Answer:
Higher voltages pose huge dangers to humans with changes in resistance however, the use of hand lotion and rubber gloves help to reduce the amount of current flowing through the human. Notwithstanding higher currents can lead to death by stopping blood flow to the heart causing a heart attack. Best way to survive voltages above 120 volts is to use a rubber glove and avoid touching bare wires.
Explanation:
Other classes of gloves that can be used include: Class 1 gloves which can be used for current up to 7,500 volts of AC, Class 2 up to 17,000 volts AC, Class 3 up to 26,500 volts AC, and Class 4 up to 36,000 volts AC. However, cotton gloves can be used inside to absorb perspiration and to improve the comfort of the user.
You are trying to overhear a juicy conversation, but from your distance of 20.0 m, it sounds like only an average whisper of 20.0 dB. So you decide to move closer to give the conversation a sound level of 70.0 dB instead. How close should you come?
Given that,
Distance = 20.0 m
Average whisper = 20.0 dB
Sound level = 70.0 dB
We know that,
The minimum intensity is
[tex]I_{o}=10^{-12}\ W/m^2[/tex]
We need to calculate the sound intensity in the distance of 20 m
Using formula of sound intensity
[tex]dB=10\log(\dfrac{I_{a}}{I_{o}})[/tex]
Put the value into the formula
[tex]20=10\log(\dfrac{I_{a}}{10^{-12}})[/tex]
[tex]10^{2}=\dfrac{I_{a}}{10^{-12}}[/tex]
[tex]I_{a}=10^{-10}\ W/m^2[/tex]
If the conversation a sound level of 70.0 dB instead
We need to calculate the sound intensity
Using formula of sound intensity
[tex]dB=10\log(\dfrac{I_{b}}{I_{o}})[/tex]
Put the value into the formula
[tex]70=10\log(\dfrac{I_{a}}{10^{-12}})[/tex]
[tex]10^{7}=\dfrac{I_{b}}{10^{-12}}[/tex]
[tex]I_{b}=10^{-5}\ W/m^2[/tex]
We know that,
The intensity is inversely proportional with the square of the distance.
We need to calculate the distance
Using formula of intensity
[tex]\dfrac{I_{a}}{I_{b}}=\dfrac{R_{b}^2}{R_{a}^2}[/tex]
Put the value into the formula
[tex]\dfrac{10^{-10}}{10^{-5}}=\dfrac{R_{b}^2}{20^2}[/tex]
[tex]R_{b}^2=20^2\times\dfrac{10^{-10}}{10^{-5}}[/tex]
[tex]R_{b}=\sqrt{20^2\times\dfrac{10^{-10}}{10^{-5}}}[/tex]
[tex]R_{b}=0.063\ m[/tex]
Hence, The distance from the conversation should be 0.063 m.
The distance you should move to achieve a loudness of 70 dB is; 0.0633 m
We are given;
Average distance 1; R_1 = 20 m
Average whisper 1; dB1 = 20 dB
Average whisper 2; dB2 = 70 dB
Now, the formula for loudness in dB is;
dB = 10log(I/I_o)
Where;
I is the intensity of the sound
I_o is the minimum intensity that a human ear can detect = 10^(-12) W/m²
Thus, for dB1 = 20 dB;20 = 10log (I/10^(-12))
20/10 = log (I/10^(-12))
log (I/10^(-12)) = 2
10² = (I/10^(-12))
I = 10² × 10^(-12)
I_1 = 10^(-10) W/m²
Similarly, for dB2 = 70 dB;70 = 10log (I/10^(-12))
70/10 = log (I/10^(-12))
log (I/10^(-12)) = 7
10^(7) = (I/10^(-12))
I = 10^(7) × 10^(-12)
I_2 = 10^(-5) W/m²
The relationship between their intensities and distance is;
I_1/I_2 = (R_2/R_1)²
Where R_2 is the distance you should move to achieve a loudness of 70 dB.
Thus;
(10^(-10))/(10^(-5)) = R_2/20
(R_2)/20 = √(10^(-5))
(R_2)/20 = 0.00316227766
R_2 = 20 × 0.00316227766
R_2 = 0.0632 m
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A force of only 150 N can lift a 600 N sack of flour to a height of 0.50 m when using a lever as shown in the diagram below. a. Find the work done on the sack of flour (in J). b. Find the distance you must push with the 150 N force-on the left side (in m). c. Briefly explain the benefit of using a lever to lift a heavy object.
The disks you will be using in the lab have the following parameters. All disks are solid with outer radius R1 = 0 0632 m and inner radius R2 = 0 0079 m The masses of the disks are: MT starless steel = 1357kg, MBottom starless stee,ã¼1.344kg. Calculate the moments of inertia of the disks.
Answer:
MT disc I = 2,752 10-3 kg m²
MB disc I = 2,726 10⁻³ kg m²
Explanation:
The moment of inertia given by the expression
I = ∫ r² dm
for bodies with high symmetry it is tabulated
for a hollow disk it is
I = ½ M (R₁² + R₂²)
let's apply this equation to our case
disc MT = 1,357 kg
I = ½ 1,357 (0.0079² + 0.0632²)
I = 2,752 10-3 kg m²
disk MB = 1,344 kg
I = ½ 1,344 (0.0079² + 0.0632²)
I = 2,726 10⁻³ kg m²
A sled on ice moves in the ways described below. Friction is so small that it can be ignored. A person wearing spiked shoes standing on the ice can apply a force to the sled and push it along the ice. Choose the one force (A through G) which would keep the sled moving as described in each statement below. You may use a choice more than once or not at all but choose only one answer for each blank. If you think that none are correct, answer choice J.A. The force is toward the right and is decreasing in strength (magnitude).B. The force is toward the right and is increasing in strength (magnitude).C. The force is toward the right and is of constant strength (magnitude).D. The force is toward the left and is decreasing in strength (magnitude).E. The force is toward the left and is increasing in strength (magnitude).F. The force is toward the left and is of constant strength (magnitude).G. No applied force is needed.
Answer:
G. No applied force is needed.
Explanation:
Force is any motion which changes or intends to change the position of an object. When a force is applied to an object it causes acceleration to increase and velocity changes. The sled is moving at a constant velocity, there is no acceleration in the force therefore force is not applied nor needed.
Find the period of revolution for the planet Mercury, whose average distance from the Sun is 5.79 x 1010 m.
Answer:
T = 7.61*10^6 s
Explanation:
In order to calculate the Mercury's period. in its orbit around the sun, you take into account on the Kepler's law. You use the following formula:
[tex]T=\sqrt{\frac{4\pi^2r^3}{GM_s}}[/tex] (1)
T: period of Mercury
r: distance between Mercury and Sun
Ms: mass of the sun = 1.98*10^30 kg
G: Cavendish's constant = 6.674*10^-11 m^3 kg^-1 s^-2
You replace the values of all parameters in the equation (1):
[tex]T=\sqrt{\frac{4\pi^2(5.79*10^{10}m)^3}{(6.674*10^{-11}m^3kg^{-1}s^{-2})(1.98*10^{30}kg)}}\\\\T=7.61*10^6s*\frac{1h}{3600s}*\frac{1d}{24h}=88.13\ days[/tex]
The period of Mercury is 7.61*10^6 s, which is approximately 88.13 Earth's days
Rice has a bulk of 0.9 g/mL, but an individual grain of rice has a density of 1.1g/mL. Water has a density of 1.0g/mL. what would happen to a cup of rice when poured into a lot of water?
Since the grains are not connected together, it's every grain for himself when they're dropped into the water.
The density of each grain is greater than the density of water, so each grain on its own sinks to the bottom.
The reason why a CUP or a BAG of rice has less density is because the grains don't all fit together perfectly, and 15% or 20% of the volume in the cup or bag is air, not rice grains.
"Rice is the perfect snack, when you're hungry for 2000 of something."
Mitch Hedberg
A person is riding on a Ferris wheel of radius R. He starts at the lowest point of the wheel. When the wheel makes one complete revolution, is the net work done by the gravitational force positive, zero or negative? Do you need to know how the speed of the person changed before you can answer the question?
Answering the two questions in reverse order:
-- No. I don't need to know how the speed of the person changed before I can answer the question. I can answer it now.
-- The NET work done by the gravitational force is zero.
-- As the person and his girl-friend go up the first half of the wheel, the motor does positive work and gravity does negative work.
-- After they pass the peak at the top and come down the second half of the wheel, the motor does negative work and gravity does positive work, even though the couple may be interested in other things during that time.
-- The total work done by gravity in one complete revolution is zero.
-- The total work done by the motor in one complete revolution is only what it takes to pay back the energy robbed by friction and air resistance.
The work done by the gravitational force is zero.
Work Done:Work done by a conservative force is path independent. Which means it only depends on the initial and final position of the body. The gravitational force is a conservational force and the gravitational potential energy depends only upon the height of the body.
Let the lowest point of the body is at some height h, then the initial gravitational potential energy of the person is:
PE(initial) = mgh
The final position of the person is also at a height h, thus, the final gravitational potential energy :
PE(final) = mgh
According to the work-energy theorem:
work done = - change in potential energy
work done = -(mgh - mgh) = 0
Thus, the work done is zero in the given case.
Learn more about work done:
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The coefficient of static friction is usually
Answer:
Higher than the coefficient of kinetic friction.
Explanation:
Hope it helps u.. :)
Imagine two free electrons that collide elastically in an acidic solution where one electron was moving and the other electron was stationary. When the electrons separate the moving electron now has a velocity of 400 m/s and the stationary electron now has a velocity of 200 m/s. What was the initial kinetic energy of the moving electron
Answer: 9.1 × 10^-26 Joule
Explanation:
Since the collision is elastic. The kinetic energy will be conserved. That is, the sum of kinetic energy before collision will be the same as the sum of the
energy after collision.
Mass of an electron = 9.1 × 10^-31 kg
Given that the velocity of the moving electron = 400 m/s and the stationary electron now has a velocity = 200 m/s.
K.E = 1/2mv^2
Add the two kinetic energies
1/2mV1^2 + 1/2mV2^2
1/2m( V1^2 + V2^2 )
Since they both have common mass
Substitute m and the two velocities
1/2 × 9.1×10^-31( 400^2 + 200^2)
4.55×10^-31 ( 160000 + 40000 )
4.55×10^-31 × 200000
K.E = 9.1 × 10^-26 Joule
Therefore, the initial kinetic energy of the moving electron is 9.1×10^-26 J
lens is to be used to allow this eye to clearly focus on objects 25 cm in front of it. What should be the focal length of this lens
The nearpoint of an eye is 151 cm. A corrective lens is to be used to allow this eye to clearly focus on objects 25 cm in front of it. What should be the focal length of this lens?
Answer:
29.96cm
Explanation:
Using the corrective lens, the image should be formed at the front of the eye and be upright and virtual.
Now using the lens equation as follows;
[tex]\frac{1}{f} = \frac{1}{v} + \frac{1}{u}[/tex] -------------(i)
Where;
f = focal length of the lens
v = image distance as seen by the lens
u = object distance from the lens
From the question;
v = -151cm [-ve since the image formed is virtual]
u = 25cm
Rewrite equation (i) to have;
[tex]f = \frac{uv}{u+v}[/tex]
Substitute the values of v and u into the equation;
[tex]f = \frac{25*(-151)}{25-151}[/tex]
[tex]f = \frac{-3775}{-126}[/tex]
f = 29.96cm
The focal length should be 29.96cm
Ripple tanks are used to observe two-dimensional waves. What should be the name of the amplitude control for two-dimensional waves
Answer:
Amplitude
Explanation:
A ripple tank is a tank of liquid (mostly water) which is used to observe the behaviour of the propagation of wave. It is commonly used to describe the principles involved in the reflection, refraction and diffraction of waves.
The tank is filled at its bottom with liquid. A bar is located at one of its ends and this bar is connected to an electric motor which causes a vertical motion of the bar. When the frequency of the electric motor is changed, there is a corresponding change in the speed, amplitude and position of the oscillator. Two common controls of the ripple tank are;
i. the frequency control.
ii. the amplitude control.
For two dimensional waves, these controls are called Frequency and Amplitude respectively.
In-car temperature is controlled in a variable displacement compressor system by varying the capacity of the _______________, not by cycling the _______________ on and off.
Answer:
In-car temperature is controlled in a variable displacement compressor system by varying the capacity of the refrigeration not by cycling the compressor on and off.
Explanation:
A Variable Displacement Compressor (VDC) works in a separate and rather more efficient manner than the fixed displacement compressor (FDC). The VDC automatically regulates the cooling effect of the refrigerator by changing its pumping capacity by means of a "wobbling piston" system.
During times of peak load, the refrigeration pumping capacity is increased, delivering a large flow rate of refrigerant, thus giving a high refrigerating effect. The reverse happens during times of load refrigeration load.
The fixed compressor tries to vary the temperature of the car by simply switching off the compressor, and turning it on during times of low and peak loads.
The power lines are at a high potential relative to the ground, so there is an electric field between the power lines and the ground. To maximize the potential difference between one end of the fluorescent tube and the other, how should the tube be held
Answer:
The tube should be held vertically, perpendicular to the ground.
Explanation:
As the power lines of ground are equal, so its electrical field is perpendicular to the ground and the equipotential surface is cylindrical. Therefore, if we put the position fluorescent tube parallel to the ground so the both ends of the tube lie on the same equipotential surface and the difference is zero when its potential.
And the ends of the tube must be on separate equipotential surfaces to optimize potential. The surface near the power line has a greater potential value and the surface farther from the line has a lower potential value, so the tube must be placed perpendicular to the floor to maximize the potential difference.
A 75 kg ball carrier is running to the right at 6.5 m/s. An 80 kg defender is chasing the ball carrier running at 7.0 m/s. The defender catches the ball carrier in a completely inelastic collision. What is the final speed of the defender/ball carrier mass
Answer:
6.758V
Explanation:
The computation of the final speed of the defender/ball carrier mass is shown below:-
Data provided in the question
Measurement of ball = 75 kg
Right m/s = 6.5 m/s
Defender = 80 kg
Carrying running 7.0 m/s
Based on the above information, the final speed of the defender or ball carrier mass is
As we know that
Conservation of momentum is
= Ball in Kg × Right m/s + Defender in Kg × running m/s
= 75 × 6.5 + 80 × 7
= (75 + 80)V
Therefore, the Final speed = 6.758V
The mass of M1 = 12 Daltons and it has a speed of v1 = 200 m/s. The mass of M2 = 4 Daltons. What was the total momentum of the system consisting of both masses before the collision (in Dalton meters per second, assume positive to the right and negative to the left)?
Answer:
The total momentum is [tex]p__{T }} =(2400 -4 v_2) \ Dalton \cdot m/s[/tex]
Explanation:
The diagram illustration this system is shown on the first uploaded image (From physics animation)
From the question we are told that
The mass of the first object is [tex]M_1 = 12 \ Dalton[/tex]
The speed of the first mass is [tex]v_1 = 200 \ m/s[/tex]
The mass of the second object is [tex]M_2 = 4 \ Dalton[/tex]
The speed of the second object is assumed to be [tex]- v_2[/tex]
The total momentum of the system is the combined momentum of both object which is mathematically represented as
[tex]p__{T }} = M_1 v_1 + M_2 v_2[/tex]
substituting values
[tex]p__{T }} = 12 * 200 + 4 * (-v_2)[/tex]
[tex]p__{T }} =(2400 -4 v_2) \ Dalton \cdot m/s[/tex]
A 100 cm length of nichrome wire has a radius of 0.50 mm, a resistivity LaTeX: \rho_0ρ 0= 1.0 × 10-6 Ω ∙ m , and a temperature coefficient LaTeX: \alphaα = 0.4 × 10-3 (oC)-1. At T0 = 20 oC the wire carries current of 0.50 A. How much power does the wire dissipate at a temperature T = 350 oC? Assume the potential difference across the ends of the wire remains constant. Group of answer choices
Answer:
P₃₅₀ = 0.28 watt
Explanation:
First we find the resistance of the wire at 20°C:
R₀ = ρL/A
where,
ρ = resistivity = 1 x 10⁻⁶ Ωm
L = Length of wire = 100 cm = 1 m
A = cross-sectional area of wire = πr² = π(0.5 x 10⁻³ m)² = 0.785 x 10⁻⁶ m²
Therefore,
R₀ = (1 x 10⁻⁶ Ωm)(1 m)/(0.785 x 10⁻⁶ m²)
R₀ = 1.27 Ω
Now, from Ohm's Law:
V = I₀R₀
where,
V = Potential Difference = ?
I₀ = Current Passing at 20°C = 0.5 A
Therefore,
V = (0.5 A)(1.27 Ω)
V = 0.64 volts
Now, we need to find the resistance at 350°C:
R₃₅₀ = R₀(1 + αΔT)
where,
R₃₅₀ = Resistance at 350°C = ?
α = temperature coefficient of resistance = 0.4 x 10⁻³ °C⁻¹
ΔT = Difference in Temperature = 350°C - 20°C = 330°C
Therefore,
R₃₅₀ = (1.27 Ω)[1 + (0.4 x 10⁻³ °C⁻¹)(330°C)]
R₃₅₀ = 1.44 Ω
Now, for power at 350°C:
P₃₅₀ = VI₃₅₀
where,
P₃₅₀ = Power dissipation at 350°C = ?
V = constant potential difference = 0.64 volts
I₃₅₀ = Current at 350°C = V/R₃₅₀ (From Ohm's Law)
Therefore,
P₃₅₀ = V²/R₃₅₉
P₃₅₀ = (0.64 volts)²/(1.44 Ω)
P₃₅₀ = 0.28 watt
two 200 pound lead balls are separated by a distance 1m. both balls have the same positive charge q. what charge will produce an electrostatic force.between the balls that is of the same order of magnitude as the weight of one ball?
Answer:
The charge is [tex]q = 3.14 *10^{-4} \ C[/tex]
Explanation:
From the question we are told that
The mass of each ball is [tex]m = 200 \ lb = \frac{200}{2.205} = 90.70 \ kg[/tex]
The distance of separation is [tex]d = 1 \ m[/tex]
Generally the weight of the each ball is mathematically represented as
[tex]W = m * g[/tex]
where g is the acceleration due to gravity with a value [tex]g = 9.8 m/s^2[/tex]
substituting values
[tex]W = 90.70 * 9.8[/tex]
[tex]W = 889 \ N[/tex]
Generally the electrostatic force between this balls is mathematically represented as
[tex]F_e = \frac{k * q_1* q_2 }{d^2}[/tex]
given that the the charges are equal we have
[tex]q_1= q_2 = q[/tex]
So
[tex]F_e = \frac{k * q^2 }{d^2}[/tex]
Now from the question we are told to find the charge when the weight of one ball is equal to the electrostatic force
So we have
[tex]889 = \frac{9*10^9 * q^2}{1^2}[/tex]
=> [tex]q = 3.14 *10^{-4} \ C[/tex]
The magnitude of charge on the balls is [tex]3.14 \times 10^{-4} \;\rm C[/tex].
Given data:
The masses of two lead balls are, m = 200 lb = 200/2.205 = 90.70 kg.
The distance of separation of two balls is, d = 1 m.
First of all we need to obtain the weight of ball. The weight of the ball is expressed as,
W = mg
Here,
g is the gravitational acceleration.
Solving as,
W = 90.70 × 9.8
W = 888.86 N
The expression for the electrostatic force between this balls is mathematically represented as,
[tex]F = \dfrac{k \times q_{1} \times q_{2}}{d^{2}}[/tex]
Since, the charges are equal then,
[tex]q_{1} =q_{2}=q[/tex]
Also, the magnitude of force between the balls is same as the weight of one ball. Then,
F = W
Solving as,
[tex]F =W= \dfrac{(9 \times 10^{9}) \times q^{2}}{1^{2}}\\\\889= \dfrac{(9 \times 10^{9}) \times q^{2}}{1^{2}}\\\\q = 3.14 \times 10^{-4} \;\rm C[/tex]
Thus, we can conclude that the magnitude of charge on the balls is [tex]3.14 \times 10^{-4} \;\rm C[/tex].
Learn more about the Coulomb's law here:
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1. In Newton’s ring experiment, the diameter of the 5th ring is 0.30 cm and diameter of 15th the ring is 0.62 cm. Find the diameter of the 25th ring.
Answer:
Diameter of Newton’s 5th ring = 0.30 cm
Diameter of Newton’s 15th ring = 0.62 cm
Diameter of Newton’s 25th ring = ?
From Newton’s rings experiment we infer that
D2n+m − D2n = 4λmR
For the 5th and 15th rings we have
D215 − D25 = 4λ * 10 * R _______ (1) (m = 10)
For 15th and 25th rings
D225 − D215 = 4λ * 10 * R _______ (2) (m = 10)
We equate the two derivatives
Equation (2) = Equation (1)
D225 − D215 = D215 − D25
D225 = 2D215 – D25
Substituting the values into the equation
D225 = 2 * 0.62 * 0.62 – 0.3 * 0.3 =0.6788 cm2
D25 = 0.8239 cm
Diameter of [tex]25^{th}[/tex] Newton Ring = 0.97 cm
Newton Rings is an experiment based on principle of thin film interference
In Newton Rings Experiment the Diameter of [tex]n^{th}[/tex] dark ring is given by equation (1)
[tex]\rm D_n= 2\sqrt{n\lambda R} ......(1)\\where \; \\D_n = Diameter\; of \; n^{th} \; dark \; ring }\\n = Number \; of \; Ring\\\lambda = Wavelength \\R = Radius \; of \; Curvature \; of\; the \; lens[/tex]
From the condition given
[tex]\rm D_5 = 0.3 \; cm \\D_{15} = 0.62 \; cm\\\\D_{25} = To \; be \; determined \\[/tex]
Putting the values in equation (1) for fifth diameter we get
[tex]\rm D_5 = 0.3=2\sqrt{5\lambda R}.......(2) \\D_{15} = 0.62 = 2\sqrt{15\lambda R}.......(3) \\\\Equation \; (3) - Equation (2) \\\\0.32 = 2\sqrt{\lambda R} ( \sqrt{15} -\sqrt{5})\\\\2\sqrt{\lambda R } = 0.1954....(4)\\[/tex]
So From equation (1) and (4)
[tex]\rm Diameter \; of \; 25^{th} Ring =D_{25} = 2\sqrt{\lambda R } \times \sqrt{25} \\\\D_{25} = 0.1954\times 5 \\\\D_{25} = 0.97 \; cm[/tex]
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If you put a total of 8.05×106×106 electrons on an intially electrically neutral wire of length 1.03 m, what is the magnitude of the electric field a perpendicular distance of 0.201 m away from the center of the wire?
Answer:
The magnitude of the electric field is 0.1108 N/C
Explanation:
Given;
number of electrons, e = 8.05 x 10⁶
length of the wire, L = 1.03 m
distance of the field from the center of the wire, r = 0.201 m
Charge of the electron;
Q = (1.602 x 10⁻¹⁹ C/e) x (8.05 x 10⁶ e)
Q = 1.2896 x 10⁻¹² C
Linear charge density;
λ = Q / L
λ = (1.2896 x 10⁻¹² C) / (1.03 m)
λ = 1.252 x 10⁻¹² C/m
The magnitude of electric field at r = 0.201 m;
[tex]E = (\frac{1}{4 \pi \epsilon_o} )\frac{ 2 \lambda}{r} \\\\E = k \frac{ 2 \lambda}{r}\\\\E = (8.89*10^9)*\frac{2*1.252*10^{-12}}{0.201} \\\\E = 0.1108 \ N/C[/tex]
Therefore, the magnitude of the electric field is 0.1108 N/C
Consider a loop of wire placed in a uniform magnetic field. Which factors affect the magnetic flux Φm through the loop?
Answer:
* The value of the magnetic field changes either in time or space
* The waxed area changes, the bow is fitting in size
* The angle between the field and the area changes
Explanation:
Magnetic flux is the scalar product of the magnetic field over the area
Ф = ∫ B. dA
where B is the magnetic field and A is the area
Let's look at stationary, for which factors affect flow
* The value of the magnetic field changes either in time or space
* The waxed area changes, the bow is fitting in size
* The angle between the field and the area changes
If you were to experimentally determine the length of the pendulum, why would you not get the same length in Iowa?
Answer:
The length of the pendulum depends on acceleration due to gravity (g) which varies in different Earth's location beacuse Earth is not perfectly spherical.
Explanation:
The period of oscillation is calculated as;
[tex]T = 2\pi\sqrt{\frac{l}{g} }[/tex]
where;
L is the length of the pendulum bob
g is acceleration due to gravity
If we make L the subject of the formula in the equation above, we will have;
[tex]T = 2\pi\sqrt{\frac{l}{g}}\\\\\sqrt{\frac{l}{g} } = \frac{T}{2\pi} \\\\\frac{l}{g} = (\frac{T}{2\pi} \)^2\\\\\frac{l}{g} =\frac{T^2}{4\pi^2}\\\\L = \frac{gT^2}{4\pi^2}[/tex]
The length of the pendulum depends on acceleration due to gravity (g).
Acceleration due to gravity is often assumed to be the same everywhere on Earth, but it varies because Earth is not perfectly spherical. The variation of acceleration due to gravity (g) as a result of Earth's geometry, will also cause the length of the pendulum to vary.
A uniform electric field stack \rightarrow Ei subscript I with rightwards arrow on top is present in the region between infinite parallel plane plates A and B and a uniform electric field stack\rightarrow Eii subscript I I end subscript with rightwards arrow on top is present in the region between infinite parallel plane plates B and C. When the plates are vertical, stack \rightarrow Ei subscript I with rightwards arrow on top is directed to the right and stack \rightarrow Eii subscript I I end subscript with rightwards arrow on top to the left. The signs of the charges on plates A, B and C may be
a. ?, ?, ?.
b. +, ?, ?.
c. +, ?, +.
d. +, +, +.
e. any one of the above
Answer:
e. any one of the above
Explanation:
What are the two types of long-term memory?
A. sensory and short-term
B. iconic and echoic
C. explicit and implicit
D. recency and primacy
Answer:
C. explicit and implicit
Explanation:
E20
I have a question regarding friction in rolling without slipping.
There is a disk that is being acted by a force at an angle. I am asked to find the acceleration of center of mass and angular acceleration. The problem specifies, “there is friction between the disk and the ground, but the disk rolls without slipping”.
What does that mean? Does it mean I can ignore friction and use F as the only force producing torque? Does friction produce torque in this case?
I’ll leave a picture here:
Explanation:
They probably put "rolls without slipping" in there to indicate that there is no loss in friction; or that the friction is constant throughout the movement of the disk. So it's more of a contingency part of the explanation of the problem.
(Remember how earlier on in Physics lessons, we see "ignore friction" written into problems; it just removes the "What about [ ]?" question for anyone who might ask.)
In this case, you can't ignore friction because the disk wouldn't roll without it.
As far as friction producing a torque... I would say that friction is a result of the torque in this case. And because the point of contact is, presumably, the ground, the friction is tangential to the disk. Meaning the friction is linear and has no angular component.
(You could probably argue that by Newton's 3rd Law there should be some opposing torque, but I think that's outside of the scope of this problem.)
Hopefully this helps clear up the misunderstanding for you.
A loaded ore car has a mass of 950 kg and rolls on rails with negligible friction. It starts from rest and is pulled up a mine shaft by a cable connected to a winch. The shaft is inclined at 28.0° above the horizontal. The car accelerates uniformly to a speed of 2.35 m/s in 14.0 s and then continues at constant speed.(A) What power must the winch motor provide when the car is moving at constant speed? kW(B) What maximum power must the motor provide? kW(C) What total energy transfers out of the motor by work by the time the car moves off the end of the track, which is of length 1,250 m?
Answer:
a) P = 10.27 kW
b) Pmax = 10.65 kW
c) E = 5.47 MJ
Explanation:
Mass of the loaded car, m = 950 kg
Angle of inclination of the shaft, θ = 28°
Acceleration due to gravity, g = 9.8 m/s²
The speed of the car, v = 2.35 m/s
Change in time, t = 14.0 s
a) The power that must be provided by the winch motor when the car is moving at constant speed.
P = Fv
The force exerted by the motor, F = mg sinθ
P = mgv sinθ
P = 950 * 9.8 *2.35* sin28°
P = 10,271.3 W
P = 10.27 kW
b) Maximum power that the motor must provide:
[tex]P = mv\frac{dv}{dt} + mgvsin \theta\\dv/dt = \frac{2.35 - 0}{14} \\dv/dt = 0.168 m/s^2\\P = (950*2.35*0.168) + (950*9.8*2.35* sin28)\\P = 374.74 + 10271.3\\P = 10646.04 W\\10.65 kW[/tex]
c) Total energy transferred:
Length of the track, d = 1250 m
[tex]E = 0.5 mv^2 + mgd sin \theta\\E = (0.5 * 950 * 2.35^2) + (950 * 9.8 * 1250 * sin 28)\\E = 2623.19 + 5463475.31\\E = 5466098.50 J\\E = 5.47 MJ[/tex]
In an RC circuit, how many time constants must elapse if an initially uncharged capacitor is to reach 80% of its final potential difference
Answer:
1.6 time constants must elapse
Explanation:
voltage on a cap, charging is given as
v = v₀[1–e^(–t/τ)]
Where R is resistance in ohms,
C is capacitance in farads
t is time in seconds
RC = τ = time constant
v = v₀[1–e^(–t/τ)]
1–e^(–t/τ) = 0.8
e^(–t/τ) = 0.2
–t/τ = –1.609
t = 1.609τ
A ball is shot at an angle of 45 degrees into the air with initial velocity of 46 ft/sec. Assuming no air resistance, how high does it go
Answer:
5.02 m
Explanation:
Applying the formula of maximum height of a projectile,
H = U²sin²Ф/2g...................... Equation 1
Where H = maximum height, U = initial velocity, Ф = angle, g = acceleration due to gravity.
Given: U = 46 ft/sec = 14.021 m/s, Ф = 45°
Constant: g = 9.8 m/s²
Substitute these values into equation 1
H = (14.021)²sin²45/(2×9.8)
H = 196.5884×0.5/19.6
H = 5.02 m.
Hence the ball goes 5.02 m high
The ball reaches the maximum height of 54 feet
The question is about projectile motion,
the ball is shot at an angle α = 45°, and
the initial velocity u = 46 ft/s.
Under the projectile motion, the maximum height H is given by:
[tex]H=\frac{u^2sin^2\alpha }{2g} [/tex]
where, g = 9.8 m/s²
substituting the given values we get:
[tex]H=\frac{46^2sin^{2}(45)}{2*9.8}\\ \\ H=\frac{46*46*(1/2)}{2*9.8}\\ \\ H=54 feet[/tex]
Hence, the maximum height is 54 feet.
Learn more:
https://brainly.com/question/2818076?referrer=searchResults
The New England Merchants Bank Building in Boston is 152 m high. On windy days it sways with a frequency of 0.17 Hz, and the acceleration of the top of the building can reach 2.0% of the free-fall acceleration, enough to cause discomfort for occupants. What is the total distance, side to side, that the top of the building moves during such an oscillation?
Answer:
d = 8.4 cm
Explanation:
In order to calculate the amplitude of oscillation of the top of the building, you use the following formula for the max acceleration of as simple harmonic motion:
[tex]a_{max}=A\omega^2[/tex] (1)
A: amplitude of the oscillation
w: angular speed of the oscillation = 2[tex]\pi[/tex]f
f: frequency = 0.17Hz
The maximum acceleration of the top of the building is a 2.0% of the free-fall acceleration. Then, you have:
[tex]a_{max}=0.02(9.8m/s^2)=0.196\frac{m}{s^2}[/tex]
Then, you solve for A in the equation (1) and replace the values of the parameters:
[tex]A=\frac{a_{max}}{\omega^2}=\frac{a_{max}}{4\p^2i f^2}\\\\A=\frac{0.196m/s^2}{16\pi^2(0.17Hz)^2}\\\\A=0.042m=4.2cm[/tex]
The total distance, side to side, of the oscilation of the top of the building is twice the amplitude A. Then you obtain:
d = 2A = 2(4.2cm) = 8.4cm
The total side to side distance at the top of the building, covered during such an oscillation is 8.4 cm.
Given data:
The height of building is, h = 152 m.
The frequency on windy days is, f = 0.17 Hz.
The acceleration on the top of building is, a = 2/100g (Here g is gravitational acceleration).
This problem can be resolved using the concept of amplitude and angular frequency. The expression for the magnitude of acceleration at the top pf building is given as,
[tex]a = A \times \omega^{2}\\\\a = A \times (2 \pi f)^{2}\\\\\dfrac{2}{100} \times g=A \times (2 \pi \times 0.17)^{2}\\\\\dfrac{2}{100} \times 9.8=A \times (2 \pi \times 0.17)^{2}\\\\A =0.042 \;\rm m[/tex]
And, the total distance, side to side, of the oscillation of the top of the building is twice the amplitude A. Then,
]s = 2A
s = 2 (0.042)
s = 0.084 m
s = 8.4 cm
Thus, we can conclude that the total side to side distance at the top of the building, covered during such an oscillation is 8.4 cm.
Learn more about the oscillatory motion here:
https://brainly.com/question/18832329
You’re standing at the highest point on the Moon, 10,786 mm above the level of the Moon’s mean radius. You’ve got a golf club and a golf ball.
Part A
(How fast would you need to hit the ball horizontally so it goes into a circular orbit?)
Express your answer with the appropriate units.
Part B
(If you hit the ball vertically with the same speed, to what height above you would it rise?)
Express your answer with the appropriate units.
Answer:
A) v = 1,675 10³ m / s , B) r₂ = 11,673 10⁶ m
Explanation:
A) This exercise we must use Newton's second law, where the forces of gravity are the Moon
F = m a
acceleration is centripetal
a = v² / r
force is the force of universal attraction
F = G m M / r²
we substitute
G m M / r² = m v² / r
v² = G M / r
distance
r = R_moon + h
r = 1.74 10⁶ +1.0786 10⁴
r = 1,750786 10⁶ m
we calculate
v = √ (6.67 10⁻¹¹ 7.36 10²² / 1.75 10⁶)
v = √ (2,8052 10⁶)
v = 1,675 10³ m / s
B) let's use energy conservation
Starting point. In the mountain
Em₀ = K + U = ½ m v² + G m M / r
Final point. Where the speed is zero
[tex]Em_{f}[/tex] = U = G mM / r₂
Em₀ = Em_{f}
½ m v² + G m M / r = G mM / r₂
1 / r₂ = (½ v₂ + G M / r) / GM
let's calculate
1 / r₂ = (½ (1,675 10³)² + 6.67 10⁻¹¹ 7.36 10²² / 1.75 10⁶) /(6.67 10⁻¹¹ 7.36 10²²)
1 / r₂ = (1,4028 10⁶ + 2,805 10⁶) / 49.12 10¹¹
1 / r₂ = 8.5664 10⁻⁷
r₂ = 11,673 10⁶ m
The vector indicates the instantaneous displacement of a projectile from the origin. At the instant when the projectile is at , its velocity and acceleration vectors are and . Which statement is correct?
Answer:
The only force acts on a projectile is gravitational force {Fg}, therefore its acceleration a=Fg/m will always directed towards the direction of force i.e. vertically downwards. Therefore it will always be perpendicular to the x direction or here we can say that a is always perpendicular to Vx}.
Explanation:
The vector r indicates the instantaneous displacement of a projectile from the origin. At the instant when the projectile is at r , its velocity and acceleration vectors are v and a . Which statement is correct?