The geologic time scale originally ordered Earth’s rocks by relative age.
Explanation:
Geologic time scale is the measure of events occurred in year wise from the starting of universe. Mostly dating of rocks and fossil fuels are doing the trends still now. In order to measure the age of rocks, geological time scale have preferred relative age mode.
In this system, the age of rocks are measured and compared layer by layer. So the lowest layer of rock will be having the maximum age. As we don’t know the starting time of universe, so this method of comparison between the layers to order the rocks is best. So, depending upon the position of the rocks, the age can be determined.
For every _______ over 50 mph, your chances of being seriously injured, disfigured, or killed are doubled.
Answer:
10 mph
Explanation:
Exceeding speed limit is one of the major causes of road accident.Crash severity increases with the speed of the vehicle at impact.The probability of death, disfigurement, or debilitating injury grows with higher speed at impact.The above consequences double for every 10 mph over
50 mph that a vehicle travels.
Answer:
10 mph
Explanation:
Exceeding speed limit is one of the major causes of road accident.
Crash severity increases with the speed of the vehicle at impact.
The probability of death, disfigurement, or debilitating injury grows with higher speed at impact.
The above consequences double for every 10 mph over
50 mph that a vehicle travels
Explanation:
When light energy hits the retina, the retinal changes from a _____ to a _____ configuration.
Answer:
Cis, Trans.
Explanation:
Rhodopsin also known as visual purple, pigment which contains sensory protein that helps to convert light into an electrical signal. Rhodopsin present in wide range of organisms from bacteria to vertebrates.
Rhodopsin is composed of opsin, and 11-cis-retinaldehyde which is derived from vitamin A. When the eye contact with light the 11-cis component converted to all trans-retinal, which results in the changes in configuration fundamental in the rhodopsin molecule.
An insulating, solid sphere has a uniform, positive charge density of [tex]\rho[/tex] =4.40*10⁻⁷C/m³. The sphere has a radius R of 0.370m.
What is the potential [tex]V_r[/tex] at a point located at r = 0.160 m from the center of the sphere?
The electric potential at a point within a uniformly charged solid sphere can be calculated using the given formula. By substituting the given values of charge density, radius of the sphere, and distance from the center of the sphere, the electric potential can be computed.
Explanation:The potential Vr within a uniformly charged solid sphere of charge density ρ, radius R, at a distance r from the center is given by the formula:
Vr = ρr²/6ε₀ + (ρR²/2ε₀) [ 1 - 3r²/R² ] for r < R
Where ρ is the charge density (4.40*10⁻⁷ C/m³), r is the distance from the center of the sphere (0.160 m), R is the radius of the sphere (0.370 m) and ε₀ is the permittivity of free space (8.854 * 10⁻¹² C²/N.m²).
Substituting the given values into the formula, we can calculate the electric potential at a point located at r = 0.160 m from the center of the sphere.
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The potential at a point located at [tex]\(r = 0.160 \, \text{m}\)[/tex] from the center of the sphere is approximately [tex]\(1.36 \times 10^{-7} \, \text{V}\).[/tex]
The potential at a point located at a distance \(r\) from the center of a uniformly charged solid sphere can be calculated using the formula:
[tex]\[ V_r = \frac{kQ_{enc}}{r} \][/tex]
where \(k\) is the Coulomb's constant ([tex]\(8.99 \times 10^9 \, \text{Nm}^2/\text{C}^2\)), \(Q_{enc}[/tex]\) is the charge enclosed within the radius \(r\), and \(r\) is the distance from the center of the sphere to the point where the potential is being calculated.
For a uniformly charged sphere, the charge enclosed within a radius \(r\) is given by:
[tex]\[ Q_{enc} = \frac{4}{3}\pi r^3 \rho \][/tex]
where [tex]\(\rho\)[/tex] is the charge density of the sphere.
Given that [tex]\(\rho = 4.40 \times 10^{-7} \, \text{C/m}^3\)[/tex] and [tex]\(r = 0.160 \, \text{m}\),[/tex] we can calculate \(Q_{enc}\) as follows:
[tex]\[ Q_{enc} = \frac{4}{3}\pi (0.160 \, \text{m})^3 (4.40 \times 10^{-7} \, \text{C/m}^3) \][/tex]
[tex]\[ Q_{enc} = \frac{4}{3}\pi (0.004096 \, \text{m}^3) (4.40 \times 10^{-7} \, \text{C/m}^3) \][/tex]
[tex]\[ Q_{enc} = \frac{4}{3}\pi (1.8192 \times 10^{-9} \, \text{C}) \][/tex]
[tex]\[ Q_{enc} = 2.42528 \times 10^{-9} \, \text{C} \][/tex]
Now, we can calculate the potential [tex]\(V_r\):[/tex]
[tex]\[ V_r = \frac{kQ_{enc}}{r} \][/tex]
[tex]\[ V_r = \frac{(8.99 \times 10^9 \, \text{Nm}^2/\text{C}^2) (2.42528 \times 10^{-9} \, \text{C})}{0.160 \, \text{m}} \][/tex]
[tex]\[ V_r = \frac{2.17836 \times 10^{-8} \, \text{Nm}^2/\text{C}}{0.160 \, \text{m}} \][/tex]
[tex]\[ V_r = 1.36148 \times 10^{-7} \, \text{V} \][/tex]
Therefore, the potential at a point located at [tex]\(r = 0.160 \, \text{m}\)[/tex] from the center of the sphere is approximately [tex]\(1.36 \times 10^{-7} \, \text{V}\).[/tex]
A 57 kg skier starts from rest at a height of H = 27 m above the end of the ski-jump ramp. As the skier leaves the ramp, his velocity makes an angle of 28° with the horizontal. Neglect the effects of air resistance and assume the ramp is frictionless.
(a) What is the maximum height h of his jump above the end of the ramp?
(b) If he increased his weight by putting on a backpack, would h then be greater, less or, the same?
Answer:
(a)[tex]h=5.95m[/tex]
(b) h is the same
Explanation:
According to the law of conservation of energy:
[tex]E_i=E_f\\U_i+K_i=U_f+K_f[/tex]
The skier starts from rest, so [tex]K_i=0[/tex] and we choose the zero point of potential energy in the end of the ramp, so [tex]U_f=0[/tex]. We calculate the final speed, that is, the speed when the skier leaves the ramp:
[tex]mgH=\frac{mv^2}{2}\\v=\sqrt{2gH}\\v=\sqrt{2(9.8\frac{m}{s^2})(27m)}\\v=23\frac{m}{s}[/tex]
Finally, we calculate the maximum height h above the end of the ramp:
[tex]v_f^2=v_i^2-2gh\\[/tex]
The initial vertical speed is given by:
[tex]v_i=vsin\theta[/tex]
and the final speed is zero, solving for h:
[tex]h=\frac{v_i^2}{2g}\\h=\frac{((23\frac{m}{s})sin(28^\circ))^2}{2(9.8\frac{m}{s^2})}\\h=5.95m[/tex]
(b) We can observe that the height reached does not depend on the mass of the skier
If you experience __________, do not use brakes, concentrate on steering, slow down gradually, brake softly when the car is under control, and pull completely off the pavement.
Answer:
Tire blowout
Explanation:
Tire Blowout
A tire experiences a blowout when it rapidly and explosively loses the inflation pressure. It can happen when some sharp object cuts or tears the surface and the air is suddenly released, making things worse.
The driver frequently loses control of the traveling direction and can be involved in serious accidents. It is recommended not to use brakes during or after the blowout because it could bring little help or make things worse.
Focusing on steering and let the vehicle softly brake.
A physics professor did daredevil stunts in his spare time. His last stunt was an attempt to jump across a river on a motorcycle. The takeoff jump was inclined at 53.0 degrees, the river was 40.0 m wide, and the far bank was 15.0m lower than the top of the ramp. The river itself was 100 m below the ramp. You can ignore air resistance.
a. What should his speed have been at the top of the ramp to have just made it to the edge of the far bank?
b. If his speed was only half the value found in (a), where did he land?
Answer:
a) 17.8 m/s
b) 28.3 m
Explanation:
Given:
angle A = 53.0°
sinA = 0.8
cosA = 0.6
width of the river,d = 40.0 m,
the far bank was 15.0 m lower than the top of the ramp h = 15.0 m,
The river itself was 100 m below the ramp H = 100 m,
(a) find speed v
vertical displacement
[tex]-h= vsinA\times t-gt^2/2[/tex]
putting values h=15 m, v=0.8
[tex]-15 = 0.8vt - 4.9t^2[/tex] ............. (1)
horizontal displacement d = vcosA×t = 0.6×v ×t
so v×t = d/0.6 = 40/0.6
plug it into (1) and get
[tex]-15 = 0.8\times40/0.6 - 4.9t^2[/tex]
solving for t we get
t = 3.734 s
also, v = (40/0.6)/t = 40/(0.6×3.734) = 17.8 m/s
(b) If his speed was only half the value found in (a), where did he land?
v = 17.8/2 = 8.9 m/s
vertical displacement = [tex]-H =v sinA t - gt^2/2[/tex]
⇒ [tex]4.9t^2 - 8.9\times0.8t - 100 = 0[/tex]
t = 5.30 s
then
d =v×cosA×t = 8.9×0.6×5.30= 28.3 m
A series combination of two resistors of resistance 12 Ω and 8 Ω is connected across a source of emf 24 V. What is the potential difference across 8 Ω resistors?
Answer:
The Potential difference across the 8 Ω resistor = 9.6 V
Explanation:
In a series combination of resistor, The total resistance
Rt = R₁ + R₂............... Equation 1
Where Rt = total resistance, R₁ = Resistance of the first resistor, R₂ = resistance of the second resistor.
Given: R₁ = 12Ω, R₂ = 8Ω.
Substituting these vales into equation 1
Rt = 12 + 8
Rt = 20 Ω.
Since both resistors are connected in series, The same current flows through both resistors.
from ohms law,
V = IR..................... Equation 3
I = V/R....................... Equation 2
where I = current flowing through both resistors, V= Emf, R = total resistance of the circuit
Also given : V = 24 V, R = 20 Ω,
Substituting these values into equation 3,
I = 24/20
I = 1.2 A.
Note: Since both resistors are connected in series, The same current flows through both resistors.
The Potential difference across the 8 Ω resistor
V = 1.2×8
V = 9.6 V
The Potential difference across the 8 Ω resistor = 9.6 V
Someone plans to float a small, totally absorbing sphere 0.500 m above an isotropic point source of light,so that the upward radiation force from the light matches the downward gravitational force on the sphere. The sphere’s density is 19.0 g/cm3, and its radius is 2.00 mm. (a) What power would be required of the light source?
Answer:
468449163762.0812 W
Explanation:
m = Mass = [tex]\rhoV[/tex]
V = Volume =[tex]\dfrac{4}{3}\pi r^3[/tex]
r = Distance of sphere from isotropic point source of light = 0.5 m
R = Radius of sphere = 2 mm
[tex]\rho[/tex] = Density = 19 g/cm³
c = Speed of light = [tex]3\times 10^8\ m/s[/tex]
A = Area = [tex]\pi R^2[/tex]
I = Intensity = [tex]\dfrac{P}{4\pi r^2}[/tex]
g = Acceleration due to gravity = 9.81 m/s²
Force due to radiation is given by
[tex]F=\dfrac{IA}{c}\\\Rightarrow F=\dfrac{\dfrac{P}{4\pi r^2}{\pi R^2}}{c}\\\Rightarrow F=\dfrac{PR^2}{4r^2c}[/tex]
According to the question
[tex]F=mg\\\Rightarrow \dfrac{PR^2}{4r^2c}=\rho \dfrac{4}{3}\pi R^3g\\\Rightarrow P=\dfrac{16r^2\rho c\pi Rg}{3}\\\Rightarrow P=\dfrac{16\times 0.002\times 19000\times \pi\times 0.5^2\times 9.81\times 3\times 10^8}{3}\\\Rightarrow P=468449163762.0812\ W[/tex]
The power required of the light source is 468449163762.0812 W
Final answer:
To calculate the power required of the light source, you need to find the upward radiation force exerted on the sphere, which should match the downward gravitational force acting on the sphere. The power is given by P = I * A.
Explanation:
To calculate the power required of the light source, we first need to find the upward radiation force exerted on the sphere, which should match the downward gravitational force acting on the sphere. The radiation force is given by the formula:
Fr = (P / c)A
where Fr is the radiation force, P is the power of the light source, c is the speed of light, and A is the cross-sectional area of the sphere.
Since the sphere is totally absorbing, all the light incident on it is absorbed, so the intensity of the light is given by:
I = P / A
where I is the intensity of the light. Rearranging the equation, we get:
P = I * A
Substituting the value of A for the area of the sphere, we can find the power required.
If Mars were the same size as Mercury (instead of its actual size), which surface features would it have?
Answer:
A surface that is full of craters, very similar to that one seen in mercury.
Explanation:
Mercury is the smaller planet in the solar system whit a very thin atmosphere. If Mars were the same size of mercury it will certainly have a surface full of craters.
When meteoroid (fragments of rock) reach the atmosphere of a planet, they get incinerate as a consequence of the friction between the object and the atmosphere. However, if the meteoroid has the necessary size it can reach the ground (at this point is known as meteorite).
In the case propose, it is most likely that bigger meteorite reach the surface of the planet, which leads to the result of a surface full of craters since the size of mars is not enough to maintain a very dense atmosphere due to the weak gravitational field.
A bullet with a mass of 8.0 g and a speed of 650 m/s is fired at a block of wood with a mass of 0.250 kg. The block rests on a frictionless surface, and is thin enough that the bullet passes completely through it. Immediately after the bullet exits the block, the speed of the block is 12 m/s. What is the speed of the bullet when it exits the block
The speed of the block after the bullet exits is 7.5 m/s. This velocity is found by using the conservation of momentum principle and plugging in the known values into the momentum equation to solve for the unknown variable.
To solve for the speed of the wooden block after the bullet exits, we will use the principle of conservation of momentum. This principle states that in an isolated system, the total momentum before an event is equal to the total momentum after the event. The formula to calculate the final velocity of the block is:
[tex]m_{bullet} \times u_{bullet} + m_{block} \times u_{block} = m_{bullet} \times v_{bullet} + m_{block} \times v_{block}[/tex]
Where:
[tex]m_{bullet}[/tex] = mass of the bullet
[tex]u_{bullet}[/tex] = initial velocity of the bullet
[tex]m_{block}[/tex] = mass of the block
[tex]u_{block}[/tex] = initial velocity of the block (0 m/s, since it's stationary)
[tex]v_{bullet}[/tex] = final velocity of the bullet after passing through the block
[tex]v_{block}[/tex] = final velocity of the block
Given:
[tex]m_{bullet}[/tex] = 0.050 kg (50 g)
[tex]u_{bullet}[/tex] = 500 m/s
[tex]m_{block}[/tex] = 2 kg
[tex]u_{block}[/tex] = 0 m/s
[tex]v_{bullet}[/tex] = 200 m/s
Plugging in the values:
0.050 × 500 + 2 × 0 = 0.050 × 200 + 2 × [tex]v_{block}[/tex]
Solving for [tex]v_{block}[/tex]:
[tex]v_{block}[/tex] = (0.050 × 500 - 0.050 × 200) / 2
[tex]v_{block}[/tex] = (25 - 10) / 2
[tex]v_{block}[/tex] = 15 / 2
[tex]v_{block}[/tex] = 7.5 m/s
Therefore, the velocity of the block after the bullet exits is 7.5 m/s.
In an experiment to determine ΔH for the reaction between HBr and NaOH , 43.6 mL of 1.08 M HBr at 19.95 °C is placed in a coffee cup calorimeter which has a heat capacity of 7.99 J/°C. 43.6 mL of 1.08 M NaOH at 19.95 °C is added to the acid solution in the calorimeter and quickly mixed. A final temperature of 26.88 °C was recorded. What is ΔH for this reaction, in kJ? Assume that the specific heats of all solutions are the same as for pure water (4.18 J g-1 °C-1), and that all solution densities are 1.00 g/mL. Don't forget to put the proper sign on your ΔH value.
Answer:
ΔH = -57.78kj/mol
Explanation:
Assumptions
These are solutions, not pure water. The specific heat of water is 4.184 J/goC. Assume that these solutions are close enough to being like water that their specific heats are also 4.1984 J/goC.
the heat evolved will be
q = mcΔt
first of all convert ml to grams
(43.6 mL + 43.6 mL ) = 87.2mL of solution.
density of water=1g/ml
87.2 mL X 1 g/ml = 87.2 grams of solution.
(mass = Volume X Density)
Find the temperature change.
Δt =tfinal - tinitial = 26.88oC - 19.95oC = 6.93oC
q = mcΔt
= 87.2 grams X 4.184 J/goC X 6.9oC
= 2.52 X 10^3 J
This is the heat gained by the water, but in fact it is the heat lost by the reacting HBr and NaOH, therefore q = -2.52 x 10^3 J.
i.e. it is an exothermic reaction, heat was lost to the water and it got warmer
to find the how much of HBr that was used in mol
43.6 mL of HBr X 1.00 mol HBr/ 1000 mL HBr = 0.0436 mol HBr
same quantity of base NaoH was used
. molar enthalpy = J/mol = -2.52 x 10^3 J / 0.0436 mol
-57.78kj/mol
Therefore, for the neutralization of HBrl and NaOH, the enthalpy change, often called the enthalpy of reaction is ΔH = -57.78kj/mol
Some substances have the same chemical composition, but their atoms are arranged differently. A classic example is carbon. A pencil lead is made of graphite, which contains only carbon atoms. A diamond also contains only carbon atoms. However, the carbon atoms in a diamond are packed more closely together. Would you expect the density of graphite and a diamond to be the same? Explain your answer.
Answer: No, because the atoms are arranged differently. Looking at 100 atoms in each sample, the volume would be smaller in a diamond, because the atoms are packed more closely together. The density of a diamond would be higher.
Explanation:
Final answer:
Graphite and diamond have different densities due to the varied arrangements of carbon atoms in their structures, with diamonds being denser and harder due to a three-dimensional bond network.
Explanation:
The difference in density between graphite and diamond can be attributed to the variation in how carbon atoms are arranged in these two allotropes of carbon. Graphite has a layered structure with weak bonds between each layer, allowing the sheets to easily slip past each other, leading to its softer structure and lower density.
In contrast, diamonds have carbon atoms bonded together in all three dimensions, creating a dense, strong lattice, which is why diamonds are very hard and have a higher density compared to graphite.
How many of following statements about the photoelectric effect are true? (a) The greater the frequency of the incident light is, the greater is the stopping potential. (b) The greater the intensity of the incident light is, the greater is the cutoff frequency. (c) The greater the work function of the target material is, the greater is the stopping potential.
Answer:
(a) The greater the frequency of the incident light is, the greater is the stopping potential.
Explanation:
The stopping potential is crucial to terminate the braking of the photoemitted electrons, stopping the current completely.
Since the kinetic energy of electrons depends on the light's incidence frequency (rather than the intensity), therefore, the stopping potential is proportional to the light's incidence frequency.
The cutoff frequency, in turn, is a limiting frequency below which no photoelectric effect occurs. The cutoff frequency depends on the material from which it is made the emitting surface (and its work function).
A Ping-Pong ball moving East at a speed of 4 m/s collides with a stationary bowling ball. The Ping-Pong ball bounces back to the West, and the bowling ball moves very slowly to the East. Which object experiences the greater magnitude impulse during the collision?A car hits another and the two bumpers lock together during the collision. Is this an elastic or inelastic collision?In a game of pool, the white cue ball hits the #5 ball and stops, while the #5 ball moves away with the same velocity as the cue ball had originally. The type of collision is...In an inelastic collision, the final total momentum is...When a cannon fires a cannonball, the cannon will recoil backward because the.....
Answer:
Explanation:
During any collision whether elastic or inelastic, force of action and reaction are generated at the point of touch for a brief period during which they remain in touch with each other. These action and reaction forces are equal and opposite. Impulse is defined by force multiplied by time duration . Hence we can say that during any collision , the impulse generated are equal and opposite .
In inelastic collision , colliding objects coalesce with each other . Hence in the given case , collision is inelastic.
In elastic collision , when an object collides with a similar object at rest , there is exchange of velocity . In the given case also , similar objects are colliding and exchange of velocity is taking place.So it is an example of elastic collision.
In an inelastic collision , momentum is conserved. So final total momentum is equal to initial total momentum.
When a cannon fires a cannonball, the cannon will recoil backward because the total momentum is conserved.
Kenneth ran a marathon (26.2 miles) in 5.5 hours. What was Kenneth’s average speed? (Round your answer to the nearest tenth.)
Answer:
Explanation:
Speed: Speed can be defined as the ratio of distance to time. The S. I unit of speed is m/s. And it is expressed mathematically as,
speed = distance/time
S = d/t............................. Equation 1
Conversion: (i) If 1 miles = 1609.344 m,
then, 26.2 miles = 1609.344× 26.2
= 42164.813 m.
(i) if 1 hours = 3600 seconds,
then, 5.5 hours = 5.5×3600
=19800 seconds.
Given: d = 26.7 miles= 42164.813 m, t = 5.5 hours = 19800 seconds
Substituting these values into equation 1
S = 42164.813/19800
S = 2.1 m/s.
Therefore Kenneth's average speed = 2.1 m/s
Answer:4.8 mph on edge
Explanation: have a great day :)
An 899-kg (1987 lb) dragster, starting from rest, attains a speed of 26.3 m/s (58.9 mph) in 0.59 s. (a) Find the average acceleration of the dragster during this time interval. 44.58 Correct: Your answer is correct. m/s2 (b) What is the magnitude of the average net force on the dragster during this time
Explanation:
Mass, m = 899 kg
Initial velocity, u = 0 m/s
Final velocity, v = 26.3 m/s
Time, t = 0.59 s
a) We have equation of motion v = u + at
Substituting
v = u + at
26.3 = 0 + a x 0.59
a = 44.58 m/s²
Acceleration of dragster = 44.58 m/s²
b) Force = Mass x Acceleration
F = ma
F = 899 x 44.58 = 40074.07 N
F = 40.07 kN
Average net force on the dragster during this time is 40.07 kN
Find the rate of change for x3. You need to work out the change in f(x)=x3 when x is increased by a small number h to x+h. So you will work out f(x+h)-f(x). Then do some algebra to simplify this. Then divide this by h to get the average rate of change of f(x) between x and x+h.
Answer:
Explanation:
Given
[tex]F(x)=x^3[/tex]
Rate of change of F(x) is given by
[tex]F'(x)=\lim_{h\rightarrow 0}\\\frac{F(x+h)-F(x)}{x+h-x}[/tex]
[tex]F'(x)=\lim_{h\rightarrow 0}\\\frac{(x+h)^3-x^3}{h}[/tex]
[tex]F'(x)=\lim_{h\rightarrow 0}\\\frac{x^3+h^3+3x^2h+3xh^2-x^3}{h}[/tex]
[tex]F'(x)=\lim_{h\rightarrow 0}\\\frac{h^3+3x^2h+3xh^2}{h}[/tex]
[tex]F'(x)=\lim_{h\rightarrow 0}\\h^2+3x^2+3xh[/tex]
Putting limits
[tex]F'(x)=3x^2[/tex]
The average rate of change will be "3x²".
Average rate of changeAccording to the question,
The function, f(x) = x³
then,
f(x + h) = (x + h)³
Now,
→ f(x + h) - f(x) = (x + h)³ - x³
= x³ + h³ + 3x²h + 3xh² - x³
= h³ + 3x²h + 3xh²
and,
→ [tex]\frac{f(x+h) -f(x)}{h}[/tex] = [tex]\frac{h^3+3x^2h+3xh^2}{h}[/tex]
= [tex]\frac{h[h^2+3x^2+3xh]}{h}[/tex]
= h² + 3x² + 3xh
By applying the limit, we get
→ [tex]\lim_{h \to 0} \frac{f(x+h)-f(x)}{h}[/tex] = [tex]\lim_{h \to 0}[/tex] h² + 3x² + 3xh
By substituting the values,
= 0² + 3x² + 3x(0)
= 3x²
Thus the above approach is correct.
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how many seconds of space should you keep between you if you're driving a 100 foot truck 30 mph?
Answer:
The space in seconds that will be kept = 2.27 seconds
Explanation:
S = d/t..................... Equation 1
making t the subject of formula in the equation above,
t = d/S.................... Equation 2
Where S = speed, d = distance, t = time.
Conversion: (i)if 1 mph = 0.44704 m/s,
then, 30 mph = 30×0.44704
= 13.41 m/s
(ii) If 1 foot = 0.3048 m
then, 100 foot = 30.48 m.
Given: S = 30 mph = 13.41 m/s, d = 100 foot = 30.48 m
Substituting these values into equation 2
t = 30.48/13.41
t = 2.27 seconds.
Therefore the space in seconds that will be kept = 2.27 seconds
The current through a 10-ohm resistor connected to a 120-v power supply is
Answer:
12 A
Explanation:
Voltage, V = 120 V
Resistance, R 10 ohm
By using ohm's law
V = i x R
where, i is the current
i = V / R
i = 120 / 10
i = 12 A
thus, the current is 12 A.
what are the charge and the charge density on the surface of a conducting sphere of radius 0.15 m whose potantial is 200v?
Answer:
The charge and the charge density on the surface of a conducting sphere is [tex]3.34\times 10^{-9}\ C[/tex] and [tex]1.18\times 10^{-8}\ C/m^2[/tex] respectively.
Explanation:
It is given that,
Radius of the conducting sphere, r = 0.15 m
Potential, V = 200 V
Potential on the surface of sphere is given by :
[tex]V=\dfrac{kq}{r}[/tex]
q is the charge on the sphere
[tex]q=\dfrac{Vr}{k}[/tex]
[tex]q=\dfrac{200\times 0.15}{9\times 10^9}[/tex]
[tex]q=3.34\times 10^{-9}\ C[/tex]
Charge per unit area is called charge density on the surface. it is given by :
[tex]\sigma=\dfrac{q}{A}[/tex]
[tex]\sigma=\dfrac{q}{4\pi r^2}[/tex]
[tex]\sigma=\dfrac{3.34\times 10^{-9}}{4\pi (0.15)^2}[/tex]
[tex]\sigma=1.18\times 10^{-8}\ C/m^2[/tex]
So, the charge and the charge density on the surface of a conducting sphere is [tex]3.34\times 10^{-9}\ C[/tex] and [tex]1.18\times 10^{-8}\ C/m^2[/tex] respectively. Hence, this is the required solution.
Final answer:
The charge on the conducting sphere with a radius of 0.15 m and a potential of 200V is approximately 3.34 × 10⁻⁶ C, and the surface charge density is approximately 1.18 × 10⁻⁵ C/m².
Explanation:
To find the charge (q) and the surface charge density (σ) on the surface of a conducting sphere with a given potential (V), we can use the relationship between potential, charge, and radius of a sphere, along with the equation that relates charge to surface charge density.
The potential (V) of a conducting sphere is given by V = (k × q)/R, where k is Coulomb's constant (k = 8.99 × 10⁹ N.m²/C²), q is the charge on the sphere, and R is the radius of the sphere.
In this case, V = 200V and R = 0.15m. Using the formula, we can find the charge q on the sphere:
V = (k × q)/R
200V = ((8.99 × 109 N.m²/C²) × q)/0.15m
q = (200V × 0.15m)/(8.99 × 10⁹ N.m²/C²)
q ≈ 3.34 × 10⁻⁶ C
Once we have q, we can use the equation q = σ(4πR2) to find the surface charge density σ:
σ = q / (4πR²)
σ ≈ 3.34 × 10⁻⁶ C / (4π × (0.15m)²)
σ ≈ 1.18 × 10⁻⁵ C/m²
a person can throw a 300 gram ball at 25 m/s the maximum height the person can throw the ball on Earth?
Answer:
Elevation =31.85[m]
Explanation:
We can solve this problem by using the principle of energy conservation. This consists of transforming kinetic energy into potential energy or vice versa. For this specific case is the transformation of kinetic energy to potential energy.
We need to first identify all the input data, and establish a condition or a point where the potential energy is zero.
The point where the ball is thrown shall be taken as a reference point of potential energy.
[tex]E_{p} = E_{k} \\where:\\E_{p}= potential energy [J]\\ E_{k}= kinetic energy [J][/tex]
m = mass of the ball = 300 [gr] = 0.3 [kg]
v = initial velocity = 25 [m/s]
[tex]E_{k}=\frac{1}{2} * m* v^{2} \\E_{k}= \frac{1}{2} * 0.3* (25)^{2} \\E_{k}= 93.75 [J][/tex]
[tex]93.75=m*g*h\\where:\\g = gravity = 9.81 [m/s^2]\\h = elevation [m]\\replacing\\h=\frac{E_{k}}{m*g} \\h=\frac{93.75}{.3*9.81} \\h=31.85[m][/tex]
A lead bullet at 27°C just melts when stopped by an obstacle. Assuming that 25 % of heat is absorbed by the obstacle, find the velocity of the bullet if its initial temperature is 27^{\circ}C. [For lead, melting point = 327^{\circ}C, specific heat = 0.03 cal/g-^{\circ}C, latent heat of fusion = 6 cal/g and J = 4.2 J/cal.]
Answer:
409.87803 m/s
Explanation:
v = Velocity of bullet
L = Latent heat of fusion = 6 cal/g
c = Specific heat of lead = 0.03 cal/g°C
[tex]\Delta T[/tex] = Change in temperature = (327-27)
m = Mass of bullet
[tex]1\ J=4.2\ J/cal[/tex]
The heat will be given by the kinetic energy of the bullet
[tex]Q=\dfrac{1}{2}mv^2[/tex]
According to the question
[tex]Q=0.75\dfrac{1}{2}mv^2[/tex]
This heat will balance the heat going into the obstacle
[tex]Q=mc\Delta T+mL\\\Rightarrow 0.75\dfrac{1}{2}mv^2=m(c\Delta T+L)\\\Rightarrow v^2=\dfrac{2}{0.75}\times (0.03\times (327-27)+6)\\\Rightarrow v^2=40\ kcal\\\Rightarrow v^2=40\times 4.2\times 10^3\\\Rightarrow v^2=168000\ m^2s^2\\\Rightarrow v=\sqrt{168000}\\\Rightarrow v=409.87803\ m/s[/tex]
The speed of the bullet is 409.87803 m/s
Transform active margins are associated with which type of boundary? Convergent Transform Divergent
Answer:
Transform active margins are associated with which type of boundary?
Transform boundary
Explanation:
The transform boundary is a boundary where one plates(crust) slides past another plate horizontally. This kind of plate movement have been detected to exist between the interaction of the North pacific plates(continental plate) and the pacific plates(oceanic plates) .
At the transform margin the crust are usually broken. But overall crust are neither created nor destroyed . The transform margin region are active as it is marked by shallow-focus earthquakes .
Along the fractured zone where this transform movement occurs is known to create extensive transform faults .Notable transform fault that exist in this kind of boundary(transform) is the San Andrea fault and Alpine Fault.
The motion of this plates can occur on a single fault or on a group of faults.
A child's toy consists of a spherical object of mass 50 g attached to a spring. One end of the spring is fixed to the side of the baby's crib so that when the baby pulls on the toy and lets go, the object oscillates horizontally with a simple harmonic motion. The amplitude of the oscillation is 6 cm and the maximum velocity achieved by the toy is 3.2 m/s . What is the kinetic energy K of the toy when the spring is compressed 5.1 cm from its equilibrium position?
Problem-Solving Strategy: Simple Harmonic Motion II: Energy
The energy equation, E=12mvx2+12kx2=12kA2, is a useful alternative relationship between velocity and position, especially when energy quantities are also required. If the problem involves a relationship among position, velocity, and acceleration without reference to time, it is usually easier to use the equation for simple harmonic motion, ax=d2xdt2=−kmx (from Newton’s second law) or the energy equation above (from energy conservation) than to use the general expressions for x, vx, and ax as functions of time. Because the energy equation involves x2 and vx2, it cannot tell you the sign of x or of vx; you have to infer the sign from the situation. For instance, if the body is moving from the equilibrium position toward the point of greatest positive displacement, then x is positive and vx is positive.
IDENTIFY the relevant concepts
Energy quantities are required in this problem, therefore it is appropriate to use the energy equation for simple harmonic motion.
SET UP the problem using the following steps
Part A
The following is a list of quantities that describe specific properties of the toy. Identify which of these quantities are known in this problem.
Select all that apply.
Select all that apply.
maximum velocity vmax
amplitude A
force constant k
mass m
total energy E
potential energy U at x
kinetic energy K at x
position x from equilibrium
Part B
What is the kinetic energy of the object on the spring when the spring is compressed 5.1 cm from its equilibrium position?
Part C
What is the potential energy U of the toy when the spring is compressed 5.1 cm from its equilibrium position?
Answer:
Part A
Mass = 50g
Vmax = 3.2m/s
Amplitude= 6cm
Position x from the equilibrium= 5.1cm
Part B
Kinetic energy = 0.185J
Part C
Potential energy = 0.185J
Explanation:
Kinetic energy = 1/2mv×2
Vmax = wa
w = angular velocity= 53.33rad/s
Kinetic energy = 1/2mv^2×r^2 = 0.185J
Part c
Total energy = 1/2m×Vmax^2= 0.256J
1/2KA^2= 0.256J
K= 142.22N/m (force constant)
Potential energy = 1/2kx^2
=1/2×142.22×0.051^2
= 0.185J
To find the kinetic energy of the toy, we need to use the energy equation for simple harmonic motion and the relationship between velocity and position. We can then substitute the known values to calculate the kinetic energy.
Explanation:In this problem, we are given the amplitude (A) of the oscillation and the maximum velocity (vmax) achieved by the toy. We need to find the kinetic energy (K) of the toy when the spring is compressed 5.1 cm from its equilibrium position.
To solve for the kinetic energy, we can use the energy equation for simple harmonic motion: K = 1/2mvx2, where m is the mass of the object and vx is the velocity of the object at position x. The mass of the object is given as 50 g, which is equal to 0.05 kg.
Since we know the maximum velocity (vmax = 3.2 m/s), we can use the relationship between velocity and position in simple harmonic motion to find the velocity (vx) at a displacement of 5.1 cm from the equilibrium position. The velocity and position in simple harmonic motion are related by vx = ±ω√(A2 - x2), where ω is the angular frequency of the motion.
Substituting the known values into the equations, we can calculate the kinetic energy of the toy.
Learn more about Simple Harmonic Motion here:https://brainly.com/question/28208332
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If the mass of the block is too large and the block is too close to the left end of the bar (near string B) then the horizontal bar may become unstable (i.e., the bar may no longer remain horizontal). What is the smallest possible value of x such that the bar remains stable (call it xcritical)
Final answer:
The question is related to determining the critical value of x, or xcritical, that ensures a bar with a heavy block remains stable by balancing torques around the pivot point.
Explanation:
The question asks about stability conditions in a physical system involving a block on a bar and requires the application of concepts such as the center of mass, force equilibrium, and tension. When the mass of the block is too large and positioned too close to the left end of the horizontal bar, the system may become unstable, potentially causing the bar to tilt. To determine the smallest value of x, denoted as xcritical, we must set up equations considering the torques about the pivot point, ensuring that the sum of torques must equal zero for stability. This involves calculating the torques due to the weight of the bar, the weight of the block, and any other forces acting on the system. The position of the center of mass of the bar is crucial as it influences the bar's propensity to rotate and the distribution of weight. Hence, xcritical is that specific distance at which the torques balance each other out, and the system remains in a stable, horizontal state.
Worker A works at a constant rate and, working alone, can complete a job in 6 hours. Worker B works at a constant rate and, working alone, can complete the same job in 5 hours. Worker C works at a constant rate and, working alone, can complete the same job in 3 hours.
Answer:
[tex]R= \frac{1job}{1.429 hours}[/tex]
So then we will have that the 3 working together will complete 1 job in approximately 1.429 hours for this case.
Explanation:
If we want to express the situation in math terms and find the number of hours that takes to complete 1 job with the 3 at the same time, we can do this.
For this case we have the following rates:
[tex]R_A =\frac{1job}{6hours}[/tex]
[tex]R_B =\frac{1 job}{5 hours}[/tex]
[tex]R_C=\frac{1job}{3 hours}[/tex]
And we know that working together the rate would be the addition of the rates like this:
[tex]R=R_A +R_B +R_C = \frac{1job}{6hours}+\frac{1job}{5hours}+\frac{1job}{3hours} =\frac{7 jobs}{10hours}[/tex]
And if we divide the numerator and denominator by 7 we got:
[tex]R= \frac{1job}{1.429 hours}[/tex]
So then we will have that the 3 working together will complete 1 job in approximately 1.429 hours for this case.
Which of the following is characteristic of proficient catching?
A. pointing the fingers upward to catch a high ball
B. pointing the fingers upward to catch a low ball
C. letting the shoulders and elbows move to give with contact of the ball
D. A and C E. B and C
Answer:
The correct answer is option D i.e. A and C
Explanation:
The correct answer is option D i.e. A and C
for proficient catching player must
- learn to absorbed the ball force
- moves the hang according to ball direction to hold the ball
- to catch ball at high height move the finger at higher position
- to catch ball at low height move the finger at lower position
if five joules were required to move a crate in 3.7 seconds, what power was applied?
Answer:
The answer to your question is 1.35 Watts
Explanation:
Data
Work = W = 5 J
time = t = 3.7 s
Power = P = ?
Formula
Power is a rate in which work is done or energy is transferred over time
P = [tex]\frac{W}{t}[/tex]
Substitution
[tex]P = \frac{5}{3.7}[/tex]
Result
P = 1.35 W
On the planet Zorb, the acceleration due to gravity is 10 meters per second squared. If you were to launch a projectile at an angle of 30 degrees with an initial velocity of 10 meters per second, in seconds, how long would it take for the projectile to fall to the ground?
Answer:
The time taken by the projectile to fall to the ground is 1 second.
Explanation:
It is given that,
Acceleration due to gravity on the planet Zorb, [tex]a=10\ m/s^2[/tex]
Angle of projectile is 30 degrees
Initial velocity of the projectile, u = 10 m/s
Let t is the time taken for the projectile to fall to the ground. Using second equation of motion to find it
[tex]h=u\ sin\theta t+\dfrac{1}{2}at^2[/tex]
When it fall to the ground, h = 0 and a = -g
[tex]10\ sin(30)t-\dfrac{1}{2}\times 10t^2=0[/tex]
On solving the above quadratic equation, we get the value of t as,
t = 1 second
So, the time taken for the projectile to fall to the ground is 1 second. Hence, this is the required solution.
Which of the following forms an interference pattern when directed toward two suitably-spaced slits?
a. electrons
b. Sound
c. light
d. all of these
enone of these
Answer:
Option d
Explanation:
Interference is the phenomenon in which two or more waves having the same frequency combines or cancel out each other to form a resulting wave with the amplitude equals the sum of the amplitudes of the waves that are combined.
Thus it is clear that the particle with the wave nature can produce interference pattern. Hence, electron, light, sound, all of these produces interference pattern when directed towards the slits that are suitably spaced.