Answer:
√3 * Gm²/d²
Explanation:
m1 = m, m2= m, distance = d. hence:
F = Gm²/d²
Let the origin be A, the point x = d be B and the point above the first two is C.
The net force acting on the third mass (point C) [tex]F_{net}[/tex] = [tex]F_A+F_B[/tex]
Let j represent the vertical component and i the horizontal component. Hence:
[tex]F_B=-F_j\\\\F_A=-F(icos\frac{\pi}{6}+jsin\frac{\pi}{6} )=-F(i\frac{\sqrt{3} }{2}+j\frac{1}{2} )\\\\F_{net} =F_A+F_B\\\\F_{net} =-F_j+{-F(i\frac{\sqrt{3} }{2}+j\frac{1}{2} )}\\\\F_{net} =-\frac{F}{2} \sqrt{3}(i+j\sqrt{3} )\\\\The\ magnitude\ of\ the\ net\ force\ is:\\\\|F_{net}|=\frac{F}{2}\sqrt{3}(\sqrt{1^2+\sqrt{3}^2 })=\frac{F}{2} \sqrt{3}(\sqrt{4})\\\\|F_{net}|=\frac{F}{2} \sqrt{3}*2=F*\sqrt{3}\\\\|F_{net}|=\sqrt{3}*\frac{Gm^2}{d^2}[/tex]
Answer:
The net gravitational force on the third mass = [tex]3^{0.5} * \frac{Gm^2}{d^2}[/tex]Explanation:
For equilateral triangle,
[tex]\theta = 30^o[/tex]
Force between masses,
[tex]F_1 = \frac{G*m*m}{d^2}\\\\F_! = \frac{Gm^2}{d^2}[/tex]
Therefore,
[tex]F_net = 2F_1cos\theta\\\\F_net = 2 * \frac{Gm^2}{d^2} * cos30\\\\F_net = 3^{0.5} * \frac{Gm^2}{d^2}[/tex]
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what would the net force be on the box in the problems shown below.( both force and direction). for all four diagrams. please explain answers.
two spheres A and B are projected off the edge of a 1.0 m high table with the same horizontal velocity . sphere A has a mass of 20.g and sphere B has a mass of 10.g.
If both spheres leave the edge of the table at the same instant, sphere A will land
a. at some time after sphere B.
b. at the same time as sphere B.
c. at some time before sphere B.
d. There is not enough information to decide.
Answer:
A lands first becuase its heavier
Wavelength is a measurement of ___________ , while period is a measurement of ___________ . *
Speed; time
Time; speed
Distance; time
Time; distance
Distance; speed
Speed; distance
distance ; time
Explanation:
wavelength is in metres [m]
period is in seconds [s]
A small, 300 g cart is moving at 1.10 m/s on an air track when it collides with a larger, 4.00 kg cart at rest. After the collision, the small cart recoils at 0.890 m/s. What is the speed of the large cart after the collision (answer in m/s please)?
Answer:
0.0158m/s
Explanation:
Using the law of conservation of energy which states that the sum of momentum before collision is equal to the sum after collision. It is expressed mathematically as;
m1u1 + m2u2 = m1v1 + m2v1
m1 and m2 are the masses of the object
u1 and u2 are the initial velocities
v1 and v2 are the final velocities
Given
m1 = 300g = 0.3kg
u1 = 1.10m/s
m2 = 4.00kg
u2 = 0m/s (at rest)
v1 = 0.890
v2 = ?
Substitute the given values into the formula;
0.3(1.10) + 0 = 0.3(0.89) + 4v2
0.33 = 0.267 + 4v2
0.33-0.267 = 4v2
0.063 = 4v2
v2 = 0.063/4
v2 = 0.0158m/s
Hence the speed of the large cart after the collision is 0.0158m/s
According to the question:
Mass,
[tex]m_1 = 300 \ g = 0.3 \ kg[/tex][tex]m_2 = 4.00 \ kg[/tex]Final velocity,
[tex]u_1 = 1.10 \ m/s[/tex][tex]u_2 = 0 \ m/s[/tex]Initial velocity,
[tex]v_1 = 0.890 \ m/s[/tex][tex]v_2 = \ ?[/tex]By using the law of conservation, we get
→ [tex]m_1 u_1 +m_2 u_2 =m_1 v_1 +m_2 v_1[/tex]
By substituting the values, we get
→ [tex]0.3(1.10)+0 = 0.3(0.89)+4v_2[/tex]
→ [tex]0.33=0.267+4 v_2[/tex]
[tex]0.33-0.267 = 4 v_2[/tex]
[tex]0.063 =4v_2[/tex]
[tex]v_2 = \frac{0.063}{4}[/tex]
[tex]= 0.0158 \ m/s[/tex]
Thus the response above is appropriate.
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A rifle is aimed at a target 40m away. The bullet hits the target 2.2cm below the horizontal plane (ignore wind and rotational effects).
(a) What is the time of flight for bullet?
(b) What is the muzzle velocity (velocity of bullet when it leaves the rifle)?
Answer:
The value is [tex]u_x = 597 \ m/s[/tex]
Explanation:
From the question we are told that
The distance of the target from the riffle is [tex]d = 40 \ m[/tex]
The height at which the bullet hit the target is [tex]y = 2.2 \ cm = 0.022 \ m[/tex]
Considering the vertical motion
Generally from kinematic equations we have that
[tex]y = u_y t + \frac{1}{2} * gt^2[/tex]
At the initial stage the velocity is zero i.e [tex]u_y = 0 \ m/s[/tex]
=> [tex]0.022 = 0 * t + \frac{1}{2} * 9.8 t^2[/tex]
=> [tex]t = 0.067 \ s[/tex]
Generally the velocity of the bullet when it leaves the riffle is mathematically represented as
[tex]u_x = \frac{ d}{t}[/tex]
=> [tex]u_x = \frac{40 }{ 0.067 }[/tex]
=> [tex]u_x = 597 \ m/s[/tex]
ASAP
What is elastic potential energy?
Answer:
Elastic potential energy is stored when materials stretch or compress. Examples of it include springs, rubber bands, and slingshots.
Explanation:
a 2kg block is attached to a horizontal ideal spring with a spring constant of 200 Newton per minute. when the spring has its equillibrium length of block is given a speed of 5mps. what is the maximum elongation of the spring
Answer:
The maximum elongation of the spring is 0.5 meters.
Explanation:
The statement is incorrect. The correct form is:
A 2-kg block is attached to a horizontal ideal spring with a spring constant of 200 newton per meter. When the spring has its equillibrium length, the block has a speed of 5 meters per second. What is the maximum elongation of the spring?
The block experiments a simple harmonic motion, where there are no non-conservative forces and the total energy is the sum of translational kinetic energy of the mass and the elastic potential energy of the spring. The maximum elongation of the spring is done when elastic potential energy reach its maximum. By the Principle of Energy Conservation, the maximum elastic potential energy is equal to the maximum translational kinetic energy, which corresponds to the instant when the mass reaches the equilibrium position. Then, the equation modelling the system is:
[tex]U_{max} = K_{max}[/tex] (1)
Where:
[tex]U_{max}[/tex] - Maximum elastic potential energy of the spring, measured in joules.
[tex]K_{max}[/tex] - Maximum translational kinetic energy of the mass, measured in joules.
By definitions of the maximum elastic potential energy of the spring and the maximum translational kinetic energy of the mass, the expression above is expanded and simplified:
[tex]\frac{1}{2}\cdot k\cdot x_{max}^{2} = \frac{1}{2}\cdot m \cdot v_{max}^{2}[/tex]
[tex]x_{max} = \sqrt{\frac{m}{k} }\cdot v_{max}[/tex] (2)
Where:
[tex]x_{max}[/tex] - Maximum elongation of the spring, measured in meters.
[tex]m[/tex] - Mass, measured in kilograms.
[tex]k[/tex] - Spring constant, measured in newtons per meter.
[tex]v_{max}[/tex] - Maximum speed of the mass, measured in meters per second.
If we know that [tex]m = 2\,kg[/tex], [tex]k = 200\,\frac{N}{m}[/tex] and [tex]v_{max} = 5\,\frac{m}{s}[/tex], then the maximum elongation of the spring is:
[tex]x_{max} = \sqrt{\frac{2\,kg}{200\,\frac{N}{m} } }\cdot \left(5\,\frac{m}{s} \right)[/tex]
[tex]x_{max} = 0.5\,m[/tex]
The maximum elongation of the spring is 0.5 meters.
types of magnetic resonance image
Answer: ur mom
Explanation:
PLEASE HELP! what kind of electricity does electrical heat burners make
Answer:
Electrical to Thermal
calculate the force necessary to keep a mass of 2 kg moving on a circular path of radius 0.2 m with a period of 0.5 second. what is the direction of this force
How large a net force is required to accelerate a 1,586-kg SUV from rest to a speed of 10.4 m/s in a distance of 212 m?
Answer:
F = 404.6 N
Explanation:
Mechanical Force
According to the second Newton's law, the net force exerted by an external agent on an object of mass m is:
F = m.a
Where a is the acceleration of the object.
On the other hand, the equations of the Kinematics describe the motion of the object by the equation:
[tex]v_f=v_o+a.t[/tex]
Where:
vf is the final speed
vo is the initial speed
a is the acceleration
t is the time
Another useful equation for speeds and distance X is:
[tex]V_f^2=V_o^2+2aX[/tex]
To calculate the acceleration, we solve for a:
[tex]\displaystyle a=\frac{V_f^2-V_o^2}{2X}[/tex]
The SUV starts from rest (vo=0) to vf=10.4 m/s traveling a distance of X=211 m, thus the acceleration is:
[tex]\displaystyle a=\frac{10.4^2-0^2}{2(212)}[/tex]
[tex]\displaystyle a=\frac{108.16}{424}[/tex]
[tex]a=0.26 \ m/s^2[/tex]
The force needed to accelerate the SUV of mass m=1,586 Kg is:
F = 1,586*0.26
F = 404.6 N
a machine
of efficiency of 70% is used to raise
a body of
mass 80 kg through
a vertical distance of 3m in
40 seconds. Calculate the power input. (Take g = 10 m/s²)
Answer:
Power_input = 85.71 [W]
Explanation:
To be able to solve this problem we must first find the work done. Work is defined as the product of force by distance.
[tex]W =F*d[/tex]
where:
W = work [J] (units of Joules)
F = force [N] (units of Newton)
d = distance [m]
We need to bear in mind that the force can be calculated by multiplying the mass by the gravity acceleration.
Now replacing:
[tex]W=m*g*d[/tex]
[tex]W=80*10*3\\W=2400[J][/tex]
Power is defined as the work done over a certain time. In this way by means of the following formula, we can calculate the required power.
[tex]P=W/t[/tex]
where:
P = power [W] (units of watts)
W = work [J]
t = time = 40 [s]
[tex]P=W/t\\P=2400/40\\P=60 [W][/tex]
The calculated power is the required power. Now as we have the efficiency of the machine, we can calculate the power that is introduced, to be able to do that work.
[tex]Effic=0.7\\Effic=P_{required}/P_{introduced}\\P_{introduced}= 60/0.7\\P_{introduced}=85.71[W][/tex]
A 2.0 kg bucket is attached to a horizontal ideal spring and rests on frictionless ice. You have a 1.0 kg mass
that you must drop into the bucket. Where should the bucket be when you drop the mass (so it is moving
purely vertically when it lands in the bucket) if your goal is to:
(a) Maximize the amplitude of the oscillation of the resulting 3.0 kg mass and spring system.
(b) Minimize the amplitude of the oscillation of the resulting 3.0 kg mass and spring system.
Answer:
x = A cos (w \sqrt{2y_{o}/g})
a) maximun Ф= \sqrt{\frac{2}{3} \frac{2 y_{o} }{g} }
b) minimun Ф = [tex]\frac{\pi }{2}[/tex] - \sqrt{\frac{2}{3} \frac{2 y_{o} }{g} }
Explanation:
For this exercise let's use kinematics to find the time it takes for the mass to reach the floor
y = y₀ + v₀ t - ½ g t²
as the mass is released from rest, its initial velocity is zero (vo = 0) and its height upon reaching the ground is zero (y = 0)
0 = y₀ - ½ g t²
t = [tex]\sqrt{2y_{o}/g}[/tex]
The bucket-spring system has a simple harmonic motion, which is described by
x = A cos wt
in this expression we assumed that the phase constant (Ф) is zero
let's replace the time
x = A cos (w \sqrt{2y_{o}/g})
this is the distance where the system must be for the mass to fall into it.
a) The new system has a total mass of m ’= 3.0 kg, so its angular velocity changes
w = [tex]\sqrt{k/m}[/tex]
In the initial state
w = \sqrt{k/2}
When the mass changes
w ’= \sqrt{k/3}
the displacement in each case is
x = A cos (wt)
for the new case
x ’= A cos (w’t + Ф)
the phase constant is included to take into account possible changes due to the collision of the mass.
we see that this maximum expressions when the cosine is maximum
cos (w´t + Ф) = 1
w’t + Ф = 0
Ф = -w ’t
Ф = - [tex]\sqrt{k/3}[/tex] [tex]\sqrt{2y_{o}/g}[/tex]
\sqrt{\frac{2}{3} \frac{2 y_{o} }{g} }
b) the function is minimun if
cos (w’t + fi) = 0
w’t + Ф = π / 2
Ф = π / 2 - w ’t
Ф = [tex]\frac{\pi }{2}[/tex] - \sqrt{\frac{2}{3} \frac{2 y_{o} }{g} }
help please Derive an equation
Ta=1.44T1/2
Answer:
To derive an equation you must indicate the variable you want to solve for.Here we have tension of an object A and Tension 1.Two variables or unknown are given hence we cannot derive any other equations.which is NOT true of the Big Bang Theory?
1. it explains what’s happening in the universe now
2. it explains what caused the big bang.
3. it explains what the universe was like in the beginning
4. it stages that the universe is expanding
Answer:
1. it explains what's happening im the universe now
What is the volume of a brick that is 30 cm long, 8 cm wide, and 10 cm tall?
A street bridge is 5.5m long if the linear expansion of steel is 0.00001 oc How much will it expand when temperatures is by 10oc? Give answer in Cm
Answer:
[tex]l_o=550.055\ cm[/tex]
Explanation:
Given that,
Length of a street bridge, l = 5.5 m
The coefficient of bridge, [tex]\alpha =0.00001 ^0 C[/tex]
We need to find how much will it expand when temperatures is by 10°C.
The change in length per unit original length is given by :
[tex]\dfrac{\Delta l}{l}=\alpha \Delta T\\\\\Delta l = l\alpha \Delta T\\\\=5.5\times 0.00001 \times 10\\\\\Delta l=0.00055\\\\(l_o-l)=0.00055\\\\l_o=0.00055+5.5\\\\=5.50055\ m\\\\l_o=550.055\ cm[/tex]
Hence, the length will expanded 550.055 cm.
Find the total charge of a system consists 2X10^4 electrons.
ANSWER: The first thing to learn is how to convert numbers back and forth between scientific notation and ordinary decimal notation. The expression "10n", where n is a whole number, simply means "10 raised to the nth power," or in other words, a number gotten by using 10 as a factor n times:
105 = 10 x 10 x 10 x 10 x 10 = 100,000 (5 zeros)
108 = 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 = 100,000,000 (8 zeros)
Notice that the number of zeros in the ordinary decimal expression is exactly equal to the power to which 10 is raised.
If the number is expressed in words, first write it down as an ordinary decimal number and then convert. Thus, "ten million" becomes 10,000,000. There are seven zeros, so in powers of ten notation ten million is written 107.
A number which is some power of 1/10 can also be expressed easily in scientific notation. By definition,
1/10 = 10-1 ("ten to the minus one power")
More generally, the expression "10-n" (where n is a whole number) means ( 1/10 )n. Thus
10-3 = ( 1 / 10 )3 = 1 / ( 10 x 10 x 10) = 1/1000
10-8 = ( 1 / 10 )8 = 1/100,000,000
Scientific notation was invented to help scientists (and science students!)deal with very large and very small numbers, without getting lost in all the zeros. Now answer the following on a separate sheet of paper and check your answers by clicking on "Answers":
Explanation:
Explain how a fish is able to enjoy 180° field of view when in a pond.
Answer:
Time
An explanation of basic pond measurements that are vital to basic pond management practices.
Save For Later Print
Pond Measurements: Area, Volume and Residence Time - Articles UPDATED: OCTOBER 29, 2015
Pond Measurements: Area, Volume and Residence Time
The importance of getting an accurate estimation of your pond surface area cannot be overestimated. The majority of pond owners visually estimate their pond area, which usually results in an overestimate of the true pond surface area. Pond area and water volume should be calculated based on some simple measurements. The effort necessary to estimate pond surface area is directly related to your pond's shape and uniformity. The simplest method--using basic equations for common shapes--can be applied if your pond closely resembles a circle, square, rectangle, or trapezoid in shape.
Pond Shapes
Circular
pond shape can be estimated by measuring the distance around the pond shoreline in feet. Square the shoreline distance and divide by 547,390 to get the pond area in acres. For example, a pond that is 450 feet around the shoreline would have an area = (450 feet)2 / 547,390 or 0.37 acres.
Explanation:
i looked this up
What is net force?
A. The sum of all the forces acting on an object
B. The difference between the largest and smallest forces acting on
an object
C. The largest force acting on an object
D. The smallest force acting on an object
Answer:
A. The sum of all the forces acting on an object.
a 9.0-kg dog runs at 4.0 m/s and jumps onto a stationary skateboard the mas of the skateboard is 1.0 m/s what speed is the speed of the skating dog?
Answer:
1.3
Explanation:
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In a step-down transformer the secondary coil is made of thick copper coil.
Explain
(2 marks)
Answer:
In a step-up transformer, the secondary coil is made of same wire as the primary coil but it is rolled with the presence of the insulated which never conduct the opposite current to the coil.
Hence it function property without meeting any fire and other problem.
Even if you want to get clear ideas, then you have to go with official website.
Based on its location on the periodic table, which metal is likely to be less reactive than both strontium and sodium?
magnesium (Mg)
barium (Ba)
cesium (Cs)
rubidium (Rb)
Answer:
Magnesium is the correct answer
Explanation:
It is magnesium because as we move down the periodic table group, atomic sizes increase which is as a result of lesser forces of attraction between nucleus and valence of an atom which makes the atom to be prone to leave as electrons so easily, elements become more reactive.
But as we move along the period , atomic sizes decrease because there is a more force of attraction between nucleus of an atom and it valence electrons making it less reactive just as the case in magnesium. Other elements are more reactive than it across the group.
Answer:
Magnesium
Explanation: Correct on edge!
Hope this Helps! :)
Which answer below most accurately describes both cells?
What power does a toaster with 15 A of current and 20 Ωof resistance consume?
Answer:
4500W
Explanation:
[tex]P = {I}^{2} R[/tex]
where P = Power consumed , I = Current & R = Resistance.
In the question it's given that
I = 15A ; R = 20Ω
So ,
[tex]P = {15}^{2}\times 20 = 225 \times 20 = 4500W[/tex]
2. Two identical spheres having charges Q and -
2Q experience a force F at a certain distance
If the spheres are kept in contact and then
placed at same initial distance, the force
between them will be
force between the two charges (+q1 and +q2),if they are at a distance 'a' is
F1=1/4pieEo q1q2/d^2................ (1)
when the metal spheres are in contact the charge flow from one sphere to another till both the sides acquires the same charge. here q1 and q2 are of same sign,hence after contact each sphere will have a charge
[tex] \binom{q1 + q2 }{2} [/tex]
now,the force between them,
f2=1/4pieEo
(q1+q2/2)^2/d^2
from eq (1)and eq (2)
f2=f1 (q1+q2)^2/4q1q2
Which is true of low resistance wires
Answer:
the cooler the wire, the less the resistance
What is the resistance of a wire with "p" of .25, a length "L" of 11.2, and an area "A" of 1.26?
Answer:
Resistance = 2.2 Ohms.
Explanation:
Given the following data;
Resistivity, P = 0.25
Length, L = 11.2
Area, A = 1.26
To find the resistance.
Resistance is given by the formula below;
Resistance = PL/A
Substituting into the equation, we have;
Resistance = (0.25*11.2)1.26
Resistance = 2.8/1.26
Resistance = 2.2 Ohms.
Therefore, the resistance of the wire is 2.2 Ohms.
A double-slit interference pattern is created by two narrow slits spaced 0.21 mm apart. The distance between the first and the fifth minimum on a screen 61 cm behind the slits is 6.2 mm What is the wavelength of the light used in this experiment?
Answer:
[tex]\lambda =533.6 nm [/tex]
Explanation:
the slits spacing, d = 0.21 mm
distance of screen, D = 61 cm
The condition for minima is given as
[tex]dsin(\theta) = \left ( n+\frac{1}{2} \right )\lambda[/tex]
So, first minima, n = 0
[tex]dsin(\theta_1) = \frac{1}{2}\lambda[/tex]
fifth minima, n = 4
[tex]dsin(\theta_5) = \frac{9}{2}\lambda[/tex]
[tex]d(sin(\theta_5) -sin(\theta_1))= 4\lambda[/tex]
For small angle
[tex]d(tan(\theta_5) -tan(\theta_1))= 4\lambda[/tex]
From the figure:
[tex]d(\frac{y_5}{D}-\frac{y_1}{D})= 4\lambda[/tex]
[tex]\frac{d}{D} (y_5-y_1) = 4\lambda[/tex]
[tex]\lambda = \frac{d}{4D} (y_5-y_1)[/tex]
[tex]\lambda = \frac{0.021}{4(61)} (6.2 \times 10^{-3})[/tex]
[tex]\lambda =533.6 nm[/tex]
"533.5 nm" would be the wavelength of the light used in this experiment.
According to the question,
Slits spacing,
d = 0.21 mmDistance of screen,
D = 61 cmThe condition for minima is given as:
→ [tex]d (sin \Theta) = (n+\frac{1}{2} ) \lambda[/tex]
So,
1st minima, n = 0
→ [tex]d sin(\Theta_1)= \frac{1}{2} \lambda[/tex]
5th minima, n = 4
→ [tex]d (sin (\Theta_5) - sin (\Theta_1) = 4 \lambda[/tex]
For small angle,
→ [tex]d (tan (\Theta_5) - tan (\Theta_1) = 4 \lambda[/tex]
From the figure, we get
→ [tex]d(\frac{y_5}{D} - \frac{y_1}{D} )= 4 \lambda[/tex]
[tex]\frac{d}{D} (y_5 -y_1) = 4 \lambda[/tex]
[tex]\lambda = \frac{d}{4D}(y_5-y_1)[/tex]
[tex]= \frac{0.021}{4\times 61} (6.2\times 10^{-3})[/tex]
[tex]= 533.6 \ nm[/tex]
Thus the above answer is correct.
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Calculate the change in entropy of 0.020 kg of ice when it melts at 0.0°C. The heat of fusion of ice is 3.36 x 10'J/kg.
Answer:
S = 2461.53 [kJ]
Explanation:
The change in entropy in a process such as melting can be calculated by means of the following expression.
[tex]S=\frac{H*m}{T}[/tex]
where:
S = entropy [kJ]
H = fusion heat = 3.36*10¹ [J/kg]
m = mass = 0.02 [kg]
T = temperature in kelvin = 273 [K]
[tex]S = \frac{0.02*3.36*10^{1} }{(273+0)}\\S = 2461.53 [kJ][/tex]