Answer:
I'm pretty sure this is not a complete question. My guess is that you are trying to add/subtract vectors. Vectors have both magnitude and direction, so vector A is pretty clear, but a magnitude of 13 (i'm guessing a resultant) without a direction is weird.
IF 13 is the magnitude of the resultant, vector B added to vector A could have any magnitude 17 ≤ B ≤ 43
It could have any direction of
θ = (225 - 180) ± arcsin(13/30)
θ = 45 ± 25.679...
70.679 ≤ θ ≤ 19.321
components of vector B would be
Bx = |B|cosθ
By = |B|sinθ
My guess is that you are trying to add/subtract vectors. Vectors have both magnitude and direction, so vector A is pretty clear, but a magnitude of 13 (i'm guessing a resultant) without a direction is weird.
What is magnitude of the resultant?IF 13 is the magnitude of the resultant, vector B added to vector A could have any magnitude 17 ≤ B ≤ 43
It could have any direction of
θ = (225 - 180) ± arcsin(13/30)
θ = 45 ± 25.679...
70.679 ≤ θ ≤ 19.321
components of vector B would be
Bx = |B|cosθ
By = |B|sinθ
My guess is that you are trying to add/subtract vectors. Vectors have both magnitude and direction, so vector A is pretty clear, but a magnitude of 13 (i'm guessing a resultant) without a direction is weird. A vector has a magnitude of 30m at an angle of 225 degrees with respect to the positive x-axis and has a magnitude of 13m.
Therefore, My guess is that you are trying to add/subtract vectors. Vectors have both magnitude and direction, so vector A is pretty clear, but a magnitude of 13 (i'm guessing a resultant) without a direction is weird.
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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
Light from the Sun takes about 8.0 min to reach Earth. How far away is the Sun? Answer in scientific notation
E5. A ball is thrown downward with an initial velocity of 12 m/s.
Using the approximate value of g 10 m/s2, what is the
velocity of the ball 1.0 seconds after it is released?
Hi there!
We know the following kinematic equation:
vf = vi + at
Where:
vf = final velocity
vi = initial velocity
a = acceleration
t = time
In this instance, the ball is experiencing a constant acceleration of that of gravity, thus:
vf = 12 + 10(1) = 22 m/s (if downward is considered positive in this instance)
Một khối khí hidro bị nén đến thể tích bằng 1/2 lúc đầu khi nhiệt độ không đổi. Nếu vận tốc trung bình của phân tử hidro lúc đầu là V thì vận tốc trung bình sau khi nén là bao nhiêu ?
answer: what language is this???
A very bouncy mushroom can be modeled as mass of 30g(the mushroom cap) on top of a spring(the mushroom stalk) when spring constant 20 N/m. A bird of mass 50g lands on the mushroom gently so that its velocity is zero when it lands. Questions are in the image.
The motion of the bouncy mushroom can be described as a simple
harmonic motion, SHM.
a) The equilibrium height of the mushroom is 0.024525 m below its initial heightb) The frequency of resulting oscillation is 0.5 Hzc) The maximum compression of the mushroom 0.03924 md) The equation that describes the oscillation of the mushroom as a function of time is; [tex]\underline{x(t) = 0.014715 \cdot cos(0.5\cdot t)}[/tex]Reasons:
The given parameters are;
Mass of the mushroom cap, m = 30 g = 0.03 kg
Mass of the bird = 50 g = 0.05 kg
The spring constant, K = 20 N/m
a) The equilibrium height of the mass spring system, is given as follows;
F = -K·x
[tex]x = \dfrac{F}{K}[/tex]
The applied force, F = The weight of the bird
∴ F = (0.05 kg) × 9.81 m/s² = 0.4905 N
[tex]x = \dfrac{0.05 \, kg \times 9.81 \ m/s^2}{20 \, N/m} = 0.024525 \, m[/tex]
The equilibrium height of the mushroom is 0.024525 m below its initial height.
b) The frequency of oscillation of a spring, ω, is given as follows;
[tex]\omega = \sqrt{\dfrac{K}{m} }[/tex]
Therefore;
[tex]\omega = \sqrt{\dfrac{20 \, N/m }{80 \, kg} } = \dfrac{1}{2} \, Hz[/tex]
The frequency of resulting oscillation, ω = [tex]\dfrac{1}{2} \, Hz[/tex] = 0.5 Hz
c) The applied force, F = The weight of the bird and the mushroom cap
F = (0.03 kg + 0.05 kg) × 9.81 m/s² = 0.7848 N
[tex]x = \dfrac{0.7848 \, N}{20 \, N/m } = 0.03924 \, m[/tex]
The maximum compression of the mushroom = 0.03924 m
d) The motion of the mushroom is a Simple Harmonic Motion, SHM.
The equation of a SHM as a function of time is; x(t) = A·cos(ω·t + Φ)
For the mushroom, we have;
The amplitude, A = 0.03924 m - 0.024525 m = 0.014715 m
Ф = The phase angle
When t = 0, cos(ω × 0 + Φ) = 1
cos(Φ) = 1
Ф = arcos(1) = 0
The equation is therefore;
x(t) = 0.014715·cos(0.5·t)
Equation of the oscillation of the mushroom is; [tex]\underline{x(t) = 0.014715 \cdot cos(0.5\cdot t)}[/tex]
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A double-pane glass window is 60.0 cm x 90.0 cm and has 3.00-mm window panes. If the temperature difference between inside and outside is 24.0 K, how far apart should the panes be to have a heat loss of 4.09 W? Assume there is air in the gap.
The distance between the glass to have the given heat loss is 2.54 m.
The given parameters:
dimension of the window, = 60 cm by 90 cmtemperature, T = 24 Kheat lost, Q = 4.09 Wthermal conductivity of glass, k = 0.8 W/mKThe area of the glass window is calculated as follows;
[tex]A = 0.6 \times 0.9\\\\A = 0.54 \ m^2[/tex]
The distance between the glass is calculated as follows;
[tex]Q = \frac{KA \Delta T}{\Delta x} \\\\\Delta x = \frac{kA \Delta T}{Q} \\\\\Delta x = \frac{0.8 \times 0.54 \times 24 }{4.09} \\\\\Delta x = 2.54 \ m[/tex]
Thus, the distance between the glass to have the given heat loss is 2.54 m.
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A motorcycle, which has an initial linear speed of 9.7 m/s, decelerates to a speed of 4.0 m/s in 4.4 s. Each wheel has a radius of 0.61 m and is rotating in a counterclockwise (positive) directions.
What is (a) the constant angular acceleration (in rad/s2) and (b) the angular displacement (in rad) of each wheel?
Hi there!
We can begin by solving for the linear acceleration as we are given sufficient values to do so.
We can use the following equation:
vf = vi + at
Plug in given values:
4 = 9.7 + 4.4a
Solve for a:
a = -1.295 m/s²
We can use the following equation to convert from linear to angular acceleration:
a = αr
a/r = α
Thus:
-1.295/0.61 = -2.124 rad/sec² ⇒ 2.124 rad/sec² since counterclockwise is positive.
Now, we can find the angular displacement using the following:
θ = ωit + 1/2αt²
We must convert the initial velocity of the tire (9.7 m/s) to angular velocity:
v = ωr
v/r = ω
9.7/0.61 = 15.9 rad/sec
Plug into the equation:
θ = 15.9(4.4) + 1/2(2.124)(4.4²) = 20.56 rad
Identify the direction of the net force acting on the object. Explain your reasoning
Answer:
See Below
Explanation:
I am not sure what you exact question is to know what direction it is but here is how you do it.
The direction of the net force is the direction of the largest force.
for example if you were to push a box forward with 100 newton's of force and someone pushed at the same time 50 newton's backwards on the box, the box would move forwards because the was a greater force on the box in a forward direction. hope this helps
Una turbina de vapor
recibe vapor con un flujo másico de 30 kg/s a 6205 kPa, 811 K, con una velocidad a la
entrara de 10 m/s. El vapor a la entrada tiene una energía interna específica de 3150.3
kJ/kg y un volumen específico de 0.05789 m3
/kg. El vapor sale de la turbina a 9.859 kPa,
318.8 K. El vapor sale a 200 m/s con una energía interna específica de 2211.8 kJ/kg y
un volumen específico de 13.36 m3
/kg. Encuentre la potencia producida por la turbina
si ésta pierde calor a una tasa de 30 kW.
Este problema está describiendo una turbina de vapor a la que entra vapor a 30 kg/s, 6.205 kPa y 811 K con una velocidad de 10 m/s y sale a 9.859 kPa, 318.8 K y con una velocidad de 200 m/s. Adicionalmente, tanto el volumen específico como la energía interna son dados para ambas corrientes.
Con lo anterior, resulta posible escribir un balance de energía para esta turbina, despreciando todo efecto por energía potencial ya que no hay diferencia significativa entre la altura de la entrada (1) y la salida (2), pues están practicamente al mismo nivel:
[tex]mh_1+\frac{1}{2} mv^2_1=mh_2+\frac{1}{2} mv^2_2+Q_2+W_2[/tex]
Aquí vemos que la incógnita es [tex]W_2[/tex] como la potencia que produce la turbina. Ahora, el primer cáculo a realizar es el de las entalpías de las corrientes de entrada y salida, dada la energía interna, presión y volumen específico:
[tex]h_1=3150.3\frac{kJ}{kg}+6205kPa*0.05789\frac{m^3}{kg} =3509.51\frac{kJ}{kg}\\\\h_2=2211.8\frac{kJ}{kg}+9.859kPa*13.36\frac{m^3}{kg} =2342.72\frac{kJ}{kg}[/tex]
Ahora, podemos reacomodar el balance de energía con el fin de resolver [tex]W_2[/tex]:
[tex]W_2=m(h_1-h_2)+\frac{1}{2} m(v^2_1-v^2_2)-Q_2[/tex]
Finalmente, reemplazamos los valores para obtener:
[tex]W_2=10\frac{kg}{s} (3509.51-2342.72)\frac{kJ}{kg} +\frac{1}{2} *10\frac{kg}{s} [(10\frac{m}{s}) ^2-(200\frac{m}{s} )^2]*\frac{1kJ}{1000J} -30\frac{kJ}{s}\\\\W_2=11438.4 kJ/s=11438.4kW[/tex]
Es de precisar que la energía cinética como 1/2 m*v² resulta en Joules, por lo que hay que convertir a kilojoules para tener unidades consistentes de kilowatts al final.
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https://brainly.com/question/21902769https://brainly.com/question/24322350A 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.
What is friction ???
Answer:
Frictional force is produced when two bodies are rubbed against each other. It is the force that oppose the motion and therefore it stops or slow down a moving body.It depends upon the roughness or smoothness of the surface of the body in contact.Rough surface have more friction that the smooth surface. Similarly, the heavier body produces more friction than a lighter body. Frictional force acts in the opposite direction of the motion of the body.
A solenoid 91.0 cm long has a radius of 1.50 cm and a winding of 1300 turns; it carries a current of 3.60 A. Calculate the magnitude of the magnetic field inside the solenoid.
The magnitude of the magnetic field inside the solenoid is [tex]6.46 \times 10^{-3} \ T[/tex].
The given parameters;
length of the solenoid, L = 91 cm = 0.91 mradius of the solenoid, r = 1.5 cm = 0.015 mnumber of turns of the solenoid, N = 1300 current in the solenoid, I = 3.6 AThe magnitude of the magnetic field inside the solenoid is calculated as;
[tex]B = \mu_0 nI\\\\B = \mu_o(\frac{ N}{L} )I\\\\[/tex]
where;
[tex]\mu_o[/tex] is the permeability of frees space = 4π x 10⁻⁷ T.m/A
[tex]B = (4\pi \times 10^{-7}) \times (\frac{1300}{0.91} ) \times 3.6\\\\B = 6.46 \times 10^{-3} \ T[/tex]
Thus, the magnitude of the magnetic field inside the solenoid is [tex]6.46 \times 10^{-3} \ T[/tex].
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What is the momentum of a 3 kg bowling ball moving at 3 m/s?
.
O 1 kg. m/s
O 3 kg. m/s
O 6 kg. m/s
O 9 kg • m/s
Explanation:
p = mvp denotes momentumm denotes massv denotes velocity→ p = 3 kg × 3 m/s
→ p = 9 kg.m/s
Option D is correct.
Check all true statements about nonmetals. Group of answer choices
All nonmetals are gaseous at room temperature.
Nonmetals have low melting and boiling points. ✅
Nonmetals are brittle and have a relatively low density. ✅
Nonmetals are good at thermal and electrical conduction.
Using the colored periodic table below, match the group or area to its color.
alkali metals red alkaline ✅
earth metals orange ✅
transition metals white ✅
halogens purple ✅
Nobel gasses teal✅
How many of each subatomic particle are in a neutral atom of Potassium-39? What is its mass number? Potassium is in group 1 and has the symbol "K".
protons 19 ✅
neutrons 20 ✅
electrons 19 ✅
mass number 39✅
Match the ion with its charge.
11 protons, 12 neutrons, 10 electrons = +1 ✅
Groups 17, period 2 = -1 ✅
31 protons, 39 neutrons, 28 electrons = +3✅
Group 2, period 5 = +2
Which best describes a metal such as Silver (Ag)?
Answer: Lustrous, malleable, and forms cations.
Determine if the "quoted" word(s) makes each statement True or False.
Elements that share the most characteristics are found on the periodic table in the same "horizontal period". False ✅
If matter has a higher temperature, that means there will be "more" molecular motion. True✅
If you put a balloon in a freezer, its volume would "increase". False ✅
If you put the balloon into a chamber where there is half the pressure, its volume will "double". True ✅
If you cool down a propane tank, there will be "less" pressure in it. True ✅
The average atomic mass of Aluminum (Al) is "16.00 amu".
False✅
Oxygen's atomic number is "8" therefore it has "8" protons in its nucleus. True✅
An "electron" has the same mass as a neutron. False✅
The nucleus contains "protons and neutrons", virtually all the mass of an atom.✅
Metals are found towards the left hand side of the periodic table while Nonmetals are found towards the right hand side of the periodic table.
Non metals are found towards the right hand side of the periodic table. They are not all gaseous at room temperature some nonmetals such as iodine are solid at room temperature.
Nonmetals usually have low melting and boiling points, are brittle, have relatively low density and are not good at thermal and electrical conduction.
Potassium - 39 contains 19 protons, 20 neutrons and 19 electrons. Recall that the number of protons and electrons are equal in a neutral atom.
The following is an accurate matching of the ions;
11 protons, 12 neutrons, 10 electrons = +1 - Na^+
31 protons, 39 neutrons, 28 electrons = +3 - Ga^2+
Group 2, period 5 = +2 - Sr^2+
Silver is a lustrous metal.
Elements that share the most characteristics are found on the periodic table in the same "horizontal period". FalseIf matter has a higher temperature, that means there will be "more" molecular motion. TrueIf you cool down a propane tank, there will be "less" pressure in it. TrueIf you put a balloon in a freezer, its volume would "increase". FalseIf you put the balloon into a chamber where there is half the pressure, its volume will "double". TrueThe average atomic mass of Aluminum (Al) is "16.00 amu". FalseOxygen's atomic number is "8" therefore it has "8" protons in its nucleus. TrueAn "electron" has the same mass as a neutron. FalseThe nucleus contains "protons and neutrons", virtually all the mass of an atom - True
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Answer:
If your answers are the green check marks all of them are correct
Explanation: I took the test
can you classify matter based on chemical properties
Answer:
Explanation:
Matter can be broken down into two categories: pure substances and mixtures. Pure substances are further broken down into elements and compounds. Mixtures are physically combined structures that can be separated into their original components. A chemical substance is composed of one type of atom or molecule.
trình bày những hiểu biết cơ bản về năng lượng
Answer:
I don't understand your language
Be able to list the three compounds that are formed as products of highly exothermic
reactions such as detonating nitrogen-based explosives?
Answer:
Ammonium perchlorate NH4ClO4
Ammonium Nitrate
Calcium Cyanamide
When detonated, the reaction products are all gases, such as water vapor, nitrogen gas, and oxides of nitrogen.
Hopefully this helps :)
Which quantity or quantities is/are increasing for the object represented by line B?
Answer:
C. Velocity and Position
Explanation:
The quantities that are increasing for the object represented by line B are velocity and position. The correct option is b.
What is velocity?The direction of a body or object's movement is defined by its velocity. In its basic form, speed is a scalar quantity. In essence, velocity is a vector quantity. It is the speed at which distance changes. It is the displacement change rate.
Velocity is the pace and direction of an object's movement, whereas speed is the time rate at which an object is traveling along a path. In other words, velocity is a vector, whereas speed is a scalar value.
The graph is given which represents the velocity and time with terms A, B, and C. As opposed to the position-time graph, which describes an object's motion over time, the velocity-time graph reveals an object's speed.
Therefore, the correct option is b. velocity and position.
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The question is incomplete. Your most probably complete question is given below:
Velocity and acceleration
velocity and position
velocity only
velocity, position, and acceleration
what is a non-economic benefit to international trade
Answer:
non economic benefits to international trade means not being too much profit not selling too much things /items in international trade
If you were told an atom was an ion, you would know the atom must have a?
A neutral charge
B charge
C negative charge
D positive charge
Answer:
its b it must be charge
Explanation:
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
An electron is held up against the force of gravity by the attraction of a fixed proton some distance above it. How far above the electron is the proton
5.08 m
Explanation:
The weight of the electron is being counteracted by the attractive electrostatic force exerted by the proton above it. We can write the force equation as follows:
[tex]m_eg = k_e\dfrac{Q_pQ_e}{r^2}[/tex]
where the Q's are the charges of the proton and electron, r is the distance between the particles, g is the acceleration due to gravity, [tex]m_e[/tex] is the mass of the electrons and [tex]k_e[/tex] is the Coulomb constant. So solving for r, we get
[tex]r^2 = k_e\dfrac{Q_pQ_e}{m_eg}[/tex]
Taking the square root of r^2, we then get the distance as
[tex]r = \sqrt{k_e\dfrac{Q_pQ_e}{m_eg}}[/tex]
The values are given as follows:
[tex]m_e = 9.11×10^{-31}\:\text{kg}[/tex]
[tex]g = 9.8\:\text{m/s}^2[/tex]
[tex]Q_p = Q_e = 1.60×10^{-19}\:\text{C}[/tex]
[tex]k_e = 8.99×10^9\:\text{N-m}^2\text{/C}^2[/tex]
Putting in all of these values in our equation for r,
[tex]r = \sqrt{\dfrac{(8.99×10^9\:\text{N-m}^2\text{/C}^2)(1.60×10^{-19}\:\text{C})^2}{(9.11×10^{-31}\:\text{kg})(9.8\:\text{m/s}^2)}}[/tex]
[tex]\:\:\:\:\:= 5.08\:\text{m}[/tex]
A 5 kg bowling ball travelling at 2 m/s hits a motionless 10 kg bowling ball. If the smaller ball bounces back at a speed of -1 m/s, what will be the speed of larger ball after the collision? Hint: Use the conservation of momentum equation to solve this problem.
Answer:
1.5 m/s
Explanation:
Conservation of momentum means the momentum of the system before the collision is the same as after.
The before, after momentum of each ball is ...
5 kg ball: (5 kg)(2 m/s), (5 kg)(-1 m/s)
10 kg ball: (10 kg)(0 m/s), (10 kg)(v)
The sum of the "before" products is the same as the sum of the "after" products:
(5 kg)(2 m/s) +0 = (5 kg)(-1 m/s) +(10 kg)v
(10 +5) kg·m/s = (10 kg)·v . . . . . add (5 kg)(1 m/s) to both sides
v = (15 kg·m/s)/(10 kg) = 1.5 m/s
The speed of the larger ball will be 1.5 m/s. Its direction of motion will be the opposite of that of the 5 kg ball after the collision.
1. It’s fall and time for the corn maze and bonfire and you just can’t wait. On your way to the farm though a turkey flies out in front of you, so you slam on the brakes and go from from 30.0 m/s to 18.0 m/s. Luckily your date brought a stop watch and told you the whole thing took place in 10.5s. What is your acceleration and how far did you go?
Acceleration = (change in velocity ( final speed - starting speed))/ (time)
Acceleration = (18-30)/10.5
Acceleration = -12/10.5
Acceleration = -1.14 m/s^2
Distance = 30m/s x 10.5s + 1/2(1.14)(10.5)^2
Distance = 252.2 meters
Sam wants to get a refrigerator into a moving truck. He chooses to use a ramp to accomplish the task that is 25 meters long. If the fridge weighs 250 N, what force will Sam need to use on the ramp to get the fridge into the 3 m high truck bed?
Answer:
Explanation:
Ignoring friction and assuming one ramp end rests on level ground.
gravity acceleration acting parallel to the ramp is gsinθ.
F = mgsinθ
mg = 250 N, sinθ = 3/25
F = 250(3/25)
F = 30 N
A 25 meter long ramp strong enough to hold a person plus a 250 N refrigerator would weigh much more than 250 N.
 A wooden box with a mass of 10.0 kg rest on a ramp that is incline at an angle of 25° to the horizontal. A rope attached to the box runs parallel to the ramp and then passes over a frictionless bully. A bucket with a mass of M hangs at the end of the rope. The coefficient of static friction between the ramp in the box is 0.50. The coefficient of Connecticut friction between the ramp in the box is 0.35.
Suppose the box remains at rest relative to the ramp. What is the maximum magnitude of the friction force exerted on the box by the ramp?
The maximum magnitude of the friction force exerted on the box by the ramp is 44.41 N.
The given parameters;
Mass of the box, m = 10 kgInclination of the ramp, θ = 25⁰Coefficient of static friction, μ = 0.5 Coefficient of kinetic friction, μk = 0.35The normal force on the wooden box is calculated as follows;
[tex]F_n = mg \times cos(\theta)\\\\F_n = 10 \times 9.8 \times cos(25)\\\\F_n = 88.8 2 \ N[/tex]
The maximum magnitude of the friction force exerted on the box by the ramp is calculated as follows;
[tex]F_f = \mu \times F_n\\\\F_f = 0.5 \times 88.82 \\\\F_f = 44.41 \ N[/tex]
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Which is NOT a function of the
cell wall?
A. Protects cell from bursting
B. Provides support for plant cells
C. Protects cell from harsh internal
environments
D. Absorbs sunlight to give energy to the cell
Answer:
The answer is D
Explanation:
The chloroplast absorbs sunlight for energy not the cell wall
A car of mass 200 kg, moving with a forward acceleration of 3 m/s-ıs acted upon by
constant resistive force of 500 N. Calculate the force exerted from the engine to
maintain this forward acceleration.
Answer:
600 N
Explanation:
The force exerted from the engine to maintain this forward acceleration is 600 N.
What is force?Force is defined in physics as: the push or pull on an object with mass that causes it to change velocity. Force is an external agent that can change the state of rest or motion of a body.
The term "force" has a specific meaning in science. At this level, it is perfectly acceptable to refer to a force as a push or a pull.
A force is not something that an object possesses or possesses. Another object applies a force to another.
We know that,
F = ma
Here, it is given that
Acceleration = 3 m/s
Mass = 200Kg.
So, by keeping the value
F = 200 x 3
F = 600 N
Thus, the force is 600 N.
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5. The net external force on a rock of mass 4.2 kg is 8.0 N forward. Find the acceleration of the rock.
Answer:
1.904
Explanation:
F= ma
8 = 4.2 a
a = 8/4.2
a = 1.904
Suppose that Hubble's constant were H0 = 51 km/s/Mly (which is not its actual value). What would the approximate age of the universe be in that case?
Given the Hubble's constant, the approximate age of the universe is 5.88 × 10⁹ Years.
Given the data in the question;
Hubble's constant; [tex]H_0 = 51km/s/Mly[/tex]
Age of the universe; [tex]t = \ ?[/tex]
We know that, the reciprocal of the Hubble's constant ( [tex]H_0[/tex] ) gives an estimate of the age of the universe ( [tex]t[/tex] ). It is expressed as:
[tex]Age\ of\ Universe; t = \frac{1}{H_0}[/tex]
Now,
Hubble's constant; [tex]H_0 = 51km/s/Mly[/tex]
We know that;
[tex]1\ light\ years = 9.46*10^{15}m[/tex]
so
[tex]1\ Million\ light\ years = [9.46 * 10^{15}m] * 10^6 = 9.46 * 10^{21}m[/tex]
Therefore;
[tex]H_0 = 51\frac{km}{\frac{s}{Mly} } = 51000\frac{m}{s\ *\ Mly} \\\\H_0 = 51000\frac{m}{s\ *\ (9.46*10^{21}m)} \\\\H_0 = 5.39 *10^{-18}s^{-1}\\[/tex]
Now, we input this Hubble's constant value into our equation;
[tex]Age\ of\ Universe; t = \frac{1}{H_0}\\\\t = \frac{1}{ 5.39 *10^{-18}s^{-1}} \\\\t = 1.855 * 10^{17}s\\\\We\ convert\ to\ years\\\\t = \frac{ 1.855 * 10^{17}}{60*60*24*365}yrs \\\\t = \frac{ 1.855 * 10^{17}}{31536000}yrs\\\\t = 5.88 *10^9 years[/tex]
Therefore, given the Hubble's constant, the approximate age of the universe is 5.88 × 10⁹ Years.
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