Elyse's scientific investigation into the migration of leatherback sea turtles on Australia's east coast is noted for its unusually precise hypothesis of "1,200" turtles migrating. Scientific hypotheses typically predict trends or relationships rather than exact figures. Understanding the broader context of sea turtle conservation is crucial.
Explanation:Elyse is conducting a scientific investigation to determine how many leatherback sea turtles migrate to their home beach on the eastern coast of Australia this season. Her approach involves observing and counting turtles as they land on the beach. An unusual aspect of her description is the formulation of a very specific hypothesis stating "1,200 leatherback sea turtles will migrate to their home beach." Typically, a hypothesis in scientific investigations is a broader statement predicting a relationship or trend, rather than a precise number. Moreover, it's essential to understand the broader context of sea turtle conservation, including the various populations of leatherback turtles in the Pacific Ocean, and why certain beaches are critical to their survival. Factors like natural selection play a vital role in why leatherback turtles favor particular nesting sites, such as the types of beaches that provide suitable conditions for hatchling survival and thus contribute to the species' overall survival.
The manometer shown in fig. 2 contains water and kerosene. with both tubes open to the atmosphere, the free-surface elevations differ by h = 20.0 mm. determine the elevation difference when a pressure of 98.0 pa (gage) is applied to the right tube. (hint: when the gage pressure is applied to the right tube, the water in the right tube is displaced downward by the same distance that the kerosene in the left tube is displaced upward)
The elevation difference is 55.3 mm when a pressure of 98.0 Pa (gage) is applied to the right tube of the manometer, displacing water downward and kerosene upward.
When a pressure of 98.0 Pa (gage) is applied to the right tube of the manometer, the water in the right tube is displaced downward by the same distance that the kerosene in the left tube is displaced upward, maintaining equilibrium.
To find the elevation difference, we can use the hydrostatic pressure formula:
P = ρgh
Where:
P = pressure difference (98.0 Pa in this case)
ρ = density of the fluid (for water, ρ_water ≈ 1000 kg/m³, for kerosene, ρ_kerosene ≈ 820 kg/m³)
g = acceleration due to gravity (approximately 9.81 m/s²)
h = elevation difference
Since the water and kerosene columns have opposite elevation changes, we can write:
98.0 Pa = (ρ_water * g * h) - (ρ_kerosene * g * h)
Now, solve for h:
98.0 Pa = (1000 kg/m³ * 9.81 m/s² * h) - (820 kg/m³ * 9.81 m/s² * h)
98.0 Pa = (9810 h) - (8034 h)
98.0 Pa = 1776 h
h = 98.0 Pa / 1776 = 0.0553 meters = 55.3 mm
So, when a pressure of 98.0 Pa (gage) is applied to the right tube, the elevation difference between the two free surfaces in the manometer is 55.3 mm.
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It is friction that provides the force for a car to accelerate, so for high-performance cars the factor that limits acceleration isn't the engine; it's the tires. you may want to review ( pages 138 - 142) . part a for typical rubber-on-concrete friction, what is the shortest time in which a car could accelerate from 0 to 60 mph? suppose that μs=1.00 and μk=0.80.
The shortest time in which a car could accelerate from 0 mph to 60 mph is [tex]13.7s[/tex].
Further explanation:
The opposite force acting on the body is known as frictional force. It always acts in the opposite direction of motion of body.
Concept used:
The force applied to a body to keep it at rest is known as the static friction force. It always acts opposite to the direction of motion of body. It is defined as the product of coefficient of friction and the normal force acting on the body.
The expression for the normal reaction of the body is given as.
[tex]N = mg[/tex]
The expression for the net force is given as.
[tex]{F_{net}} = ma[/tex] …… (1)
The expression for the static friction is given as.
[tex]{F_s} = {\mu _s}N[/tex]
The expression for the balanced forces is given as.
[tex]{F_{net}} = {F_s} - {F_r}[/tex]
Substitute[tex]{\mu _s}N[/tex] for [tex]{F_s}[/tex] and for[tex]{F_r}[/tex] in the above expression.
[tex]\begin{aligned}{F_{net}}&={\mu _s}N-{\mu _r}N\\&= \left( {{\mu _s} - {\mu _r}} \right)N \\ \end{aligned}[/tex]
Substitute [tex]mg[/tex] for [tex]N[/tex] in above expression.
[tex]{F_{net}}=\left({{\mu _s}-{\mu _r}}\right)\left( {mg}\right)[/tex] …… (2)
Compare equation (1) and (2) we get.
[tex]a=g\left({{\mu _s}-{\mu _r}}\right)[/tex] …… (3)
Here, [tex]a[/tex] is the acceleration of the body, g is the acceleration due to gravity, [tex]{\mu _s}[/tex] is the coefficient of static friction and is the coefficient of reactive force.
The expression for the first equation of motion is given as.
[tex]v = u + at[/tex]
Rearrange the above expression for time is given as.
[tex]\fbox{\begin\\t = \dfrac{{\left( {v - u} \right)}}{a}\end{minispace}}[/tex] …… (4)
Here, [tex]v[/tex] is the final velocity, [tex]u[/tex] is the initial velocity and [tex]t[/tex] is the time.
Substitute [tex]1[/tex] for [tex]{\mu _s}[/tex], [tex]0.8[/tex] for[tex]{\mu _r}[/tex] and [tex]9.8\,{\text{m/}}{{\text{s}}^{\text{2}}}[/tex] for [tex]g[/tex] in equation (3).
[tex]\begin{aligned}a&=\left( {9.8\,{\text{m/}}{{\text{s}}^{\text{2}}}}\right)\left( {1 - 0.8}\right)\\&=1.96\,{\text{m/}}{{\text{s}}^{\text{2}}} \\ \end{aligned}[/tex]
Substitute [tex]0\,{\text{mph}}[/tex] for [tex]u[/tex], [tex]60\,{\text{mph}}[/tex] for [tex]v[/tex] and [tex]1.96\,{\text{m/}}{{\text{s}}^{\text{2}}}[/tex] for [tex]a[/tex] in equation (4).
[tex]\begin{aligned}t&=\frac{{\left( {60\,{\text{mph}}-0\,{\text{mph}}}\right)}}{{1.96\,{\text{m/}}{{\text{s}}^{\text{2}}}}}\\&=\frac{{60\,{\text{mph}}\left( {\frac{{0.447\,{\text{m/s}}}}{{1\,{\text{mph}}}}} \right)}}{{1.96\,{\text{m/}}{{\text{s}}^{\text{2}}}}}\\&= 13.7\,{\text{s}} \\ \end{aligned}[/tex]
Thus, the time required to accelerate a car is [tex]\fbox{\begin\\13.7\, {\text{s}}\end{minispace}}[/tex].
Learn more:
1. Conservation of momentum https://brainly.com/question/9484203.
2. Motion under friction https://brainly.com/question/7031524.
3. Net force on the body https://brainly.com/question/4033012.
Answer Details:
Grade: College
Subject: Physics
Chapter: Kinematics
Keywords:
Force, friction, Acceleration, acceleration due to gravity, normal, weight, mass, motion, sliding, sled, hill, inclined, plane, coefficient of friction, angle of inclination, 13.68 s, 13.69 s.
Final Answer:
So, under ideal conditions with the assumption of maximum static friction being available throughout the acceleration, the shortest time in which a typical car could accelerate from 0 to 60 mph on rubber-on-concrete would be approximately 2.734 seconds.
Explanation:
To solve this problem, we need to understand how the force of friction allows a car to accelerate and then apply the equations of motion to determine the shortest time it would take for the car to accelerate from 0 to 60 miles per hour (mph).
First, we need to define our variables and constants:
- The static friction coefficient between rubber and concrete, μs, which is 1.00. This is the factor that will limit our acceleration since it represents the maximum frictional force that can be exerted before sliding begins.
- The acceleration due to gravity, g, which is 9.81 m/s^2.
- The initial speed of the car, 0 mph, which needs to be converted to meters per second (m/s).
- The final speed of the car, 60 mph, which also needs to be converted to meters per second.
The force of static friction (f_friction) responsible for the acceleration can be calculated by multiplying the static friction coefficient (μs) with the normal force (N). The normal force, in this case, is equal to the weight of the car, which is the mass (m) times the acceleration due to gravity (g). Since we are looking for the highest possible acceleration, we can assume that the force of static friction is at its maximum value.
f_friction = μs * m * g
However, we don't need to know the mass of the car because it will cancel out in the equation of motion. The maximum possible acceleration (a_max) happens when the force of static friction is at its highest, and it can be calculated using:
a_max = f_friction / m = μs * g
Now we have the maximum possible acceleration the tires can provide on concrete, which is:
a_max = 1.00 * 9.81 m/s^2
a_max = 9.81 m/s^2
Next, let's convert the final velocity from mph to m/s:
v_final(mph) = 60 mph
v_final(m/s) = v_final(mph) * 0.44704 (conversion factor from mph to m/s)
v_final(m/s) = 60 * 0.44704
v_final(m/s) ≈ 26.8224 m/s
With the initial velocity (v_initial) being 0 m/s (since we start from rest), we can use one of the kinematic equations to find the time (t):
v_final = v_initial + a_max * t
26.8224 m/s = 0 m/s + 9.81 m/s^2 * t
Solving for t, we get:
t = v_final / a_max
t ≈ 26.8224 m/s / 9.81 m/s^2
t ≈ 2.734 seconds
So, under ideal conditions with the assumption of maximum static friction being available throughout the acceleration, the shortest time in which a typical car could accelerate from 0 to 60 mph on rubber-on-concrete would be approximately 2.734 seconds.
Which statement best describes a commission system? A legislative body hires a leader to oversee city departments. An elected leader and an elected legislative body share responsibility. A leader oversees city departments, and a legislative body has limited power. A legislative body oversees city departments directly.
Answer;
A legislative body oversees city departments directly.
Explanation;
City commission government is a form of local government in the United States. In a city commission government, voters elect a small commission, typically of five to seven members, on a plurality-at-large voting basis.
The commissioners constitute the legislative body of the city and, as a group, are responsible for taxation, appropriations, ordinances, and other general functions.
Answer:
a legislative body oversees city departments directly
Explanation:
The component of the ball's velocity whose magnitude is most affected by the collisions is
If the box is a distance 1.81 m from the rear of the truck when the truck starts, how much time elapses before the box falls off the truck?
To find the time it takes for the box to fall off the truck, we use the equation of motion and the acceleration of the box relative to the truck. By setting the displacement equal to the distance from the rear of the truck and solving for time, we find that it takes approximately 1.34 seconds for the box to fall off.
Explanation:To find the time it takes for the box to fall off the truck, we need to first find the acceleration of the box relative to the truck. Given that the truck is accelerating forward at 5.0 m/s² and the crate is accelerating forward at 2.94 m/s², we can subtract these accelerations to find the crate's acceleration relative to the truck, which is -2.06 m/s². Since the box falls off when it reaches a certain distance from the rear of the truck, we can use the equation of motion to find the time it takes to travel that distance.
The equation is:
d = ut + 1/2at²
Where:
d is the displacement or distance traveledu is the initial velocity (0 m/s since the box starts from rest on the truck)t is the time takena is the accelerationPlugging in the values, we have:
1.81 m = 0 * t + 0.5 * (-2.06) * t²
Simplifying the equation:
1.81 = -1.03t²
Multiplying both sides by -1:
-1.81 = 1.03t²
Dividing both sides by 1.03:
t² = -1.81/1.03
When we take the square root of each side, we obtain:
t = ±√(-1.81/1.03)
The negative value of time implies an event before the start of motion, so we discard it. Therefore, the time it takes for the box to fall off the truck is approximately 1.34 seconds.
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Calculate the change in temperature of 20 kg of water if 30 kg of aluminum is dropped in the water and the aluminum changes temperature by 20°C (Hint: use the principle of conservation of energy.) °C
Answer:
6.6°C
Explanation:
A 55 kg person moves at the constant speed of 7 m/s along a straight stretch of track for 20s. How far does he travel in this time ?
How will sunlight most likely affect a black shirt on a hot summer day?
The sunlight is absorbed more on to the black color and no color is reflected back therefore, heat energy is released by the black color and make more hot.
What is light reflection?When a light ray is incident on a transparent or translucent material, the light ray will return back after some fraction of it is absorbed on the material and the light ray travel in the straight direction from the material make it gleams.
White light absorbs all colors and reflects back all colors. Thus, the mixing of all reflected colors make the material white in color. A black light absorbs all colors and reflect no color.
The absorbed light energy in the black body is released as heat energy instead. This make the black colored material more hotter than other colors. Hence, a black shirt is most affected by sunlight.
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A 8.40 kg mass suspended from a spring with spring constant, k = 875.0 n/m, extends it to a total length of 0.250 m. find the total length of the spring when a 17.20 kg mass is suspended from it.
The total length of the spring when a 17.20 kg mass is suspended from it is 34.85 cm, calculated using Hooke's Law and the original unstretched length of the spring.
Explanation:To find the total length of the spring when a 17.20 kg mass is suspended from it, we use Hooke's Law, which is F = kx, where F is the force applied to the spring, k is the spring constant, and x is the extension from the spring's original unstretched length. First, we need to calculate the extension caused by the 8.40 kg mass: F = (8.40 kg)(9.8 m/s2) = 82.32 N. With k = 875 N/m, the extension (x) is F/k = 82.32 N / 875 N/m = 0.0941 m or 9.41 cm. This extension is in addition to the unstretched length of the spring (L0), given by the total length minus the extension for the 8.40 kg mass: L0 = 0.250 m - 0.0941 m = 0.1559 m.
Next, for the 17.20 kg mass, the force is F = (17.20 kg)(9.8 m/s2) = 168.56 N. The additional extension caused by this mass is x = F/k = 168.56 N / 875 N/m = 0.1926 m or 19.26 cm. Therefore, the total length of the spring with the 17.20 kg mass is L = L0 + x = 0.1559 m + 0.1926 m = 0.3485 m or 34.85 cm.
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Which of the following is not a danger associated with hurricanes?
a) droughts
b) inland flooding
c) storm surges
d) tornadoes
The correct answer is a) droughts. Hurricanes are typically associated with heavy rain and high winds, which can lead to flooding.
Explanation:Hurricanes are powerful storms that cause a variety of hazards, including inland flooding, storm surges, and tornadoes.
Inland flooding can occur when the heavy rainfall associated with a hurricane overflows rivers and streams. Storm surges are the rise in sea level that happens when a hurricane's strong winds push water onshore. This can cause significant flooding and damage in coastal areas. Tornadoes can be produced by hurricanes when thunderstorms present within the hurricane system become severe.
However, droughts are not typically a direct result of hurricanes. In fact, the heavy rainfall associated with hurricanes often alleviates drought conditions.
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Some planetary scientists have suggested that the planet mars has an electric field somewhat similar to that of the earth, producing a net electric flux of 3.66×1016 n⋅m2/c into the planet's surface. part a calculate the total electric charge on the planet.
Answer:
Electric charge, q = 323910 C
Explanation:
It is given that,
Net electric flux, [tex]\phi=3.66\times 10^{16}\ Nm^2/C[/tex]
We have to find the total electric charge on the planet. We can find it using Gauss's law. It is as follows :
[tex]\phi=\dfrac{q}{\epsilon_0}[/tex]
where
q is the net electric charge
[tex]\epsilon_0[/tex] is the electric permeability
So, net electric charge is given by :
[tex]q=\phi\times \epsilon_0[/tex]
[tex]q=3.66\times 10^{16}\ Nm^2/C\times 8.85\times 10^{-12}\ F/m[/tex]
[tex]q=323910\ C[/tex]
Hence, this is the required solution.
what approximately is the percent uncertainty for the measurement 5.2
The percent uncertainty of a 5 lb bag with an uncertainty of ±0.4 lb is calculated using the formula (SA / A) x 100%, resulting in a percent uncertainty of 8%.
Explanation:To calculate the percent uncertainty for a 5 lb bag with an uncertainty of ±0.4 lb, you would use the following formula:
% uncertainty = (SA / A) × 100%
Where A is the expected value (5 lb in this case) and SA is the uncertainty in the value (0.4 lb). Plugging in the numbers:
% uncertainty = (0.4 lb / 5 lb) × 100% = 8%
This means the percent uncertainty of the bag's weight is 8%.
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The sign has a mass of 100 kg with center of mass at g. determine the x, y, z components of reaction at the ball-and-socket joint a and the tension in wires bc and bd.
The problem is about finding the reaction components at a ball-and-socket joint and the tension in wires supporting a sign. You can do this by breaking the forces and moments into their x,y,z components, setting up force balance and moment balance equations, and solving for the unknowns.
Explanation:This problem pertains to static equilibrium, where both the sum of forces and the sum of torques are zero. To begin, let's establish a coordinate system where the ball-and-socket joint 'a' is considered the origin.
Assuming the tension in the wires is T and the angles of the wires to the x, y, z axes are known, you can break the tensions into their respective components using trigonometric principles. Let's consider Tbc and Tbd to be the tensions in the strings.
For the x-component we would sum up the forces in the x-direction and set it equal to zero. This can be represented as ∑Fx = Tbc,x + Tbd,x=0
The same goes for the y and z components with the z component taking into account the force due to gravity on the sign. Hence, ∑Fy = Tbc,y + Tbd,y=0 and ∑Fz = Tbc,z + Tbd,z - weight of the sign = 0. Here, the weight of the sign is 100kg*9.81m/s2
By balancing torques about each axis, you can find the tensions in the strings. The tension force should be the same in each wire if they make the same angles with the axes and are of equal length.
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This problem in Physics requires setting up equilibrium equations based on the given mass and calculating forces at the ball-and-socket joint and cable pressures.
Explanation:The subject of this question falls under the branch of
Physics
referred to as statics, specifically the study of systems in equilibrium. Equilibrium implies that an object is neither accelerating nor rotating, meaning the sum of all forces and the sum of all torques acting on the object must both be zero. The 'x', 'y' and 'z' components referred to are likely part of a coordinate system used to describe the forces at work on the sign.
To determine these, one would need to set up equilibrium equations based on the known details about the system, which could include the known weight (provided by the mass of the sign and gravity), any relevant distances for calculating torques, and any other forces present such as tension in wires. The process of solving this system of equations would then yield the desired reaction components at the ball-and-socket joint and the tensions in the wires.
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PLZ HELP: Which food items would be appropriate for a spacecraft? Check all that apply. A< brownie, B< dry salt and pepper, C
Which instrument provides the only means of making nearly continuous measurements of surface conditions over the ocean?
What force is needed to accelerate a 55kg shopping cart 2.0m/s to the second power?
A child sets off the firecracker at a distance of 100 m from the family house. what is the sound intensity β100 at the house?
To solve this problem, we use the formula:
I100 / I1 = [P / 4π(100m)^2] / [P / 4π(1m)^2]
I100 / I1 = 1 / 100^2
I100 / I1 = 10^-4
Therefore the change in intensity from 1m to 100m in decibels is:
B100 – B1 = 10 log(10^-4) dB = -40 dB
So the intensity at 100m is calculated as:
B100 = B1 – 40 dB = 140 dB – 40 dB = 100 dB
Answer:
100 dB
The intensity of the sound produced due to the firecracker at the distance of [tex]100\text{ m}[/tex] is [tex]\fbox{\begin\\100\text{ dB}\end{minispace}}[/tex].
Further Explanation:
The vibration, produced in the medium, which travels as an wave of pressure or density through a medium is known as sound. The sound is a longitudinal wave and it requires a medium for its propagation.
Given:
The distance between the place where firecracker is set off by a child and the family house is [tex]100\text{ m}[/tex].
The intensity of the sound produced by the firecracker at the distance [tex]1\text{ m}[/tex] is[tex]140\text{ dB}[/tex].
Concept:
The intensity of the sound wave is defined as the power carried by the sound waves in the direction perpendicular to the direction of propagation per unit time.
The intensity of the sound wave is:
[tex]\fbox{\begin\\I=\dfrac{P}{4\pi r^2}\end{minispace}}[/tex] ...... (1)
Here, [tex]I[/tex] is the intesnity of the sound wave, [tex]P[/tex] is the power carried by the sound wave and [tex]r[/tex] is the distance between source and the listener.
The intensity of the sound in decible at the house is:
[tex]\fbox{\begin\\\beta _{100}=\beta _{1}+10\log \dfrac{I_{100}}{I_{1}}\end{minispace}}[/tex] ...... (2)
Here, [tex]\beta _{100}[/tex] is the intensity of the sound at the distance of [tex]100\text{ m}[/tex], [tex]\beta _{1}[/tex] is the intensity of sound produced by the firecracker at the distance of [tex]1\text{ m}[/tex], [tex]I_{100}[/tex] is the intensity of sound produced by firecracker at [tex]100\text{ m}[/tex] and [tex]I_{1}[/tex] is the intensity of sound produced by firecracker at [tex]1\text{ m}[/tex].
The ratio of the intensity of sound at distance [tex]100\text{ m}[/tex] and [tex]1\text{ m}[/tex] is:
[tex]\fbox{\begin\\\dfrac{{{I_{100}}}}{{{I_1}}}=\frac{{r_1^2}}{{r_2^2}}\end{minispace}}[/tex] …… (3)
Calculation:
Substitute the values in equation (3).
[tex]\begin{aligned}\frac{{{I_{100}}}}{{{I_1}}}&=\frac{{{{\left( 1 \right)}^2}}}{{{{\left( {100} \right)}^2}}}\\&={10^{ - 4}}\\\end{aligned}[/tex]
Substitute the values in equation (2).
[tex]\begin{aligned}{\beta _{100}}&=140\,{\text{dB}} - 40\,{\text{dB}} \\&=100\,{\text{dB}}\\\end{aligned}[/tex].
Thus, the intensity of the sound produced due to the firecracker at the distance of [tex]100\text{ m}[/tex] is [tex]\fbox{\begin\\100\text{ dB}\end{minispace}}[/tex]
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1. The motion of a body under friction brainly.com/question/4033012
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3. Conservation of energy brainly.com/question/3943029
Answer Details:
Grade: College
Subject: Physics
Chapter: Waves and Oscillation
Keywords:
Intensity of sound, sound waves, vibration, firecracker, power carried by sound waves, 100db, 100 dB, 100dB, 100 m, bursting of firecrackers, sound inside the house, 140 db, 140 dB.
A uniform disk is constrained to rotate about an axis passing through its center and perpendicular to the plane of the disk. if the disk starts from rest and is then brought in contact with a spinning rubber wheel, we observe that the disk gradually begins to rotate too. if after 22 s of contact with this spinning rubber wheel, the disk has an angular velocity of 3.0 rad/s, find the average angular acceleration that the disk experiences. (assume the positive direction is in the initial direction of rotation of the disk. indicate the direction with the sign of your answer.)
A ball is thrown horizontally from a cliff at a speed of 10 m/s. you predict that its speed 1 s later will be slightly greater than 14 m/s. your friend says it will be 10 m/s. show who is correct.
Answer:
Speed is slightly greater than 14 m/s.
Explanation:
The ball will under go projectile motion. The horizontal velocity remains constant while there would be increase in vertical velocity due to acceleration due to gravity in downward direction.
Using first equation of motion, after 1 s, vertical velocity will be:
v = u +at
v=0+(9.8 m/s²)(1 s) = 9.8 m/s
Horizontal velocity = 10 m/s
Net velocity:
[tex]V= \sqrt{(9.8)^2+(10)^2}=\sqrt{196.04}=14.001 m/s[/tex]
Speed is slightly greater than 14 m/s.
Two waves of equal magnitude but opposite phase interfere with one another to cause the waves to disappear. This is caused by:
each wave canceling the other out, causing the waves to disappear completely
refraction of waves multiplying each other's amplitude with the effect of canceling each other out
the Conner effect
destructive interference giving the effect of making the waves disappear, but each will travel on its way
For an increase in the bulk modulus of a material but without any change in the density, what happens to the speed of sound in that material?
The speed of sound within a substance depends on the bulk modulus and density of the material. An increase in bulk modulus, with density remaining constant, would increase the speed of sound within the material, given that the speed of sound is proportional to the square root of the bulk modulus.
Explanation:The speed of sound within a substance is primarily determined by two physical properties of the material, namely the bulk modulus and density. The bulk modulus measures a substance’s resistance to uniform compression. The formula used to calculate the speed of sound within a substance is v = sqrt(B/p), where 'v' represents the speed of sound, 'B' signifies the bulk modulus of the material, and 'p' indicates the density of the material.
As such, if the bulk modulus of a material were to increase while the density remained constant, the speed of sound within that material would also increase because the speed of sound is directly proportional to the square root of the bulk modulus. For instance, since solids and liquids are generally more rigid (i.e., have higher bulk modulus values) than gases, the speed of sound tends to be greater in these media than in gases, assuming density remains constant.
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Suppose a supernova occurs in a star that initially had 10 solar masses. why would the likely end result not be a black hole?
A star with an initial mass of 10 solar masses will likely form a neutron star instead of a black hole after a supernova due to the mutual repulsion between densely packed neutrons that can support the core against its own gravitational weight.
Explanation:
Given a star with initial 10 solar masses that ends its life in a supernova explosion, it is more likely to form a neutron star rather than a black hole. Most stars end their life as white dwarfs or neutron stars. In the case of a very massive star, if the core's remaining mass is more than about three times that of the Sun, it may collapse into a black hole. However, if the mass is less than about three solar masses, the mutual repulsion between densely packed neutrons can support the core against its own weight, preventing it from becoming a black hole.
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________ occurs when an object in the outer reaches of the solar system passes between earth and a far distant star, temporarily blocking light from the star.
Answer:
Occulation
Explanation:
What would most likely happen if the power lines in a pretroleum power plant were broken
A 1490-kilogram truck moving with a speed of 27.0 m/s runs into the rear end of a 1280-kilogram stationary car. if the collision is completely inelastic, how much kinetic energy is lost in the collision?
The loss in kinetic energy in an inelastic collision is equal to 251104 J.
What is law of conservation of momentum?According to the law of conservation of linear momentum, the sum of the momentum before and after an inelastic collision must be equal.
m₁u₁ + m₂ u₂ = m₁ v₁ + m₂ v₂
where m₁ and m₂ is the mass of the collided bodies, u₁ & u₂ are their initial speed while v₁ & v₂ is their final speed.
Given the initial velocity of the truck is v₁ = 27 m/s and the other car is at rest, u₂ = 0 m/s.
The mass of the truck, m₁ = 1490 Kg
The mass of the stationary car, m₂ = 1280 Kg
From the law of conservation of momentum, find the final speed after a collision:
m₁ u₁ + m₂ u₂ = (m₁ + m₂)v'
1490 × 27 + 1280 ×0 = (1490 + 1280) v'
v' = 14.52 m/s
The loss in the kinetic energy = [tex]=\frac{1}{2} mv_1^2-\frac{1}{2} (m_1+m_2)^2v^'^2[/tex]
[tex]==\frac{1}{2} \times 1490\times(27)^2-\frac{1}{2} (2770)^2(14.52)^2[/tex]
[tex]= 543105 -292000\\=251104 J[/tex]
Therefore, the loss in the kinetic energy in the collision is equal to 251104 J.
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A 200 kg weather rocket is loaded with 100 kg of fuel and fired straight up. it accelerates upward at 30.0 m/s2 for 30.0s, then runs out of fuel. ignore any air resistance effects. part a what is the rocket's maximum altitude?
A train accelerates from its metropolitan cruising speed of 22 m/s to its countryside cruising speed of 32 m/s. If it takes the train 142 seconds to do this, what is its acceleration?
10 m/s2
14.2 m/s2
4.4 m/s2
0.07 m/s2
Answer:
Acceleration of the train is 0.07 m/s²
Explanation:
Initial speed of the train, u = 22 m/s
Final speed of the train, v = 32 m/s
Time taken by the train to do the process is 142 seconds
We have to find the acceleration of the train. It is given by the rate of change of speed of the train i.e.
[tex]a=\dfrac{v-u}{t}[/tex]
[tex]a=\dfrac{32\ m/s-22\ m/s}{142\ s}[/tex]
[tex]a=0.07\ m/s^2[/tex]
So, the acceleration of the car is 0.07 m/s². Hence, the correct option is (d) 0.07 m/s²
A heat pump absorbs heat from the cold outdoors at 3°c and supplies heat to a house at 20°c at a rate of 30,000 kj/h. if the power consumed by the heat pump is 3 kw, the coefficient of performance of the heat pump is
The coefficient of performance (COP) of a heat pump is a measure of its effectiveness and is defined as the ratio of the heat transfer into the warm space to the work input required by the heat pump.
Explanation:The coefficient of performance (COP) of a heat pump is a measure of its effectiveness and is defined as the ratio of the heat transfer (Qh) into the warm space to the work input (W) required by the heat pump. Mathematically, COP = Qh/W.
In this case, the heat pump absorbs heat from the cold outdoors at 3°C and supplies heat to a house at 20°C. The rate of heat transfer is given as 30,000 kJ/h and the power consumed is 3 kW.
To find the COP, we need to convert the rate of heat transfer and power consumed to the same units. Since 1 W = 1 J/s and 1 kJ = 1000 J, we have Qh = 30,000 kJ/h x 1000 J/kJ / 3600 s/h = 8333.33 J/s and W = 3kW x 1000 J/W = 3000 J/s.
Substituting these values into the formula, we get COP = 8333.33 J/s / 3000 J/s = 2.78.
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you toss a coin into a wishing well full of liquid denser than the coin. witch of the following could be true? the coin will
A hardworking ant must supply 0.00805 n to pull a small piece of fruit at constant velocity over a distance of 9.80 cm up a small hill to the entrance of the colony. if the coefficient of kinetic friction between the piece of fruit and the 15.8° sloped hill is 0.400, calculate the work done by the ant by pulling the piece of fruit up the hill.
In physics, the definition of work is simply the product of force exerted and the displacement. Therefore in this case, work done by the ant is:
Work = 0.00805 N * 9.80 cm * (1 m / 100 cm)
Work = 7.889 x 10^-4 J