The forces doing non-zero work as the block slides down a frictionless ramp are gravity and any applied forces. The work done can be calculated using the mass, gravitational acceleration, and height. For the pushed block, the initial kinetic energy is combined with gravitational work to predict the final speed.
The forces that do non-zero work on a block as it slides down a frictionless ramp include gravity and any applied forces (like a push or a pull), but not the normal force or friction since the ramp is frictionless. Gravity does positive work as it pulls the block down the plane, increasing its kinetic energy. To calculate the total work done on the 1.85 kg wooden block after sliding down a distance of 1.85 m on the ramp at a 59.3° angle, you can use the formula:
Work = mgh, where m is mass, g is acceleration due to gravity, and h is the height.To predict the speed of the block after sliding down, you can use the conservation of energy principle or kinematic equations that relate distance, acceleration, and velocity. For the scenario where the block is given an initial push, the final speed can be predicted using the same principles by accounting for the initial kinetic energy provided by the push.
Here's an example of calculating work done:
Problem: A block of mass 10 kg slides down through a length of 10 m over an incline of 30°. If the coefficient of kinetic friction is 0.5, then find the work done by the net force on the block. In this example, work done by the net force can be calculated by finding the components of the gravitational force parallel and perpendicular to the incline and applying the kinetic friction force in the opposite direction of motion.The maximum allowed rms current in a circuit before its circuit breaker trips is 12.5 A . If a maximum emf of 180 V is connected to a device whose resistance is 8.4 ohms, will the circuit breaker trip and interrupt the flow of electricity
Explanation:
The potential difference of 180 V is applied across resistance = 8.4 ohm
Thus the maximum current flowing I₀ = [tex]\frac{V}{R}[/tex] = [tex]\frac{180}{8.4}[/tex] = 21.4 A
The rms value of the current is = [tex]\frac{I_0}{\sqrt{2} }[/tex] = I₀ x 0.7 = 15 A
This value of current is greater than the 12.5 A . Thus the circuit will break .
To identify a diatomic gas (X2), a researcher carried out the following experiment: She weighed an empty 5.1-L bulb, then filled it with the gas at 1.00 atm and 20.0 ∘C and weighed it again. The difference in mass was 5.9 g . Identify the gas.
Answer: Nitrogen gas
Explanation:
Using ideal Gas's law
PV = nRT
where
Pressure of gas, P= 1atm
Volume of gas, V= 5.1L
no of moles of gas, n=
Ideal gas constant, R= 0.0821
Temperature of gas, T= 20°C = 20+273 = 293K
also, n= (mass/molar mass)
mass of the gas m = 5.9g
Molar mass of the gas = ?
So, PV = (mRT/M)
We're looking for molar mass M, then
M = mRT/PV
M = (5.9 * 0.0821 * 293)/(1 * 5.1)
M = 141.93/5.1
M = 27.8g/mol ~ 28g/mol
Since the gas is diatomic, then we say,
Atomic mass of gas = 1/2 * molar mass
Atomic mass = 1/2 * 28
Atomic mass = 14
Therefore, the gas is nitrogen.
Toy car a is moving at a speed of 1 m/s towards toy car b that is at rest. Toy car a has a mass of 3 kg and car b has a mass of 2 kg. They couple together after collison. At what speed are they moving after the collison?
Answer:
0.6m/s
Explanation:
Totally inelastic collision
m₁v₁ + m₂v₂ = ( m₁ + m₂)v(final)
Where v (v₁ and v₂) is the initial velocity of the objects
v(final) is the final velocity of the objects stuck together
Toy car a , m₁ = 3kg, v₁ = 1m/s
Toy car b , m₂ = 2kg, v₂ = 0m/s
m₁v₁ + m₂v₂ = ( m₁ + m₂)v(final)
3(1) + 2(0) = (3 + 2) v(final)
3 = 5 v(final)
v(final) = 0.6m/s
Explanation:
Below is an attachment containing the solution.
The key difference between the binomial and hypergeometric distribution is that
Explanation:
Both distributions describe the number of times an event occurs in a givn number of trials. In the binomial distribution, the probability is the same for each trial. While in the hypergeometric distribution, each trial changes the probability of each subsequent trial, since there is no replacement.
In class we calculated the range of a projectile launched on flat ground. Consider instead, a projectile is launched down-slope of angle with an initial velocity of magnitude directed at an angle above the horizontal. How far will this projectile travel horizontally before it lands
Answer:
With an initial speed of 10m/s at an angle 30° below the horizontal, and a height of 8m, the projectile travels 7.49m horizontally before it lands.
Explanation:
Since the horizontal motion is independent from the vertical motion, we can consider them separated. The horizonal motion has a constant speed, because there is no external forces in the horizontal axis. On the other hand, the vertical motion actually is affected by the gravitational force, so the projectile will be accelerated down with a magnitude [tex]g[/tex].
If we have the initial velocity [tex]v_o[/tex] and its angle [tex]\theta[/tex], we can obtain the vertical component of the velocity [tex]v_{oy}[/tex] using trigonometry:
[tex]v_{oy}=v_osin\theta[/tex]
Therefore, if we know the height at which the projectile was launched, we can obtain the final velocity using the formula:
[tex]v_{fy}^{2} =v_{oy}^{2}+2gy\\\\ v_{fy}=\sqrt{v_{oy}^{2}+2gy }[/tex]
Next, we compute the time the projectile lasts to reach the ground using the definition of acceleration:
[tex]g=\frac{v_{fy}-v_{oy}}{\Delta t} \\\\\Delta t= \frac{v_{fy}-v_{oy}}{g}=\frac{\sqrt{v_{oy}^{2}+2gy} -v_{oy}}{g}[/tex]
Finally, from the equation of horizontal motion with constant speed, we have that:
[tex]x=v_{ox}\Delta t= v_{ox}\frac{\sqrt{v_{oy}^{2}+2gy} -v_{oy}}{g}[/tex]
For example, if the projectile is launched at an angle 30° below the horizontal with an initial speed of 10m/s and a height 8m, we compute:
[tex]v_{ox}=10\frac{m}{s} cos30=8.66\frac{m}{s}\\v_{oy}=10\frac{m}{s} sin30=5\frac{m}{s}\\\\x=8.66\frac{m}{s} \frac{\sqrt{(5\frac{m}{s}) ^{2}+2(9.8\frac{m}{s^{2}})8m}-5\frac{m}{s} }{9.8\frac{m}{s^{2} } } =7.49m[/tex]
In words, the projectile travels 7.49m horizontally before it lands.
A circular curve of highway is designed for traffic moving at 60 km/h. Assume the traffic consists of cars without negative lift. (a) If the radius of the curve is 150 m, what is the correct angle of banking of the road? (b) If the curve were not banked, what would be the minimum coefficient of friction between tires and road that would keep traffic from skidding out of the turn when traveling at 60 km/h?
Answer:
a) 10.7° ≈ 11°
b) 0.19
Explanation:
If the road is banked at an angle, without seeking the help of friction, (i.e. frictionless road), the forces acting on the car are shown in the attached free body diagram to the question
In the y - direction
mg = N cos θ (eqn 1)
mg = weight of the car.
N = normal reaction of the plane on the car
And in the direction parallel to the inclined plane,
(mv²/r) = N sin θ (eqn 2)
(mv²/r) = force keeping the car in circular motion
Divide (eqn 2) by (eqn 1)
(v²/gr) = Tan θ
v = velocity of car = 60 km/h = 16.667 m/s
g = acceleration due to gravity
r = 150 m
(16.667²/(9.8×150)) = Tan θ
θ = Tan⁻¹ (0.18896)
θ = 10.7° ≈ 11°
b) In the absence of banking, the frictional force on the road has to balance the force keeping the car in circular motion
That is,
Fr = (mv²/r)
Fr = μN = μ mg
μ mg = mv²/r
μ = (v²/gr) = (16.667²/(9.8×150)) = 0.19
Hope this Helps!!!
A person in the passenger basket of a hot-air balloon throws a ball horizontally outward from the basket with a speed of 10.0 m/s. What initial velocity (magnitude and direction) does the ball have relative to a person standing on the ground if the hot-air balloon is rising at 6.0 m/s relative to the ground during this throw?
Answer:
Explanation:
Given
balloon is rising with a speed of [tex]u_y=6\ m/s[/tex]
Person throws a ball out of basket with a horizontal velocity of [tex]u_x=10\ m/s[/tex]
Considering upward direction to be positive
When ball is thrown it has two velocity i.e. in upward direction and in horizontal direction so net velocity is
[tex]v_{net}=\sqrt{(u_x)^2+(u_y)^2}[/tex]
[tex]v_{net}=\sqrt{(6)^2+(10)^2}[/tex]
[tex]v_{net}=\sqrt{36+100}[/tex]
[tex]v_{net}=\sqrt{136}[/tex]
[tex]v_{net}=11.66\ m/s[/tex]
Direction of velocity
[tex]\tan \theta =\dfrac{u_y}{u_x}[/tex]
[tex]\tan \theta =\dfrac{6}{10}[/tex]
[tex]\theta =30.96^{\circ}[/tex]
where [tex]\theta [/tex] is angle made by net velocity with horizontal .
You are observing a binary star system and obtain a series of spectra of the light from the two stars. In this spectrum, most of the absorption lines shift back and forth as expected from the Doppler Effect. A few lines, however, do not shift at all, but remain at the same wavelength. How could we explain the behavior of the non-shifting lines?
Answer: non-shifting lines indicate non moving star.
Explanation: when a star is moving toward the detector, the wavelength will decrease - there will be a blue shift.
When it's moving away from the earth or detector, the wavelength will increase - there will be a red shift.
Identifiable patterns of absorption lines that appear shorter or longer wavelengths than normal indicate that the star is moving
In visualisation, the bottom spectrum shows the normal position of absorption line for a star that is not moving toward or away from the earth.
Non-shifting lines in the binary star system's spectrum are attributed to absorption by interstellar clouds, which unlike the stars, do not move relative to us. The narrowness of these lines indicates the low pressure of the absorbing gas.
Explanation:When observing a binary star system and noting the Doppler Effect in the spectral lines, if certain lines do not shift, it indicates they originate from something that is not moving with respect to us. Most of the absorption lines shift due to the motion of the binary stars, which causes Doppler shifts as the stars move toward or away from us. This movement results in the spectral lines being blue-shifted when the star is approaching us, and red-shifted when it's receding. However, the lines that remain constant are likely due to the absorption by interstellar clouds located between Earth and the stars. The non-shifting lines are also much narrower, suggesting that the absorbing gas is at a very low pressure. This non-movement of specific lines helped in discovering the presence of interstellar materials, as their spectral lines do not participate in the Doppler shifts associated with the stars' orbits.
Which of the following statements is true? there are only about 100 different kinds of atoms that combine to form all substances an atom is the smallest particle known to exist a large atom can be photographed with the aid of an ordinary microscope there are thousands of different kinds of atoms that account for a wide variety of substances none of the above
THE ANSWER IS : THERE ARE ONLY ABOUT 100 DIFFERENT KINDS OF ATOMS THAT COMBINE TO FORM ALL SUBSTANCES
Explanation:
A type of light bulb is labeled having an average lifetime of 1000 hours. It’s reasonable to model the probability of failure of these bulbs by an exponential density function with mean µ = 1000. Use this model to find the probability that a bulb fails within the first 300 hours.
Answer:
[tex]0.2592 \ or \ 25.92\%[/tex]
Explanation:
The exponential density function is given as
[tex]f(t)=\left \{ {{0} \atop {ce^{ct}}} \right\\0,t<0\\ce^{ct},t\geq 0[/tex]
[tex]\mu=\frac{1}{c}\\c=\frac{1}{\mu}\\\\=\frac{1}{1000}=0.001\\\\f(t)=0.001e^{-0.001t}[/tex]
To find probability that bulb fails with the first 300hrs, we integrate from o to 300:
[tex]P(0\leq X\leq 300)=\int\limits^{300}_0 {f(t)} \, dt\\\\=\int\limits^{300}_0 {0.001e^{-001t}} \, dt\\ =|-e^{-0.001t}| \ 0\leq t\leq 300[/tex]
[tex]P(0\leq X\leq 300)=-0.7408+1\\=0.2592[/tex]
Hence probability of bulb failing within 300hrs is 25.92% or 0.2592
A 12.0-g bullet is fired horizontally into a 109-g wooden block that is initially at rest on a frictionless horizontal surface and connected to a spring having spring constant 152 N/m. The bullet becomes embedded in the block. If the bullet-block system compresses the spring by a maximum of 78.0 cm, what was the speed of the bullet at impact with the block?
Answer:
v₀ = 280.6 m / s
Explanation:
we have the shock between the bullet and the block that we can work with at the moment and another part where the assembly (bullet + block) compresses a spring, which we can work with mechanical energy,
We write the mechanical energy when the shock has passed the bodies
Em₀ = K = ½ (m + M) v²
We write the mechanical energy when the spring is in maximum compression
[tex]Em_{f} = K_{e} \\= \frac{1}{2} kx^2\\ Em_0 = Em_{f}[/tex]
½ (m + M) v² = ½ k x²
Let's calculate the system speed
v = √ [k x² / (m + M)]
v = √[152 ×0.78² / (0.012 +0.109) ]
v = 27.65 m / s
This is the speed of the bullet + Block system
Now let's use the moment to solve the shock
Before the crash
p₀ = m v₀
After the crash
[tex]p_{f} = (m + M) v[/tex]
The system is formed by the bullet and block assembly, so the forces during the crash are internal and the moment is preserved
[tex]p_0 = p_{f}[/tex]
m v₀ = (m + M) v
v₀ = v (m + M) / m
let's calculate
v₀ = 27.83 (0.012 +0.109) /0.012
v₀ = 280.6 m / s
a 4kg block is attatched to a vertical sspring constant 800n/m. the spring stretches 5cm down. how much elastic potential energy is stored in the system
The Potential energy stored in the system is 1 J
Explanation:
Given-
Mass, m = 4 kg
Spring constant, k = 800 N/m
Distance, x = 5cm = 0.05m
Potential energy, U = ?
We know,
Change in potential energy is equal to the work done.
So,
[tex]U = \frac{1}{2} k (x)^2\\\\[/tex]
By plugging in the values we get,
[tex]U = \frac{1}{2} * 800 * (0.05)^2\\ \\U = 400 * 0.0025\\\\U = 1J\\[/tex]
Therefore, Potential energy stored in the system is 1 J
A telephone lineman is told to stretch the wire between two poles so the poles exert an 800Nforce on the wire. As the lineman does not have a scale to measure forces, he decides to measure the speed of a pulse created in the wire when he hits it with a wrench. The pulse travels 60m from one pole to the other and back again in 2.6s. The 60m wire has a mass of 15kg.a. Should the wire be tightened or loosened?b. Should the wire be tightened or loosened?
Answer:
The wire should be tightened because the present tension is 532.54 N where the required tension is 800 N and the higher the tension the more tightening is required.
Explanation:
To solve the question
v = [tex]\sqrt{\frac{F_t}{\mu} } = \sqrt{\frac{L*F_t}{m} }[/tex] where
v = velocity of the pulse in the string = 46.154 m/s
[tex]F_t[/tex] = Required tension force = 800 N
m = Mass of the wire = 15 kg
L = length of the wire to be tension-ed = 60 m
Since the pulse travels twice the distance of 60 m in 2.6 s the velocity is given by
v = 2×60/2.6 = 46.154 m/s
Therefore making [tex]F_t[/tex] the subject of the formula and substituting the values, we have
[tex]F_t[/tex] = [tex]\frac{v^2m}{L}[/tex] =[tex]\frac{46.154^{2*15} }{60}[/tex] = 532.54 N
This means that, as it is, the present tension in the wire is 532.54 N which is less than the required 800 N, therefore the wire should be tightened
The wire should be tightened because the current tension on the wire is less than the required tension.
The given parameters;
Required tension on the wire, T = 800 NDistance traveled by the Pulse, d = 60 mTime of motion of the pulse, t = 2.6 sMass of the wire, m = 15 kgThe speed of the wave as the pulse traveled from one pole to the other two times, is calculated as follows;
[tex]v = \frac{2d}{t} \\\\v = \frac{2 \times 60}{2.6} \\\\v = 46.154 \ m/s[/tex]
The tension created on the wire during the pulse motion is calculated as follows;
[tex]v = \sqrt{\frac{T}{m/L} } \\\\v ^2 = \frac{TL}{m} \\\\T = \frac{v^2 m}{L} \\\\T = \frac{(46.154)^2 \times 15}{60} \\\\T = 532.55 \ N[/tex]
The current tension on the wire (532.55 N) is less than the required tension of 800 N. Thus, the wire should be tightened.
Learn more about tension on wire here: https://brainly.com/question/14336853
(BRAINLIEST W/ WORK SHOWN!!!) Help with Physics question?
A sound wave traveling at 343 m/s takes 10.0 s to go from a speaker to a detector. How far apart are the two devices?
A.) 3.43 x 10^3
B.) 3.43 x 10^2
C.) 3.43 x 10^-1
D.) 3.43 x 10^4
Answer:
[tex]\boxed {3.43 x 10^{3}}[/tex]
Explanation:
We know that speed is defined as distance moved per unit time hence expressed as [tex]v=\frac {d}{t}[/tex] where v is speed in m/s, d is distance in m and t is time in seconds. Making d the subject of the above formula then
[tex]d=vt[/tex]
Substituting 343 m/s for d and 10 s for t then
[tex]d= 343\times10= 3430= 3.43 x 10^{3}[/tex]
Therefore, the distance between speaker and deter is [tex]\boxed {3.43 x 10^{3}}[/tex]
One kind of baseball pitching machine works by rotating light and stiff rigid rod about a horizontal axis until the ball is moving toward the target. Suppose a 144 gg baseball is held 81 cm from the axis of rotation and released at the major league pitching speed of 81 mph.
a. What is the ball's centripetal acceleration just before it is released?
b. What is the magnitude of the net force that is acting on the ball just before it is released?
Answer:
(a). The ball's centripetal acceleration is [tex]16.17\times10^{2}\ m/s^2[/tex]
(b). The magnitude of the net force is 232.9 N.
Explanation:
Given that,
Mass of baseball = 144 g
Speed = 81 mph = 36.2 m/s
Distance = 81 cm
(a). We need top calculate the ball's centripetal acceleration just before it is released
Using formula of centripetal acceleration
[tex]a=\dfrac{v^2}{r}[/tex]
Where, v = speed
r = radius
Put the value into the formula
[tex]a=\dfrac{(36.2)^2}{81\times10^{-2}}[/tex]
[tex]a=1617.82\ m/s^2[/tex]
[tex]a=16.17\times10^{2}\ m/s^2[/tex]
(b). We need to calculate the magnitude of the net force that is acting on the ball just before it is released
Using formula of force
[tex]F=\dfrac{mv^2}{r}[/tex]
Put the value into the formula
[tex]F=\dfrac{144\times10^{-3}\times(36.2)^2}{81\times10^{-2}}[/tex]
[tex]F=232.9\ N[/tex]
Hence, (a). The ball's centripetal acceleration is [tex]16.17\times10^{2}\ m/s^2[/tex]
(b). The magnitude of the net force is 232.9 N.
An electric vehicle starts from rest and accelerates at a rate of1.9m/s2in a straight line until it reaches a speed of23m/s. The vehicle then slows at a constant rate of 1.4m/s2 until it stops.(a)How much time elapses from start to stop
Answer: 10.53 seconds
Explanation:
in the attachment
Explanation:
Below is an attachment containing the solution.
Which of the following statements about diffusion is true?
"1. It requires integral proteins in the cell membrane. 2. It is very rapid over long distances. 3. It is a passive process in which molecules move from a region of higher concentration to a region of lower concentration. 4. It is an active process in which molecules move from a region of lower concentration to one of higher concentration. 5. It requires an expenditure of energy by the cell."
Answer:
3. It is a passive process in which molecules move from a region of higher concentration to a region of lower concentration.
Explanation:
Diffusion can be defined as the movement of solute or gaseous molecules from the region in which they have higher concentration to the regions in which they have lower concentration until they become evenly distributed across the two regions.
Diffusion is a passive process, that is, it requires no energy.
Option 3 is the correct option.
Answer:
It is a passive process in which molecules move from a region of higher concentration to a region of lower concentration
Explanation:
Diffusion is the movement of a substance from an area of high concentration to an area of low concentration. It is an important process for living things.
A single substance tends to move from an area of high concentration to an area of low concentration until the concentration is equal across a space
A 0.40-kg block is attached to the end of a horizontal ideal spring and rests on a frictionless surface. The block is pulled so that the spring stretches for 2.0 cm relative to its unstrained length. When the block is released, it moves with an acceleration of 8.0 m/s2. What is the spring constant of the spring
Answer:
160N/m
Explanation:
According to Hooke's law which states that the extension of an elastic material is directly proportional to the applied force provided that the elastic limit is not exceeded. Mathematically,
F = ke where
F is the applied force
k is the spring constant
e is the extension
From the formula k = F/e
Since the body accelerates when the block is released, F = ma according to Newton's second law of motion.
The spring constant k = ma/e where
m is the mass of the block = 0.4kg
a is the acceleration = 8.0m/s²
e is the extension of the spring = 2.0cm = 0.02m
K = 0.4×8/0.02
K = 3.2/0.02
K = 160N/m
The spring constant of the spring is therefore 160N/m
Final answer:
The spring constant of the spring is calculated using Hooke's Law and Newton's second law of motion. By multiplying the mass of the block by its acceleration, we found the force, and then divided the force by the displacement to get the spring constant, which is 160 N/m.
Explanation:
To determine the spring constant of the spring, we need to apply Hooke's Law, which states that the force exerted by an ideal spring is directly proportional to its displacement from the equilibrium position (F = -kx), where 'F' is the force, 'k' is the spring constant, and 'x' is the displacement. Since the block is on a frictionless surface and we know the acceleration (a = 8.0 m/s2) and the mass (m = 0.40 kg), we can first find the force using Newton's second law (F = ma), and then use that force to calculate the spring constant 'k'.
The force exerted by the spring can be calculated as:
F = m * a
= 0.40 kg * 8.0 m/s²
= 3.2 N
The displacement (x) from the equilibrium position is given as 2.0 cm, which is 0.020 m in SI units. Using Hooke's Law, the spring constant can be calculated:
k = F / x
= 3.2 N / 0.020 m
= 160 N/m
Two people, one of mass 78 kg and the other of mass 59 kg, sit in a rowboat of mass 88 kg. With the boat initially at rest, the two people, who have been sitting at opposite ends of the boat 2.9 m apart from each other, now exchange seats.
How far will the boat move?
Answer:
The boat moves 0.244 m towards the end where the 59 kg person was at the start of the calculations.
Explanation:
The boat only moves because the centre of mass changes a bit if the two people on opposite ends of the boat exchange seats.
The boat moves a distance of the change in centre of mass
Noting that the weight of the boat acts at the centre of the boat at 1.45m from both ends.
For convention, we call the original position of the 59 kg person as x=0
This means,
59 kg person is at x = 0 m
88 kg of the boat acts at x = 1.45 m from the end of the 59 kg person.
78 kg person is at x = 2.90 m
Centre of mass = X = (Σ mᵢxᵢ)/(Σ mᵢ)
For the initial setup,
X = [(59×0) + (88×1.45) + (78×2.90)]/(59+88+78)
X = (353.8/225)
X = 1.572 m
(Don't forget that this is 1.572 m from the end we designated x=0 m)
When the people exchange positions,
59 kg person is now at the other end of the boat with x = 2.90 m
88 kg of the boat still acts at the centre of the boat at x = 1.45 m
And 78 kg person is now at the end of the boat with x = 0 m
Centre of mass = X = (Σ mᵢxᵢ)/(Σ mᵢ)
X = [(59×2.90) + (88×1.45) + (78×0)]/(59+88+78)
X = (298.7/225)
X = 1.328 m
(This is 1.328 m from the end we designated x=0 m from the start)
So, the centre of mass moves a distance of (1.572 - 1.328) towards the end of the boat we designated x=0 m from the start.
Hence, the boat moves 0.244 m towards the end where the 59 kg person was at the start of the calculations.
Hope this Helps!!!
Which changes would cause the fusion rate in the sun’s core to increase?
Answer:
Explanation
There are two factors that can cause the fusion rate in the sun's core to increase.
1) Rise in the temperature of core:
If the temperature of the sun's core increases then it will increases the nuclear fusion reaction. The nuclear fusion reactions has such a strong dependency on temperature that even a smallest rise in temperature will results in the higher rate of reaction. That is why these reactions happen in the hottest core of the stars.
2) Reduction in the radius of the core:
Density plays a huge role in the nuclear fusion reactions. If the radius of the sun's core decrease then there will be an increase in the density of the core. Thus the gravitational pressure will also increases. In order to resist this increase in pressure the fusion reactions will speed up and their rate becomes higher.
Typically a Switch operates at layer 2 of the OSI model. However, small organizations, such as a SOHO, can purchase a switch that also interprets Layer 3 data and works much like a router. What is this device called
Answer:
Layer 3 switch.
Explanation:
A layer 3 switch carrys out both the function of a switch and a router. It acts as a switch that links all the devices that are on the same subnet or virtual LAN at lightning speeds and has IP routing intelligence built into it to carry out the function of a router. It can support routing protocols, check incoming packets, and can also carry out routing decisions based on the source and the destination addresses.
Which of the following characterizes a beta ray? Choose all that apply. is electromagnetic radiation is a product of natural radioactive decay is attracted to the positively charged plate in an electric field is attracted to the negatively charged plate in an electric field is composed of electrons
Explanation:
When a radioactive substance decays then the fast moving electrons emitted by it is known as beta ray. Basically, a number of beta particles are ejected by a beta ray.
Symbol of a beta particle is [tex]^{0}_{-1}e[/tex]. A beta ray is a natural decay of a radioactive element. As we know that opposite charges get attracted towards each other. So, a beta ray gets attracted towards a positively charged plate.
Therefore, we can conclude that following are the characterizes a beta ray:
a product of natural radioactive decay.is attracted to the positively charged plate in an electric field. is composed of electrons.Final answer:
Beta rays are negatively charged, attracted to the positively charged plate in an electric field, and are composed of electrons. They are a product of natural radioactive decay and are lighter and much less massive than alpha particles.
Explanation:
Beta rays are characterized by specific properties that distinguish them from other types of radiation produced during natural radioactive decay. According to Ernest Rutherford's research, which involved observing the behavior of radiation in magnetic and electric fields, beta particles are known to be negatively charged and relatively light. This means they are attracted to the positively charged plate in an electric field and are significantly deflected due to their lighter mass compared to alpha particles.
Beta rays are not electromagnetic radiation; this term is reserved for gamma rays, which are uncharged and therefore unaffected by electric fields. Beta rays are indeed a product of natural radioactive decay, specifically during a process known as beta decay, in which a nucleus emits an electron or a positron. Since they are composed of high-energy electrons, the identification that beta rays are composed of electrons is also correct.
Water molecules attracting other water molecules is called
Answer:
Water molecules attracting other water molecules is called cohesive attraction.
Explanation:
They are basically two forces in liquids that determine their wetting characteristics, they are cohesive and adhesive forces.
Cohesion is the attraction between molecules of same liquid example water and water, while adhesion is attraction between molecules of different liquids example alcohol and water.
Therefore, Water molecules attracting other water molecules is called cohesive attraction.
Two trains start from towns 224 mi apart and travel towards each other on parallel tracks. They pass each other 1.6 hr later. If one train travels 10 mph faster than the other, find the speed of each train.
Answer: 65mph, 75mph
Explanation:
Let us assume x to be the speed of the slower train, in mph (miles per hour).
Then the speed of the other train is (x+10) mph, according to the question.
We then would have an equation like this
1.6x + 1.6(x+10) = 224.
This is because, the first addend in the left side is the distance covered by the slower train.
The second addend in the left side is the distance covered by the faster train.
The sum is 224 miles, because they together covered all the distance to the moment when they meet each other.
1.6x + 1.6x + 16 = 224
3.2x + 16 = 224
3.2x = 224 - 16
3 2x = 208
x = 208/3.2
x = 65
Thus the speed of the slower train is 65mph, and that of the other train is 65 + 10 = 75mph
Final answer:
By setting up and solving a linear equation, the speed of the slower train is determined to be 65 mph, while the faster train's speed is 75 mph as one travels 10 mph faster than the other.
Explanation:
To solve the problem of the two trains travelling on parallel tracks towards each other, we must first define the variables for their speeds. Let's say one train travels at x mph and the other at (x + 10) mph. Since they are moving towards each other, you can add their speeds together to find how fast the distance between them is closing. The total closing speed is then x + (x + 10) mph, which simplifies to 2x + 10 mph.
The total distance to be covered by the two trains until they pass each other is 224 miles. We know they pass each other after 1.6 hours, so using the formula Distance = Speed x Time, we can set up the equation: 224 miles = (2x + 10 mph) x 1.6 hours. Solving for x gives us the speed of the slower train.
Performing the algebraic steps:
224 = (2x + 10) x 1.6
224 = 3.2x + 16
208 = 3.2x
x = 65 mph.
Therefore, the speed of the slower train is 65 mph and the speed of the faster train is 75 mph.
Block A, with a mass of 10 kg, rests on a 35 incline. The coefficient of static friction is 0.40. An attached string is parallel to the incline and passes over a massless, frictionless pulley at the top. What is the largest mass MB, attached to the dangling end, for which A remains at rest?
The maximum mass of block B, for which block A remains at rest on a 35-degree incline given a static friction coefficient of 0.40 and a 10kg mass of block A, is approximately 1.98 kg.
Explanation:To solve this question, we will need to use the principles of static friction, inclined planes, and gravitational force. Static friction is what keeps block A from sliding down the incline. It has to overcome the downward force due to gravity on block A which is a component of the weight of block A acting downwards the inclined plane.
To determine the maximum mass of block B, we can equate the static frictional force to the net force acting downward on the inclined plane. The static frictional force is [tex]\mu N[/tex] where μ is the coefficient of static friction and N is the normal force. N = [tex]mAgcos\theta[/tex] where mA and g are the mass and acceleration due to gravity respectively and θ is the angle of the inclined plane. So, static frictional force = [tex]\mu mAgcos\theta[/tex]. The downward force is the sum of components of the weight of block A acting downwards and the weight of block B. So, [tex]mAgcos\theta (\mu)[/tex] = mAg*[tex]sin\theta[/tex] + mBg.
From this equation, you can solve for the mass of Block B: MB = [tex]mA\mu cos\theta[/tex] - mA*[tex]sin\theta[/tex].
Plugging in the given numbers, we get:
MB = 10(0.4*cos(35 degrees) - sin(35 degrees)).
This gives us approximately 1.98 kg as the maximum mass of block B before block A begins to slide.
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To find the largest mass for block MB, we first need to calculate the gravitational force and the static friction on block A, using the given values for mass, gravitational acceleration, coefficient of static friction and incline angle. The tension in the string, created by block MB, has to balance these forces. By setting the net forces equal, we can calculate the value of MB.
Explanation:The problem revolves around understanding the concept of static friction and forces acting on an inclined plane. Firstly, let's calculate the force due to gravity acting on block A. This is simply F_gravity = mg sin θ where m=10 kg is the mass of the block, g=9.8 m/s² is the gravitational acceleration, and θ is 35°. Next, we need to calculate the frictional force that prevents the block from sliding. Since block A is at rest, static friction is at work here, and we can use the formula F_friction = μN, where μ=0.40 (the coefficient of static friction) and N is the normal force acting on the block, which equals mg cos θ.
To keep block A at rest, the tension T in the string due to the dangling mass MB must balance both gravity and friction. Therefore: T=F_gravity + F_friction. The weight of the dangling mass brings about the tension in the string, and hence, T=MBg. From these, we can calculate the largest mass of MB for which A remains at rest.
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Define what is vsepr theory?
Answer:
Valence shell electron pair repulsion theory
Explanation:
VSEPR stands for valence shell electron pair repulsion theory in which helps in predicting the geometric shape of a molecule based upon the number of lone pairs of electrons.It is also called the Gillespie-Nyholm theory after its two main discoverers, Ronald Gillespie and Ronald Nyholm.
In this theory the lone pair of atoms of the valence shell repel each other and attain such an angular position which minimizes the repulsion between the lone pair of electrons and the bonded pair of electrons so that it attains a stable state.This theory is however not related to the wave function and the orbital hybridization but is based only upon the electron density.[tex]\huge\mathcal{\underline{\underline{{ɑ}{\pmb{\sf{nswer \: : - }}}}}}[/tex]
[tex]\red:\implies[/tex][tex]\underline{\underline{\textbf{\pink{Valence Shell Electron Pair Repulsion \: : -}}}}[/tex]
The shape of molecule depend upon the number of valence shell electron pairs (bonded or nonbonded) around the central atom.Pairs of electron in the valence shelk repel one another since their electron clouds are negatively charged.These pairs of electron tend to occupy such positions in space that minimise repulsion and thus maximise distance between them.The valence shell is taken as sphere with the electrins pairs localising on the spherical surface at maximum distance from one another.A multiple bond is treated as if it is a single electron pair and the two or three electron pairs of a multiple bond are treated as a single auper pair.When two or more resonance structure can represent a molecule, the VSEPR model is applicable to any such structure.A small metal ball is given a negative charge, then brought near (i.e., within about 1/10 the length of the rod) to end A of the rod ) What happens to end A of the rod when the ball approaches it closely this first time?
a.It is strongly repelled.
b.It is strongly attracted.
c.It is weakly attracted.
d.It is weakly repelled.
e.It is neither attracted nor repelled.
Answer:
Option B, it is strongly attracted
Explanation:
A Test Charge Determines Charge on Insulating and Conducting Balls, and the points made regarding conductors, it can be ascertained that in conductors, the electrons are free to move about. This means that when a charge is brought near to a conductor, the opposite charges all navigate to the sharpest point closest the charge and a strong attraction is created.
This shows that the rod will be strongly attracted. The density of the charges on the rod is mostly concentrated at the sharpest point.
End A of the rod will be strongly attracted to the negatively charged metal ball because of the process of charge induction, where opposite charges attract.
Explanation:When the negatively charged metal ball is brought near to end A of the rod, end A of the rod will be strongly attracted to the negatively charged ball. This is because of the principle of charge induction. When a charged body is brought near to another body, it will cause the charges in that body to redistribute. Opposite charges attract, so the near side of the rod (end A) will have a positive charge induced on it, and this positive charge will be attracted to the negative charge on the ball. So, the correct answer is option b. It is strongly attracted.
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A ball is thrown into the air with an initial velocity of 10 m/s at an angle of 45 degrees above the horizontal, as represented above. If air resistance is negligible the time needed for the ball to return to the ground is most nearly:_______.
Answer:
1.44 s
Explanation:
Since it is a projectile motion, we use the formula for the total time of flight,t
t = 2Usinθ/g where U = initial velocity of ball = 10 m/s, θ = 45 and g = 9.8 m/s²
t = 2Usinθ/g = 2 × 10sin45/9.8 = 1.44 s
So, the time needed for the ball to return to the ground is most nearly: 1.44 s
We have that for the Question it can be said that the time needed for the ball to return to the ground is most nearly
T=1.44
From the question we are told
A ball is thrown into the air with an initial velocity of 10 m/s at an angle of 45 degrees above the horizontal, as represented above. If air resistance is negligible the time needed for the ball to return to the ground is most nearly
Generally the Newton's equation for the vertical displacement is mathematically given as
[tex]y=ut+1/2at^2\\\\Therefore\\\\T=\frac{2Usin\theta}{g}\\\\T=\frac{2*10*sin45}{9.8}\\\\[/tex]
T=1.44
Therefore
the time needed for the ball to return to the ground is most nearly
T=1.44
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Energy is measured in ___________. a. kilograms b. joules c. electron volts d. B or C Atoms and molecules are the fundamental building blocks of ___________. a. energy b. radiation c. matter d. gravity
Answer:
Energy is measured in JOULES.
Atoms and molecules are the fundamental building blocks of matter.
Explanation:
Matter is anything that has weight and occupies space. Locked within any given molecule or atom is some form of energy waiting to be activated. Energy can neither be created nor destroyed.
The wavelength of red light is about 700 nm: λred = 700 nm. Which is larger, a grain of sand (diameter roughly 0.2 mm) or a wave of red light? How many times larger?
Answer:
The grain of sand is larger by a factor of about 286.
Explanation:
We express both values in standard form in the SI unit of metre so that the comparison can be easily done:
[tex]\lambda_R = 700 \text{ nm} = 700 \times 10^{-9} \text{ m} = 7\times10^{-7} \text{ m}[/tex]
Diameter of grain of sand = [tex]0.2 \times \text{ mm} = 0.2\times10^{-3}\text{ m} = 2\times10^{-4}\text{ m}[/tex]
It is seen that the grain of sand is larger.
The ratio of the sizes is given by
[tex]\dfrac{2\times10^{-4} \text{ m}}{7\times10^{-7} \text{ m}}=286[/tex]