Answer:
b. melting
Explanation:
it is made of sediments and that is not necessary
For the following types of electromagnetic radiation, how do the wavelength, frequency, and photon energy change as one goes from the top of the list to the bottom?
a. radio waves
b. infared radiation
c. visible light
d. ultraviolet radiation
e. gamma radiation
Answer:
Wavelength, frequency and the photon energy changes as the one goes across the ranges of the electro-magnetic radiations.
Explanation:
Electro-magnetic radiations may be defined as the form of energy that is radiated or given by the electro-magnetic radiations. The visible light that we can see is the one of the electro-magnetic radiations. Other forms are the radio waves, gamma waves, UV rays, infrared radiations, etc.
The wavelength of the radiations decreases as we go from a. radio waves -- b. infrared radiation -- c. visible light -- d. ultraviolet radiation -- e. gamma radiation.
The frequency of the radiations increases when we move from a. radio waves -- b. infrared radiation -- c. visible light -- d. ultraviolet radiation -- e. gamma radiation.
The photon energy of the radiations increases when we move from a. radio waves -- b. infrared radiation -- c. visible light -- d. ultraviolet radiation -- e. gamma radiation.
Tell me the max amount you should owe on this card.
Your credit limit is $1,000.
What is the max you should ever owe on this card?
$
Your credit limit is $2,500.
What is the max you should ever owe on this card?
$
Answer:
the max is 2,500 or less
Explanation:
because you cant owe anymore
Calculate the escape velocity
the moon's surface given that a man on the moon has 1/6 his weight on earth
Answer:
v = 2.38 × 10³ m/s
Explanation:
Escape velocity, v = √(2gR) where g = acceleration due to gravity on planet and R = radius of planet.
Since it is given that the weight of the man on the moon is 1/6 his weight on earth, and g' = acceleration due to gravity on moon and g = acceleration due to gravity on earth and m = mass of man,
mg' = mg/6
g' = g/6
Since g = 9.8 m/s²,
g'= 9.8 m/s² ÷ 6
g' = 1.63 m/s²
The escape velocity of the moon is thus v = √(2g'R) where R = radius of moon = 1.737 × 10⁶ m.
Substituting these into v, we have
v = √(2g'R)
v = √(2 × 1.63 m/s² × 1.737 × 10⁶ m)
v = √[5.663 × 10⁶ (m/s)²]
v = 2.38 × 10³ m/s
A 64.1 kg runner has a speed of 3.10 m/s at one instant during a long-distance event.(a) What is the runner's kinetic energy at this instant?
KEi = _________________J
(b) If he doubles his speed to reach the finish line, by what factor does his kinetic energy change?
KEf/KEi=______________
Answer:
(a) the runner's kinetic energy at the given instant is 308 J
(b) the kinetic energy increased by a factor of 4.
Explanation:
Given;
mass of the runner, m = 64.1 kg
speed of the runner, u = 3.10 m/s
(a) the kinetic energy of the runner at this instant is calculated as;
[tex]K.E_i = \frac{1}{2} mu^2\\\\K.E_i = \frac{1}{2} \times 64.1 \times 3.1^2\\\\K.E_i = 308 \ J[/tex]
(b) when the runner doubles his speed, his final kinetic energy is calculated as;
[tex]K.E_f = \frac{1}{2} mu_f^2\\\\K.E_f = \frac{1}{2} m(2u)^2\\\\K.E_f = \frac{1}{2} \times 64.1 \ \times (2\times 3.1)^2\\\\K.E_f = 1232 \ J[/tex]
the change in the kinetic energy is calculated as;
[tex]\frac{K.E_f}{K.E_i} = \frac{1232}{308} =4[/tex]
Thus, the kinetic energy increased by a factor of 4.
An astronaut weighing 190 lbs on Earth is on a mission to the Moon and Mars.
Required:
a. What would he weigh in newtons when he is on the Moon?
b. How much would he weigh in newtons when he is on Mars, where the acceleration due to gravity is 0.38 times that on Earth?
Answer:
The weight is defined as:
W = m*g
where:
m = mass
g = gravitational acceleration.
We know that in Earth the astronaut weights 190 lb-f (this is force, not mass, the correct unit here is 190 lb*m/s^2)
then:
190 lb*m/s^2 = m*9.8m/s^2
(190 lb*m/s^2)/(9.8m/s^2) = 19.39 lb
Now we know the mass of the astronaut.
a) wieght on the moon in Newtons.
Newtons uses kilograms as the units of mass, then we need to rewrite the mass of the astronaut in kg.
we know that 1lb = 0.454 kg
Then 19.39 lb is equal to: 19.39*0.454 kg = 8.8 kg
We know that the acceleration due to gravity on the Moon is one-sixth that on Earth.
then: g = (9.8m/s^2)/6
And the weight of the astronaut in the moon will be:
W = 8.8 kg*(9.8m/s^2)/6 = 14.37 N
b) The weight on mars, where the acceleration due to gravity is 0.38 times that on Earth, we have:
g = (9.8m/s^2)*0.38
then the weight will be:
W = 8.8kg*(9.8m/s^2)*0.38 = 32.77 N
Sugar crystals enter a dryer at the rate of 1000 kg h-1 and at 20% w.b. moisture content. They leave the dryer at 3% w.b. moisture content. If the drying process requires 3000 kJ kg-1 of water removed, estimate the amount of heat required per hour and the rate of dry crystals out of the dryer.
Answer:
The dryer evaporate 200 - 24.74 kg of water per hour
To remove 1kg of water it need 3000 K J
So to remove, 175.26 Kg
it need 5.257x [tex]10^{5}[/tex] KJ of heat per hour.
Explanation:
In one hour, the amount of sugar entering = 1000 kg
w.b moisture content is defined as,
weight of water / weight of water + weight of dry
[tex]W_{w}[/tex]/[tex]W_{w}[/tex] + [tex]W_{d}[/tex] x 100
[tex]W_{w}[/tex] + [tex]W_{d}[/tex] = 1000 kg when entering
it has 20% moisture content when entering
[tex]W_{w}[/tex] = 0.2 x 1000 = 200 kg
when leaving it has 3% moisture content then weight of dry material
[tex]W_{d}[/tex] = 1000 - 200 = 800 Kg
[tex]\frac{W_{w}^{'} }{W_{w}^{'} + W_{d}^{'} }[/tex] = 0.03
[tex]\frac{W_{w}^{'} }{W_{w}^{'} + 800 }[/tex] = 0.03
[tex]W_{w} ^{'}[/tex] = 0.03 x [tex]W_{w} ^{'}[/tex] + 0.03 x 800
[tex]W_{w} ^{'}[/tex] = 24.74 kg
When leaving the dryer, the crystals has total weight of the water = 24.74 kg per hour.
The dryer evaporate 200 - 24.74 kg of water per hour
To remove 1kg of water it need 3000 K J
So to remove, 175.26 Kg
it need 5.257x [tex]10^{5}[/tex] KJ of heat per hour.
An object has a mass of 19 kg. Attached to the end of a spring, if the spring constant is 486 N/m, what is the maximum frequency?
Answer:
The frequency [tex]f = 0.8048 \ Hz[/tex]
Explanation:
From the question we are told that
The mass of the object is [tex]m = 19 \ kg[/tex]
The spring constant is [tex]k = 486 \ N/ m[/tex]
Generally the maximum frequency is mathematically represented as
[tex]f = \frac{1}{2 \pi } * \sqrt{ \frac{k}{m } }[/tex]
=> [tex]f = \frac{1}{2 * 3.142 } * \sqrt{ \frac{486}{ 19 } }[/tex]
=> [tex]f = 0.8048 \ Hz[/tex]
What us the difference in the ways objects move at a speed of a car and an object mkvinf close to the speed of light?
Answer:
The difference is in who or what is observing the speed.
Explanation:
Giving that speed is relative between the objects and the reference point from which it is being observed.
It is concluded that speed alone has no direct effect on a moving object, hence it is just a determining unit for the difference in distance between two objects.
Therefore, in this case, the difference is in who or what is observing the speed.
What was your train of thought as you navigated the picture of the candle?
Answer:
Where is the picture
Explanation:
WHERE IS THE PICTURE
A slithering snake travels once around a circle of radius 3.20m. The coefficient of friction between the ground and the snake is 0.25, and the snake's weight is 80.0N. How much work does the snake do against friction?
Answer:
The magnitude of the work done by the snake against friction is 402 J.
Explanation:
The work is given by:
[tex] W = F*d [/tex]
Where:
F: is the force
d: is the distance
The distance traveled by the snake is given by the perimeter of the circle:
[tex] P = 2\pi r = 2\pi 3.20 m = 20.1 m [/tex]
Now, the force in the direction of the movement is:
[tex] |\Sigma F| = F_{\mu} = \mu N = \mu W = 0.25*80.0 N = 20 N [/tex]
Finally, the work is:
[tex] W = F*d = 20 N*20.1 m = 402 J [/tex]
Therefore, the magnitude of the work done by the snake against friction is 402 J.
I hope it helps you!
1. Add 17.35 g, 25.6 g and 8.498 g. chaper 1 physical quantity 11class .physic
51.448 g is the required answer!
4. You’re driving a car that can climb a maximum gradient of 500m/km. The hill in front of you starts at an elevation of 20m and reaches 100m. The total distance up the hill is 1.5km? What is the gradient of the hill and will your car make it?
Answer:
Gradient of the hill is 53.33m/km which is lesser than the 500 m/km. So, he will be able to climb the hill conveniently.
Explanation: Given that
Maximum gradient = 500 m/km
Total distance = 1.5 km
Starting elevation = 20 m
Final elevation = 100 m
Gradient = change in elevation/ total distance.
Now, substitute the values into the formula.
Gradient = (100m - 20m)/1.5km
= 80m/1.5km
= 53.33m/km
Gradient of the hill is 53.33m/km which is lesser than the 500 m/km. So, he will be able to climb the hill conveniently.
what is general relativity
Explanation:
General relativity is a theory of space and time. ... The central idea of general relativity is that space and time are two aspects of spacetime. Spacetime is curved when there is matter, energy, and momentum resulting in what we perceive as gravity. The links between these forces are shown in the Einstein field equations.
What acceleration will you give a 22.4 kg box if you push it with a force of 83.1N
Answer:
mass =22.4kg
force=83.1N
a=?
f=ma
a=f/m
a=83.1/22.4
a=3.70m/s^2
The picture shows a basic diagram of an electric motor. At top left a piece of magnet labeled N and at top right a piece labeled S. Between these a square coil of wire X sits attached to a metal rod, which runs between the 2 pieces of magnet. 2 semicircular pieces Z of metal surround the rod at its opposite end from the pieces of magnet. Wires connect to each semicircular piece at terminals Y to one pole of a battery. Blue arrows superimposed on the end of the coil away from the magnets point toward S and away from N. Which labels best complete the diagram? X: Brush Y: Armature Z: Commutator X: Commutator Y: Brush Z: Armature X: Armature Y: Commutator Z: Brush X: Armature Y: Brush Z: Commutator Mark this and return
Answer:
I know this isn't much help but its not B
Explanation:
Answer:
D
Explanation:
1) Andrea and Chuck are riding on a merry-go-round. Andrea rides on a horse at the outer rim of the circular platform, twice as far from the center of the circular platform as Chuck, who rides on an inner horse. When the merry-go-round is rotating at a constant angular speed, Andrea's tangential speed is which of the following?a) twice Chuck's.b) the same as Chuck's.c) half of Chuck's.d) impossible to determine.2) When the merry go round is rotating at a constant angular speed, Andrea's tangential speed is:______.a) twice Chuck's.b) the same as Chuck's.c) half of Chuck's.d) impossible to determine.
Explanation:
The tangential speed of Andrea is given by :
[tex]v=r\omega[/tex]
Where
r is radius of the circular path
ω is angular speed
The merry-go-round is rotating at a constant angular speed. Let the new distance from the center of the circular platform is r'
r' = 2r
New angular speed,
[tex]v'=r'\omega'\\\\v'=(2r)\omega\\\\v'=2r\omega\\\\v'=2v[/tex]
New angular speed is twice that of the Chuck's speed.
A woman exerts a horizontal force of 113 N on a crate with a mass of 31.2 kg.
Required:
a. If the crate doesn't move, what's the magnitude of the static friction force (in N)?
b. What is the minimum possible value of the coefficient of static friction between the crate and the floor?
Answer:
a) 113N
b) 0.37
Explanation:
a) Using the Newton's second law:
\sum Fx =ma
Since the crate is not moving then its acceleration will be zero. The equation will become:
\sum Fx = 0
\sumFx = 0
Fm - Ff = 0.
Fm is the moving force
Ff is the frictional force
Fm = Ff
This means that the moving force is equal to the force of friction if the crate is static.
Since applied force is 113N, hence the magnitude of the static friction force will also be 113N
b) Using the formula
Ff = nR
n is the coefficient of friction
R is the reaction = mg
m is the mass of the crate = 31.2kg
g is the acceleration due to gravity = 9.8m/s²
R = 31.2 × 9.8
R = 305.76N
Recall that;
n = Ff/R
n = 113/305.76
n = 0.37
Hence the minimum possible value of the coefficient of static friction between the crate and the floor is 0.37
PLEASE ANSWER ASAP BEFORE MY TEACHER AND MY MOM KILLES ME PLEASE ASAP
The first person with the right answer gets to be a brainlest
In the attachment there is a density column where there is colour
Question: tell me why is the red at the bottom of the density column if it is the least dense
Answer:
That is not meant to be red, it is the bottom of the beaker
That is not meant to be red, it‘s the bottom of the beaker. The star is at the very bottom of the beaker. it’s just the base of the beaker.
A solid is 5 cm tall, 3 cm wide, and 2 cm thick. It has a mass of 129 g. What is its
density?
Answer:
4.3 g/cm³ or 4.3g/cc
Explanation:
Volume(V) = Height × Length × Width
= 5cm × 3cm × 2cm
= 30cm³
Mass(m) = 129gram
So,
Density = m/V
= 129g/30cm³
= 4.3g/cc or 4.3g/cm³
Plants that respond to light are responding
to an:
a. internal stimulus
b. external stimulus
please answer
Answer:
The answer would be a
Explanation:
a fisherman moves from one end of a boat to another
Answer:
That isnt a question so no one will know the answer to what you are talking about. I suggest adding a sceenshot or picture of the question.
At a distance of 10 km from a radio transmitter, the amplitude of the E-field is 0.20 volts/meter. What is the total power emitted by the radio transmitter
Answer:
The total power is [tex]P = 6.665 *10^{4} \ W[/tex]
Explanation:
From the question we are told that
The distance is [tex]r = 10 \ km = 1000 \ m[/tex]
The amplitude of the electric field is [tex]E = 0.20 \ volt/meter[/tex]
Generally the average intensity of the electromagnetic field from the radio transmitter is mathematically represented as
[tex]I = \frac{E^2}{ 2 \mu_o * c }[/tex]
Here c is the speed of light with value [tex]c = 3.0 *10^{8} \ m/s[/tex]
[tex]\mu_o[/tex] is the permeability of free space with value [tex]\mu_o = 4\pi *10^{-7} \ N/A^2[/tex]
So
[tex]I = \frac{0.2^2}{ 2 * 4\pi *10^{-7} * 3.0*10^{8} }[/tex]
=> [tex]I = 5.307 *10^{-5} \ W/m^2[/tex]
Generally this intensity can also be mathematically represented as
[tex]I = \frac{P }{ 4 \pi r^2 }[/tex]
=> [tex]P = I ( 4 \pi r^2 )[/tex]
=> [tex]P = 5.307 *10^{-5} ( 4 * 3.142 * 1000^2 )[/tex]
=> [tex]P = 6.665 *10^{4} \ W[/tex]
The total power emitted by the radio transmitter is [tex]6.67\times 10^4 \ W[/tex].
The given parameters;
amplitude of the electric field, E = 0.2 V/mdistance of the transmitter, = 10 km = 10,000 mThe intensity of the radio wave from the transmitter is calculated as follows;
[tex]I = \frac{E^2}{2\mu_0 c} \\\\I = \frac{0.2^2 }{2\times 4\pi \times 10^{-7} \times 3\times 10^8} \\\\I = 5.305 \times 10^{-5} \ W/m^2[/tex]
The total power emitted by the radio transmitter is calculated as follows;
[tex]I = \frac{P}{A} \\\\P = IA\\\\P = I \times 4\pi r^2\\\\P = (5.305\times 10^{-5} )\times 4\pi \times (10,000)^2\\\\P = 6.67\times 10^{4} \ W[/tex]
Thus, the total power emitted by the radio transmitter is [tex]6.67\times 10^4 \ W[/tex].
Learn more here:https://brainly.com/question/19340071
A thin-walled vessel of volume V contains N particles which slowly leak out of a small hole of area A. No particles enter the volume through the hole. Find the time required for the number of particles to decrease to N/2. Express your answer in terms of A, V, and v.
Answer:
[tex]\frac{V}{2av}[/tex]
Explanation:
From the question we are told that
Volume V
Contains N particles
Leaks from a small hole of area A
Generally the equation for Flow rate is given as
Volume Flow Rate [tex]V_r = A * v[/tex]
Mathematically we find the time taken to flow half way which is given by
[tex]\frac{(V/2)}{A*v}[/tex]
Therefore the time taken is
[tex]\frac{V}{2av}[/tex]
An ideal gas is confined to a container with constant volume. The number of moles is constant. By what factor will the pressure change if the absolute temperature triples
Answer:
1/3
Explanation:
Gay Lusaac's law states that "the pressure of a given mass of gas is directly proportional with the absolute temperature of the gas, provided that the volume is kept constant."
In formula, we say that
P/T = k
Where
P = pressure at different points
T = temperature at different points
k = constant of proportionality
From the stated formula, if we multiply the temperature by 3, we have
P/3T = k
P * 1/3T = k
And from this, we see the pressure will change by a value of 1/3
If the net work done on a particle is zero, which of the following statements must be true?
A. The speed is unchanged.
B. The velocity is zero.
C. The velocity is unchanged.
D. More information is needed.
E. The velocity is decreased.
Answer:
A. The speed is unchanged.
Explanation:
In the case when the work is to be done on a particle i.e. zero so the change made in KE of the particle would be zero. This represent the work energy theroem. But when the KE remains same or does not change so it should be the same and the particle speed would also the same
Therefore as per the given statement, the first option is correct
And rest of the options are wrong
What is a
physical
property of
snowflakes?
derive an expression for torque experiend by an electric dipole placed in a uniform electric field
Answer:
The torque τ on an electric dipole with dipole moment p in a uniform electric field E is given by τ = p × E where the "X" refers to the vector cross product. Ref: Wikipedia article on electric dipole moment.
Explanation:
Question 18 of 20
Which situation results in the least attraction between two magnets?
A. The south pole of one magnet is near the south pole of the other
magnet
B. The north pole of one magnet is near the south pole of the other
magnet
C. The north pole of one magnet is far away from the north pole of
the other magnet
D. The north pole of one magnet is far away from the south pole of
the other magnet.
Answer:D
Explanation:
Answer:
Explanation: the answer is D.
If a net horizontal force of 0.8 N is applied to a toy whose mass is 1.2 kg, acceleration is?
Hello!
[tex]\large\boxed{a = \frac{2}{3}m/s^{2}}[/tex]
Use the equation F = m · a to solve. We are given the force (N) and mass (kg), so we can solve for the acceleration by plugging in the given values:
0.8 = 1.2a
0.8 / 1.2 = a
a = 2/3 m/s²
My dad gifted me a calculator. I have observed that very small cells are used in a calculator. What are these cells called and what are their main advantage?