Explanation:
Magnetic fields are areas where an object exhibits a magnetic influence. The fields affect neighboring objects along things called magnetic field lines. A magnetic object can attract or push away another magnetic object. You also need to remember that magnetic forces are NOT related to gravity.
which wave has a higher frequency and why?
Explanation:
the figure in the left side has higher frequency.
because it has more nos. of wave in 1sec.
Which of the following is NOT true of a hotspot?
Given what we know about hotspots and their characteristics, we can confirm that the option that is false regarding hotspots is "A. Hotspots are constantly moving forming chains of volcanoes".
In regards to the study of geology, we refer to hotspots as spots in an oceanic mantle where magma located below causes them to become intensely hot. This heat forces magma and therefore the mantle itself upwards as a tectonic plate moves across the spot, and therefore results in the formation of volcanic chains. Some very popular examples of volcanoes and volcanic islands formed by this method include:
Yellowstone National ParkHawaiiThe country of IcelandDespite the other options being true, option A is false given that hotspots are stationary elements of the oceanic mantle. Although some scientists believe that there is evidence of the movement of hotspots, this movement is slow enough to not be taken into account.
This question was answered in regards to the complete question found online which states:
Which of the following is NOT TRUE about hotspots? a. Hotspots are constantly moving forming chains of volcanoes.
b. Hotspots are stationary.
c. Hotspots form volcanic island arcs
d. Hotspots lie along an oceanic plate
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2. What are the challenges of wind energy that causes decrease in the electrical production?
3. Give two solutions:
1. Modification in the structure of the turbine
2. Modification in the function of the turbine.
4. You must estimate the cost in each solution, and which one is more affordable and efficient for the municipality to go for.
5. You must make sure that there is minimal energy loss.
Levelized production cost is a big challenge in the production of wind energy.
The main challenge in the production of wind energy is the high production cost. It has to minimize the Levelized Production Cost (LPC) which is the ratio between energy production cost and economic lifetime. This problem can be solved by doing modification in the structure of the turbine as well as we must ensure minimal energy loss.
The modification in the structure can produce more electrical energy and reduce the energy loss also provides more profit to the company so we can conclude that levelized production cost is a big challenge in the production of wind energy.
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A crane lifts a 10000 N load through a distance of 30mins in 40s calculate work done
Given,
Force = 10,000 N
Distance = 30 m
Time = 40 s
Work done = force × distance
Work = 10,000 × 30
Work = 300,000 J
______
꧁✿ ᴿᴬᴵᴺᴮᴼᵂˢᴬᴸᵀ2222 ✬꧂
The work done on the object is equal to the power it experiences which is equal to 735 watts.
What is work done?Work done is defined as the product of force applied and the distance over which the force is applied on the object. Work is done on an object when a force is applied to an object and the object is moved through a particular distance.
The time rate of doing work is called as power, which is designated by the symbol P. It is measured in the units of watts, where one watt is equal to one joule energy applied per second time.
Work done on an object is equal to the force (F) multiplied by the distance d in the direction of the force. Where, time interval is Δt.
P = W/Δt = F × d/Δt = m × g × h/Δt = 100 × 9.8 × 30 / 40
Power = 735 watts
The power of the object is 735 watts.
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A catamaran with a mass of 5.44×10^3 kg is moving at 12 knots. How much work is required to increase the speed to 16 knots? (One knot = 0.51 m/s.)
show all steps
The work that is required to increase the speed to 16 knots is 14,176.47 Joules
If a catamaran with a mass of 5.44×10^3 kg is moving at 12 knots, hence;
5.44×10^3 kg = 12 knots
For an increased speed to 16knots, we will have:
x = 16knots
Divide both expressions
[tex]\frac{5.44 \times 10^3}{x} = \frac{12}{16}\\12x = 16 \times 5.44 \times 10^3\\x = 7.23\times 10^3kg\\[/tex]
To get the required work done, we will divide the mass by the speed of one knot to have:
[tex]w=\frac{7230}{0.51}\\w= 14,176.47Joules[/tex]
Hence the work that is required to increase the speed to 16 knots is 14,176.47 Joules
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I need help in question 7, a and b.
Answer:
The graph for 7a is shown in the attachment. For question 7b she walks a distance of 16 meters. (m)
Explanation:
A bob attached to a string of length L = 1.25 m, initially found at the equilibrium
position, is given an initial velocity v = 0.8 m/s. The maximum displacement angle,
Omax, is: (Take g=10 m/s)
17
3.2
13
10.2
6.4
Clear selection
A hoh attached to a string of length 2 m is displaced by an angle of 8º and then
The maximum displacement angle of the bob is 13⁰.
The given parameters;
Length of the pendulum, L = 1.25 mInitial velocity of the bob, v = 0.8 m/sThe maximum displacement of the bob is calculated by applying the principle of conservation of energy;
[tex]P.E = K.E\\\\mgh = \frac{1}{2} mv^2\\\\h = \frac{v^2}{2g} \\\\h = \frac{0.8^2}{2\times 10} \\\\h = 0.032 \ m[/tex]
The maximum displacement angle is calculated as follows;
[tex]cos \theta = \frac{L-h}{L} \\\\cos \theta = \frac{1.25 - 0.032}{1.25} \\\\\cos \theta = \frac{1.218}{1.25} \\\\cos \theta = 0.9744\\\\\theta = cos^{-1}(0.9744)\\\\\theta = 13\ ^0[/tex]
Thus, the maximum displacement angle of the bob is 13⁰.
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Calculate torque using angular momentum
Answer:
The equation net τ=ΔLΔt net τ = Δ L Δ t gives the relationship between torque and the angular momentum produced.
Acceration is a change in motion over time:
True False
Scientists are experimenting with a kind of gun that may eventually be used to fire payloads directly into orbit. In one test, this gun accelerates a 5.6-kg projectile from rest to a speed of 3.5 × 103 m/s. The net force accelerating the projectile is 2.5 × 105 N. How much time is required for the projectile to come up to speed?
Answer:
7.84 s
Explanation:
Ok, one step at a time. We know the mass of the projectile, we know the force. Acceleration is easily obtained from Newton second law:
[tex]\vec F = m\vec a[/tex][tex]2.5\times 10^5 = 5.6a \rightarrow a = (2.5/5.6) \times 10^5 = 4.46 \times 10^4 ms^-^2[/tex].
At this point, we know the acceleration, we know initial and final velocity, we can time the ammount of time it took to get there.
[tex]v= v_0 +at \rightarrow 3.5\times10^3 = 0 + 4.46\times10^4 t\\t= \frac{3.5\times10^3}{4.46\times10^4} = 7.84 s[/tex]
The mercury in a barometer of a cross-sectional area 1 cm² stands at 75 cm, and the space above it is 9 cm in length. What
volume of air, measured at 3. The mercury in a barometer of a cross-sectional area 1 cm² stands at 75 cm, and the space above it is 9 cm in length. What
volume of air, measured at atmospheric pressure, would have to be admitted into the space to cause the column of the mercury
to drop to 59 cm?
The ideal gas equation and the pressure in barometer allows us to find the amount of air that we must introduce into the barometer for the change in height of the mercury column is:
The variation of the volume is: ΔV = 7.67 cm³
Pressure is defined by the relationship between force and area.
P = F / A
The ideal gas equation establishes a relationship between pressure, volume, and temperature of an ideal gas.
PV = nR T
Where P is pressure, V is volume, and T is temperature.
Let's write this equation for two points assuming that the temperature has not changed.
P₀ V₀ = P₁ V₁
V₁ = [tex]\frac{P_o}{P_1} \ V_o[/tex] (1)
The subscript "o" is used for the start point and the subscript "1" for the end point.
The pressure in a barometer is:
P = ρ g y
They indicate the initial height of the barometer y₀=75 cm, the distance from empty space y'₀ = 9 cm and the final height of the barometer y₁ = 59 cm.
The volume of the cylinder is
V = π r² y
Let's calculate the initial volume.
V₀ = π 1 9
V₀ = 28.27 cm³
We substitute in equation 1.
V₁ = [tex]\frac{\rho \ g \ y_o}{\rho \ g \ y_1} \ V_o[/tex]
V₁ = [tex]\frac{y_o}{y_1} \ V_o[/tex]
Let's calculate.
V₁ = [tex]\frac{75}{59} \ 27.27[/tex]
V₁ = 35.94 cm³
The volume to be incremented is
ΔV = V₁ - V₀
ΔV = 35.94 - 28.27
ΔV = 7.67 cm³
Using the ideal gas equation and the pressure in barometer we can find the amount of air that we must introduce into the barometer for the change in height of the mercury column is:
The change of the volume is: ΔV = 7.67 cm³
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A marble is launched at by dropping it through a marble launcher. The marble took 0.22s to land 0.31m from
where it was launched. Find the horizontal velocity of the marble as it rolls off the table.
A. 1.4m/s
B. 0.53m/s
C. 0.71 m/s
D. 0.07m/s
Hi there!
For projectile motion, the horizontal and vertical components are SEPARATE.
We can use the kinematic equation to solve:
dₓ = vₓt
We can rearrange to solve for vₓ:
dₓ/t = vₓ
0.31/0.22 = vₓ
vₓ = 1.4 m/s ⇒ A
What symbols are these?
Answer:
the bottom one is wollsiegel
What is the rotational inertia of the bug-rod-spheres system about the horizontal axis through the center of the rod
Answer:
C) 5ML^2
Explanation:
2 Spheres of mass M
Bug's mass 3M
Rod length 2L, radius L
Find Rotational Inertia I
I=Σmr^2
I=(3M+M+M)L^2
I=5ML^2
The rotational inertia of the bug-rod-spheres system about the horizontal axis through the center of the rod is 5ML².
Rotational inertia of the systemThe rotational inertia of the system is determined by summing all the masses together and multiply with the radius.
I(t) = Σmr²
Where;
m is the massr is the radiusI(t) = Inertia of the 2 spheres + inertia of the bug
I(t) = (M + M)L² + (3M)L²
I(t) = 2ML² + 3ML²
I(t) = 5ML²
Thus, the rotational inertia of the bug-rod-spheres system about the horizontal axis through the center of the rod is 5ML².
Complete question is below:
There are 2 Spheres of mass M and Bug's mass 3M. The Rod length is 2L, radius L.
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A car starts from rest and accelerates at 2.5m/s^2 for 2s. What is the final velocity of the car
Answer:
5m/s ez
Explanation:
plz make me brainliest
Without friction, what is the mass of an ball accelerating at 1.8 m/sec2 to which an
an unbalanced force of 42 Newtons has been applied?
A 75.60 kg
B 00.04 kg
C 23.33 kg
D 43.80 kg
Answer:
23.33 kgExplanation:
The mass of the object can be found by using the formula
[tex]m = \frac{f}{a} \\ [/tex]
f is the force
a is the acceleration
From the question we have
[tex]m = \frac{42}{1.8} = 23.3333... \\ [/tex]
We have the final answer as
23.33 kgHope this helps you
Answer:
[tex]\boxed {\boxed {\sf C. \ 23.33 \ kg}}[/tex]
Explanation:
According to Newton's Second Law of Motion, force is the product of mass and acceleration.
[tex]F=ma[/tex]
The mass of the ball is unknown. The ball is accelerating at 1.8 meters per second squared. An unbalanced force of 42 Newtons is applied to the ball.
Convert the units of force. 1 Newton is equal to 1 kilogram meter per second squared, so our answer of 42 Newtons is equal to 42 kg*m/s².
F= 42 kg*m/s² a= 1.8 m/s²Substitute the values into the formula.
[tex]42 \ kg*m/s^1 = m * 1.8 \ m/s^2[/tex]
We are solving for the mass, so we must isolate the variable m. It is being multiplied by 1.8 meters per second squared. The inverse operation of multiplication is division. Divide both sides by 1.8 m/s².
[tex]\frac {42 \ kg*m/s^2}{1.8 \ m/s^2} = \frac{a*1.8 \ m/s^2}{1.8 \ m/s^2}[/tex]
[tex]\frac {42 \ kg*m/s^2}{1.8 \ m/s^2} =m[/tex]
The units of meters per second squared cancel.
[tex]\frac {42 \ kg}{1.8 } =m[/tex]
[tex]23.3333333 \ kg=m[/tex]
Round to the hundredth place. The 3 in the thousandth place tells us to leave the 3 in the hundredth place.
[tex]23.33 \ kg \approx m[/tex]
The mass of the ball is approximately 23.33 kilograms.
How fast is an object going if it travels from San Diego to Anaheim in 1.25 hours (hr)? The distance from San Diego to Anaheim is 93 miles (mi).
a) 74.4 mi/hr
b) 116.25 mi/hr
c) 0.013 mi/hr
d) 84.25 mi/hr
e) None of the answers
Answer:
74.4 mph
Explanation:
Since we have the distance in miles and the time it took in hours, we can divide the two to get miles per hour.
speed = mi/hr
speed = 93/1.25
speed = 74.4 mph
Can Hotspots can be found at the edges of tectonic plates?
Answer:
No.
Explanation:
Hot spots don't appear at the barriers of tectonic plates, instead, they originate at hot centres known as mantle plumes. Mantle plumes exist below the tectonic plates and may develop a string of volcanoes on the Earths surface.
easy one - giving brainly if correct.
Gas.
HOPE YOU GET 100!
Guys plz i need those answers as soon as possible
9) a) formula = h(ρ)g by default g=9.8m/s²
Pressure = 2×1000×9.8 = 19600Pa
b)
What is the mass, in kilograms, of a large dog that weighs 441 newtons?
Answer:
441 Newtons equals 44.97 kilograms.
I agree to hire you for 30 days. You can decide between two methods of payment: either (1) $1000 a day, or (2) one penny on the first day, two pennies on the second day and continue to double your daily pay each day up to day 30. Use quick estimation to make your decision, and justify it
Answer:
The second deal.
Explanation:
I had that before, if u pick the second one you will get twice as much in a year as the first one.
What electromagnetic wave is used in nuclear research and mineral exploration?.
The electromagnetic wave used in nuclear research and mineral exploration are gamma rays.
Electromagnetic waves refer to those rays that are found in the electromagnetic spectrum. They are collectively called light. Electromagnetic waves are used for different purposes.
Gamma rays are useful in nuclear research especially in identifying decaying radionuclides. Also, gamma rays are used for geologic mapping, mineral exploration, and identification of environmental contamination.
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5 a
What causes the pressure of air at the Earth's surface?
Answer:
Atmospheric pressure is caused by the gravitational attraction of the planet on the atmospheric gases above the surface and is a function of the mass of the planet, the radius of the surface, and the amount and composition of the gases and their vertical distribution in the atmosphere
Explanation:
7. A bus covers a certain distance in 60 minutes if it runs at a speed of 60 km/hr.
What must be the speed of the bus in order to reduce the time of journey by 40
minutes?
Answer:
90 Km/h
Explanation:
60 mins is 1 hr
so in an hour bus covers 60 Km
so new speed:
d/t
(60km/40mins)*60mins/h
90Km/h
if you have 3 moles of iron, how many grams of iron do you have?
Answer:
55.84 look at periodic table
so 55.84*3= 167.52gFe
Se aplican dos fuerzas concurrentes a un objeto de 4N a la derecha y 5N a la izquierda. ¿Hacia donde se movió y con cuanta fuerza?
The forces move strongly towards the left by 1N
Given the following
Force towards the right = 4N
Force towards the left = 5N
Note that the force acting towards the left is negative, hence the force acting towards the left is -5N
Take the sum of force
Resultant force = -5N + 4N
Resultant force = -1N
This shows that the forces move strongly towards the left by 1N
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A car traveling at 35.6m/s crashes into a concert barrier and comes to a stop in 0.35 seconds. Calculate the average force applied to the 75kg driver.A 3.2kg steel ball traveling at 4.1m/s strikes a second ball of a mass 2.3kg Initially at rest. Calculate the velocity of the second ball when the first one continues traveling in the same direction with a speed of 1.5m/s2 balls of putty are shot towards one another. Ball 1 has a mass of 4.3kg and is moving at 18.6m/s . Ball 2 has a mass of 5.8kg and is moving at 9.5m/s. They collide and stick together. Calculate their final combine velocity.I really appreciate those attempting the problems. I do know the answers but I’m unaware of the steps to get there. Please include all formulas in your response and steps so I can learn and understand.Check your answer:7629N3.6m/s2.46m/sThank you all!
The force on the driver is 7629 N. The velocity of the second ball is 3.6 m/s. The combined velocity of the balls is 13.37 m/s.
We have to find the acceleration using;
v = u - at
v = final velocity = 0 m/s
u = initial velocity = 35.6m/s
a = acceleration = ?
t = time = 0.35 s
u = at
a = u/t = 35.6m/s / 0.35 s
a = 101.7 ms-2
The force on the driver = 75kg × 101.7 ms-2 = 7629 N
Using the principle of conservation of momentum;
Momentum before collision = momentum after collision
m1u1 +m2u2 = m1v1 + m2v2
Hence
(3.2 × 4.1) + 0 = (3.2 × 1.5) + 2.3v2
13.12 = 4.8 + 2.3v2
13.12 - 4.8 = 2.3v2
v2 = 13.12 - 4.8/2.3
v2 = 3.6 m/s
Using the principle of conservation of linear momentum;
m1u1 + m2u2 = m1v1 + m2v2
(4.3 × 18.6) + (5.8 × 9.5) = (4.3 + 5.8) v
v = 79.98 + 55.1/10.1
v = 13.37 m/s
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A particular roller coaster has a mass of 3500 kg, a height of 4.0, and a velocity of 12m/s. What is the potential energy? If needed, use g=10.m/s^2
[tex]▪▪▪▪▪▪▪▪▪▪▪▪▪ {\huge\mathfrak{Answer}}▪▪▪▪▪▪▪▪▪▪▪▪▪▪[/tex]
Gravitational Potential Energy of an object is calculated by formula ~
[tex] \large\boxed{\sf P = mgh}[/tex]
where,
m = mass of the object = 3500 kgg = Acceleration due to gravity = 12 m/s²h = height attained by the object = 4 mNow, let's calculate its potential energy ~
[tex]3500 \times 10 \times 4[/tex][tex]140000 \: \: joules[/tex][tex]140 \: \: kj[/tex]Answer:
Potential Energy of an object is calculated by formula:
Potential Energy (P.E)=m×g×hWhere,
m=mass of bodyg=acceleration due to gravityh=height from the earth surfaceNow, let's solve the question.
Given,
mass(m)=3500 kgheight (h)=4mvelocity (v)=10m/s²Now,
We know that,
Potential Energy (P.E)=mgh
[tex] = 3500 \times 10 \times 4[/tex]
[tex] = 3500 0\times 4[/tex]
[tex] =140000 joules [/tex]
[tex]\mathfrak{\blue{DisneyPrincess29}}[/tex]
how have astronomers interpreted the unexpectedly fast rotation of galaxies
Answer:
There must be a lot of dark matter that can be felt but not seen
The astronomers interpreted the unexpectedly fast rotation of galaxies that there must be a large quantity of dark energy whose gravity is detectable yet invisible.
What is a galaxy?Any system of stars plus interstellar material that makes up the cosmos is referred to as a galaxy. Such assemblages are common, and many of them are so massive that they hold tens of trillions of stars.
A vast variety of galaxies, from dim, hazy dwarf objects to spectacular spiral-shaped giants, have been created by nature. Almost all galaxies seem to have formed shortly after the universe started, and they are everywhere in space, even at the farthest limits of the universe that can be seen by the most advanced telescopes.
The majority of galaxies are found in clusters, many of which are further organized into clusters that span hundreds of billions of light-years.
Since there are almost empty spaces between these so-called super clusters, the universe's overall structure resembles a network of sheets or chains of galaxies.
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