What happened when a lobster release a claw? a. The lobster attacks with the claw. b. The lobster let the claw go. C. The lobster bites off the claw. D. The lobster eats with the claw.
Answer:b
Explanation:
Answer:
The lobster let the claw go. so, B.
Explanation:
i hope this helps.
URGENTLY NEED HELP!!!!!!
the stretch of a spring and the force it exerts are inversely proportional.
Select one:
O True
O False
Answer:
False.
Explanation:
Another box of samples is hoisted up by the same rope. If the rope is shaken with the same frequency as before, and the wavelength is found to be 7.9 m , what is the mass of this box of samples
The motion of the rope which is perpendicular to the direction of the
propagation of the wave is a transverse wave motion.
The mass of the box is approximately 9.93 kgReasons:
The given function for the wave speed is presented as follows;
[tex]\displaystyle v = \sqrt{\frac{T}{\mu} } \\[/tex]
Where;
[tex]\displaystyle \mu = \frac{Mass \ of \ rope }{Length \ of \ rope}[/tex]
Taking the mass of the rope as, m = 2.00 kg
The length of the rope, L = 80.0 m
The mass hanging on the rope, M = 20.0 kg
We have;
T = 20.0 kg × 9.81 m/s² = 196.2 N
[tex]\displaystyle \mu = \frac{2.0 }{80.0} = \frac{2.0 }{80.0} = 0.025[/tex]
Therefore;
Taking the wavelength as, λ = 7.9 m, and the frequency as 20 Hz, we have;
v = f × λ
Therefore;
v = 7.9 Hz × 7.9 m = 62.41 m/s
Which gives;
[tex]\displaystyle 62.41 = \sqrt{\frac{T}{0.025} }[/tex]
T = 62.41² × 0.025 = 97.3752025
[tex]\displaystyle Mass, \ m = \mathbf{\frac{T}{g}}[/tex]
Where;
g = The acceleration due to gravity which is approximately 9.81 m/s²
[tex]\displaystyle Mass, \ m = \frac{97.3752025}{9.81} \approx 9.93[/tex]
Therefore;
The mass of the box, m ≈ 9.93 kg
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The parameters obtained from a similar question online are;
[tex]\displaystyle The \ equation \ applied, \, v = \sqrt{\frac{T}{\mu} } \\[/tex]
Length of the rope, L = 80.0 m
Mass of the rope, m = 2.0 kg
Frequency of a point on the rope, f = 20 Hz
A weather forecaster uses a computational model on a Monday to predict the weather on Friday. Why might that forecast change? (1 point)
A. The forecaster may have done the calculations a second time because he made a mistake in the calculation.
B. Someone may have reported the weather incorrectly before the first computation.
C. The forecaster may have had the computer model do the calculations a second time, and he found that the prediction changed.
D. An area of low pressure might move more quickly on Tuesday and Wednesday than expected.
Answer:
(d) is your answer for your question
a wave travels at a constant speed. how does the wavelength change if the frequency is reduced by a factor of 3? assume the speed of the wave remains unchanged.
A. the wavelength decreases by a factor of 3
B. the wavelength does not change
C. the wavelength increases by a factor of 3
D. the wavelength increases by a factor of 9
Answer:
b
Explanation:
when the wavelength increase it doesnt affect the frequency of a wave.
Answer: The wavelength increases by a factor of 3
Explanation:
Entra vapor a una tobera adiabática con un flujo másico de 250 kg/h. Al entrar el vapor
tiene una energía interna específica de 2510 kJ/kg, una presión de 1378 kPa, un
volumen específico de 0.147 m3
/kg y una velocidad de 5 m/s. Las condiciones de salida
son P= 138.7 kPa, volumen específico de 1.099 m3
/kg y energía interna específica de
2263 kJ/kg. Determine la velocidad de salida.
Este problema describe el funcionamiento de una tobera adiabática (sin flujo de calor), la cual tiene una corriente de entrada que difiere de la de salida espacial y energéticamente, pero que conservan el mismo flujo másico de 250 kg/h. De este modo, usamos un balance de energía con el fin the determinar la velocidad a la que sale el fluido, según es requerido en el problema:
[tex]mu_1+mP_1v_1+mgh_1+\frac{1}{2} mv_1^2=mu_2+mP_2v_2+mgh_2+\frac{1}{2} mv_2^2[/tex]
En la que se tiene que la incógnita es la velocidad de salida, [tex]v_2[/tex], y es posible simplificar el flujo másico, [tex]m[/tex], al estar como factor común en ambos lados y despreciar la energía potencial (mgh), ya que no hay diferencia de altura significativa entre la entrada y salida de la tobera.
De este modo, es posible reemplazar los valores dados para obtener la siguiente expresión:
[tex]2510\frac{kJ}{kg} +1378kPa*0.147\frac{m^3}{kg} +\frac{1}{2} (5\frac{m}{s} )^2=2263\frac{kJ}{kg}+138.7kPa*1.099\frac{m^3}{kg} +\frac{1}{2} v_2^2[/tex]
Y así, hallar la velocidad de salida como sigue:
[tex]2725.066\frac{kJ}{kg}=2415.431\frac{kJ}{kg} +\frac{1}{2} v_2^2\\\\v_2=\sqrt{2(2725.066\frac{kJ}{kg}-2415.431\frac{kJ}{kg} )} \\\\v_2=24.9\frac{m}{s}[/tex]
Para revisar:
https://brainly.com/question/23265263https://brainly.com/question/14279777Suppose that you release a small ball from rest at a depth of 0.590 m below the surface in a pool of water. If the density of the ball is 0.370 that of water and if the drag force on the ball from the water is negligible, how high above the water surface will the ball shoot as it emerges from the water? (Neglect any transfer of energy to the splashing and waves produced by the emerging ball)
Answer:
Explanation:
The work of the buoyancy force will convert to gravity potential energy
Fresh Water has a density of 1000 kg/m³
The ball has a density of 370 kg/m³
assume the ball is 1 m³
weight of the ball is 370g
The buoyancy force is 1000g
Assume the buoyancy force drops suddenly to zero when the center of the ball clears the water level.
The Work done on the ball is
W = Fd = 1000g(0.590)
the change in potential energy is
PE = mgh = 370g(0.590 + y)
where y is the height above the water level.
1000g(0.590) = 370g(0.590 + y)
1000(0.590) = 370(0.590 + y)
y = (0.590)(1000 - 370) / 370
y = 1.004594... ≈ 1.00 m
A block slides down a smooth ramp, starting from rest at a height h. When it reaches the bottom it's moving at speed v. It then continues to slide up a second smooth ramp. At what height is its speed equal to v/2
Answer:
3h/4
Explanation:
At speed v/2 height will be 3/4 h
What is equation of motion in kinematics?Equation that describes the motion of point , bodies , and system of bodies without considering the force that cause them to move is called equation of motion in kinematics
When block is at top of first ramp
u=0 ( block was at rest )
a = g ( acceleration due to gravity
using equation of motion
2as = v^2 - u^2
2gh = v^2
Then the block continued and reached a speed of v1 = v/2 on second ramp
now , final velocity = v= v1 =[tex]\sqrt{2gh}[/tex] / 2
u= [tex]\sqrt{2gh\\}[/tex]
s= h1
using equation of motion , we get
2as = v^2 - u^2
2(-g)h1 =( [tex]\sqrt{2gh}[/tex]/2)^2 - [tex]\sqrt{2gh}[/tex]
2(-g)h1 = (g h - 4 g h) / 2
h1 = 3/4 h
At speed v/2 height will be 3/4 h
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Over half the global population faces a shortage of clean water. Which example is a point source water pollutant? pesticide run-off from farmers’ fields a sewer pipe draining directly into a river road oil washing into sewer drains, and eventually reaching waterways erosion from a denuded slope after a heavy rainfall
An example of a point source water pollutant is a sewer pipe draining directly into a river.
According to the United States Environmental Protection Agency, a point source pollution is defined as; “any single identifiable source of pollution from which pollutants are discharged, such as a pipe, ditch, ship or factory smokestack.”
Hence, an example of a point source water pollutant is a sewer pipe draining directly into a river. The pollution is coming from an easily identifiable source which is a sewer pipe connected to a river directly.
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please help asap 100 POINTS and BRAINLIEST to best answer.
D. :)
lmk if im right or wrong
Answer:
D. 5ºC
Explanation:
Not B, because 1k= -457.87 fahrenheit. That means it is also not C. Since the icicles are melting, then that means it is not A either.
5ºC=about 41ºF
Find the work done by the force field F in moving an object from A to B. F(x, y) = 6y3/2i + 9x y j A(1, 1), B(3, 4)
Answer:
138
Explanation:
(since there's a blank next to "9x y j", I'm assuming the y of F(x, y) is 9x[tex]\sqrt{y}[/tex] j)
1) find the partial derivative of each:
[tex](6y^{\frac{3}{2} })i + (9x\sqrt{y} )j[/tex]
[tex]f_{x} =(6y^{\frac{3}{2} })_{x} = \int\limits{6y^{\frac{3}{2} }} \, dx = 6xy^{\frac{3}{2} } +c \\\\f_{y} = (9x\sqrt{y} )_{y} = \int\limits{9xy^{\frac{1}{2} } } \, dy = 9x(\frac{2}{3} )y^{\frac{3}{2} } +c = 6xy^{\frac{3}{2} } +c[/tex]
2) use partial integrals to make gradient of f:
take whatever you got from partial integral and add them together (if they repeat, just use it once)
[tex]F =[/tex] Vf (V = gradient of)
[tex]F(x, y) = 6xy^{\frac{3}{2} }[/tex]
3) Evaluate the integrals with given points:
Integral of F dotted with dr = F(point B) - F(point A) = F(3, 4) - F(1, 1)
[tex]F(point B) = 6(3)(4)^{\frac{3}{2} }\\F(Point A) = 6(1)(1)^{\frac{3}{2}}\\F(point B) - F(pointA) = 6(3)(4)^{\frac{3}{2} }-(6(1)(1)^{\frac{3}{2}})[/tex]
= 144 - 6 = 138 units of work
Work done by the force field F in moving an object from A to B = 138 J
Given data :
Force field F(x,y) = [tex]6y^{\frac{3}{2} }i + (9x\sqrt{y} ) j[/tex]
Step 1 : determine the partial derivatives of the vector quantity
Fx = ∫ [tex]6y^{\frac{3}{2} }i = 6xy^{\frac{3}{2} } + c[/tex]
Fy = ∫ [tex](9x \sqrt{y})_{y} = 9x(\frac{2}{3})y^{\frac{3}{2} } + c[/tex]
Equating the partial derivatives :
[tex]9x(\frac{2}{3})y^{\frac{3}{2} } + c[/tex] = [tex]6xy^{\frac{3}{2} } + c[/tex]
therefore the gradient of F i.e. F = vF = F( x,y ) = [tex]6xy^{\frac{3}{2} }[/tex]
Next step : Determine the work done
Work done ( F.dr ) = [ F(point b ) = F( 3,4 ) ] - [ F(point A) = F( 1,1 ) ]
F(3,4 ) = 6(3)(4)[tex]^{\frac{3}{2} }[/tex] = 144
F( 1,1 ) = 6(1)(1)[tex]^{\frac{3}{2} }[/tex] = 6
Therefore the work done by the force field = 144 - 6 = 138 J
Hence we can conclude that the work done by the force field F is = 138 J
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The force of earth’s gravity is 10N downward. What us the acceleration of a 15kg backpack if lifted with a a 15N force?
Answer:
F-F(gr) = ma
a= {F-F(gr)}/m =
=(15-10)/15=0.33 m/s² (upward)
While forming a 1.5kg aluminum statue, a metal smith heats the aluminum to 2700 degrees C, pours it into a mould, and then cools it to a room temperature of 23.0 degrees C. Calculate the thermal energy released by the aluminum during the process.
Answer is supposed to be: 4.7*10^6 j/kg?
The heat released by the aluminum during the process is [tex]3.6 \times 10^6 \ J[/tex].
The given parameters:
Mass of the statues, m = 1.5 kgFinal temperature of the status, t₂ = 2700 CTemperature when it is in the mould, t₁ = 23 ⁰CSpecific heat capacity of aluminum, C = 900The heat released by the aluminum during the process is calculated as follows;
[tex]Q = mc \Delta t \\\\Q = 1.5\times 900 \times (2700 - 23)\\\\Q = 3.6 \times 10^6 \ J[/tex]
Thus, the heat released by the aluminum during the process is [tex]3.6 \times 10^6 \ J[/tex].
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A 30.0-g object moving to the right at 20.5 cm/s overtakes and collides elastically with a 13.0-g object moving in the same direction at 15.0 cm/s. Find the velocity of each object after the collision. (Take the positive direction to be to the right. Indicate the direction with the sign of your answer.)
Let m₁ and m₂ be the masses of the two objects, and v₁ and v₂ their initial velocities. So
m₁ = 30.0 g = 0.0300 kg
m₂ = 13.0 g = 0.0130 kg
v₁ = + 20.5 cm/s = 0.205 m/s
v₂ = + 15.0 cm/s = 0.150 m/s
and we want to find v₁' and v₂', the final velocities of either object after their collision.
Momentum is conserved throughout the objects' collision, so that
m₁v₁ + m₂v₂ = m₁v₁' + m₂v₂'
where v₁' and v₂' are the first and second object's velocities after the collision.
Kinetic energy is also conserved, so that
1/2 m₁v₁² + 1/2 m₂v₂² = 1/2 m₁(v₁')² + 1/2 m₂(v₂')²
or
m₁v₁² + m₂v₂² = m₁(v₁')² + m₂(v₂')²
From the first equation (omitting units), we have
0.0300 • 0.205 + 0.0130 • 0.150 = 0.0300 v₁' + 0.0130 v₂'
0.0810 = 0.0300 v₁' + 0.0130 v₂'
81 = 30 v₁' + 13 v₂'
From the second equation,
0.0300 • 0.205² + 0.0130 • 0.150² = 0.0300 (v₁')² + 0.0130 (v₂')²
0.00155 ≈ 0.0300 (v₁')² + 0.0130 (v₂')²
1.55 ≈ 30 (v₁')² + 13 (v₂')²
Solving both equations simultaneously gives two solutions, one of which corresponds to the initial conditions. The other yields
v₁' ≈ + 0.172 m/s
and
v₂' ≈ + 0.227 m/s
why is cooking faster with a pressure cooker. plz help me I beg
Explanation:
The same thing happened in pressure cooker but the temperature inside is most higher.
hope you like my answer
thank you
Because cooking reactions speed up at a higher temperature also the boiling temperature of water increases and hence, the food cooks faster.
source:
https://www.vedantu.com/question-answer/a-pressure-cooker-reduces-cooking-time-for-food-class-9-chemistry-cbse-5feaba32e9fb3d3419e4a61a
)Suppose you have two insulated buckets containing the same amount of water at room temperature. You also happen to have two blocks of metal of the same mass, both at the same temperature, warmer than the water in the buckets. One block is made of aluminum and one is made of copper. You put the aluminum block into one bucket of water, and the copper block into the other. After waiting a while you measure the temperature of the water in both buckets. Which is warmer
The water in the bucket containing the aluminium block is warmer than the bucket containing the copper block.
The specific heat is the amount of heat needed or required to elevate the temperature of 1 gram of a substance by 1° C.
At standard conditions;
the specific heat of aluminium = 0.215 cal/gm/° Cthe specific heat of copper = 0.0923 cal/gm/° CWe know that:
Heat loss by the metal block = Heat gained by the water bucketTherefore, since the specific heat of aluminium is higher than that of copper, the water bucket containing aluminium block will be warmer than the bucket containing the copper block.
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A solid sphere starts from rest and rolls down a slope that is 6.4 m long. If its speed at the bottom of the slope is 5.3 m/s, what is the angle of the slope
From the relationship between acceleration a and g on an inclined plane, the angle of the slope is 13 degrees
Given that a solid sphere starts from rest and rolls down a slope that is 6.4 m long. The speed at the bottom of the slope is 5.3 m/s, the distance travelled is 6.4 m. That is,
Initial velocity U = 0 ( since it starts from rest)
Final velocity V = 5.3 m/s
distance S = 6.4 m
Let us first calculate its acceleration by using third equation of motion.
[tex]V^{2}[/tex] = [tex]U^{2}[/tex] + 2aS
[tex]5.3^{2}[/tex] = 0 + 2 x 6.4a
28.09 = 12.8a
a = 28.09 / 12.8
a = 2.2 m / [tex]s^{2}[/tex]
To calculate the angle of the slope, let us use the relationship between acceleration a and g on an inclined plane.
acceleration a = gsin∅
substitute all the relevant parameters
2.2 = 9.8 sin∅
sin∅ = 2.2/9.8
sin∅ = 0.224
∅ = [tex]Sin^{-1}[/tex](0.224)
∅ = 12.97 degrees
∅ = 13 degrees (approximately)
Therefore, the angle of the slope is 13 degrees
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Carrie is driving a car. Which factors determine the kinetic energy of her car
at this point
Answer:
The two main factors that affect kinetic energy are mass and speed.
Explanation:
Kinetic energy is the energy that is caused by the motion. The kinetic energy of an object is the energy or force that the object has due to its motion. Your moving vehicle has kinetic energy; as you increase your vehicle's speed, your vehicle's kinetic energy increases.
Have a great day! :D
A 45 kg figure skater is spinning on the toes of her skates at 1.1 rev/s. Her arms are outstretched as far as they will go. In this orientation, the skater can be modeled as a cylindrical torso (40 kg, 20 cm average diameter, 160 cm tall) plus two rod-like arms (2.5 kg each, 71 cm long) attached to the outside of the torso. The skater then raises her arms straight above her head. In this latter orientation, she can be modeled as a 45 kg, 20-cm-diameter, 200-cm-tall cylinder.
What is her new rotation frequency, in revolutions per second?
Answer: HOPE THIS HLEP YOU
Explanation:
A merry-go-round rotates from rest with an angular acceleration of 1.08 rad/s2. How long does it take to rotate through (a) the first 3.74 rev and (b) the next 3.74 rev
Answer:
Let ω1 be the initial angular speed and ω2 the final angular speed:
α = (ω2- α1) / t
corresponding to a = (v2 - v1) / t
S (distance corresponds to theta)
1 rev = 2 pi and 3.74 rev = 7.48 pi = 23.5 radians
S = 1/2 a t^2 linear or S = 1/2 α t^2 angular acceleration
23.5 = 1/2 * 1.08 t^2 and t = 6.60 sec for first 3.84 rev
b) ω1 = 1.08 * 6.6 = 7.13 rad/sec initial speed for second 3.74
23.5 = 7.13 t + .54 t^2 compare to S = v1 t + 1/2 a t^2
.54 t^2 + 7.13 t - 23.5 = 0
t^2 + 13.2 t - 43.5 = 0
t = 2.7 sec for next 3.74
Check:
7.13 * 2.7 + .54 * 2.7^2 = 23.2 rad = 3.7 rad
Is carbon a source or a sink of energy? Why?
Carbon-Based Fuels Dominate Global Energy Use
Crude oil, coal, and natural gas supply about 85% of the energy used in the world. Fossil fuels are valuable as sources of energy because they contain hydrocarbons and other carbon-based materials.
What are the two factors that affect the frictional force between objects
A magnet is located above circular current. What is the direction of the magnetic force on the magnet
The magnet is attracted to the ring since the north pole of the current loop is above the ring and the south pole is below the ring.
2. Conner flips a coin up in the air (to determine if he or his sister needs to do the dishes) at an upward velocity of 4.00 m/s. He fails to catch it on the way down and it falls down an additional 1.6 m to the floor. Calculate the maximum height and maximum magnitude (could be positive or negative) velocity of the coin.
Answer:
5.6
Explanation:
Not so sure
Easiest way to Find fahrenhiet to celsius please i need necessary 20 for the fastest correct answer
Answer:
thx for the points
Explanation:
no need brainliest
Question 3 of 20 :
Select the best answer for the question.
3. Which of the following accurately describes the behavior of water when subjected to
temperature change?
A. The volume of water will decrease if heated from 6°C to 7°C.
B. The volume of water will increase if cooled from 3°C to 2°C.
O C. A mass of water will contract if cooled from 1°C to 0°C.
D. A mass of water will expand if heated from 0°C to 2°C.
As water cools the volume expands.
Answer: B. The volume of water will increase if cooled from 3°C to 2°C.
A car C moving with a speed of 65 km/h on a straight road is ahead of motorcycle M moving with the speed of 80 km/h in the same direction. What is the velocity of M relative to A?
Answer:
The answer is 15. Step by step is you times 80 by 65 and then you divided by 2 and you call your teacher
who the football games yesterday 2021
What is the potential energy when a 100 kg object is raised 4.00 m straight up?
Answer:
100kg x 4 x 10 = 4000J
Explanation:
PLEASE HELP ME GET THIS RIGHT
Explanation:
I'm not sure, but I would go for the more than A since its orbital speed is at its fastest and the sweep occurs in about the same period of days.
