Answer:
C. the product of the power rating of the light bulb and the time that it remains lit.
Explanation:
The power rating of the light is bulb is defined as the energy supplied to the light bulb divided by the time the bulb is lit up. Therefore,
[tex]P = \frac{E}{t}\\\\E = Pt[/tex]
where,
E = Energy Supplied to the bulb = Energy stored in capacitor = ?
P = Power rating of the bulb
t = time the bulb is lit up
Hence, the correct option is:
C. the product of the power rating of the light bulb and the time that it remains lit.
why the change of the pressure and temperature affect the velocity of the sound
Air pressure has no effect at all in an ideal gas approximation. This is because pressure and density both contribute to sound velocity equally, and in an ideal gas the two effects cancel out, leaving only the effect of temperature. Sound usually travels more slowly with greater altitude, due to reduced temperature.
State what is meant by a gravitational potential at point A is -1·70 × 109 J kg-1.
Answer:
The energy stored in a body due to either it's position or change in shape is called gravitational potential energy.
Which object has the most thermal energy?
A. 2 kg of liquid oxygen at -225°C
B. 5 kg of liquid oxygen at -220°C
C. 2 kg of oxygen gas at 20°C
D. 5 kg of oxygen gas at 30°C
5 kg of liquid oxygen at -220°C will have the most thermal energy due to its higher mass and temperature.
What is thermal energy?Thermal energy is the total heat required to raise the entire mass of a given substance.
Q = mcΔθ
where;
Q is thermal energym is mass of the substancec is specific heat capacityΔθ is change in temperatureThus, 5 kg of liquid oxygen at -220°C will have the most thermal energy due to its higher mass and temperature.
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When a player's finger presses a guitar string down onto a fret, the length of the vibrating portion of the string is shortened, thereby increasing the string's fundamental frequency. The string's tension and mass per unit length remain unchanged.
If the unfingered length of the string is l=65cm, determine the positions x of the first six frets, if each fret raises the pitch of the fundamental by one musical note in comparison to the neighboring fret. On the equally tempered chromatic scale, the ratio of frequencies of neighboring notes is 21/12
x1=
x2=
x3=
x4=
x5=
x6=
Answer:
Explanation:
For frequencies n generated in a string , the expression is as follows
n = 1 /2L√ ( T/m )
n is fundamental frequency , T is tension in string , m is mass per unit length and L is length of string.
If T and m are constant , then
n x L = constant , hence n is inversely proportional to L or length of string.
Frequencies increase by 21/12 = 1.75 , length must decrease by 1 / 1.75 times
Initial length of string is 65 cm .
x1 = 65 x 1 / 1.75 = 37.14 cm
x2 = 37.14 x 1/ 1.75 = 21.22 cm
x3 = 21.22 x 1 / 1.75 = 12.12 cm
x4= 12.12 x 1 / 1.75 = 6.92 cm
x5 = 6.92 x 1 / 1.75 = 3.95 cm
x6 = 3.95 x 1 / 1.75 = 2.25 cm
How much work does a supermarket checkout attendant do on a can of soup he pushes 0.420 m horizontally with a force of 4.60 N? Express your answer in joules and kilocalories.
We know
[tex]\boxed{\sf Work\:Done=Force\times Displacement} [/tex]
[tex]\\ \rm\longmapsto Work\:done=0.420\times 4.60[/tex]
[tex]\\ \rm\longmapsto Work\:done=1.932J[/tex]
The average 8-18 year old spends how many hours per day average in front of a screen doing little physical activity
Nearly four hours every day, doing little to no physical activity.
what is the critical angle of light traveling from vegetable oil into water
56.1∘
Question: A glass is half-full of water, with a layer of vegetable oil (n = 1.47) floating on top. A ray of light traveling downward through the oil is incident on the water at an angle of 56.1∘ .
A glass is half-full of water, with a layer of vegetable oil (n ...https://study.com › academy › answer › a-glass-is-half-ful...
What is the resistance of a copper wire 200 ft long and
0.01 in. in diameter (T 20°C)?
Answer:
R = 20.21 ohms
Explanation:
Given that,
The length of the wire, l = 200 ft
The diameter of the wire, d = 0.01 in
Radius, r = 0.005 in
200 ft = 60.96 m
0.005 in = 0.000127 m
The resistivity of copper is, [tex]\rho=1.68\times 10^{-8}\ \Omega-m[/tex]
So, the resistance of a wire is given by :
[tex]R=\rho \dfrac{l}{A}\\\\R=\rho \dfrac{l}{\pi r^2}\\\\R=1.68\times 10^{-8}\times \dfrac{60.96 }{\pi \times (0.000127 )^2}\\\\R=20.21\ \Omega[/tex]
So, the resistance of the copper wire is 20.21 ohms.
If four students separately measure the density of a rock, and they all have very low percent
differences between their measurements, what can you say for certain about the accuracy of their
results?
Answer:
Their measured results are closer to the exact or true value. Hence, their measured value is considered to be more accurate.
Explanation:
Considering the situation described above, the accuracy of a measured value depicts how closely a measured value is to the accurate value.
Hence, since the students' measured values have very low percent differences, it shows the similarity of computations or estimates to the actual values, which in turn offers a smaller measurement error.
Therefore, their measured results are closer to the exact or true value, which implies that their measured value is considered to be more accurate.
A motorcycle of mass 160 kg accelerates from rest to 53 m/s in 9 seconds. Ignore air resistance. Assuming there's no slipping between the wheels and the pavement of the road.
Required:
a. What is the average horizontal component of the force that the road exerts on the wheels (total force on all two wheels, not the force on one wheel)?
b. How far does the motorcycle travel in 9 seconds?
c. In the point-particle analysis of this situation, what is the work done by this force?
d. For the real system, how much work is done by the force of the road on the wheels?
Answer:
a) [tex]F=940.8N[/tex]
b) [tex]S=234.14m[/tex]
c) [tex]W=2.2*10^5J[/tex]
d) [tex]W=0[/tex]
Explanation:
Mass [tex]m=160kg[/tex]
Velocity [tex]v=53m/s[/tex]
Time [tex]t=9seconds[/tex]
a)
Generally the Newton's equation for motion is mathematically given by
[tex]a=\frac{v}{t}[/tex]
[tex]a=\frac{53}{9}[/tex]
[tex]a=5.9m/s^2[/tex]
Therefore
F=ma
[tex]F=160*5.88[/tex]
[tex]F=940.8N[/tex]
b)
Generally the Newton's equation for motion is mathematically given by
[tex]S=0.5at^2[/tex]
[tex]S=0.5*5.9*9^2[/tex]
[tex]S=234.14m[/tex]
c)
Generally the Newton's equation for work done is mathematically given by
[tex]W=Fd[/tex]
[tex]W=940.8*238.14[/tex]
[tex]W=2.2*10^5J[/tex]
d)
Generally the Newton's equation for work done by the force of the road on the wheels is mathematically given by
[tex]W=Fdcos\theta[/tex]
[tex]W=0[/tex]
A 215 N sign is supported by two ropes. One rope pulls up and to the right 1=29.5∘ above the horizontal with a tension 1 , and the other rope pulls up and to the left 2=44.5∘ above the horizontal with a tension 2 , as shown in the figure. Find the tensions 1 and 2 .
The sign is held in equilibrium. Using Newton's second law, we set up the equations of the net forces acting on the sign in the horizontal and vertical directions:
∑ F (horizontal) = T₁ cos(29.5°) - T₂ cos(44.5°) = 0
(right is positive, left is negative)
∑ F (vertical) = T₁ sin(29.5°) + T₂ sin(44.5°) - 215 N = 0
(up is positive, down is negative)
Solve the system of equations. I use elimination here:
• Multiply the first equation by sin(29.5°) and the second by cos(29.5°):
sin(29.5°) (T₁ cos(29.5°) - T₂ cos(44.5°)) = 0
cos(29.5°) (T₁ sin(29.5°) + T₂ sin(44.5°) - 215 N) = 0
T₁ cos(29.5°) sin(29.5°) - T₂ cos(44.5°) sin(29.5°) = 0
T₁ cos(29.5°) sin(29.5°) + T₂ cos(29.5°) sin(44.5°) = (215 N) cos(29.5°)
• Subtract the first equation from the second to eliminate T₁ :
T₂ cos(29.5°) sin(44.5°) - (- T₂ cos(44.5°) sin(29.5°)) = (215 N) cos(29.5°)
• Solve for T₂ :
T₂ (cos(29.5°) sin(44.5°) + cos(44.5°) sin(29.5°)) = (215 N) cos(29.5°)
T₂ sin(74.0°) = (215 N) cos(29.5°)
… … … (using the fact that sin(x + y) = sin(x) cos(y) + cos(y) sin(x))
T₂ = (215 N) cos(29.5°) / sin(74.0°)
T₂ ≈ 195 N
• Solve for T₁ :
T₁ cos(29.5°) - T₂ cos(44.5°) = 0
T₁ cos(29.5°) = T₂ cos(44.5°)
T₁ = T₂ cos(44.5°) / cos(29.5°)
T₁ ≈ 160. N
Some copper wire has a resistance of 200 ohms at 20 degrees C . A current is then passed through the same wire and the temperature rises to 90 degrees C. Determine the resistance of the wire at 90 degrees correct to the nearest ohm assuming the coefficient of resistance is 0.004/degree C at 0 degrees
Answer:
256 ohms
Explanation:
Applying,
R = R'[1+α(T-T')]............. Equation 1
Where R = Final resistance of the wire, R' = Initial resistance of the wire, T = Final temperature, T' = Initial temperature, α = Temperature coefficient of resistance
From the question,
Given: R' = 200 ohms, T = 90 degrees, T' = 20 degrees, α = 0.004/degree
Substitute these values into equation 1
R = 200[1+0.004(90-20)]
R = 200[1+0.28]
R = 200(1.28)
R = 256 ohms
The resistance of the wire at 90 °C correct to the nearest ohm assuming the coefficient of resistance is 0.004 °C¯¹ is 256 ohm
Data obtained from the question Original resistance (R₁) = 200 ohmOriginal temperature (T₁) = 20 °C Coefficient of resistivity (α) = 0.004 °C¯¹New temperature (T₂) = 90 °C New resistance (R₂) =? How to determine the new resistanceα = R₂ – R₁ / R₁(T₂ – T₁)
0.004 = R₂ – 200 / 200(90 – 20)
0.004 = R₂ – 200 / 200(70)
0.004 = R₂ – 200 / 14000
Cross multiply
R₂ – 200 = 0.004 × 14000
R₂ – 200 = 56
Collect like terms
R₂ = 56 + 200
R₂ = 256 ohm
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Thermodynamic Processes: An ideal gas is compressed isothermally to one-third of its initial volume. The resulting pressure will be
Answer:
The resulting pressure is 3 times the initial pressure.
Explanation:
The equation of state for ideal gases is described below:
[tex]P\cdot V = n \cdot R_{u}\cdot T[/tex] (1)
Where:
[tex]P[/tex] - Pressure.
[tex]V[/tex] - Volume.
[tex]n[/tex] - Molar quantity, in moles.
[tex]R_{u}[/tex] - Ideal gas constant.
[tex]T[/tex] - Temperature.
Given that ideal gas is compressed isothermally, this is, temperature remains constant, pressure is increased and volume is decreased, then we can simplify (1) into the following relationship:
[tex]P_{1}\cdot V_{1} = P_{2}\cdot V_{2}[/tex] (2)
If we know that [tex]\frac{V_{2}}{V_{1}} = \frac{1}{3}[/tex], then the resulting pressure of the system is:
[tex]P_{2} = P_{1}\cdot \left(\frac{V_{1}}{V_{2}} \right)[/tex]
[tex]P_{2} = 3\cdot P_{1}[/tex]
The resulting pressure is 3 times the initial pressure.
When the gun fires a projectile with a mass of 0.040 kg and a speed of 380 m/s, what is the recoil velocity of the shotgun and arm–shoulder combination?
Complete question:
The recoil of a shotgun can be significant. Suppose a 3.6-kg shotgun is held tightly by an arm and shoulder with a combined mass of 15.0 kg. When the gun fires a projectile with a mass of 0.040 kg and a speed of 380 m/s, what is the recoil velocity of the shotgun and arm–shoulder combination?
Answer:
The recoil velocity of the shotgun and arm–shoulder combination is 1.013 m/s
Explanation:
Given;
combined mass of the shotgun and arm–shoulder, m₁ = 15 kg
mass of the projectile, m₂ = 0.04 kg
speed of the projectile, u₂ = 380 m/s
let the recoil velocity of the shotgun and arm–shoulder combination = u₁
Apply the principle of conservation of linear momentum;
m₁u₁ + m₂u₂ = 0
m₁u₁ = - m₂u₂
[tex]u_1 = -\frac{m_2u_2}{m_1} \\\\u_1 = - \frac{0.04\times 380}{15} \\\\u_1 =-1.013 \ m/s\\\\u_1 = 1.013 \ m/s \ \ \ in \ opposite \ direction[/tex]
Therefore, the recoil velocity of the shotgun and arm–shoulder combination is 1.013 m/s
What is your angular position 75 seconds after the wheel starts turning, measured counterclockwise from the top? Express your answer as an angle between 0∘ and 360∘. Express your answer in degrees.
Complete Question
A Ferris wheel on a California pier is 27 m high and rotates once every 32 seconds in the counterclockwise direction. When the wheel starts turning, you are at the very top.
What is your angular position 75 seconds after the wheel starts turning, measured counterclockwise from the top? Express your answer as an angle between 0∘ and 360∘. Express your answer in degrees.
Answer:
[tex]\phi=123.75[/tex]
Explanation:
From the question we are told that:
Height [tex]h=27m[/tex]
Period [tex]T=32sec[/tex]
Time [tex]t=75sec[/tex]
Generally the equation for angular velocity is mathematically given by
[tex]\omega=\frac{2 \pi}{T}[/tex]
[tex]\omega=\frac{2 \pi}{32}[/tex]
[tex]\omega=0.196rad/s[/tex]
Therefore
[tex]\theta=\omega t[/tex]
[tex]\theta=0.196rad/s*75sec[/tex]
[tex]\theta=843.75 \textdegree[/tex]
Therefore
[tex]\phi=\theta-2(360)[/tex]
[tex]\phi=123.75[/tex]
* 1a Average speed
Carl Lewis runs the 100 m sprint in about 10 s.
His average speed in units of m/s would be:
of
Answer:
Explanation:
[tex] \implies v_{av} = \dfrac{total \: displacement}{total \: time} [/tex]
[tex] \implies v_{av} = \dfrac{100}{10} [/tex]
[tex]\implies v_{av} =10 \: {ms}^{ - 1} [/tex]
A bullet is fired vertically upward a velocity of 80m/s to what height will the bullet rise above the point of projection
Answer:
The bullet will rise 320 meters above the point of projection.
Explanation:
Assuming that air friction is negligent we can use the kinematic equation:
[tex]v_{2} ^2=v_{1} ^2+2(-a)d\\0\frac{m^2}{s^2} =6400\frac{m^2}{s^2} +2(-10\frac{m}{s^2} )d\\-6400\frac{m^2}{s^2} =(-20\frac{m}{s^2}) d\\320m=d[/tex]
*acceleration is negative (-a) as it is acting in the opposite direction of the motion of the bullet.*
The bullet rises to a height of 3600 m if a bullet is fired upward with a velocity of 80 m/s.
Assume the air friction is negligible, the kinematic equation:
[tex]v_f^2 = v_i^2 +2(-a) d[/tex]
Where,
[tex]v_i^2[/tex] - iinitial velocity = 80 m/s
[tex]v_f^2[/tex]- final velocity = 0
[tex]d[/tex]- distance= ?
[tex]a[/tex]- gravitational acceleration = 9.8 m/s² = 10 m/s²
Put the values in the formula,
[tex]\begin {aligne} 0 = (80)^2 + 2 (10)^2 d\\\\d = \dfrac {6400}{ 200}\\\\d &= 3600 \rm \ m\end {aligne}\\[/tex]
Therefore, the bullet rises to a height of 3600 m if a bullet is fired upward with a velocity of 80 m/s.
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A boy at a football field practice kicked a 0.500-kg ball with a force of 100N. How fast will the ball move after reaching a distance of 7m?
Answer:
v(7) = 52.915 m/s
Explanation:
First, find the value for acceleration.
F = ma
100 = .5 * a
a = 200 m/s²
Next find the velocity at x = 7 using kinematic equations.
v² = v₀² + 2a(Δx)
v² = (0)² + 2(200)(7)
v = [tex]\sqrt{2800}[/tex]
v = 52.915 m/s
An capacitor consists of two large parallel plates of area A separated by a very small distance d. This capacitor is connected to a battery and charged until its plates carry charges Q and - Q, and then disconnected from the battery. If the separation between the plates is now doubled, the potential difference between the plates will
Answer:
Will be doubled.
Explanation:
For a capacitor of parallel plates of area A, separated by a distance d, such that the charges in the plates are Q and -Q, the capacitance is written as:
[tex]C = \frac{Q}{V} = e_0\frac{A}{d}[/tex]
where e₀ is a constant, the electric permittivity.
Now we can isolate V, the potential difference between the plates as:
[tex]V = \frac{Q}{e_0} *\frac{d}{A}[/tex]
Now, notice that the separation between the plates is in the numerator.
Thus, if we double the distance we will get a new potential difference V', such that:
[tex]V' = \frac{Q}{e_0} *\frac{2d}{A} = 2*( \frac{Q}{e_0} *\frac{d}{A}) = 2*V\\V' = 2*V[/tex]
So, if we double the distance between the plates, the potential difference will also be doubled.
a 2.00 kg object is moving east at 4.00 m/s when it collides with a 6 kg object that is initially at rest. after the collision the larger object moves east at 1 m/s. what is the final velocity of the smaller object after the collision
The final velocity of the smaller object is 1 m/s.
To calculate the final velocity of the smaller object, we use the formula below.
Formula:
mu+m'u' = mv+m'v'............. Equation 1Where:
m = mass of the bigger objectm' = mass of the smaller objectu = initial velocity of the bigger objectu' = initial velocity of the smaller objectv = final velocity of the bigger objectv' = final velocity of the smaller object.From the question,
m = 6 kgm' = 2 kgu = 0 m/s (at rest) u' = 4 m/sv = 1 m/sSubstitute these values into equation 1
6(0)+2(4) = 6(1)+2(v')8 = 6+2v'2v' = 8-62v' = 2v' = 1 m/sHence, The final velocity of the smaller object is 1 m/s.
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In a race, Usain Bolt accelerates at
1.99 m/s2 for the first 60.0 m, then
decelerates at -0.266 m/s2 for the final
40.0 m. How much time did the race take?
(Unit = s)
Answer:
65.87 s
Explanation:
For the first time,
Applying
v² = u²+2as.............. Equation 1
Where v = final velocity, u = initial velocity, a = acceleration, s = distance
From the question,
Given: u = 0 m/s (from rest), a = 1.99 m/s², s = 60 m
Substitute these values into equation 1
v² = 0²+2(1.99)(60)
v² = 238.8
v = √238.8
v = 15.45 m/s
Therefore, time taken for the first 60 m is
t = (v-u)/a............ Equation 2
t = (15.45-0)/1.99
t = 7.77 s
For the final 40 meter,
t = (v-u)/a
Given: v = 0 m/s(decelerates), u = 15.45 m/s, a = -0.266 m/s²
Substitute into the equation above
t = (0-15.45)/-0.266
t = 58.1 seconds
Hence total time taken to cover the distance
T = 7.77+58.1
T = 65.87 s
What magnitude point charges creates a 10000 N/C electric field at a distance of 0.5 m?
a. 877.2 nC
b. 287.7 nC
c. 277.8 nC
d. 872.7 nC
Answer:
c. 277.8 nC
Explanation:
applying,
E = kq/r²............. Equation 1
Where E = electric field intensity, q = charge, r = distance, k = coulomb's constant.
make q the subject of the equation
q = Er²/k............... Equation 2
From the question,
Given: E = 10000 N/C, r = 0.5 m
Constant: k = 9×10⁹ Nm²/C²
Substitute these values into equation 2
q = (10000×0.5²)/(9×10⁹)
q = 277.8×10⁹
q = 277.8 nC
Hence the right option is c. 277.8 nC
Estimate the average power of a water wave when it hits the chest of an adult standing in the water at the seashore. Assume that the amplitude of the wave is 0.56 m , the wavelength is 2.0 m , and the period is 3.4 s . Assume that the area of the chest is 0.14 m^2.
Answer:
[tex]P=45.2W[/tex]
Explanation:
From the question we are told that:
Amplitude [tex]A=0.56m[/tex]
Period [tex]T=3.4s[/tex]
Wavelength [tex]\lambda=2.0[/tex]
Area [tex]a=0.14m^2[/tex]
Generally the equation for Power is mathematically given by
[tex]Power = 2 \pi ^2 \rho a(\frac{\lambda}{T})(\frac{1}{T})*A[/tex]
[tex]P= 2 3.142^2 1025 0.14(\frac{2.0}{3.4})(\frac{1}{3.4})^2*0.56^2[/tex]
[tex]P=45.2W[/tex]
Question 8 of 20
What identifies your skills and interests to help you plan out your career
goals?
A. A self-examination
B. A self-assessment
O C. A self-help book
O D. A bit of self-knowledge
Answer:
c
Explanation: because i took it and got it correct
A uniform circular disk has a radius of 34 cm and a mass of 350 g. Its center is at the origin. Then a circular hole of radius 6.8 cm is cut out of it. The center of the hole is a distance 10.2 cm from the center of the disk. Find the moment of inertia of the modified disk about the origin.
Answer:
u can ask it to the person who give ot to u i dont no
when we jump on a concrete surface,the feet get injured.Why
Answer:
Explanation:
Bhjb
Explanation:
its because a concrete surface is a hard surface which doesn't absorb the energy of gravitation when we fall down so we get hurt more badly..
hope this helps
Help please. I don’t understand
(D)
Explanation:
That's the statement of the Pythagorean theorem.
[tex]c^2=a^2+b^2 \Rightarrow c = \sqrt{a^2+b^2}[/tex]
While standing at the edge of the roof of a building, a man throws a stone upward with an initial speed of 6.63 m/s. The stone subsequently falls to the ground, which is 14.5 m below the point where the stone leaves his hand.
At what speed does the stone impact the ground? Ignore air resistance and use =9.81 m/s2 for the acceleration due to gravity.
impact speed:
18.1151
m/s
How much time is the stone in the air?
elapsed time:
Answer:
Explanation:
You final velocity is correct but not to the correct number of significant digits. The actual answer should be 18.1 m/s. We will use that to find the total time the stone was in the air in the equation:
v = v₀ + at
18.1 = 6.63 + (-9.81)t and
11.5 = -9.81t so
t = 1.17 seconds.
We know this was how long the stone was in the air (as compared to the time that the stone took to reach its max height or some other height) because we used the velocity with which the stone hit the ground to find the total time the rock was in the air before it hit the ground going at that velocity.
hi can anyone pls answer this question. i will mark brainliest
Answer:
a.work done by man A is zero
D is best option
Explanation:
D none of them because as we know that to do work some distance should be cover with some load as both picture they are covering some distance with carrying some load so A and B option are absolutely wrong and remaining C they are not doing same amount of work because distance cover by them and load carrying by them is different so how can work be same so D is best option none of A B C is correct answer
in an experiment the following readings were observed volume of alcohol flowing per minute equals to 10 raise to power - 5 cube volume density of alcohol is 800 kg per metre cube length of cube is 0.5 m radius of tube is 0.05 cm height of alcohol is 60 cm calculate the coefficient of viscosity of alcohol
Answer:
The viscosity is 1.30 x 10^-3 deca poise.
Explanation:
Volume per minute, V = 10^-5 m^3
Volume per second, V = 1.67 x 10^-7 m^3
density, d = 800 kg/m^3
radius, r = 0.05 cm
Length, L = 0.5 m
Height, h = 60 cm
Pressure, P = h d g = 0.6 x 800 x 9.8 = 4704 Pa
Use the formula of rate of flow
[tex]V = \frac{\pi p r^4}{8\eta L}\\\\1.67\times 10^{-7}\times8\times \eta\times 0.5 = 3.14\times 4707\times (0.05\times 10^{-2})^4\\\\\eta = 1.38\times 10^{-3} deca poise[/tex]
