Pressure and temperature ______ with depth below Earth’s surface.
Answer:
Pressure increases as you move deeper below earth's surface.
Tempurature increases as you move deeper below earth's surface.
Hope this helps!
Explanation:
The plate is a model for how sunlight hits Earth’s surface. Which parts of Earth are most similar to the plate with an axis angle of 0°?
Answer: The parts closer to the equator.
Explanation:
The parts of the earth closer to the equator are similar to the plate, because they receive more sunlights than other parts. Which makes them hotter than any other regions in the earth. Example of such countries are Gabon, Uganda, Kenya, Maldives, the democratic republic of Congo, sao tome and principle e.t.c
Which would increase the speed of a sound wave?
O A wave passes from a solid to a liquid while remaining the same temperature.
The medium increases in temperature while remaining in the same phase.
The medium decreases in temperature while remaining in the same phase.
O A wave passes from a liquid to a gas while remaining the same temperature.
Answer:
The medium increases in temperature while remaining in the same phase
Explanation:
The speed of a sound wave is increased when the medium increases in temperature while remaining in the same phase.
What is meant by the temperature coefficient of sound wave?The temperature coefficient of a sound wave is defined as the measure of increase in the velocity of the sound wave per unit rise in its the temperature.
Here,
The speed of a sound wave is affected by various factors in the medium through which it propagates. Among them, speed of the wave is mostly influenced by the temperature of the medium.
The speed of a sound wave in a medium is directly proportional to the square root of its absolute temperature. So,
v [tex]\alpha[/tex] √T
where v is thee speed of the sound wave and T is the absolute temperature.
Therefore,
As the temperature of the medium increases, the kinetic energy of the wave particles increases. Thus the speed of the sound wave is increased. As a result, the sound waves will move faster.
Hence,
The speed of a sound wave is increased when the medium increases in temperature while remaining in the same phase.
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The power that a student generates when walking at a steady pace of vw is the same as when the student is riding a bike at vb = 3vw. The student is going to travel a distance d. The energy the student uses when walking is Ew. The energy the student uses when biking is Eb. The ratio EwEb is
Answer:
3
Explanation:
a man weighing 490 n on earth weighs only 81.7 n on the moon. His mass on the moon is__kg. (Use g=9.8 m/s2
Answer:
m = 50 [kg]
Explanation:
In order to solve this problem we must be clear about the difference between weight and mass. Weight is the product of mass by the acceleration of the planet or the star. While the mass is always preserved it never changes regardless of where it is located.
So for the earth we have:
g = gravity acceleration = 9.8 [m/s^2]
m = mass [kg]
W = weigth = 490 [N]
therefore the mass will be:
m = W/g
m = 490/9.8
m = 50 [kg]
Now it is important to remember that the mass will be the same on the moon or on the earth, but the weight will be different, because the gravity acceleration of the moon is different from the gravity acceleration on earth
So the gravity on the moon is equal to:
81.7 = 50 * gm
gm = 1.634 [m/s^2]
Two boxes of masses 3M and 5M are attached by a massless rope. They are being pulled to the right with a constant force of P = 800 N, which allows them to just overcome static friction, with a μs= 0.70 between the floor and the boxes.
a. Find M.
b. Find the Tension in the rope between the two boxes.
Answer:
a) about 14.577 kg
b) 300 N
Explanation:
b) In order for the acceleration to be the same for each mass, the 800 N force must be divided between the boxes in proportion to their mass. That is, the net force on the 5M mass must be 5/8 of the total force, or 500 N. Then the tension in the rope is 800 N -500 N = 300 N, which is 3/8 of 800 N.
Tension: 300 N
__
a) The total mass is 8M, and the total normal force on the floor is ...
F = ma = (8M)(9.8 m/s^2)
The friction force is 0.7 times this, and is equal to the 800 N force pulling on the boxes.
800 N = (8M)(9.8 m/s^2)(0.7)
M = 800/(8·9.8·0.7) kg ≈ 14.577 kg
What is the answer to this ?
A block of mass m1 = 18.5 kg slides along a horizontal surface (with friction, μk = 0.22) a distance d = 2.3 m before striking a second block of mass m2 = 7.25 kg. The first block has an initial velocity of v = 8.25 m/s.
Assuming that block one stops after it collides with block two, what is block two's velocity after impact in m/s?
How far does block two travel, d2 in meters, before coming to rest after the collision?
Answer:
19.5 m/s
87.8 m
Explanation:
The acceleration of block one is:
∑F = ma
-m₁gμ = m₁a
a = -gμ
a = -(9.8 m/s²) (0.22)
a = -2.16 m/s²
The velocity of block one just before the collision is:
v² = v₀² + 2aΔx
v² = (8.25 m/s)² + 2 (-2.16 m/s²) (2.3 m)
v = 7.63 m/s
Momentum is conserved, so the velocity of block two just after the collision is:
m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂
m₁u₁ = m₂v₂
(18.5 kg) (7.63 m/s) = (7.25 kg) v
v = 19.5 m/s
The acceleration of block two is also -2.16 m/s², so the distance is:
v² = v₀² + 2aΔx
(0 m/s)² = (19.5 m/s)² + 2 (-2.16 m/s²) Δx
Δx = 87.8 m
The velocity of block 2 and the distance traveled by it prior to being at rest post-collision are 19.5 m/s and 87.8 m. Check the calculations below:
FrictionGiven that,
[tex]m_{1}[/tex] = 18.5 kg
d = 2.3m
To find,
Acceleration of block 1:
∑[tex]F = ma[/tex]
⇒ -m₁gμ = m₁a
⇒ a = -gμ
⇒ a [tex]= -(9.8 m/s^2) (0.22)[/tex]
∵ a [tex]= -2.16 m/s^2[/tex]
Now,
To determine the velocity of block one prior to the collision:
We know,
The initial velocity of block 1 = 8.25 m/s
⇒ [tex]v^2 = v_{o}^2 + 2[/tex]aΔx
⇒ [tex]v^2 = (8.25 m/s)^2 + 2 (-2.16 m/s^2) (2.3 m)[/tex]
∵ [tex]v = 7.63 m/s[/tex]
We also know,
[tex]m_{2}[/tex] = 7.25 kg
Now,
The velocity of block 2 post collision:
⇒ [tex]m_{1} u_{1} + m_{1} u_{1} = m_{1} v_{1} + m_{2} v_{2}[/tex]post-collision
Through this,
⇒ [tex](18.5 kg) (7.63 m/s) = (7.25 kg) v[/tex]
∵[tex]v = 19.5 m/s[/tex]
The distance can be found through:
⇒ [tex]v^2 = v_{o} ^{2} + 2[/tex][tex]a[/tex]Δ[tex]x[/tex]
⇒ [tex](0 m/s)^2 = (19.5 m/s)^2 + 2 (-2.16 m/s^2)[/tex]Δ[tex]x[/tex]
∵ Δ[tex]x = 87.8 m[/tex]
Thus, 19.5 m/s and 87.8 m are the correct answers.
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The first ionization potential for calcium (Z = 20, A = 40) is 6.11 eV. Singly-ionized calcium (Ca+) produces two very strong absorption lines in the Sun’s spectrum discovered by Joseph Fraunhofer in 1814, who named them "H" and "K" (he didn’t know they were from calcium, as this was >100 years before the development of quantum mechanics). Both lines always appear together, with lambda subscript H equals 3968 end subscript Å and lambda subscript K equals 3933 Å; hence they are called a "doublet
A. What is the speed of an electron that has just barely enough kinetic energy to collisionally ionize a neutral calcium atom? What is the speed of a calcium ion with this same kinetic energy?
B. What is the temperature T of a gas in which the average particle energy is just barely sufficient to ionize a neutral calcium atom?
C. The lower energy level of both lines is the ground state of Cat. What is the difference in energy in eV) between the two states that correspond to the upper energy levels of the Hand Klines, respectively? How does this compare to the energy of a calcium K photon? Can these two lines can be formed by transitions to upper energy levels with different principal quantum numbers (different n), or do they represent transitions with the same n but some different higher-order quantum number? Explain your reasoning based on your understanding of the general behavior of atomic energy levels (En).
Answer:
A) v = 1.47 10⁶ m / s, v = 0.5426 10⁴ m / s , B) T = 4.7 10⁴ K, C) n₂ = 42
Explanation:
A) For this part, let's calculate the speed of an electron that has an energy of 6.11 eV.
Let's reduce the units to the SI system
E₀ = 6.11eV (1.6 10⁻¹⁹ J / 1eV) = 9.776 10⁻¹⁹ J
The kinetic energy of the electron is
K = ½ m v²
E₀ = K
v = √ 2E₀ / m
v = √ (2 9.776 10⁻¹⁹ / 9.1 10⁻³¹)
v = √ (2.14857 10¹²)
v = 1.47 10⁶ m / s
now the speed of a calcium ion is asked, let's find sum
m = 40 1.66 10⁻²⁷ = 66.4 10⁻²⁷ kg
v = √ (2E₀ / M)
v = √ (2 9.776 10⁻¹⁹ / 66.4 10⁻²⁷)
v = √ (0.2994457 10⁸)
v = 0.5426 10⁴ m / s
B) the terminal energy of an ideal gas is
E = 3/2 kT
T = ⅔ E / k
T = ⅔ (9,776 10-19 / 1,381 10-23)
T = 4.7 10⁴ K
C) To calculate the energy of these lines we use the Planck expression
E = h f
where wavelength and frequency are related
c =λ f
f = c /λ
let's substitute
E = h c /λ
let's look for the energies
λ = 396.8 nm
E₁ = 6.63 10⁻³⁴ 3 10⁸ / 396.8 10⁻⁹
E₁ = 5.0126 10⁻¹⁹ J
λ = 393.3 nm
E₂ = 6.63 10⁻³⁴ 3 10⁸ / 3.93.3 10⁻⁹
E₂ = 5.0572 10⁻¹⁹ J
The difference in energy between these two states is
ΔE = E₂ -E₁
ΔE = (5.0572 - 5.0126) 10⁻¹⁹ J
ΔE = 0.0446 10⁻¹⁹ J
let's reduce eV
ΔE = 0.0446 10⁻¹⁹ J (1 eV / 1.6 10⁻¹⁹ J)
ΔE = 2.787 10⁻² eV
Now let's use Bohr's atomic model for atoms with one electron,
E = -13.606 Z² / n²
where 13,606 eV is the energy of the base state of the Hydrogen atom, Z is the atomic number of Calcium
n = √ (13.606 Z² / E)
λ = 396.8 nm
E₁ = 5.0126 10⁻¹⁹ J (1 eV / 1.6 10⁻¹⁹J) = 3.132875 eV
n₁ = √ (13.606 20² / 3.132875)
n₁ = 41.7
since n must be an integer we take
n₁ = 42
λ = 393.3 nm
E₂ = 5.0572 10⁻¹⁹ J (1eV / 1.6 10⁻¹⁹ J) = 3.16075 eV
n₂ = √ (13.606 20² / 3.16075)
n₂ = 41.5
Again we take n as an integer
n₂ = 42
We can see that the two lines have the same principal quantum number, so for the difference of these energies there must be other quantum numbers, which are not in the Bohr model, because of the small difference they are possibly due to small numbers of the moment angular orbital or spin
N₂ + H₂
NH3
how do i balance this equation?
Answer:
N2 + 3H2 -----> 2NH3
Explanation:
Reactants side:
2 Nitrogen
5 Hydrogen
Products Side:
2 Nitrogen
5 Hydrogen
A car which is traveling at a velocity of 15 m/s undergoes an acceleration of 6.5 m/s2 over a distance of 340 m. How fast is it going after that acceleration? (68.15 m/s)
v² - u² = 2 a ∆x
where u = initial velocity, v = final velocity, a = acceleration, and ∆x = distance traveled.
So
v² - (15 m/s)² = 2 (6.5 m/s²) (340 m)
v² = 4645 m²/s²
v ≈ 68.15 m/s
Question 4
Which of the following is unique for any given element?
O the mass of a neutron
o the number of neutrons
o the charge on the electons
O the number of protons
Hi, Please help.. I have assignments due tonight and I need someone to help me when a question I have generally..
Okay so if Density = Mass divided by Volume and I put that information into a calculator it comes out as
ex. 1.938773646 how do I make it look like something like this 1.4?
Answer:
did you tried putting it in standard form
Answer:
It may help to round all of the given numbers up to at least 1 or 2 decimal points or you could round up the number you get to 1 or 2 decimal places. For example, for this question round up your answer to 1.9 or 1.94
Explanation:
hope this helps!!
The coefficient of static friction between m1 and the horizontal surface is 0.50, and the coefficient of kinetic friction is 0.30. (a) If the system is released from rest, what will its acceleration be
This question is incomplete
Complete Question
m1 is 10kg, m2 is 4.0kg. The coefficient of static friction between m1 and the horizontal surface is 0.50. and the Coefficient of kinetic friction is 0.30.
a) if the system is released from rest what will be its acceleration
Answer:
0.7 m/s²
Explanation:
The coefficient of static friction between m1 and the horizontal surface is 0.50. and the coefficient of kinetic friction is 0.30.
(a) if the system is released from rest what will be its acceleration
g = acceleration due to gravity = 9.81 m/s²
Coefficient of Kinetic Friction = μk = 0.30
m1 = 10kg
m2 = 4.0kg
The formula to solve question a is given as:
a = acceleration at rest
m2g- μk m1g = (m1+ m2) a
Making a the subject of the formula:
a = (m2g- μk×m1g )/(m1+ m2)
a = [(4.0 kg × 9.81m/s²) – (0.30 ×9.81 × 10) ]/(10+4)
a = 0.7 m/s²
What do mammoths and tigers need energy for
A crane uses a single cable to lower a steel girder into place. The girder moves with constant speed. The cable tension does work WT and gravity does work WG. Which statement is true
Explanation:
Work done by a force is given by :
[tex]W=Fd\cos\theta[/tex]
Where
F is force, d is displacement and [tex]\theta[/tex] is the angle between F and d.
In this problem, a crane is moving in downward direction, the force gravity is in downward direction and the tension is in upward direction.
We know that if force and displacement is in same direction, work is positive while if force and displacement is in oposite direction, work is negative.
I would mean that, [tex]W_g[/tex] is positive, because gravity is parallel to the displacement and [tex]W_t[/tex] is negative, because the tension is opposite to the displacement.
If a rock is skipped into a lake at 24 m/s2, with that what force was the rock thrown if it was 1.75kg?
Answer: f= M×A
1.75kg×24= 42N
Explanation:
Because to find force you do Mass times acceleration so I did 1.75 kg times 24 would equal 42 Newtons!
(A) Electricity and Magnetism
A). Three point charges are aligned along the x axis as shown in
Fig. Find the electric field at (a) the position (2, 0) and (b) the
position (0, 2).
electricity
Explanation:
the position (2,o
At an amusement park, a swimmer uses a water slide to enter the main pool. You may want to review (Pages 234 - 241) . Part A If the swimmer starts at rest, slides without friction, and descends through a vertical height of 2.81 m , what is her speed at the bottom of the slide
Answer:
Her speed at the bottom of the slide is 7.42 m/s
Explanation:
From the question,
The swimmer starts at rest, that is, her initial speed, u is 0 m/s.
Since she slides without friction and descends through a vertical height, then it is a free fall motion (due to gravity).
Also, from the question,
She descends through a vertical height of 2.81 m.
To determine her speed at the bottom of the slide, that is her final speed,
From one of the equations of motion for freely falling bodies
v² = u² + 2gh
Where v is the final speed
u is the initial speed
g is acceleration due to gravity (g = 9.8 m/s²)
and h is height
From the question,
u = 0 m/s
h = 2.81 m
Putting the values into the equation
v² = u² + 2gh
v² = 0² + 2×9.8×2.81
v² = 55.076
v =√55.076
v = 7.42 m/s
Hence, her speed at the bottom of the slide is 7.42 m/s.
The precision value of measuring tape is
1)0.1cm
2)0.1mm
3)1cm
4)0.01cm
C.1cm
Explanation:
precision is how close two or more measurements are to each other
4?
Explanation:
sorry im not sure but
You can always take a metre ruler as a starting point. Your metre ruler has the same precision as your 15.0cm or 30.0cm ruler, so bring a ruler during exams as they'll come in handy ;)
The order goes like this:
rulers: 0.1cm or 1mm
measuring tape: 0.01cm or 0.1mm
vernier calipers: 0.001cm or 0.01mm
micrometer screw gauge: 0.0001cm or 0.001mm
((if i'm not wrong))
Help me Please!!!!!!!
The smokestack of a stationary toy tra in consists
of a vertical spring gun chat shoots a steel ball a meter
or so straight into the air-----so straight that the ball
always falls back into the smokescack. Suppose the
train moves at constant speed along the straight track. Do you think the ball will still return to the smoke-
stack if shot from the moving train? What if the train
gains speed along the straight track? What if it moves at a constant speed on a circular track? Discuss why your answers differ,
Answer:
i)The ball shot out of the smokestack of a train moving in a straight line at constant speed will fall back into the smokestack
ii)The ball shot out of the smokestack of a train moving in a straight track ( gaining speed ) will fall behind the smoke stack
iii) The ball shot out of the smokestack of a train moving in a circular track at constant speed will fall away from the smokestack in a direction that is away from the middle of the circular track
Explanation:
The ball shot out of the smokestack of a train moving in a straight line at constant speed will fall back into the smokestack
The ball shot out of the smokestack of a train moving in a straight track ( gaining speed ) will fall behind the smoke stack
The ball shot out of the smokestack of a train moving in a circular track at constant speed will fall away from the smokestack in a direction that is away from the middle of the circular track
Is the answer clockwise (CW) or counter clockwise (CCW) ?
Please help me with this!
hope this helps good luck
5. A car advertisement states that a certain car can accelerate from rest to 70 m/s in 7
seconds. Find the car's average acceleration.
O-0.10 m/s^2
10 m/s^2
-10 m/s^2
O 0.10 m/s^2
Answer:
The car's average acceleration is [tex]10\ m/s^2[/tex].
Explanation:
Constant Acceleration Motion
It's a type of motion in which the velocity of an object changes by an equal amount in every equal period of time.
Being vo the initial speed, a the constant acceleration, vf the final speed, and t the time, the following relation applies:
[tex]v_f=v_o+at[/tex]
If we need to find the acceleration, we solve the above equation for a:
[tex]\displaystyle a=\frac{v_f-v_o}{t}[/tex]
The car accelerates from rest (vo=0) to vf=70 m/s in t=7 seconds. Substitute the values into the formula:
[tex]\displaystyle a=\frac{70-0}{7}=\frac{70}{7}=10[/tex]
[tex]a=10\ m/s^2[/tex]
The car's average acceleration is [tex]10\ m/s^2[/tex].
Note: The choices are not very clear, but the second choice seems to be the correct answer.
A block of mass m begins at rest at the top of a ramp at elevation h with whatever PE is associated with that height. The block slides down the ramp over a distance d until it reaches the bottom of the ramp. How much of its original total energy (in J) survives as KE when it reaches the ground
This question is incomplete, the complete question is;
A block of mass m begins at rest at the top of a ramp at elevation h with whatever PE is associated with that height. The block slides down the ramp over a distance d until it reaches the bottom of the ramp.
How much of its original total energy (in J) survives as KE when it reaches the ground? m = 9.9 kg h = 4.9 m d = 5 m μ = 0.3 θ = 36.87°
Answer:
the amount of its original total energy (in J) that survives as KE when it reaches the ground will is 358.975 J
Explanation:
Given that;
m = 9.9 kg
h = 4.9 m
d = 5 m
μ = 0.3
θ = 36.87°
Now from conservation of energy, the energy is;
Et = mgh
we substitute
Et = 9.9 × 9.8 × 4.9
= 475.398 J
Also the loss of energy i
E_loss = (umg cosθ) d
we substitute
E_loss = 0.3 × 9.9 × 9.8 × cos36.87° × 5
= 116.423 J
so the amount of its original total energy (in J) that survives as KE when it reaches the ground will be
E = Et - E_loss
E = 475.398 J - 116.423 J
E = 358.975 J
Which two types of energy does a book have as it falls to the floor
Answer:
kinetic and potential energy
Explanation:
If the body with a mass of 4kg is moved by a force of 20 N, what is the rate of its acceleration?
Answer:
The answer is 5 m/s²Explanation:
The acceleration of an object given it's mass and the force acting on it can be found by using the formula
[tex]acceleration = \frac{force}{mass} \\[/tex]
From the question
force = 20 N
mass = 4 kg
We have
[tex]a = \frac{20}{4} \\ [/tex]
We have the final answer as
5 m/s²Hope this helps you
An electron moving in the direction of the x-axis enters a magnetic field. If the electron experiences a magnetic deflection in the -y direction, the direction of the magnetic field in this region points in the direction of the
Answer:
-z
Explanation:
The force on a moving charge due to a magnetic field follows the right hand rule, so a positive charge, experiencing a magnetic deflection in the -y direction, while it moves in the direction of the x-axis, will do it due to a magnetic field pointing in the +z direction.
As the electron has a negative charge, the magnetic field will point in the opposite direction, i.e., in the -z direction.
The boys are finally old enough to compete in the box car derby race at the local fair. They have been working on their cars since the conclusion of the race last year. One boy's car raced down the track and placed 2nd in his race. However, the other boy's car started well but half-way through the race a wheel came off and his car came to a complete stop. The boy was very disappointed and the other boy felt horrible for his friend. Which of the following graphs best represents the motion of boy's car that stopped?