are the same on the equator as in Canada
Answer:
Option E is correct.
none of the listed options are the same on the equator as at the celestial north pole; the region where Canada is.
Explanation:
Firstly, let's explain some terms.
The earth's rotation is not straight forward as a sphere just rotating. The shape of the earth (better described as geoid; spherical, but a bit flattened at the poles), some gravitational forces on the earth and the existence of seasons mean the earth's rotation is tilted at an angle of 23.5° relative to our orbital plane; the plane of the earth's orbit round the sun.
When the tilted earth is considered, the celestial north and south poles are the north and south poles on the tilted earth's rotational axis.
And the celestial equator is when the equator on the tilted earth, equidistant from the celestial north and south poles is projected into space. A portion of the celestial equator passes through the real equator.
Now, the options one by one,
A) the celestial poles are on the north and south points of the horizon
At the celestial north and south poles, the poles are observable at the north and south points of the horizon.
At the celestial north pole, it is usually located by locating the north star, Polaris, and at the celestial South poles, there are about 4 different methods of locating the poles on the horizon.
But these poles aren't noticeable on the horizon at the equator. The northern one can only be noticed at regions around the celestial north pole.
Hence, the statement A isn't correct.
B) the celestial equator is overhead and passes through the zenith.
Zenith describes the the point in the sky directly above one's head.
At some points on the equator, the celestial equator is truly overhead, but the celestial equator doesn't pass overhead at the north pole, So, this isn't something common to the equator and the north pole.
C) all stars rise and set (none remain in the sky all night long)
Stars remaining in the sky all night long only occurs at the poles (north pole especially). Depending on the seasons, there are times and/or places that have 24 hour sunlight or darkness.
But, this is usually not the case at the equator. At the equator, all stars rise and set all year round. Only small noticeable longer days, shorter nights or longet nights, shorter days are experienced at the equator.
Hence, the statement given isn't true for the north pole and the equator.
D) all-stars are above the horizon exactly half a day.
The paths of the stars are vertical and are cut exactly in half by the horizon at the equator. Each star is up half the time and down half the time.
Unlike at the poles, as explained in the previous statement, they enjoy more varied daytime and night times.
Again, the statement described isn't true for the north pole and the equator.
Tech A says that the solenoid on the starter motor switches the high current on and off. Tech B says that the solenoid engages the starter motor with the ring gear. Who is correct
Tech B is correct.
Explanation:
Generally the electromagnet for a coil of wire, which allow the system to transform electric energy into mechanical energy is understood as a solenoid, by generating magnetic field from electric current and induces linear motion by utilizing the magnetic field. Every modern starters depend on the solenoid to use the flywheel ring gear to enable the start drive. When the solenoid is rejuvenated, a plunger or lever is controlled which pushes the ring gear into mesh with the pinion. A standard starter solenoid has one small adapter for the starter control wire and two major terminals: one for the positive battery cable, and another one for the thick wire that drive the starter motor by its own.
The correct answer is that both Tech A and Tech B are correct.
In a vehicle's starting system, the solenoid serves two main functions:
1. Switching High Current: When the ignition key is turned to start the engine, a small current flows through the solenoid coil, creating an electromagnetic field that pulls the solenoid plunger inward. This action closes a pair of heavy contacts, allowing a high current to flow from the battery to the starter motor. This is what Tech A is referring to when they mention that the solenoid switches the high current on and off.
2. Engaging the Starter Motor with the Ring Gear: The movement of the solenoid plunger also pushes a lever that engages the starter motor's pinion gear with the ring gear on the engine's flywheel. This engagement is necessary to turn the engine over during the starting process. This is the function that Tech B is describing.
Therefore, the solenoid is responsible for both switching the high current to the starter motor and engaging the starter motor with the ring gear. Both technicians are correct in their statements about the functions of the solenoid in the starting system.
Tito knows that one of the reasons people do not return to his electronics store is because of the slow service. How would a SWOT analysis classify the slow service at Tito's electronic store?
It is classified as Weakness in SWOT analysis.
Explanation:
SWOT analysis is performed to understand the characteristics of any start ups. It is the abbreviation of Strength, Weakness, Opportunities and Threats. It helps in self analyzing and to plan strategically for improvements. So in this different loopholes, advantages, benefits, profits and loss attained by any organized is classified in these four options like Strength, Weakness, Opportunities and Threats.
In the present case, the problem of slow service speed by Tito's company lead to reduction in customers. Also many of the customers are not returning back due to its slow speed in servicing any instrument. So this loophole is placed in the box of weakness in SWOT analysis. So it is classified as Weakness in SWOT analysis.
f we apply an electrical stimulus to a muscle cell to cause it to contract, the magnitude of that stimulus must be strong enough to reach a critical value that is essential to initiate contraction. This critical value is known as __________.
Answer:
critical value is known as threshold
Explanation:
solution
when we apply electrical stimulus to the cell of muscle
then magnitude of stimulus will be strong enough for reaching the critical value
and this critical value is threshold because when the size of an stimulus change in requirement for the detection, this is called the threshold
but if change is done by without stimulation then there threshold stimulus is absolute threshold
so correct answer is threshold
An open container holds ice of mass 0.500kg at a temperature of -16.1?C . The mass of the container can be ignored. Heat is supplied to the container at the constant rate of 850J/minute .
The specific heat of ice to is 2100 J/kg?K and the heat of fusion for ice is 334
Answer: Tmelt = 19.89mins
Explanation: (complete question- 334×10^3j/kg. How much time, Tmelt passes before the ice starts to melt?)
For ice to melt, its temperature must be 0* C.
Let
Q1 = heat required to raise the ice temp from -16.1 °C. to 0* C.
Q2 = heat required to melt the ice
Q1 = MCp(delta T)
Q2 = m(Hf)
where
M = mass of the ice = 0.500 kg (given)
Cp = specific heat of ice = 2100 J/kg K (given)
delta T = temperature change = 0 - -16.1= 16.1 °C.
Hf = heat of fusion of ice = 334 x 10^3 J/jg.
Substituting values,
Q1 = 0.500× 2100× 16.1 = 16905J
Q2 = 0.500 × 334×10^3 = 167,000J
Heat required to melt the ice = Q1 + Q2 = 16905 + 167000 = 183905 J
How much time Tmelts = Q1/ 850 = 16905/850 = 19.89mins.
A bucket of water is being raised from a well using a rope. If the bucket of water has a mass of 6.2 kg, how much force (in N) must the rope exert on the bucket to accelerate it upward at 1 m/s2
Answer:
6.2N force
Explanation:
According to Newton's second law of motion, force is equal to the product of the mass of a body and its acceleration. Mathematically,
Force = mass × acceleration
Given mass of bucket of water = 6.2kg
acceleration of the bucket = 1m/s²
Force exerted on the rope = 6.2 × 1
= 6.2N
Consider a steam power plant that operates between the pressure limits of 5 MPa and 10 kPa. Steam enters the pump as saturated liquid and leaves the turbine as saturated vapor. Determine the ratio of the work delivered by the turbine to the work consumed by the pump. Assume the entire cycle to be reversible and the heat losses from the pump and the turbine to be negligible
The ratio of the work delivered by the turbine to the work consumed by the pump in the steam power plant is 1.
To determine the ratio of the work delivered by the turbine to the work consumed by the pump in the steam power plant, we need to consider the energy conversions that occur in the cycle. The work done by the turbine is equal to the area under the pressure-volume curve in the turbine phase, while the work consumed by the pump is equal to the area under the pressure-volume curve in the pump phase. Since the process is reversible and heat losses are negligible, the amount of work done in the turbine is equal to the amount of work consumed by the pump. Therefore, the ratio of the work delivered by the turbine to the work consumed by the pump is 1.
Learn more about Steam power plant cycle here:https://brainly.com/question/31731141
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The muzzle velocity of a gun is the velocity of the bullet when it leaves the barrel. The muzzle velocity of one rifle with a short barrel is greater than the muzzle velocity of another rifle that has a longer barrel. In which rifle is the acceleration of the bullet larger
Answer:small barrel gun
Explanation:
Given
Muzzle velocity of bullet is greater in short barrel gun as compared to larger barrel gun
acceleration is given by change in velocity with respect to time
[tex]a=\dfrac{\Delta v}{\Delta t}[/tex]
In case of short barrel bullet time taken by bullet to reach its muzzle velocity is less therefore acceleration of small barrel bullet is more compared to long barrel bullet.
Final answer:
The muzzle velocity of a bullet can be calculated using the formula v = u + at. Using the given acceleration of 6.20 x 10⁵ m/s² and time of 8.10 x 10⁻⁴ s, the muzzle velocity is 502 m/s.
Explanation:
To calculate the muzzle velocity of a bullet, which is the velocity of the bullet as it leaves the barrel of a gun, we can use the formula for uniform acceleration: v = u + at, where v is the final velocity, u is the initial velocity (which is 0 m/s in this case, as the bullet starts from rest), a is the acceleration, and t is the time. Here, we are given an acceleration of 6.20 × 10⁵ m/s² and a time of 8.10 × 10⁻⁴ s. Plugging these values into the formula gives us the muzzle velocity.
Using the formula, we have: v = 0 m/s + (6.20 × 10⁵ m/s²)(8.10 × 10⁻⁴ s), which simplifies to v = 502 m/s. Therefore, the muzzle velocity of the bullet is 502 meters per second. This value represents how fast the bullet is traveling when it exits the gun barrel.
What enables humans to see light in the infrared range of the electromagnetic spectrum
Answer: night vision goggles
Explanation: These goggles use technology to enhance all light, including the infrared rays, which aren't normally visible to the naked eye. This way you can see any type of light that is surrounding you.
please give brainliest!
Nothing does. That's why we can't see it without some sort of external imaging device.
Now suppose the crate is lifted so rapidly that air resistance was significant during the raising. How much work was done by the lifting force as the box was raised 1.5 m?
Answer:
A
Explanation:
- The complete question is:
" A crate is lifted vertically 1.5 m and then held at rest. The crate has weight 100 N (i.e., it is supported by an upward force of 100 N).Now suppose the crate is lifted so rapidly that air resistance was significant during the raising. How much work was done by the lifting force as the box was raised 1.5 m?"
Options:
1. More than 150 J
2. A bit less than 150 J because the air partially supported the crate.
3. No work was done.
4. Still 150 J
5. None of these
Solution:
- As the box is raised the work is done against gravity and air resistance opposes the motion of box. Hence, both the force of gravity ( Weight and air resistance act downward).
- The amount of work done is the sum of both work done against gravity and against air resistance.
- W = W_g + W_r
W > W_g
W > 100*1.5
W > 150 J
- Hence, the work done is greater than 150 J i.e work is also done against ai r resistance.
human beings can typically detect a difference in sound level of 2.0 dB. What is the ratio of the amplitudes of two sounds whose levels differ by this amount
Answer: 1.25
Explanation:
The ratio of the amplitude of the sound can be calculated using the wave intensity by squaring the amplitude of the two wave particles.
Let the sound level be S2 and S1 and intensities be I1 and I2
Difference in sound level(S) = S2 - S1 = 2dB
Using the relationship between difference in sound level and Intensity
S2 - S1 = 10log(I2 ÷ I1)
Note : Intensity is the square of the amplitude of sound
Therefore,
I1 = A1^2 and I2 = A2^2
Substituting the values into the equation:
S2 - S1 = 10log (I2 ÷ I1)
2dB = 10log (A2^2 ÷ A1^2)
2dB = 10log(A2 ÷ A1)^2
2 dB= 20log(A2 ÷ A1)
2÷20 = log (A2 ÷ A1)
0.1 = log (A2 ÷ A1)
(A2 ÷ A1) = 10^0.1
A2/A1 = 1.25
A satellite system that has unique properties is the ____ satellite, which is used by governments for spying and by scientific agencies for observing celestial bodies
Answer: Highly-elliptical-earth-orbit (heo)
Explanation: Highly-elliptical-earth-orbit (heo) satellite system has unique properties
which is used by governments for spying and by scientific agencies for observing celestial bodies. It is an extremely elongated orbit that is useful for communication satellites which creates signals between a source transmitter and a receiver at different locations on earth. They are used by government for spying, scientific agencies for observing celestial bodies, for the internet and telephone communications.
Answer:
Highly elliptical satellite orbit.
Explanation:
Highly elliptical satellite orbit is used to provide coverage over any point on the globe. The HEO is not limited to equatorial orbits like the geostationary orbit and the resulting lack of high latitude and polar coverage.
As a result of its ability to provide high latitude and polar coverage, countries such as Russia which need coverage over polar and near polar areas make significant use of highly elliptical orbits, HEO. With two satellites in any orbit, they are able to provide continuous coverage.
What should be the speed for this to happen? Assume that the radius of the Moon is rMrM = 1.74×106m×106m, and the mass of the Moon is mMmM = 7.35×1022kg.
Answer:
a. v = 1.679 × 10⁹ m/s b. The answer of v = 1.679 × 10⁹ m/s is not reasonable
Explanation:
Here is the complete question
You read in a science magazine that on the Moon, the speed of a shell leaving the barrel of a modern tank is enough to put the shell in a circular orbit above the surface of the Moon (there is no atmosphere to slow the shell).
A.....What should be the speed for this to happen? Assume that the radius of the Moon is rM = 1.74×106m, and the mass of the Moon is mM = 7.35×1022kg.(Express your answer to two significant figures and include the appropriate units.)
B.....Is this number reasonable?
Solution
From the question, the centripetal force in orbit for the shell = gravitational force of moon on shell.
So, mv²/r = GMm/r²
So v = √(GM/r) where v = velocity of shell, G = universal gravitational constant = 6.67 × 10⁻¹¹ Nm²/kg², M = mass of moon = 7.35 × 10²² kg and r = radius of moon = 1.74 × 10⁶ m.
v = √(GM/r) = v = √(6.67 × 10⁻¹¹ Nm²/kg² × 7.35 × 10²² kg/1.74 × 10⁶ m)
v = √(49.0245/1.74 × 10¹⁷) m/s
v = √28.175 × 10¹⁷ m/s
v = √2.8175 × 10¹⁸ m/s
v = 1.679 × 10⁹ m/s
b. The answer of v = 1.679 × 10⁹ m/s is not reasonable because, it is over 1000000 km/s which is greater than the speed of light which is 300000 km/s