Answer:
[tex]p.d' = 37.6 V[/tex]
Explanation:
From the question we are told that:
Potential difference [tex]p.d=18.8V[/tex]
New Capacitor [tex]C_1=C_2/2[/tex]
Generally the equation for Capacitor capacitance is mathematically given by
[tex]C=\frac{eA}{d}[/tex]
Generally the equation for New p.d' is mathematically given by
[tex]C_2V=C_1*p.d'[/tex]
[tex]p.d' = 2V[/tex]
[tex]p.d'= 2 * 18.8[/tex]
[tex]p.d' = 37.6 V[/tex]
A T-shirt cannon launches a shirt at 5.30 m/s from a platform height of 4.00 m from ground level. How fast (in m/s) will the shirt be traveling if it is caught by someone whose hands are at 5.20 m from ground level (b) 4.00 m from ground level?
Answer:
(a) the velocity of the shirt is 2.14 m/s
(b) the velocity of the shirt is 5.3 m/s
Explanation:
Given;
initial velocity of the shirt, u = 5.3 m/s
height of the platform above the ground, h = 4.00 m
(a) When the shirt is caught by someone whose hand is 5.20 m from the ground level, the height traveled by the shirt = 5.2 m - 4.0 m = 1.2 m
The velocity at this position is calculated as;
[tex]v^2 = u^2 + 2(-g)h\\\\v^2 = u^2 - 2gh\\\\v^2 = 5.3^2 - (2\times 9.8 \times 1.2)\\\\v^2 = 4.57\\\\v= \sqrt{4.57} \\\\v = 2.14 \ m/s[/tex]
(b) When the shirt is caught by someone whose hand is 4.00 m from the ground level, the height traveled by the shirt = 4.00 m - 4.00 m = 0 m
The velocity at this position is calculated as;
[tex]v^2 = u^2 + 2(-g)h\\\\v^2 = u^2 - 2gh\\\\v^2 = 5.3^2 - (2\times 9.8 \times 0)\\\\v^2 = 28.09\\\\v= \sqrt{28.09} \\\\v = 5.3 \ m/s[/tex]
The elastic extensibility of a piece of string is .08. If the string is 100 cm long, how long will the string be when it is stretched to the point where it becomes plastic?
Answer:
The elastic extensibility of a piece of string is .08. If the string is 100 cm long, how long will the string be when it is stretched to the point where it becomes plastic? is your ansewer dont take tension
The string will be 108 cm long when it is stretched to the point where it becomes plastic.
What is elasticity?Elasticity in physics and materials science refers to a body's capacity to withstand a force that causes distortion and to recover its original dimensions once the force has been withdrawn.
When sufficient loads are applied, solid objects will deform; if the material is elastic, the object will return to its original size and shape after the weights have been removed. Unlike plasticity, which prevents this from happening and causes the item to stay deformed,
Given parameters:
The elastic extensibility of a piece of string is 0.08.
The string is 100 cm long.
Hence, it becomes plastic, after it is stretched up to = 100 × 0.08 cm = 8 cm. The string will be 108 cm long.
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A baseball pitcher brings his arm forward during a pitch, rotating the forearm about the elbow. If the velocity of the ball in the pitcher's hand is 34.0 m/s and the ball is 0.310 m from the elbow joint, what is the angular velocity (in rad/s) of the forearm
Answer:
[tex]\omega=109.67\ rad/s[/tex]
Explanation:
Given that,
The speed of the ball, u = 34 m/s
The ball is 0.310 m from the elbow joint.
We need to find the angular velocity (in rad/s) of the forearm.
We know that,
[tex]v=r\omega\\\\\omega=\dfrac{v}{r}\\\\\omega=\dfrac{34}{0.31}\\\\\omega=109.67\ rad/s[/tex]
So, the required angular velocity of the forearm is 109.67 rad/s.
A positive statement is:________. a. reflects oneâs opinions. b. can be shown to be correct or incorrect. c. a value judgment. d. based upon an optimistic judgment.
Answer:
b
Explanation:
At a distance of 14,000 km from the center of Planet Z-99, the acceleration due to gravity is 32 m/s2. What is the acceleration due to gravity at a point 28,000 km from the center of this planet
A body of mass m feels a gravitational force due to the planet of
F = GmM/R ² = ma
where
• G = 6.67 × 10⁻¹¹ N•m²/kg² is the universal gravitational constant
• M is the mass of the planet
• R is the distance between the body and the planet's center
• a is the acceleration due to gravity
Solving for a gives
a = GM/R ²
Notice that 28,000 km is twice 14,000 km. The equation says that the acceleration varies inversely with the square of the distance. So if R is changed to 2R, we have a new acceleration of
GM/(2R)² = 1/4 × GM/R ² = a/4
so the acceleration of the body at 28,000 km from the planet's center would be (32 m/s²)/4 = 8 m/s².
A solid conducting sphere of radius ra is placed concentrically inside a conducting spherical shell of inner radius rb1 and outer radius rb2. The inner sphere carries a charge Q while the outer sphere does not carry any net charge. The potential for rb1 < r < rb2 is:________
Answer:
The right answer is "[tex]\frac{KQ}{r_b_2}[/tex]".
Explanation:
As the outer spherical shell is conducting, so there is no electric field in side from
⇒ [tex]r_b_1 < r < r_b_2[/tex].
So the electric potential at all points inside the conducting shell that from
⇒ [tex]r_b_1<r<r_b_2[/tex]
and will be similar as well as equivalent to the potential on the outer surface of the shell that will be:
⇒ [tex]v=\frac{KQ}{r_b_2}[/tex]
Thus the above is the right solution.
What bet force is required to stop a 2250 kg car if the decelerates at a rate of -4.3 m/s^2 please answer fast
Answer:
Force = Mass × Acceleration
[tex]{ \tt{force = 2250 \times 4.3}} \\ = { \tt{9675 \: newtons}}[/tex]
A boy of mass 50 kg on a motor bike is moveny coith 20m/see what is hio k.E
Which quantities below of a solid object on this planet are NOT the same as on Earth?
Choose all
possible answers.
Weight
Mass
Volume
Density
Acceleration when it falls vertically.
Color
Answer:
Weight, acceleration when it falls vertically, are not same as that of earth.
Explanation:
Weight of the object is given by the product of mass of the object and the acceleration due to gravity of the planet.
So, the weight of object is not same as that on earth.
The mass is defined as the amount of matter contained in the object.
So, the mass of the object is same as that of earth.
The volume of the object is defined as the space occupied by the object.
So, the volume of the object is same as that of earth.
The density is defined as the ratio of mass of the object to its volume.
So, the density of the object is same as that of earth.
The acceleration due to gravity on a planet depends on the mass of planet and radius of planet.
So, the acceleration is not same as that of earth.
The color of the object is its characteristic.
It is same as that of earth.
A penny of mass 3.10 g rests on a small 20.0 g block supported by a spinning disk with radius of 12.0 cm. The coefficients of friction between block and disk are 0.850 (static) and 0.575 (kinetic) while those for the penny and block are 0.395 (kinetic) and 0.495 (static). What is the maximum rate of rotation in revolutions per minute that the disk can have, without the block or penny sliding on the disk
Answer:
do this Q yourself because i havent read the chapter
The maximum rate of rotation in revolutions per minute that the disk can have, without the block or penny sliding on the disk is 63 rpm.
How to solveThis is calculated using the coefficient of static friction between the penny and block, which is 0.495.
The maximum angular velocity of the disk is when the force of static friction is just sufficient to prevent the penny from sliding.
This force is equal to the mass of the penny multiplied by the acceleration due to gravity, multiplied by the coefficient of static friction.
The angular velocity of the disk is then calculated from this force and the radius of the disk.
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An object moves in a direction parallel to its length with a velocity that approaches the velocity of light. The length of this object, as measured by a stationary observer:________
a. approaches infinity.
b. approaches zero.
c. increases slightly.
d. does not change.
Answer:
b. approaches zero.
Explanation:
The phenomenon is known as length contraction.
Length contraction is a result of Einstein's special theory of relativity. This theory states that an observer in an inertial frame of reference will observe a decrease in the length of any moving object placed at another inertial frame of reference.
let the length of the train = L
Let the length observed when the train is in motion = L₀
Apply Einstein's special theory of relativity;
[tex]L_0 = L \times \sqrt{1 - \frac{v^2}{c^2} } \\\\where;\\\\v \ is \ the \ velocity \ of \ the \ train\\\\c \ is \ the \ speed \ of \ light\\\\[/tex]
from the equation above, when v = 0, the length observed is equal to the initial length of the train. (L₀ = L)
As the velocity of the train (v) approaches the speed of light (c), the length of the train observed (L₀) becomes smaller than the initial length of the train (L). (L₀ < L)
Eventually, when v equals c, we will have a square root of zero (0), and the length observed will become zero. (L₀ = 0)
Thus, the length of this object, as measured by a stationary observer approaches zero
Question 1 of 10
Which nucleus completes the following equation?
239UHe+?
A. 228 Th
B. 2220
c. 23. Pu
D. 78Th
SUBMIT
Answer:
Option D. ²²²₉₀Th
Explanation:
Let the unknown be ⁿₘZ. Thus, the equation becomes:
²²⁶₉₂U —> ⁴₂He + ⁿₘZ
Next, we shall determine n, m and Z. This can be obtained as follow:
For n:
226 = 4 + n
Collect like terms
226 – 4 = n
222 = n
n = 222
For m:
92 = 2 + m
Collect like terms
92 – 2 = m
90 = m
m = 90
For Z:
ⁿₘZ => ²²²₉₀Z => ²²²₉₀Th
Therefore, the complete equation becomes:
²²⁶₉₂U —> ⁴₂He + ⁿₘZ
²²⁶₉₂U —> ⁴₂He + ²²²₉₀Th
Thus, the unknown is ²²²₉₀Th
4. Which of the following statements best describes the relationship
between mechanical, kinetic and potential energies of an object of mass
m kg that is thrown vertically upwards with in initial velocity of v. m/s.
A. Kinetic energy increases while potential energy decrease and mechanical
energy remains constant.
B. Kinetic energy decreases, while potential energy increases and mechanical
energy remains constant.
C. Both kinetic and potential energies decrease while mechanical energy
increases.
D. Both kinetic and potential energies increase while mechanical energy
remains constant.
(1)
21
Answer:
D
Explanation:
increase while mechanical energy remains constant
friction between two flat surfaces can be divided into two categories. what are the two most common kinds of friction?
Answer:
kinetic and static
Explanation:
hope it helps! ^w^
Use the following information to answer the next question.
Environmental Concerns
1. release of greenhouse gases
2. release of gases that cause acid rain
3. release of excess heat
4. depletion of solar energy
5. depletion of geothermal energy
6. flooding of land
Which of the above environmental concerns are associated with the production of electricity?
Select one:
O A. 2, 3, and 4
O B. 1, 2, 3, and 6
O C. 1, 2, 3, 5 and 6
O D. 1, 3, and 5
Answer:
1.Emitted primarily through the burning of fossil fuels (oil, natural gas, and coal), solid waste, and trees and wood products. Changes in land use also play a role. Deforestation and soil degradation add carbon dioxide to the atmosphere, while forest regrowth takes it out of the atmosphere.
2.Acid rain is caused by a chemical reaction that begins when compounds like sulfur dioxide and nitrogen oxides are released into the air. These substances can rise very high into the atmosphere, where they mix and react with water, oxygen, and other chemicals to form more acidic pollutants, known as acid rain.
3.Untreated, heat exhaustion can lead to heatstroke, a life-threatening condition that occurs when your core body temperature reaches 104 F (40 C) or higher. Heatstroke requires immediate medical attention to prevent permanent damage to your brain and other vital organs that can result in death.
4.The loss of solar energy in passing through the atmospheric layers is called the atmospheric deflection. ... The longer the path traversed, the greater the amount of radiant energy depleted. Various processes whereby heat energy is lost through the atmosphere are known as scattering, diffusion, absorption, and reflection.
5.Geothermal energy is renewable because the Earth has retained a huge amount of the heat energy that was generated during formation of the planet. In addition, heat is continuously produced by decay of radioactive elements within the Earth. The amount of heat within the Earth, and the amount that is lost though natural processes (e.g. volcanic activity, conduction/radiation to the atmosphere), are much, much more than the amount of heat lost through geothermal energy production. At any one geothermal field, however, the temperature of the geothermal reservoir or the fluid levels/fluid pressure in the reservoir may decrease over time as fluids are produced and energy is extracted. Produced fluids can be re-injected to maintain pressures, although this may further cool down the reservoir if care is not taken. Over time, it is commonly necessary to drill additional wells in order to maintain energy production as temperatures and/or reservoir fluid pressures decline.
6.Floods, Floodplains, and Flood-Prone Areas. ... Flooding is a result of heavy or continuous rainfall exceeding the absorptive capacity of soil and the flow capacity of rivers, streams, and coastal areas. This causes a watercourse to overflow its banks onto adjacent lands.
The environmental concerns associated with the production of electricity are the release of greenhouse gases, the release of gases that cause acid rain, the release of excess heat, the flooding of land, and the depletion of geothermal energy so, option C is correct.
What is electricity?The presence or movement of charged particles is electricity. The movement of electrons through a circuit is known as an electric current. The accumulation of electrons on an insulator causes static electricity.
Mostly released when solid trash, trees, and wood products are burned, along with fossil fuels (coal, gas, and oil).
Land use changes also have an impact. Carbon dioxide is released into the atmosphere by deforestation and soil erosion, while it is removed from it by forest regeneration.
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A spherical conductor of radius = 1.5 cm with a charge of 3.9 pC is within a concentric hollow spherical conductor of inner radius = 3 cm, and outer radius = 4 cm, which has a total charge of 0 pC. What is the magnitude of the electric field 2.3 cm from the center of these conductors?
Answer:
The answer is "66.351 N/C"
Explanation:
Given:
[tex]a=1.5\ cm= 1.5 \times 10^{-2}\ m\\\\q_1=3.9\ pc\\\\b=3\ cm\\\\c= 4\ cm\\\\q_2=0 \ pc\\\\[/tex]
Using Gauss Law:
[tex]\oint \vec{E} \cdot \vex{dA}= \frac{Q_{enc}}{\varepsilon_0 }[/tex]
[tex]E \times 4 \pi\ r^2=\frac{Q_{enc}}{\varepsilon_0}\\\\E= \frac{Q_{enc}}{4 \pi\ r^2 \varepsilon_0}= \frac{1}{4 \pi \varepsilon_0} \frac{Q_{enc}}{r^2}= \frac{k_e\ Q_{enc}}{r^2}\\\\[/tex]
[tex]=\frac{9\times 10^{9} \times 3.9 \times 10^{-12}}{(2.3\times 10^{-2})^2}\\\\=\frac{35.1\times 10^{-3}\ }{(2.3\times 10^{-2})^2}\\\\=\frac{35.1\times 10^{-3}\ }{5.29 \times 10^{-4}}\\\\=\frac{35.1\times 10 }{5.29 }\\\\=\frac{351}{5.29 }\\\\=66.351\ \frac{N}{C}[/tex]
A Man has 5o kg mass man in the earth and find his weight
Answer:
49 N
Explanation:
Given,
Mass ( m ) = 50 kg
To find : Weight ( W ) = ?
Take the value of acceleration due to gravity as 9.8 m/s^2
Formula : -
W = mg
W = 50 x 9.8
W = 49 N
A spinning wheel having a mass of 20 kg and a diameter of 0.5 m is positioned to rotate about its vertical axis with a constant angular acceleration, a of 6 rad/s If the initial angular velocity is 1.5 rad/s, determine The maximum angular velocity and linear velocity of the wheel after 1 complete revolution.
Answer:
ωf = 8.8 rad/s
v = 2.2 m/s
Explanation:
We will use the third equation of motion to find the maximum angular velocity of the wheel:
[tex]2\alpha \theta = \omega_f^2 -\omega_I^2[/tex]
where,
α = angular acceleration = 6 rad/s²
θ = angular displacemnt = 1 rev = 2π rad
ωf = max. final angular velocity = ?
ωi = initial angular velocity = 1.5 rad/s
Therefore,
[tex]2(6\ rad/s^2)(2\pi\ rad)=\omega_f^2-(1.5\ rad/s)^2\\\omega_f^2=75.4\ rad/s^2+2.25\ rad/s^2\\\omega_f = \sqrt{77.65\ rad/s^2}[/tex]
ωf = 8.8 rad/s
Now, for linear velocity:
v = rω = (0.25 m)(8.8 rad/s)
v = 2.2 m/s
The patellar tendon attaches to the tibia at a 20 deg angle 3 cm from the axis of rotation at the knee. If the force generated in the patellar tendon is 400 N, what is the resulting angular acceleration, in rad/s2), if the lower leg and foot have a combined mass of 4.2kg and a given radius of gyration of 25 cm
Answer:
the resulting angular acceleration is 15.65 rad/s²
Explanation:
Given the data in the question;
force generated in the patellar tendon F = 400 N
patellar tendon attaches to the tibia at a 20° angle 3 cm( 0.03 m ) from the axis of rotation at the knee.
so Torque produced by the knee will be;
T = F × d⊥
T = 400 N × 0.03 m × sin( 20° )
T = 400 N × 0.03 m × 0.342
T = 4.104 N.m
Now, we determine the moment of inertia of the knee
I = mk²
given that; the lower leg and foot have a combined mass of 4.2kg and a given radius of gyration of 25 cm ( 0.25 m )
we substitute
I = 4.2 kg × ( 0.25 m )²
I = 4.2 kg × 0.0626 m²
I = 0.2625 kg.m²
So from the relation of Moment of inertia, Torque and angular acceleration;
T = I∝
we make angular acceleration ∝, subject of the formula
∝ = T / I
we substitute
∝ = 4.104 / 0.2625
∝ = 15.65 rad/s²
Therefore, the resulting angular acceleration is 15.65 rad/s²
B.F.Skinner emphesized the importance of-----?
Answer:
BFSkinner enfatizó la importancia de creía en la importancia de desarrollar la psicología experimental y dejar atrás el psicoanálisis y las teorías acerca de la mente basadas en el simple sentido común.
Explanation:
The acceleration vector of a particle in uniform circular motion:___________
a) points outward from the center of the circle.
b) points toward the center of the circle.
c) is zero.
d) points along the circular path of the particle and opposite the direction of motion.
e) points along the circular path of the particle and in the direction of motion.
(B)
Explanation:
Centripetal means "towards the center" so the acceleration vector of an object undergoing UCM is always pointed towards the center.
The acceleration vector of a particle in a uniform circular motion points toward the center of the circle, The correct option is option (b).
Centripetal force is the force acting on an object in curvilinear motion directed towards the axis of rotation or center of curvature. The unit of centripetal force is Newton.
Centripetal means "towards the center" so the acceleration vector of an object undergoing circular motion is always pointed towards the center.
Therefore, The acceleration vector of a particle in a uniform circular motion points toward the center of the circle, The correct option is option (b).
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the product 17.10 ✕
Explanation:
pls write the full question
An electric field E⃗ =5.00×105ı^N/C causes the point charge in the figure to hang at an angle. What is θ?
We have that the angle is
[tex]\theta=32.53[/tex]
From the Question we are told that
E⃗ =5.00×105ı^N/C
Generally the equation for Tension is mathematically given
[tex]W=Tcos\theta[/tex]
Where
[tex]tan\theta=\frac{2.5*10^{-9}(5*10{5})}{2*10^{-3}(9.8)}[/tex]
[tex]\theta=32.53[/tex]
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A ten loop coil of area 0.23 m2 is in a 0.047 T uniform magnetic field oriented so that the maximum flux goes through the coil. The average emf induced in the coil is
Answer:
Explanation:
From the question we are told that:
Number of turns [tex]N=10[/tex]
Area [tex]a=0.23m^2[/tex]
Magnetic field [tex]B=0.947T[/tex]
Generally the equation for maximum flux is mathematically given by
[tex]\phi=NBa[/tex]
[tex]\phi=10*0.047*0.23[/tex]
[tex]\phi=0.1081wbi[/tex]
Therefore induced emf
[tex]e= \frac{d\phi}{dt}[/tex]
Since
[tex]t=0[/tex]
Therefore
[tex]e=0[/tex]
A 3.00-kg ball swings rapidly in a complete vertical circle of radius 2.00 m by a light string that is fixed at one end. The ball moves so fast that the string is always taut and perpendicular to the velocity of the ball. As the ball swings from its lowest point to its highest point Group of answer choices the work done on it by gravity is -118 J and the work done on it by the tension in the string is zero. the work done on it by gravity is -118 J and the work done on it by the tension in the string is 118 J. the work done on it by gravity and the work done on it by the tension in the string are both equal to -118 J. the work done on it by gravity is 118 J and the work done on it by the tension in the string is -118 J. the work done on it by gravity and the work done on it by the tension in the string are both equal to zero.
Answer:
The ball moves from lowest to highest point:
W = M g h = 3 * 9.8 * 4 = 118 J
This is work done "against" gravity so work done by gravity is -118 J
The tension of the string does no work because the tension does not
move thru any distance W = T * x = 0 because the length of the string is fixed.
A 12.0 g sample of gas occupies 19.2 L at STP. what is the of moles and molecular weight of this gas?
At STP, 1 mole of an ideal gas occupies a volume of about 22.4 L. So if n is the number of moles of this gas, then
n / (19.2 L) = (1 mole) / (22.4 L) ==> n = (19.2 L•mole) / (22.4 L) ≈ 0.857 mol
If the sample has a mass of 12.0 g, then its molecular weight is
(12.0 g) / n ≈ 14.0 g/mol
Do all substances conduct heat ?Why/ Why not ?
Answer:
no, all substances doesnot conduct heat
Answer:
No, all substances do not conduct heat easily because it depends on the nature of the substance. Some are good conductors of heat and some are bad. Therefore, it depends on their characteristics and their ability to conduct heat.
The bad conductors of heat are water, air, plastic, wood, etc.
Gold, Silver, Copper, Aluminium, Iron, etc. are good heat conductors as well as electrical conductors.
A wire carrying a 30.0-A current passes between the poles of a strong magnet that is perpendicular to its field and experiences a 2.15-N force on the 4.00 cm of wire in the field. What is the average field strenth?
Answer:
1.79 T
Explanation:
Applying,
F = BILsin∅................ Equation 1
Where F = Force, B = magnetic field, I = current flowing through the wire, L = length of the wire, ∅ = angle between the magntic field and the force
make B the subject of the equation
B = F/ILsin∅............. Equation 2
From the question,
Given: F = 2.15 N, I = 30 A, L = 4.00 cm = 0.04 m, ∅ = 90° (perpendicular to the field)
Substitute these values into equation 2
B = 2.15/(30×0.04×sin90°)
B = 2.15/1.2
B = 1.79 T
Hence the average field strength is 1.79 T
In a large chemical factory, a feed pipe carries a liquid at a speed of 5.5 m/s. A pump pushes the liquid along at a gauge pressure of 140,000 Pa. The liquid travels upward 6.0 m and enters a tank at a gauge pressure of 2,000 Pa. The diameter of the pipe remains constant. At what speed does the liquid enter the tank
Answer:
v₂ = 15.24 m / s
Explanation:
This is an exercise in fluid mechanics
Let's write Bernoulli's equation, where the subscript 1 is for the factory pipe and the subscript 2 is for the tank.
P₁ + ½ ρ v₁² + ρ g y₁ = P₂ + ½ ρ v₂² + ρ g y₂
They indicate the pressure in the factory P₁ = 140000 Pa, the velocity
v₁ = 5.5 m / s and the initial height is zero y₁ = 0
the tank is at a pressure of P2 = 2000 Pa and a height of y₂ = 6.0 m
P₁ -P₂ + ρ g (y₁ -y₂) + ½ ρ v₁² = ½ ρ v₂²
let's calculate
140,000 - 2000 + ρ 9.8 (0- 6) + ½ ρ 5.5² = ½ ρ v₂²
138000 - ρ 58.8 + ρ 15.125 = ½ ρ v2²
v₂² = 2 (138000 /ρ - 58.8 + 15.125)
v₂ = [tex]\sqrt{\frac{276000}{\rho } - 43.675 }[/tex]
In the exercise they do not indicate what type of liquid is being used, suppose it is water with
ρ = 1000 kg / m³
v₂ = [tex]\sqrt{\frac{276000}{1000} - 43.675}[/tex]
v₂ = 15.24 m / s
What is the approximate radius of an equipotential spherical surface of 30 V about a point charge of +15 nC if the potential at an infinite distance from the surface is zero?
Answer:
V = k Q / R potential at distance R for a charge Q
R = k Q / V
R = 9 * 10E9 * 15 * 10E-9 / 30 = 9 * 15 / 30 = 4.5 m
Note: Our equation says that if R if infinite then V must be zero.
