Determine whether the following statements are true and give an explanation or counterexample.(A) If the acceleration of an object remains constant, its velocity is constant.
(B) If the acceleration of object moving along a line is always 0, then its velocity is constant.
(C) It is impossible for the instantaneous velocity at all times a(D) A moving object can have negative acceleration and increasing speed.
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Answer:
Explanation:(A)if a body is accelerating then it's velocity can't be constant since an object is said to be accelerating if it is changing velocity (B)if the acceleration of an object moving along a line is 0 then it's velocity will be constant since there is no change in direction or speed(C)No.it is not possible for a moving body to have an instantaneous velocity at all times since instantaneous velocity is the velocity of a body at a certain instant of time..(D)Yes a moving object can have a negative acceleration and increasing speed,it can also have a positive acceleration with decreasing speed.
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A 13.0-Ω resistor, 13.5-mH inductor, and 50.0-µF capacitor are connected in series to a 55.0-V (rms) source having variable frequency. If the operating frequency is twice the resonance frequency, find the energy delivered to the circuit during one period.
Consider an astronomical telescope with a 48 centimeter focal-length objective lens and a 10 centimeter focal-length eyepiece. Approximately how many centimeters apart should the lenses be placed
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Two large parallel metal plates are 1.6 cm apart and have charges of equal magnitude but opposite signs on their facing surfaces. Take the potential of the negative plate to be zero. If the potential halfway between the plates is then +3.8 V, what is the electric field in the region between the plates?
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Answer:
Force that the output plunger applies to the car; F2 = 3888N
Explanation:
For a hydraulic device, the relationship between the force and the area using Pascal's principle is;
F1/A1 = F2/A2
Where;
F1 is force applied to the input piston
F2 is force that the output plunger applies to the car
A1 is Area of input piston
A2 is area of larger piston
We are given;
R2/R1 = 9
So,R2 = 9R1
F1 = 48N
Area of input piston;
A1 = π(R1)²
Area of output piston;
A2 = π(9R1)²
Since, (F1/A1) = (F2/A2)
Thus;
F1/(π(R1)²) = F2/(π(9R1)²)
If we simplify, π(R1)² will cancel out to give;
F1 = F2/9²
Thus;
F2 = 9² x F1
Plugging in 48N for F1, we have;
F2 = 9² x 48
F2 = 81 x 48
F2 = 3888N
Final answer:
Using the principle of Pascal's law and ignoring the height difference, the output force is found by the formula F2 = F1*(r2/r1)^2. Given F1 is 48N and r2/r1 is 9.0, the output force F2 equates to 3888N.
Explanation:
In the case of a hydraulic jack, the principle of Pascal's law is applied. According to this law, pressure applied at one point of the fluid is transmitted equally in all directions. Therefore, if we ignore the height difference between pistons, the pressure exerted on both pistons would be the same.
Pressure is equal to the force divided by the area, where area equals π times the radius squared (π*r^2). So, the pressure at the input piston (P1) is the force at the input piston (F1) divided by its area (A1): P1 = F1/A1, where A1 = π*(r1)^2.
For the output plunger(P2 = F2/A2), where F2 = force at the output plunger and A2 = π*(r2)^2. By equating the pressures (P1=P2) and simplifying, we find that F2 = F1*(r2/r1)^2, where r2/r1 is given as 9.0. So, the output force F2 would be 48N*(9.0)^2 = 3888N.
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Hope this helps!
Answer:
At a higher velocity.
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Answer:
The boiling point of Acetone is 329K (in 3 significant figures)
Explanation:
Boiling point of Acetone = 56°C = 56 + 273K = 329K (in 3 significant figures)
Answer: using the formula 0°C + 273.15 = 273.15K the boiling point in units of kelvin to significant figures is 329.15k.
Explanation: The boiling point of a substance ( acetone) is the temperature at which the vapour pressure of the liquid substance equals the pressure surrounding it. The boiling point of acetone serves as it's characteristic physical properties. This is measured in degree Celsius (°C ) which can be converted to units of Fahrenheit or kelvin. To convert degree Celsius to kelvin this formula is used: 0°C + 273.15 = 273.15K . Given that acetone has boiling point of 56°C,from the formula 0°C is substituted for 56°C. This gives us:
56°C + 273.15= 319.15k.
Also,measurements given in Kelvin will always be larger numbers than in Celsius and the Kelvin temperature scale does not use the degree (°) symbol because Kelvin is an absolute scale, based on absolute zero, while the zero on the Celsius scale is based on the properties of water. I hope this helps. Thanks
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Answer:
U/U₀ = 2
(factor of 2 i.e U = 2U₀)
Therefore, the energy stored in the capacitor is doubled when the plate separation is doubled while the capacitor has been disconnected
Explanation:
Energy stored in a capacitor can be expressed as;
U = 0.5CV^2 = Q^2/2C
And
C = ε₀ A/d
Where
C = capacitance
V = potential difference
Q = charge
A = Area of plates
d = distance between plates
So
U = Q^2/2C = dQ^2/2ε₀ A
The initial energy of the capacitor at d = d₀ is
U₀ = Q^2/2C = d₀Q^2/2ε₀ A ....1
When the plate separation is increased after the capacitor has been disconnected, the charge Q of the capacitor remain constant.
The final energy stored in the capacitor at d = 2d₀ is
U = 2d₀Q^2/2ε₀ A ...2
The factor U/U₀ can be derived by substituting equation 1 and 2
U/U₀ = (2d₀Q^2/2ε₀ A)/( d₀Q^2/2ε₀ A )
Simplifying we have;
U/U₀ = 2
U = 2U₀
Therefore, the energy stored in the capacitor is doubled when the plate separation is doubled while the capacitor has been disconnected.
Answers
Answer:
160 Hz , 240 Hz , 400 Hz
Explanation:
Given that
Frequency of forth harmonic is 320 Hz.
Lets take fundamental frequency = f₁
f₁=80 Hz
Frequency of first harmonic = f₂
f₂=2 f₁
f₂ =2 x 80 = 160 Hz
Frequency of second harmonic = f₃
f₃= 3 f₁=3 x 80 = 240 Hz
Frequency of fifth harmonic = f₅
f₅= 5 f₁= 5 x 80 = 400 Hz
Three frequencies are as follows
160 Hz , 240 Hz , 400 Hz
Final answer:
The resonant frequencies of a string depend on its length, tension, and linear mass density. For a string resonating in four loops at 320 Hz, three possible smaller frequencies could be 160 Hz, 106.7 Hz, and 80 Hz.
Explanation:
When a string resonates, it vibrates at certain frequencies called its resonant frequencies. The resonant frequencies of a string depend on factors such as its length, tension, and linear mass density. In this case, the string resonates in four loops at a frequency of 320 Hz.
Three other possible resonant frequencies at which the string could vibrate with smaller loops include:
- 160 Hz: This is half the frequency of the given resonant frequency, which means the string vibrates with twice the number of loops.
- 106.7 Hz: This is one third of the given resonant frequency, which means the string vibrates with three times the number of loops.
- 80 Hz: This is one fourth of the given resonant frequency, which means the string vibrates with four times the number of loops.
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Answer:
The wavelength of that tone in air at standard condition is 0.96 m.
Explanation:
Given that, a trombone can produce pitches ranging from 85 Hz to 660 Hz approximately. We need to find the wavelength of that tone in air when the trombone is producing a 357 Hz tone.
We know that the speed of sound in air is approximately 343 m/s. Speed of a wave is given by :
So, the wavelength of that tone in air at standard condition is 0.96 m. Hence, this is the required solution.