Answer: 400 m
Explanation:
Vf= 20 + (4*10)
Vf= 60 [m/s]
x= (60^2 - 20^2) / (2*4)
x= 400 m
If a red star and a blue star both have the same radius and both are the same distance from Earth, which one looks brighter in the night sky? Explain why.
Answer:
The blue one.
Explanation:
Im pretty sure its because the blue is hotter.
A car start moving from the rest.If the acceleration of the car is 2m/2 for 10 seconds what will be it final velocity
Answer:
20 m/s
Explanation:
Applying,
a = (v-u)/t.................... Equation 1
Where a = acceleration of the car, v = final velocity of the car, u = initial velocity of the car, t = time.
make v the subject of the equation
v = u+at.............. Equation 2
From the question,
Given: u = 0 m/s(start from rest), a = 2 m/s², t = 10 seconds
Substitute these values into equation 2
v = 0+(2×10)
v = 20 m/s
What types of changes occur during the erosion and deposition of sediments in a river?
Answer:
Slow-moving rivers generate extensive floodplains and meanders through erosion and deposition. Stream and river deposition can result in the formation of alluvial fans and deltas. Natural levees may be formed by floodwaters. Caves and sinkholes can arise as a result of groundwater erosion and deposition.
Explanation:
s
The erosion and deposition of sediments in a river creates broad floodplains and meanders.
What is Erosion and Deposition?Deposition occurs when sediment, a combination of soil and rock produced by weathering, is eroded and transported to a new area.
Deposition is the act of depositing silt that has been transported by the wind, water, sea, or ice.
Earthen materials are worn away during erosion, a geological process in which they are moved by wind or water.
The removal of soil, rock, or dissolved material from one area on the Earth's crust and subsequent transport to another region for deposition are known as erosional processes. Erosion differs from weathering, which is a static process.
Given data ,
Let the erosion and deposition of sediments be deposited in a river
Now , Pebbles, sand, mud, and salts that have been dissolved in water can all be used to convey sediment. Afterwards, salts may be left behind by organic action.
Now , floodplain is caused when erosion happens
And , A floodplain is a broad, level or nearly level area of land where the stream flows.
Meandering streams that wander from side to side broaden the plain by eroding it during the formation of the plain. Flooding can occur when stream flows overflow from their channel due to very excessive rainfall or quick snow-melt.
Hence , floodplain and meandering occurs due to erosion and deposition
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please help
a girl pulls a wheeled suitcase with a force of 3N. If the suitcase has a mass of 6 kg, what is the acceleration?
Explanation:
Start with what you know and list your knowns and unknowns
F = ma
F= 3N
m = 6kg
a =?
3N = 6kg x a
solve for a
3N / 6kg = a
plsss plsss plsss helppppp ASAP thank u ❤️
Answer:
7. (D) uniformly accelerated vertical motion
8. (A) zero
9. (A) zero
10. (C) parabolic
Answer:
7.Uniformly accelerated vertical motion
8.0m/s²
9.9.8m/s
10.parabolic
11.vertical component.
A 0.2 kg hockey park is sliding along the eyes with an initial velocity of -10 m/s when a player strikes it with his stick, causing it to reverse its direction and giving it a velocity of +25 m/s the impulse the stick applies to the park is most nearly
Answer:
The impulse applied by the stick to the hockey park is approximately 7 kilogram-meters per second.
Explanation:
The Impulse Theorem states that the impulse experimented by the hockey park is equal to the vectorial change in its linear momentum, that is:
[tex]I = m\cdot (\vec{v}_{2} - \vec{v_{1}})[/tex] (1)
Where:
[tex]I[/tex] - Impulse, in kilogram-meters per second.
[tex]m[/tex] - Mass, in kilograms.
[tex]\vec{v_{1}}[/tex] - Initial velocity of the hockey park, in meters per second.
[tex]\vec{v_{2}}[/tex] - Final velocity of the hockey park, in meters per second.
If we know that [tex]m = 0.2\,kg[/tex], [tex]\vec{v}_{1} = -10\,\hat{i}\,\left[\frac{m}{s}\right][/tex] and [tex]\vec {v_{2}} = 25\,\hat{i}\,\left[\frac{m}{s} \right][/tex], then the impulse applied by the stick to the park is approximately:
[tex]I = (0.2\,kg)\cdot \left(35\,\hat{i}\right)\,\left[\frac{m}{s} \right][/tex]
[tex]I = 7\,\hat{i}\,\left[\frac{kg\cdot m}{s} \right][/tex]
The impulse applied by the stick to the hockey park is approximately 7 kilogram-meters per second.
If you shine a light of frequency 375hz on a double slit setup , and you measure the slit separation to be 950 nm and the screen distance to be 4030 nm away , what is the distance from the zero order fringe to the first order fringe ?
Answer:
y = 33.93 10⁵ m
Explanation:
This is an interference exercise, for the contributory interference is described by the expression
d sin θ = m λ
let's use trigonometry for the angle
tan θ = y / L
how the angles are small
tan θ = sin θ / cos tea = sin θ
we substitute
sin θ = y / L
d y / L = m λ
y = m λ L / d
the light fulfills the relation of the waves
c = λ f
λ = c / f
λ = 3 10⁸ /375
λ = 8 10⁵ m
first order m = 1
let's calculate
y = 1 8 10⁵ 4030 10-9 / 950 10-9
y = 33.93 10⁵ m
Q:
The kinetic energy of an object can be determined
using the equation KE = 1/2mv^2, where ke is the
kinetic energy, m is the mass, and v is the speed of
the object. What is the mass of an object moving at
a speed of 12m/s that has 36 J of kinetic energy?
A:
0
2.0 kg
8.0 kg
0.13 kg
0.50 kg
Answer:
Explanation:
We will use the KE equation you wrote here and fill in what we are given:
[tex]36=\frac{1}{2}m(12)^2[/tex] and isolating the m:
[tex]m=\frac{2(36)}{12^2}[/tex] which gives us
m = .50 kg
A student applies a 10 N force to a wood block with a mass of 5 kg. The block is pushed across four different surfaces. The accelerations of the block are recorded. Which surface showed the least friction?
The complete question is as follows: A student is subjected to a reaction force of 10 N northward from a 5 kg block while pushing the block over a smooth, level surface. Ignoring friction, what is the acceleration of the block?
Answer: The acceleration of the block is [tex]2 m/s^{2}[/tex].
Explanation:
Given: Force = 10 N
Mass = 5 kg
It is known that force applied on an object is the product of mass and acceleration.
Mathematically, [tex]F = m \times a[/tex]
where,
F = force
m = mass
a = acceleration
Substitute the values into above formula as follows.
[tex]F = m \times a\\10 N = 5 kg \times a\\a = \frac{10}{5}\\= 2 m/s^{2}[/tex]
Thus, we can conclude that the acceleration of block is [tex]2 m/s^{2}[/tex].
A tank has the shape of an inverted circular cone with height 16m and base radius 3m. The tank is filled with water to a height of 9m. Find the work required to empty the tank by pumping all of the water over the top of the tank. Use the fact that acceleration due to gravity is 9.8 m/sec2 and the density of water is 1000kg/m3. Round your answer to the nearest kilojoule.
Answer:
[tex]W=17085KJ[/tex]
Explanation:
From the question we are told that:
Height [tex]H=16m[/tex]
Radius [tex]R=3[/tex]
Height of water [tex]H_w=9m[/tex]
Gravity [tex]g=9.8m/s[/tex]
Density of water [tex]\rho=1000kg/m^3[/tex]
Generally the equation for Volume of water is mathematically given by
[tex]dv=\pi*r^2dy[/tex]
[tex]dv=\frac{\piR^2}{H^2}(H-y)^2dy[/tex]
Where
y is a random height taken to define dv
Generally the equation for Work done to pump water is mathematically given by
[tex]dw=(pdv)g (H-y)[/tex]
Substituting dv
[tex]dw=(p(=\frac{\piR^2}{H^2}(H-y)^2dy))g (H-y)[/tex]
[tex]dw=\frac{\rho*g*R^2}{H^2}(H-y)^3dy[/tex]
Therefore
[tex]W=\int dw[/tex]
[tex]W=\int(\frac{\rho*g*R^2}{H^2}(H-y)^3)dy[/tex]
[tex]W=\rho*g*R^2}{H^2}\int((H-y)^3)dy)[/tex]
[tex]W=\frac{1000*9.8*3.142*3^2}{9^2}[((9-y)^3)}^9_0[/tex]
[tex]W=3420.84*0.25[2401-65536][/tex]
[tex]W=17084965.5J[/tex]
[tex]W=17085KJ[/tex]
'
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Numa máquina térmica uma parte da energia térmica fornecida ao sistema(Q1) é transformada em trabalho mecânico (τ) e o restante (Q2) é dissipado, perdido para o ambiente.
sendo:
τ: trabalho realizado (J) [Joule]
Q1: energia fornecida (J)
Q2: energia dissipada (J)
temos: τ = Q1 - Q2
O rendimento (η) é a razão do trabalho realizado pela energia fornecida:
η= τ/Q1
Exercícior resolvido:
Uma máquina térmica cíclica recebe 5000 J de calor de uma fonte quente e realiza trabalho de 3500 J. Calcule o rendimento dessa máquina térmica.
solução:
τ=3500 J
Q1=5000J
η= τ/Q1
η= 3500/5000
η= 0,7 ou seja 70%
Energia dissipada será:
τ = Q1 - Q2
Q2 = Q1- τ
Q2=5000-3500
Q2= 1500 J
Exercicio: Qual seria o rendimento se a máquina do exercicio anterior realizasse 4000J de trabalho com a mesma quantidade de calor fornecida ? Quanta energia seria dissipada agora?
obs: Entregar foto da resolução ou o cálculo passo a passo na mensagem
A 6.93*10-4 C charge has a
potential energy U = -3.09 J at a
point in space. What is the electric
potential V at that point?
Include the sign, + or -
(Unit = V)
Answer:
P = V * Q potential energy = potential * charge
V = =3.09 J / 6.93 * 10E-4 C = 4460 Joules / Coulomb
The electric potential, V at the point given the data from the question is –4458.87 V
What is electric potential?The electric potential or electromotive force (EMF) is defined as the energy supplied by a battery per unit charge. Mathematically, it can be expressed as:
Electromotive force (EMF) = Work (W) / charge (Q)
V = EMF = W / Q
How to determine the Electric potentialwork (W) = –3.09 JCharge on electron = 6.93×10⁻⁴ CElectric potential (V) =?V = W / Q
V = –3.09 / 6.93×10⁻⁴
V = –4458.87 V
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3 - An object is being pushed with a net force of 15 N. If the net force is cut in third to 5 N, how will the acceleration be changed?
Answer:
Explanation:
F = ma is a linear equation. This means that the Force change as the accleration changes. And vice versa. If the Force is cut in thirds, then the acceleration is also cut in thirds. Let's do some math on this just to prove it, shall we?
We know that at first, the F = 15. Let's give this object a mass of 5kg. That means that
15 = 5a so
a = 3
Then the F is cut into thirds, so
5 = 5a so
a = 1
The second acceleration is one-third of the first one, where the Force is 3 times greater.
The blank
is the organ responsible for cleaning a person's blood.
A beam of light shines on a plane mirror with an angle of incidence of 40°. The reflected light then hits another parallel plane mirror. What is the angle of reflection from the second mirror? A. 40° B. 50° C. 80° D. 90° E. 140°
Answer:
40degrees
Explanation:
Since the mirrors are parallel, the angle of incidence on the first mirror will be the same as the angle of incidence on the second mirror.
Angle of incidence on the second mirror = 40degrees
According to the second law of reflection, the angle of incidence is equal to the angle of reflection. Hence;
angle of incidence = angle of reflection = 40degrees
This shows that the angle of reflection from the second mirror is 40degrees
Who's Buzz Aldrin?
______________
Answer:
Buzz Aldrin was American astronaut, engineer and fighter pilot. he did three spacewalks as pilot of the 1966 Gemini 12 mission, and as the lunar modulbe pilot on the 1969 Apollo 11 mission, he and mission commander Neil Armstrong were the first two people to land on the Moon.
I hope it helps
have have a great day
If a sled has a mass of 4kg what is the force of gravity on the sled?
Answer:
Explanation:
Gravity pulls everything down at the same rate of 9.8 m/s/s. If you're looking for the normal force, which is the same as the weight of the object, we'll find that, just in case.
w = mg which says that the normal force/weight of an object is equal to its mass times the pull of gravity:
w = 4.0(9.8) so
w = 39N
particles that are found in the sun's plasma
Answer:
This plasma mostly consists of electrons
ayuda porfa es de actividades fisicas
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1. If you use an applied force of 45N to slide a 12Kg wooden crate across a floor at a constant velocity, what is the coefficient of kinetic friction between the crate and the floor?
Answer:
Coefficient of kinetic friction = 0.38 (Approx.)
Explanation:
Given:
Applied force = 45 N
Mass of wooden crate = 12 kg
Find:
Coefficient of kinetic friction
Computation:
Coefficient of kinetic friction = Applied force / (Mass)(Acceleration due to gravity)
Coefficient of kinetic friction = 45 / (12)(9.8)
Coefficient of kinetic friction = 45 / 117.6
Coefficient of kinetic friction = 0.3826
Coefficient of kinetic friction = 0.38 (Approx.)
In a warehouse, the workers sometimes slide boxes along the floor to move them. Two boxes were sliding toward each other and crashed. The crash caused both boxes to change speed. Based on the information in the diagram, which statement is correct? In your answer, explain what the forces were like and why the boxes changed speed.
Box 1 has more mass than Box 2.
Box 1 and Box 2 are the same mass.
Box 1 has less mass than Box 2
Answer:
The second one.
Explanation:
It caused both to change speed because they have both the same mass.
Determine the applied force required to accelerate a 2.25 kg object rightward with a
constant acceleration of 1.50 m/s2 if the force of friction opposing the motion is 18.2 N.
(Neglect air resistance.)
Answer:
Explanation:
Im going to be using the rules for significant digits properly so I hope you're quite familiar with them. The equation we need for this is
F - f = ma where F is the applied force (our unknown), f is the frictional force, m is the mass, and a is the acceleration. Filling in:
F - 18.2 = 2.25(1.50) and
F = 2.25(1.50) + 18.2 Do the multiplication first and round to get
F = 3.38 + 18.2 The addition rules tell us that we will be rounding to the tenths place after we add to get
F = 21.6 N
2. In any energy transformation, energy is _____. A created B conserved C destroyed
Answer:
B energy can't be created or destroyed
Planets in the solar system revolve around the Sun in an orbital path and can vary from rocky and terrestrial objects to gaseous and liquid object. Which set of characteristics is unique to the outer planets
Answer:
Gaseous; liquid.
Explanation:
A planet can be defined as a large celestial body having sufficient mass to allow for self-gravity and make it assume a nearly circular shape (hydrostatic equilibrium), revolves in an orbit around the Sun in the solar system and has a cleared neighborhood.
Some examples of the planet are Mars, Venus, Earth, Mercury, Neptune, Jupiter, Saturn, Uranus, Pluto, etc.
Basically, the planets are divided into two (2) main categories and these includes;
I. Inner planets: these planets are the closest to the sun and comprises of mercury, venus, earth and mars.
II. Outer planets: these planets are beyond the asteroid belt and comprises of jupiter, saturn, uranus and neptune, from left to right of the solar system.
These outer planets are made mostly of gases (hydrogen and helium) causing them to be less dense than the solid inner planets. These gases are generally known to be less dense in terms of physical properties.
In conclusion, Planets in the solar system can vary from rocky and terrestrial objects to gaseous and liquid object.
However, the set of characteristics which is unique to the outer planets are gaseous and liquid.
Nitrogen gas at standard atmospheric pressure 101.3 kPa, has a volume of 0.080 m3. If there are 3.6 mol of the gas, what is the temperature?
Answer:
271 K
Explanation:
Answer:
The temperature of Nitrogen is 270.85 K.
Explanation:
What is perfect gas ?A gas that follows Boyle's law , Charle's law and law of pressure , is called perfect gas.
Given data;
[tex]P =101.3 KPa , V = 0.08 m^{3} , \\ n = 3.6 mol , R = 8.31\\T = ?[/tex]
By the formula,
P.V = n .R.T
On putting values
[tex]1.013[/tex] x [tex]10^{5}[/tex] x [tex]0.08[/tex] = [tex]3.6[/tex] x [tex]8.31[/tex] x [tex]T[/tex]
So [tex]T =\frac{1.013*10^{5} * 0.08}{3.6 * 8.31}[/tex]
[tex]T =\frac{8104}{29.92}[/tex]
[tex]T = 270.85 K[/tex]
Hence the temperature of Nitrogen is 270.85 K
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television set changes electrical energy to sound and light energy. In this process, some energy is *
Answer :light
Explanation:
A television set is another device that operates by the transformation of energy. An electrical beam from the back of the television tube strikes a thin layer of chemicals on the television screen, causing them to glow. In this case, electrical energy is converted into light.
A solid cylinder has a mass of 5 kg and radius of 2 m and is fixed so that it is able to rotate freely around its center without friction. A 0.02 kg bullet is moving from right to left with an angular momentum of 9 kgm2s just before it strikes the cylinder near its bottom and gets stuck at the outer radius. What is the angular velocity (magnitude and direction) of the cylinder bullet system after the impact
Answer:
0.893 rad/s in the clockwise direction
Explanation:
From the law of conservation of angular momentum,
angular momentum before impact = angular momentum after impact
L₁ = L₂
L₁ = angular momentum of bullet = + 9 kgm²/s (it is positive since the bullet tends to rotate in a clockwise direction from left to right)
L₂ = angular momentum of cylinder and angular momentum of bullet after collision.
L₂ = (I₁ + I₂)ω where I₁ = rotational inertia of cylinder = 1/2MR² where M = mass of cylinder = 5 kg and R = radius of cylinder = 2 m, I₂ = rotational inertia of bullet about axis of cylinder after collision = mR² where m = mass of bullet = 0.02 kg and R = radius of cylinder = 2m and ω = angular velocity of system after collision
So,
L₁ = L₂
L₁ = (I₁ + I₂)ω
ω = L₁/(I₁ + I₂)
ω = L₁/(1/2MR² + mR²)
ω = L₁/(1/2M + m)R²
substituting the values of the variables into the equation, we have
ω = L₁/(1/2M + m)R²
ω = + 9 kgm²/s/(1/2 × 5 kg + 0.02 kg)(2 m)²
ω = + 9 kgm²/s/(2.5 kg + 0.02 kg)(4 m²)
ω = + 9 kgm²/s/(2.52 kg)(4 m²)
ω = +9 kgm²/s/10.08 kgm²
ω = + 0.893 rad/s
The angular velocity of the cylinder bullet system is 0.893 rad/s in the clockwise direction-since it is positive.
(reply with an answer to the question or get reported) Please solve it with the steps ty
Answer:
vertical force=50
horizontal force=35
A circuit has a voltage of 10 V and a current of 5 A. What must the resistance be?
0.5 ohms
50 ohms
2 ohms
15 ohms
Answer:
R=V/I
R= 2
Explanation:
R = 10V/5A
R = 2ohms
If a 70 kg man is standing on the surface of the earth, then what is his weight on the moon
Answer:
P = 113.4 N
Explanation:
We know that if it is on the Moon it has another gravity. Therefore, it is known through the International System that the gravity of the Moon is 1.62 m/s², so..
Data:
m = 70 kgg = 1.62 m/s²P = ?Use formula:
[tex]\boxed{\bold{P=m*g}}[/tex]Replace and solve:
[tex]\boxed{\bold{P=70\ kg*1.62\frac{m}{s^{2}}}}[/tex][tex]\boxed{\boxed{\bold{P=113.4\ N}}}[/tex]The man's weight is 113.4 Newtons.
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