Answer:
they have the same mass
Explanation:
The force applied by the field is a function of the charge and velocity, so the acceleration experienced by a particle will be dependent upon its mass. Particles in orbits with the same radius are exhibiting the same acceleration, so must have the same mass.
what tools use cut wood
Answer:
hand saws
power saws
Circular Saw
Explanation:
that is all that i know
A uniform electric field exists in the region between two oppositely charged plane parallel plates. A proton is released from rest at the surface of the positively charged plate and strikes the surface of the opposite plate, 1.60 cm distant from the first, in a time interval of 3.20×10−6s3.20×10 −6 s. (a) Find the magnitude of the electric field. (b) Find the speed of the proton when it strikes the negatively charged plate.
Answer:
E = 326.17 N/C
Explanation:
(a) In order to calculate the magnitude of the electric field between the parallel plates you first calculate the acceleration of the proton. You use the following formula:
[tex]x=v_ot+\frac{1}{2}at^2[/tex] (1)
vo: initial speed of the proton = 0m/s
t: time that the proton takes to cross the space between the plates = 3.20*10^-6 s
a: acceleration of the proton = ?
x: distance traveled by the proton = 1.60cm = 0.016m
You solve the equation (1) for a, and replace the values of all parameters:
[tex]a=\frac{2x}{t^2}=\frac{2(0.016m)}{(3.20*10^{-6}s)^2}=3.125*10^{10}\frac{m}{s^2}[/tex]
Next, you use the Newton second law for the electric force, to find the magnitude of the electric field:
[tex]F_e=qE=ma[/tex] (2)
q: charge of the proton = 1.6*10^-19C
m: mass of the proton = 1.77*10^-27kg
You solve the equation (2) for E:
[tex]E=\frac{ma}{q}=\frac{(1.67*10^{-27}kg)(3.125*10^{10}m/s^2)}{1.6*10^{-19}C}\\\\E=326.17\frac{N}{C}[/tex]
The magnitude of the electric field in between the parallel plates is 326.17N/C
The Huka Falls on the Waikato River is one of New Zealand's most visited natural tourist attractions. On average, the river has a flow rate of about 300,000 L/s. At the gorge, the river narrows to 20-m wide and averages 20-m deep.
(a) What is the average speed of the river in the gorge?
(b) What is the average speed of the water in the river downstream of the falls when it widens to 60 m and its depth increases to an average of 40 m?
Answer:
(a) V = 0.75 m/s
(b) V = 0.125 m/s
Explanation:
The speed of the flow of the river can be given by following formula:
V = Q/A
V = Q/w d
where,
V = Speed of Flow of River
Q = Volume Flow Rate of River
w = width of river
d = depth of river
A = Area of Cross-Section of River = w d
(a)
Here,
Q = (300,000 L/s)(0.001 m³/1 L) = 300 m³/s
w = 20 m
d = 20 m
Therefore,
V = (300 m³/s)/(20 m)(20 m)
V = 0.75 m/s
(b)
Here,
Q = (300,000 L/s)(0.001 m³/1 L) = 300 m³/s
w = 60 m
d = 40 m
Therefore,
V = (300 m³/s)/(60 m)(40 m)
V = 0.125 m/s
Stress is a factor that contributes to heart disease risk.true or false
The resonance tube used in this experiment produced only one resonance tone. What length of tube would be required to produce a second tone under the same experimental conditions? Explain your answer.
Answer:
the length that would produce a sound tone under the same experimental contditions must be increased by Δl = [tex]\frac{v}{2f}[/tex]
Explanation:
Recall
V = f ×λ
where λ is ⁴/₃l₂ for second resonance
f = [tex]\frac{3v}{4l_{2} }[/tex]
l₂ = [tex]\frac{3v}{4f}[/tex]
where λ is 4l₁ for 1st resonance
f = [tex]\frac{v}{4l_{1} }[/tex]
l₁ = [tex]\frac{v}{4f}[/tex]
∴ Δl = l₂ - l₁ = [tex]\frac{3v}{4f}[/tex] ⁻ [tex]\frac{v}{4f}[/tex]
Δl= [tex]\frac{2v}{4f}[/tex]
Δl = [tex]\frac{v}{2f}[/tex]
Therefore, the length should increase by [tex]\frac{v}{2f}[/tex]
A passenger jet flies from one airport to another 1,233 miles away in 2.4 h. Find its average speed. = ____ m/s
Speed = (distance) / (time)
Speed = (1,233 mile) / (2.4 hour)
Speed = 513.75 mile/hour
Speed = (513.75 mi/hr) x (1609.344 meter/mi) x (1 hr / 3600 sec)
Speed = (513.75 x 1609.344 / 3600) (mile-meter-hour/hour-mile-second)
Speed = 229.7 meter/second
Two identical loudspeakers 2.30 m apart are emitting sound waves into a room where the speed of sound is 340 m/s. Abby is standing 3.00 m in front of one of the speakers, perpendicular to the line joining the speakers, and hears a maximum in the intensity of the sound. Part A What is the lowest possible frequency of sound for which this is possible
Answer:
1089.74 Hz
Explanation:
Using Pythagoras theorem, we can find the distance from the point to the second speaker.
Thus;
d2 = √(3² + 2.3²)
d2 = √(9 + 5.29)
d2 = √14.29
d2 = 3.78 m
Then, the path distance which is the extra distance travelled would be;
Δd = d2 - d1
Δd = 3.78 - 3
Δd = 0.78 m
Now, the destructive interference condition is given by the formula;
Δd = (m + ½)λ
λ is the wavelength
m is a non - negative integer.
In this case, m = 2
Thus;
0.78 = (2 + ½)λ
λ = 0.78/(2½)
λ = 0.312 m
Now the formula for frequency of a wave is given by;
f = v/λ
Where v is speed of sound.
Thus;
f = 340/0.312
f = 1089.74 Hz
In a ballistic pendulum experiment, a small marble is fired into a cup attached to the end of a pendulum. If the mass of the marble is 0.0215 kg and the mass of the pendulum is 0.250 kg, how high h will the pendulum swing if the marble has an initial speed of 5.15 m/s? Assume that the mass of the pendulum is concentrated at its end so that linear momentum is conserved during this collision.
Answer:
h = 8.48*10^-3m
Explanation:
In order to calculate the height reached by the pendulum with the marble, you first take into account the momentum conservation law, to calculate the speed of both pendulum and marble just after the collision.
The total momentum of the system before the collision is equal to the total momentum after:
[tex]m_1v_1+m_2v_2=(m_1+m_2)v[/tex] (1)
Here you used the fact that the pendulum has its total mass concentrated at the end of the pendulum.
m1: mass of the marble = 0.0215kg
m2: mass of the pendulum concentrated at its end = 0.250kg
v1: horizontal speed of the arble before the collision = 5.15m/s
v2: horizontal speed of the pendulum before the collision = 0m/s
v: horizontal speed of both marble and pendulum after the collision = ?
You solve the equation (1) for v, and replace the values of the other parameters:
[tex]v=\frac{m_1v_1+m_2v_2}{m_1+m_2}\\\\v=\frac{(0.0215kg)(5.15m/s)+(0.250kg)(0m/s)}{0.0215kg+0.250kg}=0.40\frac{m}{s}[/tex]
Next, you use the energy conservation law. In this case the kinetic energy of both marble and pendulum (just after the collision) is equal to the potential energy of the system when both marble and pendulum reache a height h:
[tex]U=K\\\\(m_1+m_2)gh=\frac{1}{2}(m_1+m_2)v^2\\\\h=\frac{v^2}{2g}[/tex]
v = 0.40m/s
g: gravitational acceleration = 9,8m/s^2
[tex]h=\frac{(0.40m/s)^2}{2(9.8m/s^2)}=8.48*10^{-3}m[/tex]
Then, the height reached by marble and pendulum is 8.48*10^-3m
3. Which is a general chemical equation for an endothermic, double-replacement reaction?
✓
O AB+ CD + energy-> AD+ CB
O AB+ CD AD + CB + energy
O AB+ C + energy - A+ CB
O AB+C- A+ CB + energy
The correct answer is A. AB+ CD + energy-> AD+ CB
Explanation:
In chemistry, a reaction is endothermic if the reaction involves absorption of heat or energy and this is necessary for the reaction to start. In terms of the chemical equation, this implies energy is part of the reactants or initial substances. Besides this, if the reaction is a double-replacement reaction this means two ions of the original substances are swapped or replaced, which means new substances in the products.
According to this, option A is the correct chemical equation because energy is part of the reactants, which shows the reaction is endothermic and the reactants AB + CD lead to the products AD + CB which shows two ions of the compounds were replaced (double replacement).
Answer:
The proper Answer is A) AB+CD + Energy --> AD + CB
Explanation:
The molecules of a gas are in constant random motion. This means that they have energy in what type of energy store?
Answer:
Heat causes the molecules to move faster, (heat energy is converted to kinetic energy ) which means that the volume of a gas increases more than the volume of a solid or liquid.
Explanation:
Electronic flash units for cameras contain a capacitor for storing the energy used to produce the flash. In one such unit, the flash lasts for a time interval of 1.50×10^−3 s with an average light power output of 2.90×10^5 W
a) If the conversion of electrical energy to light has an efficiency of 90.0 % (the rest of the energy goes to thermal energy), how much energy must be stored in the capacitor for one flash?
b) The capacitor has a potential difference between its plates of 125 V when the stored energy equals the value calculated in part A. What is the capacitance?
Answer:
(a) E = 483.33 J
(b) C = 0.062 F = 62 mF
Explanation:
(a)
First we need to calculate the energy output for one flash. For that purpose we have the formula:
E₀ = Pt
where,
E₀ = Energy output for one flash = ?
P = Light Output Power for one flash= 2.9 x 10⁵ W
t = time interval for one flash = 1.5 x 10⁻³ s
Therefore,
E₀ = (2.9 x 10⁵ W)(1.5 x 10⁻³ s)
E₀ = 435 J
Now, for energy to be stored in capacitor, we use the following formula:
Efficiency = E₀/E
where,
E = Energy required to be stored in capacitor for one flash = ?
Efficiency = 90% = 0.9
Therefore,
0.9 = 435 J/E
E = 435 J/0.9
E = 483.33 J
(b)
The energy stored in the capacitor is given by the formula:
E = (1/2)(CV²)
where,
C = Capacitance = ?
V = Voltage = 125 V
Therefore,
483.33 J = (1/2)(C)(125 V)²
C = (483.33 J)(2)/(125 V)²
C = 0.062 F = 62 mF
Light from a helium-neon laser (? = 633 nm) is used to illuminate two narrow slits. The interference pattern is observed on a screen 3.2m behind the slits. Eleven bright fringes are seen, spanning a distance of 60mm .
What is the spacing (in mm) between the slits?
Answer:
0.3376 mm
Explanation:
The computation of the spacing in mm between the slits is shown below:
As we know that
[tex]d = \frac{m\lambda L}{\Delta y}[/tex]
where,
[tex]\lambda[/tex] = wavelength
L = distance from the scrren
[tex]\Delta y[/tex] = spanning distance
As there are 11 bright fingers seen so m would be
= 11 - 1
= 10
Now placing these values to the above formula
So, the spacing is
[tex]= \frac{(10)(633 \times 10^{-9})(3.2m)}{60 \times 10^{-3}}[/tex]
= 0.3376 mm
We simply applied the above formula.
Answer:
Explanation:
Maximum occurs when the path difference is an integral multiple of wavelength
Here [tex]\lambda[/tex] - Wavelength, [tex]d-[/tex] slit separation and [tex]m-[/tex] Order of pattern
Rearrange the equation for
[tex]\begin{aligned}d &=\frac{m \lambda}{\sin \theta} \\
\text { Here, } \sin \theta &=\frac{y}{L} \quad\left(\begin{array}{l}
\text { Here, } L-\text { separation between slit and screen } \\
y-\text { Distance between respective fringe from center on screen }\end{array}\right)[/tex]
[tex]d=\frac{m \lambda}{\left(\frac{y}{L}\right)} \\
&=\frac{m \lambda L}{y}[/tex]
Here, order
Due to the fact that there are 11 bright fringes seen, you take [tex]11-1=10[/tex]
since starts from 0,1,2,3
Substitute given values
[tex]\begin{aligned}d &=\frac{(10)\left(633 \times 10^{-9} \mathrm{m}\right)(3.2 \mathrm{m})}{60 \times 10^{-3} \mathrm{m}} \\&=\left(3.376 \times 10^{-4} \mathrm{m}\right)\left(\frac{1 \mathrm{mm}}{10^{-3} \mathrm{m}}\right) \\&=0.3376 \mathrm{mm}\end{aligned}[/tex]
A hollow conducting spherical shell has radii of 0.80 m and 1.20 m, The radial component of the electric field at a point that is 0.60 m from the center is closest to
Complete Question
The complete question is shown on the first uploaded image
Answer:
The electric field at that point is [tex]E = 7500 \ N/C[/tex]
Explanation:
From the question we are told that
The radius of the inner circle is [tex]r_i = 0.80 \ m[/tex]
The radius of the outer circle is [tex]r_o = 1.20 \ m[/tex]
The charge on the spherical shell [tex]q_n = -500nC = -500*10^{-9} \ C[/tex]
The magnitude of the point charge at the center is [tex]q_c = + 300 nC = + 300 * 10^{-9} \ C[/tex]
The position we are considering is x = 0.60 m from the center
Generally the electric field at the distance x = 0.60 m from the center is mathematically represented as
[tex]E = \frac{k * q_c }{x^2}[/tex]
substituting values
[tex]E = \frac{k * q_c }{x^2}[/tex]
where k is the coulomb constant with value [tex]k = 9*10^{9} \ kg\cdot m^3\cdot s^{-4} \cdot A^{-2}.[/tex]
substituting values
[tex]E = \frac{9*10^9 * 300 *10^{-9}}{0.6^2}[/tex]
[tex]E = 7500 \ N/C[/tex]
A stunt man whose mass is 70 kg swings from the end ofa 4.0 m long rope along thearc of a vertical circle. Assuming that he starts from rest whenthe rope is horizontal, find the tensions in the rope that arerequired to make him follow his circular path at each of thefollowing points.
(a) at the beginning of his motion
N
(b) at a height of 1.5 m above the bottom of the circular arc
N
(c) at the bottom of the arc
N
Answer:
a. T= 0
b. T = 1286N
c. T= 2058N
Explanation:
C.
(11) in parallel
A potentiometer circuit consists of a
battery of e.m.f. 5 V and internal
resistance 1.0 12 connected in series with a
3.0 12 resistor and a potentiometer wire
AB of length 1.0 m and resistance 2.0 12.
Calculate:
(i) The total resistance of the circuit
The current flowing in the circuit
(iii) The lost volt from the internal
resistance of battery across the
battery terminals
(iv) The p.d. across the wire AB
(v) The e.m.f. of a dry cell which can be
balanced across 60 cm of the wire
AB.
Assume the wire has a uniform cross-
sectional area.
Answer:
fggdfddvdghyhhhhggghh
When moving to a new apartment, you rent a truck and create a ramp with a 244 cm long piece of plywood. The top of the moving ramp lies on the edge of the truck bed at a height of 115 cm. You load your textbooks into a wooden box at the bottom of the ramp (the coefficient of kinetic friction between the box and ramp is = 0.2). Then you and a few friends give the box a quick push and it starts to slide up the ramp. A) What angle is made by the ramp and the ground?B) Unfortunately, after letting go, the box only tables 80cm up the ramp before it starts coming back down! What speed was the box initially traveling with just after you stopped pushing it?
Answer:
A) θ = 28.1º , B) v = 2.47 m / s
Explanation:
A) The angle of the ramp can be found using trigonometry
sin θ = y / L
Φ = sin⁻¹ y / L
θ = sin⁻¹ (115/244)
θ = 28.1º
B) For this pate we can use the relationship between work and kinetic energy
W =ΔK
where the work is
W = -fr x
the negative sign is due to the fact that the friction force closes against the movement
Lavariacion of energy cineta is
ΔEm = ½ m v² - mgh
-fr x = ½ m v² - m gh
the friction force has the equation
fr = very N
at the highest part there is no speed and we take the origin from the lowest part of the ramp
To find the friction force we use Newton's second law. Where one axis is parallel to the ramp and the other is perpendicular
Axis y . perpendicular
N- Wy = 0
cos tea = Wy / W
Wy = W cos treaa
N = mg cos tea
we substitute
- (very mg cos tea) x = ½ m v²2 - mgh
v2 = m (gh- very g cos tea x)
let's calculate
v = Ra (9.8 0.80 - 0.2 9.8 0.0 cos 28.1)
v = RA (7.84 -1.729)
v = 2.47 m / s
When the charges in the rod are in equilibrium, what is the magnitude of the electric field within the rod?
Answer: If we have equilibrium, the magnitude must be zero.
Explanation:
If the charges are in equilibrium, this means that the total charge is equal to zero.
And as the charges must be homogeneously distributed in the rod, we can conclude that the electric field within the rod must be zero, so the magnitude of the electric field must be zero
A particle of charge = 50 µC moves in a region where the only force on it is an electric force. As the particle moves 25 cm, its kinetic energy increases by 1.5 mJ. Determine the electric potential difference acting on the partice
Answer:
nvbnncbmkghbbbvvvvvvbvbhgggghhhhb
A student has made the statement that the electric flux through one half of a Gaussian surface is always equal and opposite to the flux through the other half of the Gaussian surface. This is:_______.
a. never true.
b. never false.
c. true whenever enclosed charge is symmetrically located at a center point, or on a center line or centrally placed plane
d. true whenever no charge is enclosed within the Gaussian surface.
e. true only when no charge is enclosed within the Gaussian surface.
Answer:
E.true only when no charge is enclosed within the Gaussian surface.
Explanation:
Because Gauss’s law states that the net flux of an electric field in a closed surface is directly proportional to the enclosed electric charge.
gg The sound source of a ship’s sonar system operates at a frequency of 22.0 kHzkHz . The speed of sound in water (assumed to be at a uniform 20∘C∘C) is 1482 m/sm/s . What is the difference in frequency between the directly radiated waves and the waves reflected from a whale traveling straight toward the ship at 4.95 m/sm/s ? Assume that the ship is at rest in the water.
Answer:
Δf = 73.72Hz
Explanation:
In order to calculate the difference in frequency between the direct waves and the reflected waves, you first take into account the Doppler's effect for an observer getting closer to the source:
[tex]f'=f(\frac{v+v_o}{v-v_s})[/tex] (1)
You can assume that the reflected waves come from a source "the whale". Then you have:
f': frequency of the reflected waves = ?
f: frequency of the source = 22.0*kHz = 22.0*10^3 Hz
v: speed of the sound in water = 1482m/s
vs: speed of the source = 4.95m/s
vo: speed of the observer = 0m/s
You replace the values of the parameters in the equation (1):
[tex]f'=(22.0*10^3Hz)(\frac{1482m/s}{1482m/s-4.95m/s})=22073.72Hz[/tex]
Then, the difference in frequency is:
[tex]\Delta f = f'-f=22000Hz-22073.72Hz=73.72Hz[/tex]
Bromine, a liquid at room temperature, has a boiling point
Yes it does ! The so-called "boiling point" is the temperature at which Bromine liquid can change state and become Bromine vapor, if enough additional thermal energy is provided. The boiling point is higher than room temperature.
In the far future, astronauts travel to the planet Saturn and land on Mimas, one of its 62 moons. Mimas is small compared with the Earth's moon, with mass Mm = 3.75 ✕ 1019 kg and radius Rm = 1.98 ✕ 105 m, giving it a free-fall acceleration of g = 0.0636 m/s2. One astronaut, being a baseball fan and having a strong arm, decides to see how high she can throw a ball in this reduced gravity. She throws the ball straight up from the surface of Mimas at a speed of 43 m/s (about 96 mph, the speed of a good major league fastball)."
Required:
a. Predict the maximum height of the ball assuming g is constant and using energy conservation. Mimas has no atmosphere, so there is no air resistance.
b. Now calculate the maximum height using universal gravitation.
c. How far off is your estimate of part (a)? Express your answer as a percent difference and indicate if the estimate is too high or too low.
Answer:
a)[tex]h_{max}=14536.16 m[/tex]
b)[tex]h = 15687.9 m[/tex]
c)[tex]PD=7.62\%[/tex] The estimate is low.
Explanation:
a) Using the energy conservation we have:
[tex]E_{initial}=E_{final}[/tex]
we have kinetic energy intially and gravitational potential energy at the maximum height.
[tex]\frac{1}{2}mv^{2}=mgh_{max}[/tex]
[tex]h_{max}=\frac{v^{2}}{2g}[/tex]
[tex]h_{max}=\frac{43^{2}}{2*0.0636}[/tex]
[tex]h_{max}=14536.16 m[/tex]
b) We can use the equation of the gravitational force
[tex]F=G\frac{mM}{R^{2}}[/tex] (1)
We have that:
[tex] F = ma [/tex] (2)
at the surface G will be:
[tex]G=\frac{gR^{2}}{M}[/tex]
Now the equation of an object at a distance x from the surface.
is:
[tex]F=\frac{mgR^{2}}{(R+x)^{2}}[/tex]
[tex]m\frac{dv}{dt}=\frac{mgR^{2}}{(R+x)^{2}}[/tex]
Using that dv/dt is vdx/dt and integrating in both sides we have:
[tex]v_{0}=\sqrt{\frac{2gRh}{R+h}}[/tex]
[tex]h=\frac{v_{0}^{2}R}{2gR-v_{0}^{2}}[/tex]
[tex]h=15687.9[/tex]
c) The difference is:
So the percent difference will be:
[tex]PD=|\frac{14536.16-15687.9}{(14536.16+15687.9)/2}*100%[/tex]
[tex]PD=7.62\%[/tex]
The estimate is low.
I hope it helps you!
A d'Arsonal meter with an internal resistance of 1 kohm requires 10 mA to produce full-scale deflection. Calculate thew value of a series
A d’Arsonval meter with an internal resistance of 1 kΩ requires 10 mA to produce full-scale deflection. Calculate the value of a series resistance needed to measure 50 V of full scale.
Answer:4kΩ
Explanation:Given;
internal resistance, r = 1kΩ
current, I = 10mA = 0.01A
Voltage of full scale, V = 50V
Since there is full scale voltage of 50V, then the combined or total resistance (R) of the circuit is given as follows;
From Ohm's law
V = IR
R = [tex]\frac{V}{I}[/tex] [substitute the values of V and I]
R = [tex]\frac{50}{0.01}[/tex]
R = 5000Ω = 5kΩ
The combined resistance (R) is actually the total resistance of the series arrangement of the series resistance([tex]R_{S}[/tex]) and the internal resistance (r) in the circuit. i.e
R = [tex]R_{S}[/tex] + r
[tex]R_{S}[/tex] = R - r [Substitute the values of R and r]
[tex]R_{S}[/tex] = 5kΩ - 1kΩ
[tex]R_{S}[/tex] = 4kΩ
Therefore the series resistance is 4kΩ
Based on what you know about electricity, hypothesize about how series resistors would affect current flow. What would you expect the effective resistance of two equal resistors in series to be, compared to the resistance of a single resistor?
Answer:
Effective resistance of two equal resistors in series is twice that of a single resistor and in essence will reduce the amount of current flowing in the circuit.
Explanation:
When two resistors are connected in series, their effective resistance is the sum of their individual resistances. For example, given two resistors of resistance values R₁ and R₂, their effective resistance, Rₓ is given by;
Rₓ = R₁ + R₂ --------------(1)
If these resistors have equal resistance values, say R, then equation 1 becomes;
Rₓ = R + R
Rₓ = 2R
This means that their effective resistance is twice of their individual resistances. In other words, when two equal resistors are in series, their effective resistance is twice the resistance of each single one of those resistors.
Now, according to Ohm's law, voltage(V) is the product of current (I) and resistance (R). i.e
V = IR
I = [tex]\frac{V}{R}[/tex]
We can deduce that current increases as resistance decreases and vice-versa.
So, if the two equal resistors described above are connected in series, the amount of current flowing will be reduced compared to having just a single resistor.
If the person is shaking her hand up-and-down 12 times per second, what is the wave speed?
Welllll, first of all, it would take incredible muscular development and control to be able to do that, and I don't believe it's actually humanly possible.
But for Math and Physics problems, that's OK. We don't mind suspending our disbelief, accepting a temporary alternate reality, and working with the hand that is dealt.
The speed of a wave doesn't depend on how the wave is created. A puppy wagging its tail, a fly batting its wings, or a person shaking her hand up and down, are moving the air. The wave that travels away from the vibration is a sound wave in air. Its speed depends only on the characteristics of the air it travels through.
For some typical combination of temperature, pressure, and humidity, this speed (of sound) is taken to be 343 meters per second.
Notice that the 'sound' of shaking her hand up and down will not be 'heard' by anyone, no matter how close she stands to them. 12 Hz (12 per second) is not a fast-enough wiggle to be sensed as sound by human ears. If the person senses the wave at all, it will only be as some kind of pulsating breeze.
A 25.0 kg block is initially at rest on a horizontal surface. A horizontal force of 75 N is required to set the block in motion, after which a horizontal force of 60 N is required to keep the block in moving with constant speed. Find the coefficient of static and kinetic friction between the block and the surface.
Answer:
μs = 0.30
μk = 0.24
Explanation:
In order to calculate the kinetic friction and static friction between the block and the surface, you take into account that the kinetic friction is important when the block is moving and the static friction when the block is at rest.
You use the following formula to find the coefficient of static friction:
[tex]F_1=\mu_s Mg[/tex] (1)
F1 = 75N
μs: coefficient of static friction = ?
M: mass of the block = 25.0kg
g: gravitational acceleration = 9.8m/s^2
You solve for μs in the equation (1):
[tex]\mu_s=\frac{F_1}{Mg}=\frac{75N}{(25.0kg)(9.8m/s^2)}=0.30[/tex]
For the coefficient of kinetic friction you have:
[tex]F_2=\mu_k Mg[/tex] (2)
F2 = 60N
μk: coefficient of kinetic friction = ?
You solve for μk in the equation (2):
[tex]\mu_k=\frac{F_2}{Mg}=\frac{60N}{(25.0kg)(9.8m/s^2)}=0.24[/tex]
Then, you have:
coefficient of static friction = 0.30
coefficient of kinetic friction = 0.24
The lightest and heaviest flying birds are the bee hummingbird of Cuba, which weighs about 1.6 grams, and the great bustard of Europe and Asia, which can weigh as much as 21 kilograms. Show that the bee hummingbird produces about 0.016 newton of lift when it flies, whereas the great bustard produces about 205.8 newtons of lift. Which species would you expect to have proportionally larger wings? Why?
Answer:
for the birds to be able to stay vertical in flight without falling down to earth, they must produce a lift that will counteract their weight
for the small bee humming bird,
mass = 1.6 g = 1.6 x [tex]10^{-3}[/tex] kg
weight of the bird under acceleration due to gravity = mg
where g = acceleration due to gravity = 9.81 m/s^2
weight of the bird = 1.6 x [tex]10^{-3}[/tex] x 9.806 = 0.0156 ≅ 0.016 N
for this bird to maintain flight, the least lift upward, it must generate must be equal to its weight downwards, i.e
lift = weight
therefore,
lift = 0.016 N
For the bustard of Europe and Asia,
mass = 21 kg
weight of the bird under acceleration due to gravity = mg
weight of the bird = 21 x 9.806 = 205.9 N
lift = weight = 205.9 N
lift generated is proportional to the wing surface area according to the lift equation
L = Cs x p x [tex]\frac{v}{2}[/tex] x S
where L = lift
C = lift coefficient
p = density of air
v = relative velocity of bird and air
S = surface are of the wing.
The great bustard will have a proportionally larger wing area to hold its weight in flight
A sunbather stands waist deep in the ocean and observes that six crests of periodic surface waves pass each minute. The crests are 16.00 meters apart. What is the wavelength, frequency, period, and speed of the waves
Answer:
Wavelength = 16 m
Frequency = 0.1 Hz
Period = 10 s^-1
speed of the wave = 1.6 m/s
Explanation:
The crests of the wave is 16.00 m apart
Also, 6 crests pass per minute
The wavelength of this wave is the distance between consecutive corresponding troughs or crests. This means that the wavelength λ is 16 m
Frequency is defined as a number of cycles per seconds.
A minute has 60 sec, therefore, the frequency of this wave is
==> f = 6/60 = 0.1 Hz
Period is the inverse of the frequency, therefore period of the wave is
==> T = 1/0.1 = 10 s^-1
Speed of the wave is the frequency times the wavelength
v = λf = 16 x 0.1 = 1.6 m/s
An ac circuit consist of a pure resistance of 10ohms is connected across an ae supply
230V 50Hz Calculate the:
(i)Current flowing in the circuit.
(ii)Power dissipated
Plz check attachment for answer.
Hope it's helpful
Which of the following biotic organisms makes its own energy from inorganic substances?
producers
consumers
decomposers
minerals
Answer:
producers make its own energy frominorganic substances.