the answer is "all organisms evolve at the same rate" :)
Answer:
answer is b
Explanation:
PLS ANSWER FAST WILL GIVE BRAINLY!!!
how do scientist measure mass of an atom?
An object with a mass of 2.0 kg accelerates 2.0 m/s 2when an unknown force is applied to it. What is the amount of force?
Answer:
4 N
Explanation:
mass = 2 kg
acceleration = 2 m/s^2
Force = mass * acceleration
= 2 *2
= 4 N
Determine the torque
produced by a perpendicular force of 75
N at the end of a 0.2 m wrench.
Answer:
Explanation:
the answer is C.)
The torque produced by a perpendicular force of 75 N at the end of a 0.2 m wrench is 15 Nm.
What is torque?The force which causes the object to rotate about any axis is called torque. In math form, it is equivalent to the product of force and perpendicular distance.
Torque is a twisting or turning force that frequently results in rotation around an axis, which may be a fixed point or the center of mass. The ability of something rotating, such as a gear or a shaft, to overcome turning resistance is another way to think of torque.
Given:
The perpendicular force on the wrench, P = 75 N,
The length of the wrench, r = 0.2 m,
Calculate the torque of the wrench as shown below,
The torque of the wrench = r × P × sin a
Here, the force is perpendicular hence, a = 90°
The torque of the wrench = 0.2 × 75 × sin90
The torque of the wrench = 15 × 1
The torque of the wrench = 15
Thus, the torque on the wrench is 15 Nm.
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The law of universal gravitation describes the relationships between
Group of answer choices
A. unbalanced force, balanced force, and net force.
B. speed, distance, and time.
C. speed, velocity, and acceleration.
D .force, mass, and distance.
Answer:
D force mass and distance
PLEASE PLEASE HELPPP!
Select all that apply!
Answer:
1 AND 3
Explanation:
which system of units is used by only a small number of countries in the world, including the u.s
Answer:
imperial system
Explanation:
Answer:
british system
Explanation:
A P E X
What is the charge on an object that experiences a force of 5 Newtons in an electric field of 50 Newtons per coulomb?
Answer:
Explanation:
F = qE
F is the force in Newtons
q is the test charge
E is the electrical field produced by the source charge
[tex]5=q(50)\\q=1*10^-^1Coulombs[/tex]
The magnitude of charge on the object is of 0.1 C.
Given Data:
The magnitude of electric field is, E = 50 N/C.
The magnitude of Electric force is, F = 5 N.
The region where any particle feels the effect of electric force on itself, due to other charged entities, is known as Electric field. And the relation between the electric force and electric field is given as,
F = E × q
Here,
q is the magnitude of charge on the object.
Solving as,
q = F/E
q = 5 / 50
q = 0.1 C
Thus, we can conclude that the magnitude of charge on the object is of 0.1 C.
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An old lorry is 20% efficient at converting the
chemical energy in the diesel into useful
mechanical energy. How many joules of
mechanical energy are produced when 600
joules of chemical energy have been used by
the engine?
Answer:
The amount of mechanical energy produced by the old lorry when 600 Joules of chemical energy have been used by the engine, M.E. is 120 Joules
Explanation:
The question is a word problem on the relationship between energy efficiency and energy consumption
The given parameters of the old lorry are;
The chemical energy to mechanical energy conversion efficiency of the old lorry = 20%
The amount of chemical energy used by the engine = 600 joules
Let 'M.E.' represent the amount of mechanical energy produced by the old lorry, we have;
[tex]Energy \ Efficiency = \dfrac{Energy \ produced }{Total \ Energy \ Taken \ In \ or \ Used} \times 100[/tex]
Therefore, the for the old lorry, we have;
[tex]20\% = \dfrac{M.E. \ Produced}{600 \ Joules} \times 100[/tex]
∴ M.E. Produced = 20/100 × 600 J = 120 J
The amount of mechanical energy produced by the old lorry when 600 Joules of chemical energy have been used by the engine, M.E. = 120 Joules
2. A large collection of stars, gas, and dust is called a _____?
Whats the word for when "velocity equals 0 and direction changes" Its 14 letters, and I have _ _ _ _ i _ _ _ _ _ h_ i _ _ t
Answer:
Maximum Height
Explanation:
The maximum height is the highest point reached by a projected body. At this point, final velocity, v of the body is equal to 0; because upward motion is at its peak and the body start to fall again.
Hence, final velocity v is always 0 at this maximum point and the direction of motion changes from upward to downward.
Final velocity at this maximum height is zero because of the directional change experied by the object and the fact that upward motion of the body terminates at this point.
A 0.225 kg sample of tin initially at 97.5°c is dropped into 0.115 kg of water. the initial temperature of the water is 10.0°c. if the specific heat capacity of tin is 230 j/kg • °c, what is the final equilibrium temperature of the tin-water mixture
Answer:
The final equilibrium temperature of the tin-water mixture is approximately 18.468 °C
Explanation:
The parameters of heat energy transfer from the tin to the water are given as follows;
The mass of the sample of tin, m₁ = 0.225 kg
The initial temperature of the tin, T₁ = 97.5 °C
The mass of the water into which the tin is dropped, m₂ = 0.115 kg
The initial temperature of the water, T₂ = 10.0 °C
The specific heat capacity of tin, c₁ = 230 J/(kg·°C)
The specific heat capacity of water, c₂ = 4,200 J/(kg·°C)
Let 'T' represent the final equilibrium temperature of the tin-water mixture, we have;
The heat lost by the tin, ΔQ[tex]_{tin}[/tex] = The heat gained by the water ΔQ[tex]_{water}[/tex]
∴ ΔQ[tex]_{tin}[/tex] = ΔQ[tex]_{water}[/tex]
Where;
ΔQ[tex]_{tin}[/tex] = m₁·c₁·(T₁ - T)
ΔQ[tex]_{water}[/tex] = m₂·c₂·(T - T₂)
By substitution, we have;
ΔQ[tex]_{tin}[/tex] = 0.225 kg × 230 J/(kg·°C) × (97.5°C - T)
ΔQ[tex]_{water}[/tex] = 0.115 kg × 4,200 J/(kg·°C) × (T - 10.0°C)
From ΔQ[tex]_{tin}[/tex] = ΔQ[tex]_{water}[/tex], we have;
0.225 kg × 230 J/(kg·°C) × (97.5°C - T) = 0.115 kg × 4,200 J/(kg·°C) × (T - 10.0°C)
∴ 5,045.625 J - 51.75 J/°C × T = 483 J/°C × T - 4,830 J
5,045.625 J + 4,830 J = 534.75 J/°C × T
∴ 534.75 J/°C × T = 9,875.625 J
T = 9,875.625 J/(534.75 J/°C) = 18.4677419 °C ≈ 18.468 °C
The final equilibrium temperature of the tin-water mixture, T ≈ 18.468 °C.
A horizontal net force F is exerted on an object at rest. The object starts at x=0 m and has a speed of 4.0 m/s after moving 4.0 m along a horizontal frictionless surface. The net force F as a function of the object's position x is shown below
Explanation:
The solution is be found in the attachment.
A horizontal net force F is exerted on an object at rest. The object starts at x = 0 m and has a speed of 4.0 m/s after moving 8.0 m along a horizontal frictionless surface. The net force F as a function of the object's position x is shown below.
What is the mass of the object?
Answer: 30 kg
please help me out here
Answer:
Explanation:
it's blocked for me...it was unblocked but you know how everything else is blocked
A rock has 750 J of potential energy as it sits on a ledge. If the rock were pushed off the ledge, how much would it have just before it hit the ground?
Answer:
the 750 j will have potential energy is 375
Explanation:
750/2 is 375
a clay ball with a mass of 0.35 kg hits another 0.35 kg ball at rest and the two stick together. The first balls intial speed us 4.2m/s.
What are the balls final speed?
Answer:
4.2 m/s
the balls final speed is V (m/s)
because the clay ball hits another and the two stick together
=> 0,35.4,2 + 0,35.4,2 = (0,35 + 0,35).V
<=> 2,94 = 0,7V
<=> V = 2,94/0,7 = 4,2
For a clay ball with a mass of 0.35 kg, the ball's final speed is mathematically given as
V2= 4.2
What is the balls final speed?Question Parameter(s):
A clay ball with a mass of 0.35 kg
Hits another 0.35 kg ball at rest and the two stick together.
The first ball's initial speed is 4.2m/s.
Generally, the equation for the conservation of momentum is mathematically given as
M1v1=m2v2
Therefore
0.35*4.2 + 0.3*5.4.2 = (0.35 + 0.35)*V
2,94 = 0,.7V
V2 = 2.94/0.7
V2 = 4.2
In conclusion
V2= 4.2
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The molecules of a substance in a particular state of matter move freely with random motion. The average speed of the molecules is increasing. What is being described?
Answer:
A gas
Explanation:
A student attaches a block to a vertical spring of unknown spring constant so that the block-spring system will oscillate if the block-spring system is released from rest at a vertical position that is not the system’s equilibrium position. The student varies the object’s mass and uses a stopwatch to determine the time it takes the object to make one oscillation. The student creates the graph that is shown. The slope of the line of best fit is equal to which of the following quantities?
Answer:
D: 2π/√k0
Explanation:
I got it right
The slope of the line of best fit is the quantity 2π/√ko.
The line of best fits is used to show the relationship between two variables. The nature of the graph shows the relationship between the variables.
We can see that this is a graph of the period of oscillation against the square root of the mass of the body. The slope of the line of best fit is the quantity 2π/√ko.
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A steel ball bearing is released from a height H and
rebounds after hitting a steel plate to a height H.
What is true about the collision with the steel plate?
Answer:
ELASTIC collision
kinetic energy is conservate
Explanation:
As the ball bounces to the same height, it can be stated that the impact with the floor is ELASTIC.
As the floor does not move the conservation of the moment
po = pf
-mv1 = m v2
- v1 = v2
So the speed with which it descends is equal to the speed with which it rises
Therefore the kinetic energy of the ball before and after the collision is the same
Answer:
CORRECT (SELECTED)
It is elastic since kinetic energy was conserved.
Explanation:
How can I skip class:
A: Fake sick.
B: Have a heating pad on my heat to have a fake fever.
C: Fake barf
D: Act fine until a test the neb sick
Answer:
c gl
Explanation:
::::::::::::::::::::::::
The outside diameter of your teacher's rear bicycle tire is 16 inches. How far will he travel if the rear wheel makes 1200 revolutions on the road?
Answer:
241,274.32 inches
Explanation:
How far will he travel if the rear wheel makes 1200 revolutions on the road?
Since the rear wheel makes one revolution in the distance of a circumference of a circle, C with diameter, d = 16 inches
C = πd²/4
So, the distance, travelled in 1200 revolutions is D = 1200 × C = 1200πd²/4
Substituting d = 16 into D, we have
D = 1200πd²/4
D = 1200π(16)²/4
D = 76800π
D = 241,274.32 inches
Which statement describes the difference between ionic bonds and Van der
Waals forces?
A. Ionic bonds involve the sharing of electrons between atoms, while
Van der Waals forces involve the attraction of nonpolar molecules.
B. Ionic bonds involve the transfer of electrons between ions, while
Van der Waals forces involve the sharing of electrons between
atoms.
C. lonic bonds involve the transfer of electrons between ions, while
Van der Waals forces involve the attraction of nonpolar molecules.
D. Ionic bonds involve the attraction of nonpolar molecules, while
Van der Waals forces involve the transfer of electrons between
ions.
Answer:
C. lonic bonds involve the transfer of electrons between ions, while
Van der Waals forces involve the attraction of nonpolar molecules.
Explanation:
Just did it
Please answer quick! Thanks guys!
Answer:
wire By please mark my answer in brain list
Answer:
Remove wire G, if the wire after it is F. Remove wire A as C is connecting to it already.
a car travels straight for 20 miles on a road that is 30 north of east. what is the easiest component of the cars displacement to the nearest tenth of a mile
A -17.3 miles
B -10.0miles
C 10.0 miles
D 17.3 miles
Answer:
17.3 miles
Explanation:
!!!URGENT!!! Worth 100 points!!
A snail travels down the sidewalk at 25 mm/min for 4 minutes How far does the snail travel?
The snail would travel 100mm
Each minutes is 25mm and it took 4 minutes
4 x 25 = 100
or
25+25+25+25=100
Is the potential energy on the surface of Earth zero?If not,what will be it's value?
Answer:
Potential energy does not have an absolute measurement. It is always relative to some reference point. Gravitational potential always increases when you go up and decreases when you go down. But the choice of a zero point is arbitrary.
If you’re dropping objects onto the ground, then choosing the ground as a zero point makes the calculations easier. But you could just as easily make the zero point the top of Mt. Everest, and all the answers would turn out the same. Up still means more energy and down means less, but now the PE of an object at sea level would have a negative sign, but not as negative as an object ten meters above sea level.
So everything would still work fine.
In fact, planetary astronomers take this idea to extremes. Instead of the top of Mt. Everest, they set the zero point for potential energy as infinitely high—so far “up” that gravity is so weak that going “up” another kilometer doesn’t gain you any energy. Then the value for potential energy everywhere else in the universe anywhere near a planet has a negative sign, but just as before, all the answers in relative terms turn out fine.
Explanation:
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calculate the ground pressure of a 90kg gas cylinder the diameter of the cylinder is 1.1 m
Answer:
P = 928.09 Pa
Explanation:
Given that,
Mass of a gas cylinder, m = 90 kg
The diameter of the cylinder, d = 1.1 m
Radius, r = 0.55 m
We need to find the ground pressure. The pressure is equal to force per unit area. So,
[tex]P=\dfrac{F}{A}\\\\P=\dfrac{mg}{\pi r^2}\ (A\ is\ area\ of\ base)\\\\P=\dfrac{90\times 9.8}{\pi \times (0.55)^2}\\\\P=928.09\ Pa[/tex]
So, the ground pressure is 928.09 Pa.
Someone plz help I’m on a time limit :(
What patterns occur in the Earth-Sun-Moon system that allow us to identify the cause of
eclipses?
Answer:
Earth's rotation and orbit and the Moon's orbit cause observable patterns (Day and night, seasons, phases of the Moon, tides).
Why does the Moon appear red during a lunar eclipse?
1.)Because the sunlight bends as it passes through the Earth's atmosphere
2.)Because the sunlight bends as it passes through the Moon's atmosphere
Answer:
Image result for Why does the Moon appear red during a lunar eclipse?
A lunar eclipse takes place when the sun, Earth and moon line up in space. The moon passes through Earth's shadow. ... Bottom line: The moon can look red during a total lunar eclipse because of sunlight that's filtered and refracted by Earth's atmosphere
Rain and wind place which type of load on structures?
A) Dynamic load
B) Soft load
C) Gradual load
D) Static load
Un bloque de 3 kg en reposo se deja libre a una altura de 5 m sobre una rampa curva y sin rozamiento. Al pie de la rampa se encuentra un resorte de constante k = 400 N/m, como se muestra en la fig. El objeto desliza por la rampa y llega a chocar contra el resorte comprimiéndolo una distancia x antes de que quede en reposo momentáneamente. Determinar: a) La velocidad con la que el bloque alcanza al resorte. ____________________ b) La distancia x que el bloque comprime al resorte. __________________ c) La velocidad con la que el bloque es expulsado por el resorte. ____________________ d) La altura que alcanza sobre la parte curva. ________________ e) ¿Alcanzará la misma altura si la rampa no está libre de rozamiento? ___________________
Answer:
a) La velocidad del bloque cuando llega al resorte es de aproximadamente 9,9 m / s
b) La distancia a la que se comprime el resorte es de aproximadamente 0,86 m
c) La velocidad con la que el resorte expulsa el bloque es de aproximadamente 9,9 m / s
d) La altura que alcanza el bloque es de 5 metros.
e) El bloque no alcanzará la misma altura si la rampa no está libre de fricción
Explanation:
a) Los parámetros dados del bloque son;
La masa del bloque, m = 3 kg
La altura a la que se coloca el bloque, h = 5 m
La constante de resorte, k = 400 N / m
La aceleración debida a la gravedad, g = 9,8 m / s²
La energía potencial de un cuerpo, P.E. = m · g · h
Por tanto, la energía potencial inicial del bloque, P.E. se da como sigue;
P.E. = 3 kg × 9,8 m / s² × 5 m = 147 julios
P.E. = 147 julios
La energía cinética del bloque al pie de la rampa, K.E. = 1/2 · m · v²
Dónde;
v = La velocidad del bloque cuando llega al resorte
Por lo tanto, para el bloque dado tenemos;
K.E. = 1/2 · m · v² = 1/2 × 3 kg × v²
Por el principio de conservación de la energía, tenemos;
El PE. del bloque en reposo a una altura de 5 m = La energía cinética al pie de la rampa. K.E.
∴ P.E. = K.E.
147 J = 1/2 × 3 kg × v²
v² = 147 J / (1/2 × 3 kg) = 98 m² / s²
v = √ (98 m² / s²) = 7 · √2 m / s
v = 7 · √2 m / s ≈ 9,9 m / s
b) La energía recibida por el resorte comprimido, E = 1/2 · k · x²
Dónde;
k = La constante del resorte = 400 N / m
x = La distancia a la que se comprime el resorte
Por el principio de conservación de la energía, tenemos;
La energía recibida por el resorte comprimido, E = La energía potencial inicial del resorte, P.E.
∴ E = 1/2 · k · x² = P.E.
De lo que tenemos;
E = 1/2 × 400 N / m × x² = 147 julios
x² = 147 Julios / (1/2 × 400 N / m) = 0,735 m²
x = √ (0,735 m²) = 0,7 · √ (3/2) m ≈ 0,86 m
La distancia a la que se comprime el resorte = x ≈ 0.86 m
c) La velocidad con la que el resorte expulsa el bloque se indica a continuación;
La energía en el resorte = 1/2 · k · x² = La energía cinética dada al bloque, 1/2 · m · v²
∴ 1/2 · k · x² = 1/2 · m · v²
∴ La velocidad con la que el bloque es expulsado por el resorte, v = La velocidad con la que el bloque llega al resorte = 7 · √2 m / s
La velocidad con la que el resorte expulsa el bloque, v = 7 · √2 m / s ≈ 9,9 m / s
d) La altura que alcanza el bloque también viene dada por la siguiente relación anterior;
P.E. = K.E.
∴ m · g · h = 1/2 · m · v²
v = 7 · √2 m / s
De donde tenemos h = La altura inicial del bloque en la rampa = 5 metros
e) El bloque no alcanzará la misma altura si la rampa no está libre de fricción porque se utilizará energía para superar la fuerza de fricción
a) La velocidad final del bloque es aproximadamente 9.903 metros por segundo.
b) El resorte se deforma 0.858 metros.
c) Por el principio de la conservación de energía y sabiendo la ausencia de fuerzas disipativas, la velocidad del objeto expulsado del resorte es aproximadamente 9.903 metros por segundo.
d) Por el principio de la conservación de energía y si existieran fuerzas disipativas, la altura máxima sería menor a la hallada en el punto a).
a) Conforme a la situación de este problema, la energía cinética traslacional final ([tex]K[/tex]), en joules, es igual a la energía potencial gravitacional inicial ([tex]U[/tex]), en joules.
[tex]U = K[/tex] (1)
Por las definiciones de las energías cinética traslacional y potencial gravitacional expandimos la ecuación anterior:
[tex]m\cdot g\cdot h = \frac{1}{2}\cdot m\cdot v^{2}[/tex] (1)
Ahora despejamos la velocidad de esa ecuación:
[tex]v = \sqrt{2\cdot g\cdot h}[/tex]
Donde:
[tex]m[/tex] - Masa del bloque, en kilogramos.[tex]g[/tex] - Aceleración gravitacional, en metros por segundo al cuadrado.[tex]h[/tex] - Altura inicial del bloque, en metros.[tex]v[/tex] - Velocidad final del bloque, en metros por segundo.Si sabemos que [tex]g = 9.807\,\frac{m}{s^{2}}[/tex] y [tex]h = 5\,m[/tex], entonces la velocidad final del bloque es:
[tex]v = \sqrt{2\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (5\,m)}[/tex]
[tex]v\approx 9.903\,\frac{m}{s}[/tex]
La velocidad final del bloque es aproximadamente 9.903 metros por segundo.
b) Por el principio de conservación de la energía, la energía cinética traslacional inicial es igual a la energía potencial elástica final, cuyas fórmula es la siguiente:
[tex]\frac{1}{2}\cdot k\cdot x^{2} = \frac{1}{2}\cdot m \cdot v^{2}[/tex] (2)
Where:
[tex]k[/tex] - Constante de resorte, en newtons por metro.[tex]x[/tex] - Deformación del resorte, en metros.Ahora despejamos la deformación del resorte:
[tex]x = \sqrt{\frac{m}{k} }\cdot v[/tex] (3)
Si sabemos con [tex]k = 400\,\frac{N}{m}[/tex], [tex]m = 3\,kg[/tex] y [tex]v \approx 9.903\,\frac{m}{s}[/tex], entonces la deformación del resorte es:
[tex]x = \sqrt{\frac{3\,kg }{400\,\frac{N}{m} } }\cdot \left(9.903\,\frac{m}{s} \right)[/tex]
[tex]x \approx 0.858\,m[/tex]
El resorte se deforma 0.858 metros.
c) Por el principio de la conservación de energía y sabiendo la ausencia de fuerzas disipativas, la velocidad del objeto expulsado del resorte es aproximadamente 9.903 metros por segundo.
d) Por el principio de la conservación de energía y si existieran fuerzas disipativas, la altura máxima sería menor a la hallada en el punto a).
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