Answer:
The models are larger versions of atoms/molecules. It is a visualization so we can understand the super small particles.
Explanation:
Atoms and molecules are very, very, small. They average 100 picometers across.
They definitely cannot be seen with the naked human eye. We can only see them with special microscopes (electron, optical, etc). that magnify hundreds of thousands of times.
So, we use models of atoms and molecules. It allows us to study, visualize, understand, and "see" the particles on a much larger scale.
Surface water is replaced by the blank cycle.
Answer: Surface water is replaced by the water cycle.
Explanation: The water cycle is a cycle that describes the movement of water.
Answer:
it is water cycle when it's one of the process occurs and process is precipation this replaces surface water
Block X and block Y travel toward each other along a horizontal surface with block X traveling in the positive direction. Block X has a mass of 4kg and a speed of 2ms. Block Y has a mass of 1kg and a speed of 1 ms. A completely inelastic collision occurs in which momentum is conserved. What is the approximate speed of block X after the collision
Answer:
v = 1.4 m / s
Explanation:
To solve this problem we use the law of conservation of momentum, for this we define a system formed by the two blocks in such a way that the force during the collision have been internal
initial instant. Just before the crash
p₀ = M v₁ - m v₂
final instant. Right after the crash
p_f = (M + m) v
the moment is preserved
p₀ = p_f
M v₁ - m v₂ = (M + m) v
v = [tex]\frac{1}{M+m}[/tex] (M v₁ - mv₂)
let's calculate
v = [tex]\frac{1}{4+1}[/tex] (4 2 - 1 1)
v = 1.4 m / s
in the same direction as the largest block (M)
The approximate speed of block X after the collision is 1.4 m/s.
Given information:
Block X and block Y travel toward each other along a horizontal surface with block X traveling in the positive direction.
Block X has a mass of [tex]m_1=[/tex] 4 kg and a speed of [tex]u_1=[/tex] 2 m/s.
Block Y has a mass of [tex]m_2=[/tex] 1 kg and a speed of [tex]u_2=[/tex] -1 m/s.
The collision between them is perfectly inelastic. So, the final velocity of both the objects will be the same.
The momentum of the system will be conserved. Let v be the final velocity of both the blocks.
The final velocity v can be calculated as,
[tex]m_1u_1+m_2u_2=(m_1+m_2)v\\4\times 2-1\times1=(4+1)v\\5v=7\\v=1.4\rm\;m/s[/tex]
Therefore, the approximate speed of block X after the collision is 1.4 m/s.
For more details about collision, refer to the link:
https://brainly.com/question/12941951
What is the torque on a bolt applied with a wrench that has a
lever arm of 30 cm with a force of 30 N?
Answer:
9 Nm
Explanation:
The formula for torque is;
τ = lever arm * force applied
τ = 30/100 * 30
τ = 9 Nm
The torque on a bolt will be "9 Nm".
Given values are:
Force applied,
30 NLever arm,
30 cmAs we know the formula,
→ [tex]\tau = Lever \ arm\times force \ applied[/tex]
By substituting the values, we get
→ [tex]= \frac{30}{100}\times 30[/tex]
→ [tex]= 9 \ Nm[/tex]
Thus the above response is correct.
Learn more:
https://brainly.com/question/20372081
Of the charge Q initially on a tiny sphere, a portion q is to be transferred to a second, nearby sphere. Both spheres can be treated as particles. For what value of q/Q>0.5 will the electrostatic force between the two parts have 1/3 of the maximum possible value?
Answer:
Explanation:
Maximum value of force will be possible when both the sphere will have same charge . In that case charge on each sphere = Q / 2 =.5Q
F( max ) = k .5Q x .5Q / R²
=.25kQ² /R²
For the second case
F = k q ( Q-q)/ R²
F = .25kQ² /3R²
.25kQ² /3R² = k q ( Q-q)/ R²
.25 Q² = 3qQ - 3q²
3q² - 3qQ + .25 Q² = 0
q =
In which part of a lab report would be the following sentence most likely occur? “Since the data showed that the
Answer:
most likely be included in the analysis section of a lab report
Explanation:
In the picture shown below A represents a characteristic of only geocentric model, B represents a characteristic common to both geocentric and heliocentric models, C represents a characteristic of only heliocentric model, and D represents a characteristic which the geocentric and heliocentric models do not have.
Under which label will the characteristic, "The sun and planets revolve around a central moon in the solar system" fall?
A
B
C
D
A frictionless cart of mass M is attached to a spring with spring constant k. When the cart is displaced 6 cm from its rest position and released, it oscillates with a period of 2 seconds.Four English majors are discussing what would happen to the period of oscillation if the cart was displaced 12 cm from its rest position instead of 6 cm and again released.With which, if any, of these students do you agree?
Answer:
Time period of horizontal Oscillation = T = 2[tex]\pi[/tex][tex]\sqrt{\frac{m}{k} }[/tex]
As you can see, from the equation, Period of oscillation depends upon the mass and the spring constant. Not on the displacement.
Explanation:
Solution:
As we know that:
F = Kx
Here,
K = Spring constant
x = displacement.
First, they are displacing it with 6 cm from its rest position for which Time period of the oscillation is T = 2 seconds.
But next, they want to know the effect on the time period of the oscillation if the displacement x is doubled from 6cm to 12 cm.
First of all, let us see the equation of the time period of the oscillation.
We need to check, if time period does depend on the displacement or not.
As we know,
Time period of horizontal Oscillation = T = 2[tex]\pi[/tex][tex]\sqrt{\frac{m}{k} }[/tex]
As you can see, from the equation, Period of oscillation depends upon the mass and the spring constant. Not on the displacement.
Since, K is the constant for a particular spring, we need to change the mass of the cart to change the time period.
Hence the Time period will remain same.
The time period will remain the same in both conditions. The time period of horizontal Oscillation will be [tex]2 \pi \sqrt{\frac{m}{k} }[/tex].
What is the time period of oscillation?The period is the amount of time it takes for a particle to perform one full oscillation. T is the symbol for it. Taking the reciprocal of the frequency yields the frequency of the oscillation.
From Hooke's law;
F = Kx
Where,
K is the spring constant
x is the displacement
First, they move it 6 cm from its rest position, with a T = 2 second oscillation period.
However, they want to know what influence doubling the displacement x from 6 cm to 12 cm has on the oscillation's time period.
Let's start by looking at the oscillation's time period equation. We need to see if the time period is affected by the shift.
The time period of the horizontal oscillation is given by;
[tex]\rm T = 2 \pi \sqrt{\frac{m}{k} }[/tex]
As the equation shows, the period of oscillation is determined by the mass and the spring constant. On the displacement, no.
We must modify the mass of the cart to change the time period since K is the constant for each spring.
Hence the time period will not change.
To learn more about the time period of oscillation refer to the link;
https://brainly.com/question/20070798
How does the geosphere interact with the biosphere during a fire?
Answer: A forest fire could affect the geosphere by ash and the heat from the fire, but the fire would affect the biosphere by less oxygen in the air and would make it hard for human to breath affecting climate change. Also many homes to wildlife that live in the forest would lose their homes. This would result in lose of money from no hunting and that affect the economy.
What happens to Average Speed if distance decreased & time stayed the same?
Answer:
I think you are trying to ask 'what happens to the average speed (of humans or animals.) if the distance decreased and time stayed the same (the time such as what hour it is or what second?)' If that is the case then the average speed will be more since the distance decreased. And the time staying the same might affect it I am not fully sure so I will not say anything about that. I hope this helps!
Various amplifier and load combinations are measured as listed below using rms values. For each, find the voltage, current, and power gains ( A v , Ai , and Ap, respectively) both as ratios and in dB:
(a) vI= 100 mV, iI = 100 μA, vO = 10 V, RL = 100Ω
(b) vI = 10 μV, iI = 100 nA, vO =1 V, RL= 10 kΩ
(c) vI =1 V, iI = 1 mA, vO =5 V, RL = 10Ω
Answer:
The solution to this question can be defined as follows:
Explanation:
In point (a):
[tex]v_i= 100 \ mV\\\\ i_I = 100 \mu \ A\\\\ v_O = 10 \ V\\\\ R_L = 100 \ \Omega \\\\i_L = \frac{V_0}{R_L} = \frac{10}{100} = 100 \ MA \\\\A_v = \frac{V_0}{V_i} = \frac{10}{100 \times 10^{-3}} =100 \\\\A_v(db) = 20 \lag (100) =40 \ db \\\\ A_i= \frac{i_L}{i_i} = \frac{100 \times 10^{-3}}{100 \times 10^{-6}} =1000 \\\\A_i(db) = 20 \lag (100) =60 \ db \\\\[/tex]
[tex]A_p= \frac{P_0}{p_i} =\frac{v_0 i_L}{v_i i_i} = \frac{ 10(100 \times 10^{-3})}{100 \times 10^{-6} \times 100 \times 10^{-3}} =100000\\\\ A_p(db) =10 \log (100000) =50 \ db \\\\[/tex]
In point (b):
[tex]v_i = 10 \mu V\\\\ i_i = 100 \ nA \\\\ v_O =1 \ V \\\\ R_L= 10 \ k \Omega \\\\i_0 = \frac{V_0}{R_L} = \frac{1}{10 \ K} = 100 \ \muA \\\\A_v = \frac{V_0}{V_i} = \frac{10}{10 \times 10^{-6}} =100000 \\\\A_v(db) = 20 \lag (100000) =100 \ db \\\\ A_i= \frac{i_0}{i_i} = \frac{100 \times 10^{-6}}{100 \times 10^{-9}} =1000 \\\\A_i(db) = 20 \lag (1000) =60 \ db \\\\[/tex]
[tex]A_p= \frac{P_0}{p_i} =\frac{v_0 i_0}{v_i i_i} = \frac{ 1 \times 100 \times 10^{-6})}{10 \times 10^{-6} \times 100 \times 10^{-9}} =100000000\\\\A_p(db) =10 \log (100000000) =80 \ db \\\\[/tex]
In point (C):
[tex]v_i =1\ V\\\\ i_I = 1 \ mA\\\\ v_O =5\ V \\\\ R_L = 10 \ \Omega \\\\i_0 = \frac{V_0}{R_L} = \frac{5}{10 } = 0.5 \ A \\\\A_v = \frac{V_0}{V_i} = \frac{5}{1} =5 \\\\A_v(db) = 20 \log 5 =13.97 \ db = 14 \db \\\\ A_i= \frac{i_0}{i_i} = \frac{0.5}{1\times 10^{-3}} =500 \\\\A_i(db) = 20 \log (500) =53.97 \ db = 54 \db \\\\[/tex]
[tex]A_p= \frac{P_0}{p_i} =\frac{v_0 i_0}{v_i i_i} = \frac{ 5 \times 0.5 }{1 \times 1 \times 10^{-3}} =2500\\\\A_p(db) =10 \log (2500) = 33.97 \ db = 34 \db\\\\[/tex]
Let w(x)=3x-7.If w(x)=14, find x
Answer:
7
Explanation:
We are given:
w(x) = 3x - 7
w(x) = 14
The problem here entails us to solve for x;
To solve for x; equate the two expressions:
3x - 7 = 14
3x = 14 + 7
3x = 21
x = 7
So the value of x = 7
Some giant ocean waves have a wavelength of 25 m and travel at 6.5 m/s with a frequency of 0.26 HZ. What is the period of such a wave ?
Answer:
3.85s
Explanation:
Given parameters:
Wavelength = 25m
Velocity = 6.5m/s
Frequency = 0.26Hz
Unknown:
Period of the wave = ?
Solution:
The period of a wave is the inverse of the frequency of the wave.
Period = [tex]\frac{1}{frequency}[/tex]
Period = [tex]\frac{1}{0.26}[/tex] = 3.85s
When the sum of all the forces acting on a block on an inclined plane is zero, the block
A) must be at rest
B) must be accelerating
C) may be slowing down
D) may be moving at constant speed
Answer:
hmmm thats too hard for me.
Explanation:
What is the gravitational force between two students, John and Mike, if John has a mass of 81.0 kg, Mike has a mass of 93.0 kg, and their centers are separated by a distance of .620 m?
Answer:
1.31×10¯⁶ N
Explanation:
From the question given above, the following data were obtained:
Mass of John (M₁) = 81 Kg
Mass of Mike (M₂) = 93 Kg
Distance apart (r) = 0.620 m
Gravitational constant (G) = 6.67×10¯¹¹ Nm²/Kg²
Force (F) =?
The gravitational force between the two students, John and Mike, can be obtained as follow:
F = GM₁M₂ / r²
F = 6.67×10¯¹¹ × 81 × 93 / 0.62²
F = 6.67×10¯¹¹ × 7533 / 0.3844
F = 1.31×10¯⁶ N
Therefore, the gravitational force between the two students, John and Mike, is 1.31×10¯⁶ N
Suppose Isaac Newton can swim with a velocity of 35 m/sec. If Isaac Newton swims straight
across a river with a current of 22 m/sec, then what is the magnitude of Newton's resulting
velocity as he swims across the river. Note: Sir Isaac Newton swims like Aquaman.
Answer:
Let Vx = 35 m/s speed of swimmer across river
Vy = 22 m/s speed of river dowstream
V = (Vx^2 + Vy^2)^1/2 = 41.3 m/s net speed of swimmer
tan theta = Vy / Vx = .629 theta = 32.2 deg
(Theta would be zero if speed of river was zero)\
PLEASE HELP IM HEING TIMED
Answer:
Uh, I could be wrong but doesn’t it mean that the wave and particle are reacting together to make light? I think it’s something like that... I hope this helps!
I will give brainly
Defend Democritus' work on the atom and its contribution to the modern atomic model.
The speed limit on some segments of interstate 4 is 70 mph. What is this in km/h?
Answer:
112.63km/hr
Explanation:
The given dimension is :
70mph
We are to convert this to km/hr
1 mile = 1.609km
so;
70mph x 1.609 = 112.63km/hr
So,
The solution is 112.63km/hr
Artificial satellites in space can help you find locations on
Earth. True or false?
•What is the gravitational potential energy of a girl
who has a mass of 40 kg and is standing on the
edge of a diving board that is 5 m above the water?
Answer:
1960 joule
Explanation:
Plz help this is so confusing
The correct answer is 5 km/h
Explanation:
The speed at which the duck travels can be found by using the equation that is given (speed= distance/ time). The first step to do this is to replace distance and time using the values given. Here is the process:
speed = distance / time
speed = 10 km / 2 hours
Now, solve this equation
speed = 10 km/ 2 hours
10 / 2 or 10 divided 2 = 5
Finally, use the units, in this case, the correct units are km/h
a spring stretches from an initial height of 5 cm to a final stretch of 10 cm. the spring constant is 800 n/m.How much work was done on the spring?
what is the final force on the spring when it is at its 10 cm stretch?
explain why it is not appropriate to use the equation w=f//d when considering springs.
Answer:
.
Explanation:
F = kx so k = 800/((10-5)/100) = 16000 N/m
W = 1/2 kx^2 = 1/2 * 16000 * .05^2 = 20 J.
(sorry if it's wrong)
Heat traveling through a pan to warm food in the pan is an example of what kind of heat transfer?
A. Convection
B. Radiation
C. Insulation
D. conduction
Answer:
A
Explanation:
Answer:
D. Conduction
Explanation:
Convection heat transfer occurs in fluids, while conduction heat transfer occurs in solids.
A 62 kg student, starting from rest, slide down an 10.6 m high water slide. How fast is he going at the bottom of the slide? Use g = 10 m/s2
Answer:
14.6m/s
Explanation:
Given parameters:
Mass of the student = 62kg
Initial velocity = 0m/s
Height of slide = 10.6m
g = 10m/s²
Unknown:
Speed at the bottom of the slide = ?
Solution:
The speed at the bottom of the slide is the final velocity;
v ² = u² + 2gh
v is the final velocity
u is the initial velocity
g is the acceleration due to gravity
h is the height
v² = 0² + 2x 10 x 10.6
v² = 212
v = 14.6m/s
Please answer this question I don't know how to do it.
A safety plug is designed to melt when the pressure inside a metal tank becomes too high. A gas
at 51.0 atm and a temperature of 23.0°C is contained in the tank, but the plug melts when the
pressure reaches 75.0 atm. What temperature did the gas reach?
To review the solution to a similar problem, consult Interactive Solution 1.43. The magnitude of a force vector is 87.4 newtons (N). The x component of this vector is directed along the x axis and has a magnitude of 72.1 N. The y component points along the y axis. (a) Find the angle between and the x axis. (b) Find the component of along the y axis.
Answer:
(a) 34.4°
(b) 49.4 N
Explanation:
(a) From the diagram,
Amgle between the x axis can be calculated as,
cosΘ = adjacent/hypoteneous
cosΘ = 72.1/87.4
cosΘ = 0.8249
Θ = cos⁻¹(0.8249)
Θ = 34.4°.
Hence the angle between the x axis is 34.4°
(b) To find the component along the y axis we make use of pythagoras theorem.
a² = b²+c²................... Equation 1
Where a = 87.4 N, b = 72.1 N, c = y.
Substitute these values into equation 1
87.4² = 72.1² + y²
y² = 87.4²-72.1²
y² = 2440.35
y = √(2440.35)
y = 49.4 N
A person's speed around the Earth is faster at the poles than it is at the equator.
True or False?
Answer:
no
Explanation:
it is faster at the equator
A uniform disk with mass 35.2 kg and radius 0.200 m is pivoted at its center about a horizontal, frictionless axle that is stationary. The disk is initially at rest, and then a constant force 34.5 N is applied tangent to the rim of the disk.
a) What is the magnitude v of the tangential velocity of a point on the rim of the disk after the disk has turned through .200 revolution?
b) What is the magnitude a of the resultant acceleration of a point on the rim of the disk after the disk has turned through .200 evolution?
Answer:
a) v = 1.01 m/s
b) a = 5.6 m/s²
Explanation:
a)
If the disk is initially at rest, and it is applied a constant force tangential to the rim, we can apply the following expression (that resembles Newton's 2nd law, applying to rigid bodies instead of point masses) as follows:[tex]\tau = I * \alpha (1)[/tex]
Where τ is the external torque applied to the body, I is the rotational inertia of the body regarding the axis of rotation, and α is the angular acceleration as a consequence of the torque.Since the force is applied tangentially to the rim of the disk, it's perpendicular to the radius, so the torque can be calculated simply as follows:τ = F*r (2)For a solid uniform disk, the rotational inertia regarding an axle passing through its center is just I = m*r²/2 (3).Replacing (2) and (3) in (1), we can solve for α, as follows:[tex]\alpha = \frac{2*F}{m*r} = \frac{2*34.5N}{35.2kg*0.2m} = 9.8 rad/s2 (4)[/tex]
Since the angular acceleration is constant, we can use the following kinematic equation:[tex]\omega_{f}^{2} - \omega_{o}^{2} = 2*\Delta \theta * \alpha (5)[/tex]
Prior to solve it, we need to convert the angle rotated from revs to radians, as follows:[tex]0.2 rev*\frac{2*\pi rad}{1 rev} = 1.3 rad (6)[/tex]
Replacing (6) in (5), taking into account that ω₀ = 0 (due to the disk starts from rest), we can solve for ωf, as follows:[tex]\omega_{f} = \sqrt{2*\alpha *\Delta\theta} = \sqrt{2*1.3rad*9.8rad/s2} = 5.1 rad/sec (7)[/tex]
Now, we know that there exists a fixed relationship the tangential speed and the angular speed, as follows:[tex]v = \omega * r (8)[/tex]
where r is the radius of the circular movement. If we want to know the tangential speed of a point located on the rim of the disk, r becomes the radius of the disk, 0.200 m.Replacing this value and (7) in (8), we get:[tex]v= 5.1 rad/sec* 0.2 m = 1.01 m/s (9)[/tex]
b)
There exists a fixed relationship between the tangential and the angular acceleration in a circular movement, as follows:[tex]a_{t} = \alpha * r (9)[/tex]
where r is the radius of the circular movement. In this case the point is located on the rim of the disk, so r becomes the radius of the disk.Replacing this value and (4), in (9), we get:[tex]a_{t} = 9.8 rad/s2 * 0.200 m = 1.96 m/s2 (10)[/tex]
Now, the resultant acceleration of a point of the rim, in magnitude, is the vector sum of the tangential acceleration and the radial acceleration.The radial acceleration is just the centripetal acceleration, that can be expressed as follows:[tex]a_{c} = \omega^{2} * r (11)[/tex]
Since we are asked to get the acceleration after the disk has rotated 0.2 rev, and we have just got the value of the angular speed after rotating this same angle, we can replace (7) in (11).Since the point is located on the rim of the disk, r becomes simply the radius of the disk,, 0.200 m.Replacing this value and (7) in (11) we get:[tex]a_{c} = \omega^{2} * r = (5.1 rad/sec)^{2} * 0.200 m = 5.2 m/s2 (12)[/tex]
The magnitude of the resultant acceleration will be simply the vector sum of the tangential and the radial acceleration.Since both are perpendicular each other, we can find the resultant acceleration applying the Pythagorean Theorem to both perpendicular components, as follows:[tex]a = \sqrt{a_{t} ^{2} + a_{c} ^{2} } = \sqrt{(1.96m/s2)^{2} +(5.2m/s2)^{2} } = 5.6 m/s2 (13)[/tex]
need help pleaseee !
Answer:
08 and 09 with be 67
Explanation:
i know this because it was the vase major