The braking distance is the distance traveled by a car experiencing a braking force until it comes to rest.
Our initial energy is solely kinetic:
[tex]E_i = \frac{1}{2}mv^2[/tex]
And, since the car goes to rest, it is no longer in motion. It will have no kinetic energy.
[tex]E_f = 0[/tex]
Therefore, there was work done by the braking force.
[tex]W_B = E_f - E_i = -\frac{1}{2}mv^2[/tex]
Recall the definition of work:
[tex]W = F\cdot \Delta x[/tex]
Or in this case, since the displacement and breaking force are antiparallel:
[tex]W = -F_B\Delta x[/tex]
This is equivalent to the dissipation of kinetic energy:
[tex]W = -F_B\Delta x = -\frac{1}{2}mv^2[/tex]
Now, to visualize this, let's rearrange the equation to solve for displacement.
[tex]\Delta x =\frac{mv^2}{2F_B}[/tex]
There is a direct, SQUARE relationship between necessary braking distance speed.
If the speed was reduced by 10.3 percent, its new speed is only 89.7% percent of the original, so:
[tex]\Delta x' =\frac{m(0.897v)^2}{2F_B}[/tex]
[tex]\Delta x' = 0.8046\Delta x[/tex]
The reduction by a percentage is:
[tex]1 - 0.8046 = 0.1954 \\\\\boxed{= 19.54\%}[/tex]
If you place a hot block of metal in room temperature water is it convection or conduction?
Answer:
convection
Explanation:
Convection is a process by which heat is transferred by the movement of a heated fluid such as air or water.
The answer is Conduction: the process of heat is directly transmitted through a substance when there is a difference in temperature.
After brushing, Fluffy's fur has a charge of +8.0 × 10-9 coulombs and her plastic brush has a charge of –1.4 × 10-8 coulombs. If the distance between the fur and brush is roughly 5.0 × 10-1 meters, what is the approximate magnitude of the force between them?
(k = 9.0 × 109 newton·meters2/coulomb2)
The approximate magnitude of the force between them is [tex]-4.032*10^-^6 N[/tex]
What is columbo's law?According to Coulomb, the electric force for charges at rest has the following properties:
Like charges repel each other; unlike charges attract.
Thus, two negative charges repel one another, while a positive charge attracts a negative charge.
The equation in Coulomb's law is :
[tex]F =\frac{{kq_1q_2}}{r^2}[/tex]
Where F is the force between charges, k is a constant, q₁ is the charge of the first particle, q₂ is the charge of the second particle and r is the distance between the two particle charges.
According to the question,
The given values are:-
q₁ = +8.0 × 10⁻⁹
q₂ = –1.4 × 10⁻⁸
Substituting to the equation:
[tex]F =\frac{{kq_1q_2}}{r^2}[/tex]
=[tex]\frac{ { 9*10^{9} * 8* 10^{-9} *-1.4*10^{-8} }}{(5x10^-1)^2}[/tex]
= -4.032 x 10⁻⁶ N
Therefore,
-4.032 x 10⁻⁶ N is the approximate magnitude of the force between them.
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What frequency (in Hz) of sound at 20.1 degrees Celsius has a wavelength of 1.71 m?
The frequency of sound at the given temperature is 200.58 Hz.
Frequency of sound at the given temperatureThe frequency of sound at the given temperature is calculated as follows;
v = fλ
f = v/λ
where;
v is speed of sound at 20 ⁰C = 343 m/sλ is wavelength of the sound = 1.71 mf = 343/1.71
f = 200.58 Hz
Thus, the frequency of sound at the given temperature is 200.58 Hz.
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A hockey puck of mass 150 g is sliding south along the ice and slows at a rate of 1.2 m/s². What is the net force acting on the puck?
Answer:
0.18 N or 180 N
Explanation:
Mass,m = 150g
acceleration,a = 1.2m/s2
CHANGE FROM g TO kg
1kg =1000g
so 150g =150/1000 = 0.15kg
But Force, F = mass,m × acceleration,a
F=ma
F=0.15×1.2
F= 0.18 N
(b)) Suggest a reason why pail manufacturers prefer the shape shown to other shapes. 30 cm .... ( b ) ) Suggest a reason why pail manufacturers prefer the shape shown to other shapes . 30 cm ....
Manufacturers of pails prefer a cone shape because it has a suitable base as well as more space for its contents.
What is the most common shape of pails?The most common shape of pails is the shape of the frustum of a cone.
Pails are useful to contain liquids as well as some grainy solids like sand.
Because of their unique shapes, pails are able to rest on their bases and also has enough space to contain enough liquid as their top part spreads outwards.
In conclusion, the choice of shape of pails is to ensure they rest on the base and also have enough space for their contents.
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Three ropes A, B and C are tied together in one single knot K. (See figure.)
If the tension in rope A is 65.3 N, then what is the tension in rope B?
The tension in rope B is determined as 46.17 N.
Tension in rope A
Tension in rope A is the resultant tension of the rope B and C.
T(C) = T(B) = T
where;
T(C) is tension in rope CT(B) is tension in rope BT² + T² = 65.3²
2T² = 4,264.09
T² = 4,264.09/2
T² = 2,132.045
T = √2,132.045
T = 46.17 N
Thus, the tension in rope B is determined as 46.17 N.
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Find the components of vtot along the x and y axes in Figure 3.25, where = 23.0° and vtot = 7.56 m/s.
vtot, x =
m/s
vtot, y =
m/s
The component of the total velocity in the x - direction is 6.96 m/s.
The component of the total velocity in the y - direction is 2.95 m/s.
Component of the velocity in x direction
The component of the total velocity in the x - direction is calculated as follows;
v(x) = vtot cosθ
where;
vtot is total velocityv(x) is velocity in x directionv(x) = 7.56 x cos(23)
v(x) = 6.96 m/s
Component of the velocity in y - directionv(y) = vtot sinθ
v(y) = 7.56 x sin(23)
v(y) = 2.95 m/s
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A gas in a cylinder expands from a volume of 0.110 m³ to 0.320 m³. heat flows into the gas just rapidly enough to keep the pressure constant at 1.65×10⁵ during the expansion. the total heat added is 1.15×10⁵ j. find the change in internal energy of the gas.
80000 Joule is the change in the internal energy of the gas.
In Thermodynamics, work done by the gas during expansion at constant pressure:ΔW = -pdV
ΔW = -pd (V₂ -V₁)
ΔW = - 1.65×10⁵ pa (0.320m³ - 0.110m³)
= - 0.35×10⁵ pa.m³
= - 35000 (N/m³)(m³)
= -35000 Nm
ΔW = -35000 Joule
Therefore, work done by the system = -35000 Joule
Change in the internal energy of the gas,ΔV = ΔQ + ΔW
Given:
ΔQ = 1.15×10⁵ Joule
ΔW = -35000 Joule
ΔU = 1.15×10⁵ Joule - 35000 Joule
= 80000 Joule.
Therefore, the change in the internal energy of the gas= 80000 Joule.
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A big rock falls from a height towards water below, and hits water at a speed of 12 m/s. Rock was at a height of 15 m before falling and weighs 65 kg.
a) What was the average force of air resistance (e.g., friction) acting on the rock?
b) What is the force of friction underwater if the rock reaches a depth of 2.5 m before stopping and there is a buoyant force of 650 N [upward] acting on the rock once underwater.
In a resonance tube experiment, a tuning fork of frequency 256Hz gave a resonance when the water level was 30cm below the open end of the tube. If the next position of resonance was 100cm, what is the velocity of sound in air?
The velocity of sound in air is 358m/s
Calculations and ParametersGiven;
Frequency= 256 HzWater level= 30cmPosition of resonance= 100cmThe velocity of sound in air= ?v= 2f(L2-L1)
= 2.2.56.(1- 0.3)
=358 m/s
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Discuss how oxygen is used in spacecraft's air supplies in at least 3 paragraphs.
Short-duration spacecraft typically have one backup system and carry their own supply of oxygen. A large portion of the required oxygen is produced on long-duration missions, such as the International Space Station (ISS), which has been in orbit since 1998. Different sources provide the oxygen utilized on the ISS. The water electrolyzer is the primary source of metabolic oxygen. As an alternative to the electrolyzer, oxygen candles (also known as SFOGs) can produce metabolic oxygen. Additionally, oxygen is carried up whenever a cargo ship docks and stored in two tanks on the ISS Airlock. The electrolyzer electrolyzes water to create oxygen by running an electric current through it. Since water is a poor electrical conductor by itself, a little quantity of common salt is dissolved in the water to improve its electrical conductivity. Water is split into hydrogen and oxygen throughout the process.
We must keep in mind that oxygen by itself cannot be inhaled; it must be combined in the proper ratio with nitrogen to make it breathable. Two tanks aboard the ISS are used to store nitrogen, and the cargo ships that travel by from time to time also transport nitrogen cylinders. Through the electrical grid of the station, the solar panels on the station supply the necessary electricity for the oxygen generators. The majority of the required water is transported to the station by cargo supply ships. Condensers, which draw water vapor even from the station's air, ensure that not a drop of water is wasted. Using the proper equipment, water is also recycled from the astronauts' urine.
Through a suitable vent, the hydrogen gas produced during the electrolysis process is released into space. Pressurized tanks at the airlock nodes at the space station are pumped with oxygen when the cargo vehicles arrive there. Pressurized tanks there are also pumped with nitrogen. It goes without saying that the station's atmospheric controls combine the gases in the right amounts for the atmosphere of Earth and then distribute the combination throughout the cabin. The production of oxygen in space is impossible.
A 0.17 kg steel ball is tied to the end of a string and then whirled in a vertical circle at a constant speed v . The length of the string is 0.62 m , and the tension in the string when the ball is at the top of the circle is 4.0 N . What is v ?
What is the magnitude of the centripetal acceleration of the weight at this speed?
At the top of the circle, the net force on the ball is pointing downward, so that by Newton's second law
[tex]\sum F = -ma \implies F_{\rm tension} + F_{\rm weight} = \dfrac{mv^2}R[/tex]
where [tex]a=\frac{v^2}R[/tex] because the ball is undergoing circular motion with constant speed.
Plug in everything you know and solve for [tex]v[/tex].
[tex]4.0\,\mathrm N + (0.17\,\mathrm{kg}) g = \dfrac{(0.17\,\mathrm{kg})v^2}{0.62\,\rm m} \\\\ \implies v^2 = \dfrac{0.62\,\rm m}{0.17\,\rm kg} \left(4.0\,\mathrm N + (0.17\,\mathrm{kg}) g\right) \\\\ \implies v^2 \approx 21 \dfrac{\mathrm m^2}{\mathrm s^2} \implies \boxed{v \approx 4.5\dfrac{\rm m}{\rm s}}[/tex]
Plug this into the acceleration equation and solve for [tex]a[/tex].
[tex]a = \dfrac{21\frac{\mathrm m^2}{\mathrm s^2}}{0.62\,\rm m} \implies \boxed{a \approx 33\dfrac{\rm m}{\mathrm s^2}}[/tex]
1. Refer to the diagram. A magnet is moved away from a coil that is connected to a galvanometer. Explain how applying Lenz’s law helps determine the direction of the induced emf in the coil. What is the direction of the induced emf in the coil? Explain how the direction can be verified using the second right-hand rule.
The Lenz's law gives the direction of induced current. According to this law, the induced current opposes the cause that produces it.
What is Lenz law ?Lenz's law states that: The current induced in a circuit due to a change in a magnetic field is directed to oppose the change in flux and to exert a mechanical force which opposes the motion.
According to the question,
If we keep the magnet stationary and moved the coil back and forth within the magnetic field, an electric current would be induced in the coil.
Then by either moving the wire or changing the magnetic field we can induce a voltage and current within the coil and this process is known as Electromagnetic Induction.
When the coil AB is approaching towards north pole of the magnet, the magnetic flux linking to the coil increases.
Now according to Lenz's law,
This magnetic field created will oppose its own or we can say opposes the decrease in flux through the coil and this is possible only if approaching coil side attains south polarity, as we know dissimilar poles attract each other.
Once we know the magnetic polarity of the coil side, we can easily determine the direction of the induced current by applying right hand rule. In this case, the current flows in the A to B direction.
The right-hand rule. If you point your pointer finger in the direction the positive charge is moving, and then your middle finger in the direction of the magnetic field, your thumb points in the direction of the magnetic force pushing on the moving charge.
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Three resistors with resistances of R1=12 ohms (Ω), R2=8 ohms, and R3=24 ohms are in a series circuit with a 12-volt battery. (A) What is the total resistance of the resistors? (B) How much current can move through the circuit? (C) What is the current through each resistor?
The total resistance of the resistors is R=44Ω. And the current will be I= 0.27 A.
What is the function of a resistor?Resistors are electronic components whose main function is to limit the flow of electrical charges by converting electrical energy into thermal energy.
A)The resistances connected in series in an arm of a circuit, total resistance of that arm of is given by algebric sum individual resistances.
R = R1 + R2 + R3 = 12 + 8 + 24 = 44Ω
B) Ohm's law gives that:
V = IR
Where,
Voltage applied to circuit (V)current flowing through the circuit (I) total resistance of the circuit (R)(I) = V/R = 12 / 44 = 3/11 = 0.27 A
C) In series combination, all the resistances are connected one after other in a linear fashion. So, current flowing through first resistor passes to second, and then third and so on. Hence, the current flowing through all the resistance in series combination is same as the current flowing through that arm of circuit.
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The velocity of a 0.25kg model rocket changes from 15m/s [up] to 40m/s [up] in
0.60s. The gravitational field intensity is 9.8N/kg (acceleration due gravity). Calculate
the force the escaping gas exerts of the rocket.
a) 12.14 N [up]
b) 41. N [up]
c) 12.9 N [down]
d) 13 N [up]
The force the escaping gas exerts of the rocket is 10.42 N.
Force escaping gas exerts
The force the escaping gas exerts of the rocket is calculated as follows;
F = m(v - u)/t
where;
m is mass of the rocketv is the final velocity of the rocketu is the initial velocity of the rockett is time of motionF = (0.25)(40 - 15)/0.6
F = 10.42 N
Thus, the force the escaping gas exerts of the rocket is 10.42 N.
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exlapin the working of an electric bell
A point charge Q1 = +6.0 nC is at the point (0.30 m, 0.00 m); a charge Q2 = -1.0 nC is at (0.00 m, 0.10 m), and a charge Q3 = +5.0 nC is at (0.00 m, 0.00 m). What are the magnitude and direction of the net force on the +5.0-nC charge due to the other two charges? (k = 1/4πε0 = 9.0 × 109 N ∙ m2/C2)
Answer:
My Answer is Answer
Explanation:
We can use Coulomb's equation, which states that the force between two point charges is imparted: due to the other two charges ([tex]\rm Q_1[/tex] and [tex]\rm Q_2[/tex]) to obtain a net force on the +5.0 nC charge ([tex]\rm Q_3[/tex]).
[tex]\rm F = k * |Q_1 * Q_2| / r^2[/tex]
where:
F = force between the charges
k = Coulomb's constant ≈ 9.0 x [tex]\rm 10^9 N.m^2/C^2[/tex]
[tex]\rm Q_1[/tex] and [tex]\rm Q_2[/tex] = magnitudes of the charges
r = distance between the charges
Calculating the forces separately for each pair of charges and then find the net force:
Force between [tex]\rm Q_3[/tex] and [tex]\rm Q_1[/tex]:
[tex]\rm Q_3[/tex] = +5.0 nC = +5.0 x [tex]10^-^9[/tex] C
[tex]\rm Q_1[/tex] = +6.0 nC = +6.0 x [tex]10^-^9[/tex] C
[tex]\rm r_1[/tex] = distance between [tex]\rm Q_3[/tex] and [tex]\rm Q_1[/tex] = √[(0.00 m - 0.30 m[tex])^2[/tex] + (0.00 m - 0.00 m[tex])^2[/tex]] ≈ 0.30 m
[tex]\rm F_1[/tex]= k * |[tex]\rm Q_3[/tex] * [tex]\rm Q_1[/tex]| / [tex]\rm r_1^2[/tex]
[tex]\rm F_1[/tex] = (9.0 x [tex]10^9[/tex] N·[tex]\rm m^2/C^2[/tex]) * |(5.0 x [tex]10^-^9[/tex] C) * (6.0 x [tex]10^-^9[/tex] C)| / (0.30 [tex]\rm m)^2[/tex]
[tex]\rm F_1[/tex]≈ 6.0 N (towards the left)
Force between [tex]\rm Q_3[/tex] and [tex]\rm Q_2[/tex]:
[tex]\rm Q_3[/tex] = +5.0 nC = +5.0 x [tex]10^-^9[/tex] C
[tex]\rm Q_2[/tex] = -1.0 nC = -1.0 x [tex]10^-^9[/tex] C
r2 = distance between Q3 and Q2 = √[(0.00 m - 0.00 m)^2 + (0.00 m - 0.10 m)^2] ≈ 0.10 m
[tex]\rm F_2[/tex] =[tex]\rm k * |Q_3 * Q_2| / r2^2[/tex]
[tex]\rm F_2[/tex] = (9.0 x [tex]\rm 10^9 N.m^2/C^2[/tex]) * |(5.0 x [tex]10^-^9[/tex] C) * (-1.0 x [tex]10^-^9[/tex] C)| / (0.10 [tex]\rm m)^2[/tex]
[tex]\rm F_2[/tex] ≈ 45.0 N (towards the top)
Now we can sum the forces in the x and y directions to determine the net force:
Net force in the x-direction = [tex]\rm F_1[/tex] = 6.0 N (to the left)
Net force in the y-direction = [tex]\rm F_2[/tex] = 45.0 N (upwards)
The Pythagorean theorem can be used to determine the magnitude and direction of the net force.
Magnitude of net force = √(Net force in the x-direction[tex])^2[/tex]+ (Net force in the y-direction[tex])^2[/tex]
Magnitude of net force ≈ √(6.0 N[tex])^2[/tex] + (45.0 N[tex])^2[/tex]≈ √(36 [tex]\rm N^2[/tex] + 2025 [tex]\rm N^2[/tex]) ≈ √(2061 [tex]\rm N^2[/tex]) ≈ 45.4 N
The direction of the net force is the angle θ:
θ = arctan(Net force in the y-direction / Net force in the x-direction)
θ = arctan(45.0 N / 6.0 N) ≈ arctan(7.5) ≈ 82.9°
As a result, the net force exerted by the other two charges on the +5.0 nC charge has a strength of about 45.4 N and is directed about 82.9° above the negative x-axis (in the upper left direction).
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A crate with a mass of M = 62.5 kg is suspended by a rope from the endpoint of a uniform boom. The boom has a mass of m = 116 kg and a length of l = 7.65 m. The midpoint of the boom is supported by another rope which is horizontal and is attached to the wall as shown in the figure.
1. The boom makes an angle of θ = 57.7° with the vertical wall. Calculate the tension in the vertical rope.
2. What is the tension in the horizontal rope?
The boom makes an angle of θ = 60.2° with the vertical wall and the tension in the horizontal rope is mathematically given as
[tex]T_1=730.85 \mathrm{~N}[/tex]
[tex]T_1'=\frac{1980.51 \mathrm{~N}}{}[/tex]
What is the tension in the vertical rope.?Generally, the equation for is mathematically given as
1) Tension in vertical rope,
[tex]T_{1}=74.5 \times 9.81\\T_1=730.85 \mathrm{~N}[/tex]
2) Tension in horizontal rope,
[tex]\sum{Mg} =0\\Mg\frac{l}{2} \sin \theta+T_{1} \frac{l}{2} \sin \theta &=T_{2} l \cos \theta \\[/tex]
[tex]T_{2} &=\frac{M g(l / 2) \sin \theta+T_{1}(l / 2) \sin \theta}{l \cos \theta} \\\\&=\frac{M g \tan \theta}{2}+\frac{T_{1} \tan \theta}{2} \\\\T_{2} &=\frac{142 \times 9.81 \tan 61.8}{2}+\frac{730.85 \times \tan 61.8}{2} \\\\=& 1299+681.51[/tex]
[tex]T_1=\frac{1980.51 \mathrm{~N}}{}[/tex]
In conclusion, the tension in the horizontal rope is
[tex]T_1=\frac{1980.51 \mathrm{~N}}{}[/tex]
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In Bolt’s fastest 100 meter, he accelerated from the starting block to a speed of 27.8 mi/hr in 9.58 s. What was his acceleration?
O 9.58 mi/hr/s
O 3.60 mi/hr/s
O 2.90 mi/hr/s
O 2780 mi/hr/s
Answer: 2.90mi/hr/s
Explanation:
Acceleration = Final Velocity - Initial over time
Acceleration = 27.8mi/hr over 9.58s.
Answer is 2.90mi/hr/s
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Calculate the reversible work done in compressing 1 ft3 of mercury at a constant temp. of 0 o C from 1 atm to 3000 atm. The isothermal compressibility of mercury at 0 o C is : κ = 3.9 x 10-6 – 0.1 x 10-9 P , where P (atm) and κ (atm-1 ).
Compressibility of a substance is defined as the ratio of volumetric strain to the change in pressure. It is also equal to the reciprocal of the bulk modulus of the substance. It is denoted by K (kappa).
, for an isothermal process.
As we know that, work done for isothermal reversible expansion is given by-
Answer : C) Electrons are many thousand times smaller than the nucleus and negatively charged.
Reasoning : Edguinity
The mass of a proton and neutron and the positive charge of the proton accounts for why the electrons are thousands of times smaller than the nucleus and negatively charged.
What are electrons?Electrons are one of the three fundamental particles that make up an atom.
Electrons are negatively-charged particles which are found in certain regions of space around the nucleus of an atom at a particular instant.
The nucleus is composed of protons and neutrons. The protons are positively-charged while the neutrons are neutral. Thu, the negative charge of the electron balances the positive charge of the nucleus.
The mass of a proton is about 1840 times heavier than the electron. Thus, the nucleus is many thousand times heavier than the electron.
In conclusion, the negative charge of the electron ensures an atom is neutral.
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Magnets attract objects made up of _______;_______and _______
A device that uses electricity and magnetism to create motion is called a
Answer:
electric motors is the answer
An extended body (not shown in the figure) has its center of mass at the origin of the reference frame. In the case below give the direction for the torque τ with respect to the center of mass of the body due to force F acting on the body at a location indicated by the vector r.
Options are:
X
-Y
Z
-Z
The direction of torque τ this method is mathematically given as
D=X
Option A is correct.
What is the direction of torque?Generally, the equation for torque is mathematically given as
τ = r X F
Since F is the force exerted on the body at r, we need to know the direction of this torque with regard to the body's center of mass.
The dominant hand (the right) is used.Just run the fingers of our right hand along the line r.Our right palm should be placed on the letter F. Then, we'll gradually move r into F. The direction of the thumb's motion will indicate the torque's vector.In conclusion, the direction of this method is
D=X Option A.
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HELP ASAP PLEASE!!!
In the circuit, the resistance of C is 24-Ohms, the resistance of A is 4-Ohms and the resistance of B is 8-Ohms. If the current through resistor A is 2A, what is the current through the resistor C?
If resistor A, B and C are arranged in series, same current will flow in all the resistors and 2A will flow is resistor C.
What is series circuit?
This is the type of circuit arrangement in which the resistors are arranged in series order.
In this type of circuit arrangement, same current flows in every component of the circuit.
Thus, if resistor A, B and C are arranged in series, same current will flow in all the resistors and 2A will flow is resistor C.
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A 5.0 kg hammer strikes a 0.25 kg nail with a force of 10.0N, causing the nail to accelerate at 40.0m/s2. What is the acceleration of the hammer?
According to Newton's second law
F=maa=F/ma=10/5.25a=1.9m/s²Calculate the focal length of a spherical mirror whose radius of curvature is 25 cm
Answer:
The focal length is R/2 = 25 cm / 2 = 12.5 cm
A ray from the center of curvature is focused back on itself (R cm)
As the angle of the ray is changed to being parallel to the axis the ray from the mirror will be brought closer to the mirror
f(parallel) < focus (back upon itself)
compare angles of incidence and reflection
Calculate the work done by pulling an object through a distance of 30metres by a force of 220 newton
Answer:
6600J
Explanation:
the formula for work is: W= F x d
F- force, d- distance
mass is the amount of matter in an object what describes the amount of space the object takes up?
A.Color
B. Temperature
C. Volume
D. Texture
answer asap this is a exam
What do we call fixed point around which a lever pivots?
A. A fulcrum
B. A whell and axle
c. A wedge
D. A cylinder
Answer asap for beainlist