Given,
The initial pressure of the gas, P₁=8.5 kPa
The initial volume of the gas, V₁=2.4 m³
The initial temperature of the gas, T₁=10 °C=282.15 K
The temperature of the building, i.e., the temperature of the gas after entering the building, T₂=35 °C=308.15 K
The pressure of the gas after entering the building, P₂=2 kPa
Let us assume the new volume of the gas is V₂
From the combined gas law,
[tex]\begin{gathered} \frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2} \\ \Rightarrow V_2=\frac{P_1V_1T_2}{T_1P_2} \end{gathered}[/tex]On substituting the known values,
[tex]undefined[/tex]PLEASE HELPPPPPPPPPPPPPPPPPPPPPPP
Answer:
right answer is death valley
Explanation:
because it is close to surface gravitational field
mexico
cus its the farthest
In order to hear a sound, even though there is an obstacle between you and the source, the sound wave must:A.diffract.B.refract.C.shorten.D.reflect.
We will have the following:
The sound must diffract. [Option A]
Will mark as Brainlist!
Which of the following best represents R= A - B ?
Please help, it’s due soon!
The option (C) best represents the R = A- B
What is vector in mathematics?A vector in mathematics is a quantity that not only expresses magnitude but also motion or position of an object in relation to another point or object. Euclidean vector, geometric vector, and spatial vector are other names for it.
In mathematics, a vector's magnitude is defined as the length of a segment of a directed line, and the vector's direction is indicated by the angle at which the vector is inclined.
What are the components in vector quantity?A vector primarily consists of two elements, the horizontal component and the vertical component. The horizontal component's value is cosθ, and the vertical component's value is sinθ.
There are two types of vector multiplication, they are dot products and cross products.
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What is the kinetic energy of the ocean liner ?
In order to calculate the kinetic energy, we can use the formula:
[tex]E_k=\frac{mv^2}{2}[/tex]Where m is the mass in kg and v is the speed in m/s.
So, for the bullet, we have:
[tex]E_k=\frac{0.0014\cdot408^2}{2}=116.5\text{ J}[/tex]And for the liner, we have:
[tex]E_k=\frac{59000000\cdot12^2}{2}=4248000000\text{ J}[/tex]So the ocean liner has greater kinetic energy.
Which of the following is an example of Newton's third law of motion?A. A skydiver slows down when her parachute opens.B. A grocery cart moves forward when it is pushed.C. A cannon recoils backwards when it is fired.D. A rolling rock slows down due to friction.
Explanation:
The third law of Newton says that when an object exerts a force on a second object, the first object experiences an equal and opposite force that is exerted by the second object.
So, the example that shows this law is:
C. A cannon recoils backward when it is fired.
Because the cannon e
A power company transmits current through a 240,000 V transmission line. This voltage is stepped down at an area substation to 40,000 V by a transformer that has 940 turns on the primary coil. How many turns are on the secondary of the transformer? _________turns
Given:
The voltage in the transmission line is: V = 240000 v
The stepped-down voltage is: Vs = 40000 v
The turns of the primary coil of the transformer are: Np = 940
To find:
The turns of the secondary coil of the transformer.
Explanation:
The voltage in a transmission line is used to step down by using the transformer. Thus, the primary voltage of the transformer will be the voltage in the transmission line.
Thus, Vp = V = 240000 v
The primary voltage Vp, the secondary voltage Vs, the primary turns on the coil Np and the secondary turns on the coil Ns are related as:
[tex]\frac{V_p}{V_s}=\frac{N_p}{N_s}[/tex]Rearranging the above equation, we get:
[tex]\begin{gathered} N_s=N_p\frac{V_s}{V_p} \\ \\ N_s=940\times\frac{40000\text{ v}}{240000\text{ v}} \\ \\ N_s=940\times\frac{4}{24} \\ \\ N_s=156.67 \\ \\ N_s\approx157 \end{gathered}[/tex]Final answer:
The number of turns on the secondary coil of the transformer are approximately 157.
If a bus is traveling at 12m/s and a passenger on the bus is walking to the back of the bus at a velocity of 5m/s, what is the relative velocity of the passenger relative to the ground?
The relative velocity is 17m/s.
The relative velocity of the passenger relative to the ground can be found by applying the concept of relative motion.
speed of bus (vb)=12m/s
speed of passenger inside the bus(vp)= 5m/s opposite to the speed of bus
speed of passenger relative to the ground = v
v= vb+vp
v= 12+(-5), since passenger is traveling in opposite direction
v=7m/
Therefore, the velocity of passenger relative to the ground is 7m/s.
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You turn a corner and are driving up asteep hill. Suddenly, a small boy runs out on the street chasing a ball. You slam on the brakes and skid to astop leaving a 50-foot-long skid mark on the street. The boy calmly walks away but a policemen watching fromthe sidewalk walks over and gives you a speeding ticket. He points out that the speed limit on this street is 25mph. After you recover your wits, you begin to examine the situation. You determine that the street makes anangle of 25◦with the horizontal and that the coefficient of static friction between your tires and the street is0.80. You also find that the coefficient of kinetic friction between your tires and the street is 0.60. Your car’sinformation book tells you that the mass of your car is 1600 kg. You weigh 140 lbs.
Given data:
Total displacement of the car;
[tex]s=50\text{ ft}[/tex]Speed limit;
[tex]v_m=25\text{ mph}[/tex]The angle of street from horizontal;
[tex]\theta=25\degree[/tex]Coefficient of static friction;
[tex]\mu_s=0.80[/tex]Coefficient of kinetic friction;
[tex]\mu_k=0.60[/tex]Mass of the car;
[tex]M=1600\text{ kg}[/tex]Weight of the man;
[tex]W=140\text{ lbs}[/tex]The kinetic friction force is given as,
[tex]F_k=\mu_k(M+m)g\cos \theta[/tex]Here, m is the mass of the man and g is the acceleration due to gravity.
The acceleration of the car driving up a steep hill is given as,
[tex]\begin{gathered} (M+m)g\sin \theta+F_k=(M+m)a \\ (M+m)g\sin \theta+\mu_k(M+m)g\cos \theta=(M+m)a \\ g\sin \theta+\mu_kg\cos \theta=a \end{gathered}[/tex]Substituting all known values,
[tex]\begin{gathered} (32\text{ ft/s}^2)\times\sin (25\degree)+0.6\times(32\text{ ft/s}^2)\times\cos (25\degree)=a \\ \approx30.92\text{ ft/s}^2 \end{gathered}[/tex]The velocity of the car is given as,
[tex]v^2=u^2-2as[/tex]Here, v is the final velocity (v=0, as the car stops), and u is the initial velocity.
The initial velocity of the car is given as,
[tex]u=\sqrt[]{v^2+2as}[/tex]Substituting all known values,
[tex]\begin{gathered} u=\sqrt[]{0^2+2\times(30.92\text{ ft/s}^2)\times(50\text{ ft})} \\ \approx55.61\text{ ft/s} \\ \approx37.91\text{ mph} \end{gathered}[/tex]Therefore, your speed is greater than the speed limit. Thus, you can not fight the ticket in the court.
In one or two sentences, explain how you would draw the particles in a gas.
Explanation
In gases the particles move rapidly in all directions, frequently colliding with each other and the side of the container
so
I would draw it with the particles separated moving and colliding with the others
I hope this helps you
The figure shows a 100 W light bulb 1 meter away from my finger. If my finger tip has an area of 1 cm2 and if the wavelength of the light from the bulb is λ = 588 nm = 588 × 10−9 m, then show that the number of photons hitting my finger per second is about 1015γ/second.1 Watt of power is 1 Joule/second.Number of photons per second?
We are given the following information
Energy of bulb = 100 W = 100 Joules/second
Area of fingertip = 1 cm² = 0.0001 m²
Wavelength of light = 588×10⁻⁹ m
Number of photons per second = ?
Let us first convert the wavelength into energy
[tex]E=\frac{h\cdot c}{\lambda}[/tex]Where h is the plank's constant (6.626x10⁻³⁴J.s), c is the speed of light (3×10⁸ m/s) and λ is the wavelength.
[tex]\begin{gathered} E=\frac{6.626\times10^{-34}\cdot3\times10^8}{588\times10^{-9}} \\ E=3.3806\times10^{-19}\; \frac{J}{s} \end{gathered}[/tex]A total of 8.0 joules of work is done when a constant horizontal force of 2.0 newtons to the left is used to push a 3.0-kilogram box acrossa counter top. Determine the total horizontal distance the box moves.
ANSWER
4 meters
EXPLANATION
Given:
• Work = 8.0 J
,• Force = 2.0 N
,• Mass of the box = 3.0 kg
Unknown:
• Distance the box moves
Work is the product of the applied force in the direction of motion and the distance an object moves,
[tex]W=F\cdot d[/tex]In this case, a box is pushed horizontally with a force of 2N producing a work of 8J. Solve the equation above for d,
[tex]d=\frac{W}{N}=\frac{8.0J}{2.0N}=4m[/tex]Hence, the horizontal distance the box moves is 4 meters.
You decide to roll a 0.10 kg ball across the floor so slowly that it will have a small momentum and a large de Broglie wavelength. If you roll it at 1.4x1^-3 m/s, what is its wavelength? Express your answer to two significant figures and include the appropriate units. How will the answer from above compare with the de Broglie wavelength of the high-speed electron that strikes the back face of one of the early models of a TV screen at 1/10 the speed of light (2x10^-11 m) ?
ANSWER:
(a) 4.73*10^-30 m
(b) 2.37*10^-19 times smaller
STEP-BY-STEP EXPLANATION:
Given:
Mass of ball = m = 0.10 kg
Speed of ball = v = 1.4x10^-3 m/s
(a)
Since, de Broglie wavelength is given by:
[tex]\lambda=\frac{h}{mv}[/tex]Where, h is the Plank's Constant ( h = 6.626x10^-34 kg m^2/s). Therefore, de Broglie wavelength of the ball will be:
[tex]\begin{gathered} \lambda=\frac{6.626\cdot10^{-34}}{0.10\cdot1.4\cdot10^{-3}} \\ \lambda_{\text{ball}}=4.73\cdot10^{-30} \end{gathered}[/tex](b)
[tex]\begin{gathered} \lambda_{electron}=2\cdot10^{-11} \\ \frac{\lambda_{\text{ball}}}{\lambda_{electron}}=\frac{4.73\cdot10^{-30}}{2\cdot10^{-11}} \\ \frac{\lambda_{\text{ball}}}{\lambda_{electron}}=2.37\cdot10^{-19} \end{gathered}[/tex]It means that the wavelength of the ball is 2.37*10^-19 times smaller
Given a DC battery of voltage, V = 4.00 V connected to a resistor R with a current I = 3.00 A through the resistor. What power is in this circuit? 15.5 W 12.0 W 39.4 W 45.5 W 8.88 W
12.0 W
Explanation
Electric power is the rate, per unit time, at which electrical energy is transferred by an electric circuit. to find the power in the circuit we need to use the expression:
[tex]P=IV[/tex]where P is the powe I is the current and V is the voltage
Step 1
a)Let
[tex]\begin{gathered} I=\text{ 3 Amperes} \\ V=4.0\text{ volts} \end{gathered}[/tex]b) now,replace
[tex]\begin{gathered} P=IV \\ P=3\text{ A*4 V} \\ P=12\text{ W} \end{gathered}[/tex]therefore, the answer is
12.0 W
I hope this helps you
Solve: What work is done when 3.0 C is moved through an electric potential difference of 1.5 V?1) 0.5 J2) 2.0 J3) 4.0 J4) 4.5 J
We know that the work done in an electric potential difference is
[tex]\begin{gathered} W=V\cdot Q \\ W=1.5V\cdot3.0C \\ W=4.5J \end{gathered}[/tex]Therefore, the work done is 4.5 J.I’m confused about which electromagnetic waves have the lowest frequency
The eletromagnetic wave that has the highest frequency is the gamma rays. It also has the highest energy and shortest wavelengths.
On the other hand, the type of eletromagnetic wave that has the lowest frequency, lowest energy and longest wavelength is radio waves.
b. Andrea has a particularly old bike, and she has not lubricated the chain orgears. This causes friction inside the gears as she rides. She begins coasting onflat ground at a speed of 10 m/s, and after a few seconds is only going 8 m/s.How much energy has been converted to heat? (1 point) (The combined mass is 50kg)c. Andrea is going 10 m/s toward a hill. She coasts up the hill without pedaling. Iffriction causes 10% of her energy to be converted to heat inside the gears, howhigh up the hill will she be able to coast? (2 points)
b.
The energy converted in heat is the change in kinetic energy, that is, the kinetic energy lost is the same as the heat.
The kinetic energy is given by:
[tex]K=\frac{1}{2}mv^2[/tex]The change in kinetic energy is given by:
[tex]K_f-K_0=\frac{1}{2}mv^2_f-\frac{1}{2}mv^2_0[/tex]Plugging the values given we have that:
[tex]\begin{gathered} K_f-K_0=\frac{1}{2}(50)(8)^2-\frac{1}{2}(50)(10)^2 \\ =-900 \end{gathered}[/tex]Hence Andrea has lost 900 J of energy and therefore 900 J is converted to heat.
c.
We know that the energy is conserved, in this case this means that the initial kinetic energy has to be equal to the energy lost by heat in the gears and the potential energy gained by climbing up the hill, that is:
[tex]K=Q+U[/tex]where K is the kinetic energy, Q is the heat and U is the potential energy.
We know that 10% of her energy is converted in heat, this means that:
[tex]Q=0.1K[/tex]and hece we have:
[tex]\begin{gathered} K=0.1K+U \\ U=0.9K \end{gathered}[/tex]the potential energy is given by:
[tex]U=mgh[/tex]then we have that:
[tex]mgh=0.9(\frac{1}{2}mv^2)[/tex]Plugging the values given and solving for h we have:
[tex]\begin{gathered} (50)(9.8)h=0.9(\frac{1}{2}(50)(10^2)) \\ 490h=2250 \\ h=\frac{2250}{490} \\ h=4.59 \end{gathered}[/tex]Therefore Andrea will be able to climp up to 4.59 meters
carts, bricks, and bands
4. Which two trials demonstrate the effect of a doubling of force upon the acceleration of a cart of constant mass?
a. Trials 2 and 4
b. Trials 2 and 6
c. Trials 4 and 7
d. Trials 6 and 7
A. The two trials that demonstrate the effect of a doubling of force upon the acceleration of a cart of constant mass are Trials 2 and 4.
What is the applied force on an object?The force applied on object is obtained by multiplying the mass and acceleration of the object.
According to Newton's second law of motion, the force exerted on an object is directly proportional to the product of mass and acceleration of the object.
Also, the applied force is directly proportional to the change in the momentum of the object.
Mathematically, the force acting on object is given as;
F = ma
a = F/m
where;
a is the acceleration of the objectm is the mass of the objectF is the applied forceAt a constant mass;
F₁/a₁ = F₂/a₂
When the force is doubled, the acceleration of the object is given as;
a₂ = F₂a₁/F₁
a₂ = (2F₁ x a₁) / F₁
a₂ = 2a₁
From the trials,
acceleration of trial 2 = 0.51 m/s²
acceleration of trial 4 = 1 m/s²
Thus, the two trials that demonstrate the effect of a doubling of force upon the acceleration of a cart of constant mass are Trials 2 and 4.
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Please help me with this! (Sadly my previous tutor couldn't help me with this)
a.
The free body diagram (not at scale) for each crate is shown below:
b.
In this case we know that the force is just sufficient to keep the crates from sliding, this means that the acceleration of the system is zero.
From the free body diagram and Newton's second law we have that for the 45 kg crate that:
[tex]\begin{gathered} T-W=0 \\ T=W \\ T=(45)(9.8) \\ T=441 \end{gathered}[/tex]For the 35 kg crate the equations of motion would be:
[tex]\begin{gathered} F+f_f-T=0 \\ N-W^{\prime}=0 \end{gathered}[/tex]but we know that the force of friction is given by:
[tex]F_f=\mu N[/tex]and from the second equation of motion we have that:
[tex]\begin{gathered} N=W^{\prime} \\ N=(35)(9.8) \\ N=343 \end{gathered}[/tex]Then we have that:
[tex]\begin{gathered} F+F_f-T=0 \\ F+343\mu-441=0 \end{gathered}[/tex]Since the crates are not moving we need to use the static coefficient of friction, then:
[tex]\begin{gathered} F+343\mu-441=0 \\ F+343(0.5)-441=0 \\ F+171.5-441=0 \\ F-269.5=0 \\ F=269.5 \end{gathered}[/tex]Therefore the force applied is 269.5 N
c.
The diagram in this case is:
d.
In this case we know that the 35 kg is sliding to the right at constant velocity, this means that the acceleration for the system is zero. (Notice that the difference with the previous case is that the friction points to the left)
From the discussion in part b we know that for the 45 kg block:
[tex]T-W=0[/tex]and then:
[tex]T=441[/tex]For the 35 kg we have that:
[tex]\begin{gathered} F-F_f-T=0 \\ N-W^{\prime}=0 \end{gathered}[/tex]from the previos discussion we know that:
[tex]N=343[/tex]and since in this case the crates are moving we need to use the kinetic coefficient of friction, then we have:
[tex]\begin{gathered} F-F_f-T=0 \\ F-(0.3)(343)-441=0 \\ F-102.9-441=0 \\ F-543.9=0 \\ F=543.9 \end{gathered}[/tex]Therefore in this case the force applied is 543.9 N
e.
In this case the free body diagram is:
f.
Since the crates are moving with an accelearion of 0.5 m/s^2 we have for the 45 kg crate that:
[tex]T-W=ma[/tex]from where:
[tex]\begin{gathered} T-(9.8)(45)=(45)(0.5) \\ T-441=22.5 \\ T=441+22.5 \\ T=463.5 \end{gathered}[/tex]For the 35 kg crate we have that:
[tex]\begin{gathered} F-F_f-T=m^{\prime}a \\ N-W^{\prime}=0 \end{gathered}[/tex]from the previous discussion we know that N=343, plugging this in the first equation we have:
[tex]\begin{gathered} F-(0.3)(343)-463.5=(35)(0.5) \\ F-102.9-463.5=17.5 \\ F-566.4=17.5 \\ F=566.4+17.5 \\ F=583.9 \end{gathered}[/tex]Therefore the force applied in this case is 583.9 N
PLEASE HELP :((
A particular cookie provides 54.0 kcal of energy. An athlete does an exercise that involves repeatedly lifting (without acceleration) a 103-kilogram weight 2.45-decimeters above the ground with an energy efficiency of 25%. How many repetitions of this exercise can the athlete do with the energy supplied from one of these cookies?
Energy effectiveness would be a term that refers to the proportion of input power over output. Power generation, as well as simply energy utilization, is the process of reducing the amount of energy used to produce goods and services.
Considering that,
A cookie contains 54.0 kcal of energy. An athlete utilizes the 54.0 kcal inside this cookie from input energy.
The following diagram illustrates the relationship among input as well as output energy:
Efficiency = output energy / input energy...(i)
Output energy = efficiency × input energy
By using equation (i)
⇒ output energy = 0.25 × 54 kcal = 13.5 kcal.
The lifting exercise has been performed n times for the output energy.
In terms of potential energy, such output energy could be written as follows:
Mass × gravity ×height.
So, energy per repetition = mgh = 103 kg × 9.8 m/ × (2.45 × 0.1m) = 247.303 J = 0.059 kcal.
So, Count of repetitions = sim of output energy / energy per repetition..(ii)
By using equation (ii)
Count of repetitions = 13.5 kcal / 0.059 kcal =229 repetitions.
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55
3. A ball is rolling at a rate of -2.4 m/s.
The ball then rolls down a hill, causing it to accelerate to a velocity of
8.6 m/s in 4.5 seconds time. What is the acceleration of the ball?
An object initially at rest experiences an acceleration of 9.8 m/s2. How much time will it take it to achieve a velocity of 49 m/s? v = vo + at
Given data:
* The initial velocity of the object is 0 m/s.
* The final velocity of the object is 49 m/s.
* The acceleration of the object is,
[tex]a=9.8ms^{-2}[/tex]Solution:
The time taken by the object to achieve the final velocity is,
[tex]v=v_o+at[/tex]where v is the final velocity, v_o is an initial velocity, a is an acceleration, and t is the time taken by the object to acquire the final velocity,
Substituting the known values,
[tex]\begin{gathered} 49=0+9.8\times t \\ 49=9.8t \\ t=\frac{49}{9.8} \\ t=5\text{ s} \end{gathered}[/tex]Thus, the time taken by the object to acqurie the final velocity is 5 seconds.
What the difference between velocity and speed
Answer:
Speed is the time rate at which an object is moving along a path, while velocity is the rate and direction of an object's movement.
how far away is a star if it takes light 12.5 years to reach the earth?
What would the separation between two identical objects, one carrying
2 C
of positive charge and the other
2 C
of negative charge, have to be if the electrical force on each was precisely
2 N?
Please Help
The distance between the two charges is 134,164.1 m.
What is the distance between the two identical charges?
The distance between the two identical charges is determined by applying Coulomb's law as shown below.
F = kq²/r²
where;
K is Coulomb's constantq is the magnitude of the chargesr is the distance between the chargesF is the electric force between the two chargesr = √(kq²/F)
r = √(9 x 10⁹ x 2²) / 2)
r = 134,164.1 m
Thus, the distance between the two charges is determined by applying Coulomb's law.
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I need help with some questions to study them for midterms!
The correct answer is option C.
The acceleration of an object is defined as the time rate of change of velocity. Velocity is a vector and has magnitude and direction. The change velocity may mean either a change in magnitude or change in direction or both.
Thus an object can have acceleration even when the speed is constant. For example, an object in a circular motion.
Thus the correct answer is option C.
Identify the kinematic equation which relates the velocity and time.
The kinematic equation which relates velocity and time is
[tex]v=v_0+at[/tex]As when the acceleratio
The picture below shows a person swinging a toy plane attached to a string in
uniform circular motion.
Which vector points in the direction of the centripetal acceleration of the
plane?
According to the image A vector points in the direction of the centripetal acceleration of the plane.
The correct option is C.
What is centripetal acceleration?Centripetal acceleration is a property of an object's motion along a circular path. Centripetal force refers to any item travelling in a circle with an acceleration vector pointing in the direction of the circle's center.
Briefing:A person is seen in the image swinging a toy plane on a string in a smooth, circular motion. The velocity direction is tangent to the circular orbit and perpendicular to the direction of the position.
The direction of the velocity and the speed may both vary as an object travels in a circular orbit. The velocity's direction is continually shifting. While acceleration is constantly moving uniformly in a circular path toward the recent, tangent at each point indicates the direction of velocity at that place.
As a result, vector A is pointed in the plane's centripetal acceleration's direction.
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The complete question is -
The picture below shows a person swinging a toy plane attached to a string in uniform circular motion.
Which vector points in the direction of the centripetal acceleration of the
plane?
A-B
B-D
C-A
D-C
carts, bricks, and bands
3. Which statement describes the effect of a doubling of force upon the acceleration of a cart of constant mass?
a. Doubling the force will cause the acceleration to be twice the original value.
b. Doubling the force will cause the acceleration to be one-half the original value.
c. Doubling the force will cause the acceleration to be four times the original value.
d. Doubling the force will cause the acceleration to be one-fourth the original value.
The statement that describes the effect of a doubling a force at a constant mass is "doubling the force will cause the acceleration to be twice the original value.
The correct answer is option A.
What is the applied force on an object?
The force applied on object is obtained by multiplying the mass and acceleration of the object.
According to Newton's second law of motion, the force exerted on an object is directly proportional to the product of mass and acceleration of the object.
Also, the applied force is directly proportional to the change in the momentum of the object.
Mathematically, the force acting on object is given as;
F = ma
a = F/m
where;
a is the acceleration of the objectm is the mass of the objectF is the applied forceAt a constant mass;
F₁/a₁ = F₂/a₂
When the force is doubled, the acceleration of the object is given as;
a₂ = F₂a₁/F₁
a₂ = (2F₁ x a₁) / F₁
a₂ = 2a₁
Thus, when the force on the cart is doubled and the mass is constant, the acceleration of the cart will double as well.
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The motor of a table saw is rotating at 3450 rev/min. A pulley attached to the motor shaft drives a second pulley of half the diameter by means of a V-belt. A circular saw blade of diameter 0.208 m is mounted on the same rotating shaft as the second pulley.
Answer: It should be the answer beginning like this
The linear spread is that
The radial acceleration of locations along the blade's outer edge is approximately 17580 [tex]m/s^2[/tex].
What is Radial acceleration?Radial acceleration describes the acceleration of an object travelling on a circular path towards the circle's center. It can be defined as the rate of change of tangential velocity with regard to time and is also known as centripetal acceleration.
Given:
A table saw's engine rotates at 3450 revolutions per minute.A V-belt connects a pulley that's attached to the motor shaft to a second pulley half the diameter.A 0.208 m circular saw blade is installed on the same rotating shaft as the second pulley.We know that the motor is rotating at 3450 rev/min. One revolution is equal to 2π radians, so we can convert the motor speed to radians per minute:
ω₁ = (3450 rev/min) x (2π rad/rev) = 21675π rad/min
The second pulley is half the diameter of the first pulley, so its angular speed, ω₂, is twice that of ω₁:
ω₂ = 2ω₁ = 43350π rad/min
The circular saw blade is mounted on the same shaft as the second pulley, so it also rotates at the same angular speed:
ω = ω₂ = 43350π rad/min
We can now calculate the linear speed of the small piece of wood moving at the same rate as the rim of the circular saw blade, indicated by v. The circumference of the circle is supplied by the rim of the circular saw blade:
C = πd = π(0.208 m) = 0.6548 m
The linear speed of the little piece of wood is equal to the tangential speed of the circular saw blade's rim:
v = ωr
where r is the circular saw blade's radius, given by half its diameter:
r = d/2 = 0.208/2 = 0.104 m
By substituting the values, we obtain:
v = r = (43350 rad/min) x (0.104 m) x (1/60) = approx. 23.0 m/s
As a result, the linear speed of the little piece of wood moving at the same rate as the rim of the circular saw blade is about 23.0 m/s.
Next, compute the radial acceleration of locations on the blade's outer edge, represented by. The radial acceleration is calculated as follows:
α = rω²
By substituting the values, we obtain:
r2 = (0.104 m) x (43350 rad/min)2 x (1/602) = 17580 m/s2 (approximate)
Therefore, the radial acceleration of points on the outer edge of the blade is approximately 17580 m/s². This high radial acceleration explains why sawdust doesn't stick to the teeth of the saw blade.
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Your question is incomplete, most probably the complete question is:
The motor of a table saw is rotating at 3450{\rm rev/min}. A pulley attached to the motor shaft drives a second pulley of half the diameter by means of a V-belt. A circular saw blade of diameter 0.208{\rm m}is mounted on the same rotating shaft as the second pulley.
The operator is careless and the blade catches and throws back a small piece of wood. This piece of wood moves with linear speed equal to the tangential speed of the rim of the blade. What is this speed?
v =_______________________ m/s
Calculate the radial acceleration of points on the outer edge of the blade to see why sawdust doesn't stick to its teeth.
\alpha=______________________m/s2
Two points ___________ create a line.A. sometimesB. neverC. alwaysD. not enough information
According to the Euclidean Postulates, a straight line segment can be drawn joining any two points.
Therefore, the answer is:
[tex]C)\text{ Always}[/tex]