originlal wire :
conductance = 0.9
conductivity = n
length = l
area = A
New wire -
conductance = ?
conductivity = n
length = l
area = 3A
Conductance of original wire :
C = (nA)/l = 0.9 s
new conductance :
C' = (n3A)/l = 3× (nA)/l = 3 × 0.9 = 2.7 s
Typical value for the magnitude of the electric field inside the atom is
a. 10-11N/C
b. 1011N/C
c. 10-9N/C
d. 109N/C
Answer:d. 109N/C
Explanation: The atomic electric field, the field between the atomic nucleus and the surrounding electron cloud, should possess information about the atomic species, local chemical bonding, and charge redistributions between bonded atoms.
It takes 80 pounds of force to
stretch a particular spring 2
inches. How much work is done
in stretching it from its relaxed
state a total of 4 inches?
[?] inch - pounds
Answer: 240 inch-lbs.
Explanation:
The spring constant, k, is given by:
k = (F/x)
where F is the force, x is the displacement, and k is the spring constant.
Using the given information, we can calculate the spring constant as:
k = (80 lbs / 2 in) = 40 lbs/in
To find the work done in stretching the spring from its relaxed state to a total of 4 inches, we need to integrate the force over the displacement:
W = ∫F dx
Since the force required to stretch the spring varies with displacement, we need to break up the integration into two parts: one from 0 to 2 inches, and another from 2 to 4 inches.
W = ∫0^2 F dx + ∫2^4 F dx
The first integral is:
∫0^2 F dx = ∫0^2 (kx) dx = (1/2)kx^2 = (1/2)(40 lbs/in)(2 in)^2 = 80 inch-lbs
The second integral is:
∫2^4 F dx = ∫2^4 (2k) dx = 2kx = 2(40 lbs/in)(4 in - 2 in) = 160 inch-lbs
Therefore, the total work done in stretching the spring from its relaxed state to a total of 4 inches is:
W = ∫0^2 F dx + ∫2^4 F dx = 80 inch-lbs + 160 inch-lbs = 240 inch-lbs
So, the work done in stretching the spring from its relaxed state a total of 4 inches is 240 inch-lbs.
A 32.9 kg child on a sled slides down a hill, reaching a speed of 9.94 m/s. How high was the hill (in m)?
Answer:
To find the height of the hill, we can use the conservation of energy principle, which states that the initial potential energy of the child and sled at the top of the hill is equal to their final kinetic energy at the bottom of the hill. The potential energy is given by:
PE = mgh
where m is the mass of the child and sled, g is the acceleration due to gravity (9.81 m/s^2), and h is the height of the hill.
The kinetic energy is given by:
KE = (1/2)mv^2
where v is the speed of the child and sled at the bottom of the hill.
Equating the potential and kinetic energies, we have:
mgh = (1/2)mv^2
Canceling the mass, we get:
gh = (1/2)v^2
Solving for h, we have:
h = (1/2) v^2 / g
Substituting the given values, we get:
h = (1/2) (9.94 m/s)^2 / 9.81 m/s^2
h = 5.06 m
Therefore, the height of the hill is 5.06 meters.
A ball rolls along flat ground with a speed of 5.2 m/s when it encounters a hill. What vertical height (in m) above the ground does the ball reach?
a ring of aluminium bronze alloy has internal diameter 300mm and 50mm wide .the coefficient of cubical expansion of alloy is 51×10^-6 /degree celcius .for a temperature rise of 600 degree celsius find the final internal diameter?
According to the given statement The ring's final internal diameter would be 1218 mm.
What is cubical expansion explanation?Cubical expansion is the term for the phenomenon wherein the volume of a solid increases as it is heated. Also called volumetric expansion. The coefficient of volumetric expansion measures how much a material's volume expands as its temperature rises by one degree.
The following formula can be used to get the ring's ultimate interior diameter:
ΔL = αLΔT
where ΔL is the length increase, is the cubical growth coefficient, L is the starting length, and ΔT is the temp change.
In this instance, the change in width, which is double the change in length, is what we are searching for. Thus, the formula may be rewritten as follows:
ΔD = 2αDLΔT
where D represents the starting diameter and ΔD represents the diameter change.
By putting in the indicated values, we get:
ΔD = 2(51×10⁻⁶ /degree celsius)(300 mm)(600 degree celsius)
ΔD = 918 mm
As a result, the ring's final internal diameter would be:
Dfinal = Dinitial + ΔD = 300 mm + 918 mm = 1218 mm
Hence, the ring's final internal diameter would be 1218 mm.
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A horizontal pipe of diameter 1.11 m has a
smooth constriction to a section of diameter
0.666 m . The density of oil flowing in the pipe
is 821 kg/m3
.
If the pressure in the pipe is 8130 N/m
2
and in the constricted section is 6097.5 N/m2
,
what is the rate at which oil is flowing
PLEASE ANSWER THISSSSSSSSS!!!!!
The rate at which οil is flοwing is 0.494 m³/s. This means that the prοduct οf the fluid's density (ρ), crοss-sectiοnal area (A), and velοcity (v) is cοnstant.
What is Density?Density is a physical prοperty οf matter that represents hοw much mass is cοntained within a given vοlume οf a substance. It is defined as the amοunt οf mass per unit vοlume and is typically expressed in units οf kilοgrams per cubic meter (kg/m³) οr grams per cubic centimeter (g/cm³).
We can use the principle οf cοntinuity tο sοlve this prοblem. Accοrding tο this principle, the mass flοw rate οf a fluid remains cοnstant as it flοws thrοugh a pipe οf varying diameter.
Therefore, we can write:
[tex]\rho_1A_1v_1 = \rho_2A_2v_2[/tex]
where the subscripts 1 and 2 refer to the sections of the pipe before and after the constriction, respectively.
We can rearrange this equation to solve for the velocity of the oil in the pipe:
[tex]v_2 = (A_1/A_2) \times (v_1 \times (\rho_1/\rho_2))[/tex]
where A1 and A2 are the cross-sectional areas of the pipe before and after the constriction, respectively.
Using the given values, we get:
[tex]v_2 = (\pi/4) \times (1.11 m)^2 \times (8130 N/m^2 / 821 kg/m^3) / [(\pi/4) \times (0.666 m)^2] \times (6097.5 N/m^2 / 821 kg/m^3)[/tex]
v₂ = 1.74 m/s
Finally, we can calculate the rate at which oil is flowing using the formula:
Q = A₂ × v₂
Using the given values, we get:
Q = (π/4) × (0.666 m)² × 1.74 m/s
Q = 0.494 m³/s
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Question #1. 500 pages of a book has a total mass of 2.5 kg. What is the mass of each page in: (i) kg: (ii) mg: (iii). µg:
Answer:
i) 2.5kg ii)2.5x10^6mg III) 2.5x10^9μgHow much energy is produced in J when the sun converts 2 kg of mass into energy?
Answer:
The energy produced by the sun when it converts 2 kg of mass into energy is given by Einstein's famous equation E = mc², where E is the energy produced, m is the mass converted, and c is the speed of light.
Substituting the values, we have:
E = (2 kg) x (299,792,458 m/s)²
E = 2 x 89,875,517,873,681,764 J
E ≈ 1.7975 x 10²⁰ J
Therefore, the sun produces approximately 1.7975 x 10²⁰ joules of energy when it converts 2 kg of mass into energy.
Two resistors of 5.0 and 9.0 are connected in parallel. A 4.0- resistor is then connected in series with the parallel combination. A 6.0-V battery is then connected to the series-parallel combination. What is the current through the 5.0- resistor?
Answer:
First, we need to find the equivalent resistance of the parallel combination of 5.0 and 9.0 resistors:
1/R = 1/5.0 + 1/9.0
1/R = 0.4 + 0.1111
1/R = 0.5111
R = 1/0.5111
R ≈ 1.955 ohms
The equivalent resistance of the parallel combination is approximately 1.955 ohms.
Next, we need to find the total resistance of the circuit:
R_total = 4.0 + 1.955
R_total = 5.955 ohms
The total resistance of the circuit is approximately 5.955 ohms.
Using Ohm's Law, we can find the current through the circuit:
I = V/R_total
I = 6.0/5.955
I ≈ 1.006 A
The current through the circuit is approximately 1.006 A.
Finally, we can use the current divider rule to find the current through the 5.0-ohm resistor:
I_5 = (R_parallel / (R_parallel + R_series)) * I_total
I_5 = (1.955 / (1.955 + 4.0)) * 1.006
I_5 ≈ 0.383 A
The current through the 5.0-ohm resistor is approximately 0.383 A.
A container with a height of 7.7 inches with an open top has a 5.6 inch diameter and is open to the atmosphere. The container is filled with water. The bottom of the container has a 0.87 inch diameter hole. Calculate ρgh at the top of the container if the datum is set at the bottom of the container.
This is the pressure at the top of the container due to the height of the water. The pressure at the bottom of the container due to the height of the water is 0 Pa since the datum was set at the bottom.
What is pressure?Pressure is a physical quantity used to measure the force applied by an object to another object or by an object to a surface. It can be expressed as the force per unit area and is typically measured in units such as pounds per square inch (psi) or pascals (Pa). Pressure can be applied to gases, liquids, and solids, and is generated by the weight of the atmosphere, the force of gravity, and by the movement of air or liquids. Pressure can also be created by mechanical devices such as pumps and compressors, as well as by chemical reactions.
ρ = density of water = 1000 kg/m3
g = acceleration due to gravity = 9.81 m/s2
h = height of the container = 7.7 in = 0.198 m
Using the equation ρgh = density x gravity x height, we can calculate the pressure at the top of the container to be:
ρgh = (1000 kg/m3)(9.81 m/s2)(0.198 m) = 1960.38 Pa
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Which of the following thermometers responds best to changing temperature? A Mercury thermometer. BAlcohol thermometer. C Resistance thermometer. D Thermoelectric thermometer. E Gas thermometer.
Answer:
D. Thermoelectric thermometer
Explanation:
It preferred for rapidly changing temperature
- An object in equilibrium has three forces exerted on it. A 33-N force act at 90° from the x-axis and a 46-N force act at 60°. What are the magnitude and direction of the third force
Answer:
Explanation:
The 33 N force is at a 90 degree angle, whereas the 44 N force is at a 60 degree angle with the x-axis.
Assume that the third force makes a theta-angle contact with the x-axis.
Since the object is in balance, the total force acting on it will equal zero.
Find the accumulation of the x-axis forces.
[tex]\begin{aligned} 33\cos 90{}^\circ +44\cos 60{}^\circ +{{F}_{3}}\cos \theta &=0 \\ 0+22\text{ N}+{{F}_{3}}\cos \theta &=0 \\ {{F}_{3}}\cos \theta &=-22\text{ N }......\text{ }\left( 1 \right) \end{aligned}[/tex]
Find accumulation of the y-axis forces.
[tex]\begin{aligned} 33\sin 90{}^\circ +44\sin 60{}^\circ +{{F}_{3}}\sin \theta &=0 \\ 33\text{ N}+38.11\text{ N}+{{F}_{3}}\sin \theta &=0 \\ {{F}_{3}}\sin \theta &=-71.11\text{ N }......\text{ }\left( 2 \right) \end{aligned}[/tex]
Identify the magnitude.
[tex]\begin{aligned} F&=\sqrt{{{\left( {{F}_{3}}\cos \theta \right)}^{2}}+{{\left( {{F}_{3}}\sin \theta \right)}^{2}}} \\ &=\sqrt{{{\left( -22\text{ N} \right)}^{2}}+{{\left( -71.11\text{ N} \right)}^{2}}} \\ &=74.43\text{ N} \end{aligned}[/tex]
Identify the direction.
[tex]\begin{aligned} \tan \theta &=\left( \frac{{{F}_{3}}\sin \theta }{{{F}_{3}}\cos \theta } \right) \\ \theta &={{\tan }^{-1}}\left( \frac{{{F}_{3}}\sin \theta }{{{F}_{3}}\cos \theta } \right) \\ \theta &={{\tan }^{-1}}\left( \frac{-71.11\text{ N}}{-22\text{ N}} \right) \\ \theta &=72.8{}^\circ \end{aligned}[/tex]
A faulty model rocket moves in the xy-plane (the positive y-direction is vertically upward). The rocket's acceleration has components ax(t)=αt2
and ay(t)=β−γt
, where α
= 2.50 m/s4
, β
= 9.00 m/s2
, and γ
= 1.40 m/s3
. At t=0
the rocket is at the origin and has velocity v⃗ 0=v0xi^+v0yj^
with v0x
= 1.00 m/s
and v0y
= 7.00 m/s
.
The rocket travels a horizontal distance of 57.4 m before hitting the ground.
What is the initial speed of the rocket?The initial speed of the rocket is v0=√(v0x^2+v0y^2)=7.28 m/s.
What is the rocket's velocity at the maximum height?The rocket's velocity at the maximum height is zero, as it momentarily stops moving vertically and starts falling back down.
To solve this problem, we can use the kinematic equations of motion. Let's first find the velocity and position as a function of time:
vx(t) = v0x + ∫ax(t) dt = v0x + (1/3)αt^3
vy(t) = v0y + ∫ay(t) dt = v0y + βt - (1/2)γt^2
x(t) = ∫vx(t) dt = v0x t + (1/12)αt^4
y(t) = ∫vy(t) dt = v0y t + (1/2)βt^2 - (1/6)γt^3
Now, let's find the time t1 when the rocket reaches its maximum height:
ay(t1) = 0
β - γt1 = 0
t1 = β/γ = 6.43 s
At t1, the rocket's height is:
y(t1) = v0y t1 + (1/2)βt1^2 - (1/6)γt1^3
y(t1) = 7.00 m/s × 6.43 s + (1/2) × 9.00 m/s2 × (6.43 s)^2 - (1/6) × 1.40 m/s3 × (6.43 s)^3
y(t1) = 92.5 m
Now, let's find the time t2 when the rocket hits the ground. We can do this by solving for the positive root of the quadratic equation:
y(t) = 0
(1/2)γt^2 - βt - v0y = 0
Using the quadratic formula, we get:
t2 = (β + √(β^2 + 2γv0y))/γ = 8.01 s
Finally, let's find the horizontal distance traveled by the rocket:
x(t2) = v0x t2 + (1/12)αt2^4
x(t2) = 1.00 m/s × 8.01 s + (1/12) × 2.50 m/s4 × (8.01 s)^4
x(t2) = 57.4 m
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A man stands on the roof of a building of height 14.0 m and throws a rock with a velocity of magnitude 32.0 m/s at an angle of 29.0 ∘ above the horizontal. You can ignore air resistance.
A) Calculate the maximum height above the roof reached by the rock.
Express your answer in meters
B) Calculate the magnitude of the velocity of the rock just before it strikes the ground.
Express your answer in meters per second.
C) Calculate the horizontal distance from the base of the building to the point where the rock strikes the ground.
Express your answer in meters.
Answer:
A) The maximum height above the roof reached by the rock can be found using the formula:
h = (v₀²sin²θ)/(2g)
where v₀ is the initial velocity (32.0 m/s), θ is the angle of the initial velocity (29.0°), and g is the acceleration due to gravity (9.81 m/s²).
Plugging in the values, we get:
h = (32.0²sin²29.0)/(2(9.81)) = 31.1 m
Therefore, the maximum height above the roof reached by the rock is 31.1 meters.
B) The vertical component of the velocity just before the rock strikes the ground is:
vᵥ = v₀sinθ - gt
where t is the time it takes for the rock to reach the ground.
We can find t by using the formula:
h = v₀sinθt - (1/2)gt²
where h is the height of the building (14.0 m). Rearranging this formula and solving for t, we get:
t = (v₀sinθ + sqrt((v₀sinθ)² + 2gh))/g
Plugging in the values, we get:
t = (32.0sin29.0 + sqrt((32.0sin29.0)² + 2(9.81)(14.0)))/9.81 = 4.01 s
Therefore, the vertical component of the velocity just before the rock strikes the ground is:
vᵥ = 32.0sin29.0 - 9.81(4.01) = -14.3 m/s
Note that the negative sign indicates that the velocity is directed downwards.
C) The horizontal distance from the base of the building to the point where the rock strikes the ground can be found using the formula:
d = v₀cosθt
Plugging in the values, we get:
d = 32.0cos29.0(4.01) = 96.4 m
Therefore, the horizontal distance from the base of the building to the point where the rock strikes the ground is 96.4 meters.
A body of mass 1 kg is pushed at a constant speed against a vertically placed wall, with a constant force F, which forms an angle of 45 degrees with the horizontal, as in the picture. The sliding friction coefficient is 0.2. What is the intensity of the force F? For the acceleration of the heavier force, take g=10 m/s2.
The intensity of the force F is calculated to be approximately 2.83 N.
What is sliding friction coefficient?Coefficient of sliding friction is a value that measures force of sliding friction for particular surface type.
Weight = mg = 1 kg x 10 m/s² = 10 N (acting downwards)
F_vertical - Weight = 0
F_vertical = 10 N
As Frictional force = coefficient of friction x normal force
Frictional force = 0.2 x 10 N = 2 N (acting to the left)
F_horizontal = F x cos(45) = F / √2
F_horizontal - frictional force = 0
F_horizontal = frictional force = 2 N
F / √2 = 2 N
F = 2 N x √2
F ≈ 2.83 N
Therefore, the intensity of the force F is approximately 2.83 N.
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A horizontal ruler from 0 to 100. An image of an upward pointing arrow in a frame, at 0. A double convex lens at 100. X marked in the center of the ruler.
What does the “X” on the horizontal line represent?
Draw the ray diagram in your notes, showing only the principal rays. Explain why you don’t need to draw more rays.
Where will the image appear? On the left or on the right of the lens? At which mark on the ruler?
How will the image look? Upright or inverted? Same size, larger, or smaller?
Horizontal lines are referred to as being parallel to a x-axis in coordinate geometry. A line is referred to as horizontal if two points on the line share the same y-coordinate points.
What does a graph's x-axis / horizontal line represent?The intersection of the vertical and horizontal (X axis) real number lines is shown on the axis graph (Y axis). The Y axis is known as the dependent variable of the data set, whereas the X axis is typically used to refer to the independent variable of the data set.
In the line graph, how does X show up?The horizontal x-axis and vertical y-axis are the two axes of a line graph (vertical). A different type of data is indicated at each of the points where the axes connect, and (0,0). The x-axis is referred to as an independent axis since the numbers it represents are not reliant on any of the variables being assessed.
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5. While driving a car Rahul saw one sign board and suddenly, he reduced the speed. Which of
the following signboard he has seen?
The signboard Rahul saw was likely a speed limit sign. This sign is used to inform drivers of the maximum speed they should be travelling at in that area.
What is limit?Limit is a mathematical concept that describes the highest or lowest value that a given function or expression can reach. It is used as a tool to measure how a function or expression behaves as its variables approach a certain value. For example, the limit of a function as x approaches infinity is the highest value that the function can take on. It is often used to calculate the area under a curve, the rate of change of a function, and the slope of a line at a given point. Limits are also used to compare the relative sizes of different functions, and to determine the continuity of a function.
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Drag each tile to the correct box. Not all tiles will be used. A chemical reaction takes place in which energy is released. Arrange the reaction’s characteristics in order from start to finish. lower energy of reactants higher energy of products higher energy of reactants transition state
Answer:
Start:
Higher energy of reactants -> Transition state
Finish:
Lower energy of reactants -> Higher energy of products
Explanation:
4. A 25,000kg asteroid is flying towards the Earth at a speed of 1500.0m/s. How much Silly Putty would it take to stop the asteroid if we launched it at a velocity of -100.0m/s? Assume the Silly Putty sticks to the asteroid.
To solve this problem, we can use the principle of conservation of momentum, which states that the total momentum of a system remains constant if there are no external forces acting on it. We can set the momentum of the asteroid before the collision equal to the momentum of the asteroid-Silly Putty system after the collision.
The momentum of the asteroid before the collision is:
P_before = m_ast * v_ast
where m_ast is the mass of the asteroid and v_ast is its velocity.
The momentum of the asteroid-Silly Putty system after the collision is:
P_after = (m_ast + m_sp) * v_final
where m_sp is the mass of the Silly Putty, v_final is the velocity of the asteroid-Silly Putty system after the collision, which we assume to be zero.
We can equate these two expressions for momentum and solve for the mass of the Silly Putty:
m_sp = (m_ast * v_ast) / (-v_final)
Substituting the given values:
m_ast = 25,000 kg
v_ast = 1500.0 m/s
v_final = -100.0 m/s
m_sp = (25,000 kg * 1500.0 m/s) / (-(-100.0 m/s))
m_sp = 375,000 kg m/s / 100.0 m/s
m_sp = 3750 kg
Therefore, we would need 3750 kg of Silly Putty to stop the asteroid if we launched it at a velocity of -100.0 m/s.
PLEASE HELPPP!!!! i really need help with this report if anyone can!!!!
The U.S. Army is planning to drop supplies from a plane at a refugee camp. The supplies are divided into 700-kilogram parcels, and the parachutes have an area of 100 square meters. The only problem is that the parcels cannot hit the ground at a velocity of more than 5 meters per second without damaging the contents. Are these parachutes suitable for this task?
For the purposes of this exercise, assume that the for the drag coefficient of the parachute is 1.5 and that the air density is 1.22 kilograms per cubic meter. Write a report detailing why these parachutes are or are not suitable and determining the minimum size parachute that can be used in this situation.
Answer:
search it up
Explanation:
If two balls have the same volume, but ball A has twice as much mass as ball B, which one will have the greater density? If ball C is 3 times the volume of ball D and ball D has 1/3 the mass of ball C, which has the greater density? If two balls have the same mass, but ball P is twice as large as ball Q, which one will have the greater density? If ball X is twice as big as ball Y and weighs only half as much as ball Y, then which one will have the greater densitv? Previous Activity
Ball A will have the greater density because it has twice as much mass as ball B for the same volume. Ball C will have the greater density because it has 3 times the volume of ball D and only 1/3 the mass.
What is volume?Volume is the quantity of three-dimensional space occupied by an object or a substance. It is measured in cubic units, such as liters or gallons. Volume is an important concept in mathematics, physics, chemistry, and engineering, and is often used to calculate the amount of material needed for a certain project. For example, in architecture, engineers may use volume to determine the amount of concrete needed to build a bridge. In cooking, cooks use volume to measure the amount of ingredients needed for a recipe. In physics, volume is used to measure the amount of space an object occupies, or the amount of space within an object, such as a liquid or gas.
Ball P will have the greater density because it is twice as large as ball Q for the same mass. Ball X will have the greater density because it is twice as big as ball Y but weighs only half as much.
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in the summer people often wear light colored clothing to stay cool this is a good idea because light colors tend to_____electromagnetic waves
1. absorb
2. refract
3. reflect
The fastest recorded pitch in Nippon Professional Baseball, thrown by Shohei Otani in 2016, was clocked at 102.5 mi/h. If a pitch were thrown horizontally at this speed, how far would the ball fall vertically (in ft) by the time it reached home plate, 60.5 ft away?
Incorrect: Your answer is incorrect.
ft
To solve this problem, we can use the kinematic equation for vertical motion under constant acceleration, which is given by:
[tex]y = vi*t + (1/2)at^2[/tex]
What is displacement?
Displacement is a vector quantity that refers to the overall change in position of an object from its initial position to its final position.
It is a straight line distance between the initial and final position, in a specific direction.
where y is the displacement (in this case, the vertical distance the ball falls), vi is the initial velοcity (0 in this case since the ball is thrοwn hοrizοntally), a is the acceleratiοn due tο gravity[tex](-32.2 ft/s^2)[/tex], and t is the time it takes fοr the ball tο reach hοme plate. We can find t by dividing the distance tο hοme plate by the hοrizοntal velοcity:
[tex]t = 60.5 ft / (102.5 mi/h * 5280 ft/mi * 1 h/3600 s) = 0.400 s[/tex]
Now we can use the kinematic equation to find y:
[tex]y = 0 + (1/2)(-32.2 ft/s^2)(0.400 s)^2 = -2.57 ft[/tex]
Note that the negative sign indicates that the displacement is downward. Therefore, the ball falls about 2.57 feet vertically by the time it reaches home plate.
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Which statement correctly identifies and interprets the figurative
language in this excerpt from the passage?
Other times, he seemed to saw himself in two, his legs driving him one way
while his head and torso faked another, his body rejoining at the rim to lay the
ball in with a knowing smirk.
The metaphor "sawing himself in two" highlights the basketball player's agility as his legs and upper body move in opposing directions before coming together at the rim to hit a shot.
Which one best defines figurative language?Essentially, employing figurative language involves distorting the meaning of words to convey a point, sound clever, or create a joke. Figurative language is a common technique used in narrative writing when the author wants to make the reader feel strongly about something.
What circumstance might figurative language be used in?Literature forms like poetry, drama, prose, and even speeches use figurative language. Figurative language is a literary element that is employed throughout
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Since the moon rotates on its own axis while revolving around the Earth, shouldn’t we be able to see all sides of it? So, why do we see only one side of the moon?
A block of mass m is supported by two identical parallel vertical springs, each with spring stiffness constant k. What will be the frequency of vibration? The answer is not a number, but an equation.
f=*****
The frequency of vibration of the block supported by two identical parallel vertical springs with spring stiffness constant k and mass m is [tex](1 / 2\pi) * \sqrt(2k / m).[/tex]
What does physics mean by vibrational frequency?In physics, frequency is the number of waves that pass a fixed point in a unit of time as well as the number of cycles or vibrations that a body in periodic motion experiences in a unit of time.
The frequency of vibration of a mass-spring system is given by the formua:
[tex]f = (1 / 2\pi) * \sqrt(k / m)[/tex]
where f is the frequency of vibration, k is the spring constant, and m is the mass of the object.
In this case, the block is supported by two identical parallel vertical springs, each with spring stiffness constant k. So, the effective spring constant is the sum of the individual spring constants:
k_eff = 2k
The mass of the block is given as m.
So, the frequency of vibration of the block supported by two identical parallel vertical springs can be calculated as:
[tex]f = (1 / 2\pi) * \sqrt(k_eff / m) = (1 / 2\pi) * \sqrt((2k) / m)\\f = (1 / 2\pi) * \sqrt(2k / m)[/tex]
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What is the tension on a stone of mass 50 g, tied to a string of length 50 cm and rotated at a speed of 1 m/s?
0.1
100
10
Answer: 0.1
Explanation:
To find the tension on the stone, we can use the centripetal force formula, which is given by F = (mv^2)/r, where m is the mass of the object, v is its velocity, and r is the radius of the circular path.
In this case, the stone is tied to a string and is moving in a circle of radius 50 cm (or 0.5 m), so we have:
F = (0.05 kg) x (1 m/s)^2 / 0.5 m
F = 0.1 N
Therefore, the tension on the string is 0.1 N.
So, the correct answer is 0.1.
A block of mass m = 1.9 kg is attached to a string that is wrapped around the circumference of a wheel of radius R = 8.1 cm. The wheel rotates freely about its axis and the string wraps around its circumference without slipping. Initially, the wheel rotates with an angular speed ω, causing the block to rise with a linear speed v = 0.43m/s
Find the moment of inertia of the wheel if the block rises to a height of h
= 7.5 cm before momentarily coming to rest.
The moment of inertia of the wheel if the block rises to a height of h is 7.5 cm before momentarily coming to rest is 0.068 kg m².
We can use the conservation of mechanical energy to solve for the moment of inertia of the wheel. Initially, the system has kinetic energy, which is converted to potential energy at the highest point of the block's trajectory. Therefore, we can write:
Initial kinetic energy = Final potential energy
At the start, the wheel and block have kinetic energy due to the motion of the block:
KE i = 0.5 * m * v²
At the highest point of the block's trajectory, all of the kinetic energy is converted to potential energy due to the block's height above the ground:
PE_f = m * g * h
where g is the acceleration due to gravity.
Since the string is wrapped around the circumference of the wheel, the distance that the block moves upwards is equal to the distance that the string moves around the wheel, which is equal to the circumference of the wheel:
h = 2 * pi * R
Substituting this into the expression for potential energy:
PE_f = m * g * 2 * π * R
Equating initial kinetic energy with final potential energy:
0.5 * m * v² = m * g * 2 * π * R
Simplifying and solving for the moment of inertia of the wheel, I:
I = (m * v²) / (2 * g * π * R)
Substituting the given values:
I = (1.9 kg * 0.43 m/s)² / (2 * 9.81 m/s² * π * 0.081 m)
I = 0.068 kg m²
Therefore, the moment of inertia of the wheel is 0.068 kg m².
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We measure the intensity of a sound source in open air to be 0.3 W/m2 when we are located a distance of 15m away from the source. If we were to move to a distance of 25 m, what would be the intensity of the sound? How about at a distance of 45 m?
The intensity of sound are- For distance of 25 m: I = 0.108 W/m² and For distance of 45 m: I = 0.034 W/m².
Explain about the intensity of a sound?The sound becomes softer the further you get from the sound source, particularly when you are outside. This is not shocking at all. Rather like light, sound expands out as it moves away from where it originated. The strength of the sound decreases as you move further away from the source if there aren't surfaces for it to reflect from.Your comprehension of the inverse square law, which states that a sound's intensity varies inversely to the square of its distance from its source, will be put to the test in this challenge.
I = k • (1/R²)
when, distance R is 15m , intensity of a sound source in open air to be 0.3 W/m².
So,
0.3 = k • (1/15²)
k = 225 * 0.3
k = 67.5
For distance of 25 m:
I = k • (1/R²)
I = 67.5 • (1/25²)
I = 0.108 W/m².
For distance of 45 m:
I = k • (1/R²)
I = 67.5 • (1/45²)
I = 0.034 W/m².
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A car is moving at 32 miles per hour. The kinetic energy of that car is 5 × 10^5 J.
How much energy does the same car have when it moves at 101 miles per hour?
Answer in units of J.
Answer:
The car has approximately 1.42 × 10^6 J of energy when it moves at 101 miles per hour.
Explanation:
First, we need to convert the initial velocity and kinetic energy to SI units:
Initial velocity: 32 miles per hour = 14.3 meters per second (rounded to 2 decimal places)
Kinetic energy: 5 × 10^5 J (given)
Next, we can use the formula for kinetic energy:
KE = (1/2)mv^2
where KE is the kinetic energy, m is the mass of the car, and v is the velocity of the car.
Solving for mass:
m = 2KE/v^2
Substituting the given values:
m = 2(5 × 10^5 J) / (14.3 m/s)^2 ≈ 1569.93 kg (rounded to 2 decimal places)
Now, we can use the same formula to calculate the kinetic energy of the car when it moves at 101 miles per hour (rounded to 2 decimal places):
KE = (1/2)mv^2 = (1/2)(1569.93 kg)(45.06 m/s)^2 ≈ 1.42 × 10^6 J
Therefore, the car has approximately 1.42 × 10^6 J of energy when it moves at 101 miles per hour.