Answer:
Explanation:
F = ma
3(9.81) - 3.8(9.81sin36) = (3 + 3.8)a
a = 1.10566...
a = 1.1 m/s²
the 3.8 kg mass will move up slope and the 3 kg mass will fall at that acceleration.
Which of the following best describes an achievement test that is given at the end of a learning segment to evaluate mastery of objectives?
Answer
An Achievement test is an assessment of developed knowledge or skill. ... Achievement tests are developed to measure skills and knowledge learned in a given grade level, usually through planned instruction, such as training or classroom instruction. Achievement tests are often contrasted with aptitude tests.
Explanation:
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Good morning dear...
Have a beautiful and joyful day ahead.
Using Electrostatic Concept Explain The Classification of substances in Terms of Their ability to Conduct Electric Charges.
Answer:
please mark me as brainliest I'm begging u
Explanation:
The behavior of an object that has been charged is dependent upon whether the object is made of a conductive or a nonconductive material. Conductors are materials that permit electrons to flow freely from particle to particle. An object made of a conducting material will permit charge to be transferred across the entire surface of the object. If charge is transferred to the object at a given location, that charge is quickly distributed across the entire surface of the object. The distribution of charge is the result of electron movement. Since conductors allow for electrons to be transported from particle to particle, a charged object will always distribute its charge until the overall repulsive forces between excess electrons is minimized. If a charged conductor is touched to another object, the conductor can even transfer its charge to that object. The transfer of charge between objects occurs more readily if the second object is made of a conducting material. Conductors allow for charge transfer through the free movement of electrons
Eric drops a 2.20 kg water balloon that falls a distance of 45.08 m off the top of a
building. What is the kinetic energy at the bottom?
Answer:
972 J
Explanation:
At the bottom, all the gravitational potential energy was converted into kinetic energy. If you calculate the GPE, its value will be the same that the KE at the bottom. The GPE can be calculated this way:
GPE = mass×gravity×heigth
GPE = 2.2×9.8×45.08 ≈ 972
Although genes contribute to whatever IQ scores measure, IQ can change radically due to changes in the __________.
Answer: Even though the genetic susceptibility plays a crucial role on the IQ of the individual, various modifiable environmental factors like education, premature birth, nutrition, pollution, drug and alcohol abuse, mental illnesses, and diseases can have an influence on an individual's IQ.
Some amount of ideal gas with internal energy U was heated from 100^0C to 200^0C. We can predict that internal energy after heating in terms of U is:
The internal energy after heating in terms of U is 100U.
The given parameters;
initial temperature of the gas, T₁ = 100 ⁰Cfinal temperature of the gas, T₂ = 200 ⁰CAssuming a constant pressure, the internal energy of the ideal gas is equal to the change in the enthalpy of the ideal gas.
[tex]\Delta H = U \times \Delta T\\\\\Delta H = U (200 - 100)\\\\\Delta H = 100 U[/tex]
Thus, we can conclude that the internal energy after heating in terms of U is 100U.
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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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Two parallel plates, separated by 0.20 m, are connected to a 12-V battery. An electron released from rest at a location 0.10 m from the negative plate. When the electron arrives at a distance 0.050 m from the positive plate, what is the potential difference between the initial and final points
The potential difference between the initial and final point is 3.0 V.
The given parameters:
distance between the plates, d = 0.2 mvoltage across the plates, V = 12 Vposition of the electron from negative plate, x₁ = 0.1 mposition of the electron from the positive plate, x₂ = 0.0 5mThe potential difference between the initial and final point is calculated as follows;
[tex]E = \frac{V}{d} \\\\\frac{V_1}{d_1} = \frac{V_2}{d_2}[/tex]
where:
[tex]d_2[/tex] is the distance of the electron between the positive and negative plate
[tex]0.1 + d_2 + 0.05 = 0.2\\\\d_2 + 0.15 = 0.2\\\\d_2 = 0.2 - 0.15\\\\d_2 = 0.05 \ m[/tex]
[tex]V_2 = \frac{V_1d_2}{d_1} \\\\V_2 = \frac{12 \times 0.05}{0.2} \\\\V_2 = 3.0 \ V[/tex]
Thus, the potential difference between the initial and final point is 3.0 V.
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The current in the long wire is decreasing. What is the direction of the current induced in the conducting loop below the wire
Answer:
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lutfsn...
Another nervous papa bear paces at 1.0 m/s north for 3.0 s, then at 1.6 m/s south for 5.0 s, and then
at 1.4 m/s north for 4.0 s. The average speed of this bear is:
a) 1.3 m/s
b) 1.4 m/s
c) 4.0 m/s
d) 8.0 m/s
Answer:
Explanation:
Average speed is total distance traveled over the time needed to do so.
d = vt
v = (1.0(3.0) + 1.6(5.0) + 1.4(4.0) / (3.0 + 5.0 + 4.0)
v = 16.6 / 12
v = 1.383333333...
v = 1.4 m/s
How much heat must be removed from 1.61 kg of water at 0 ∘C to make ice cubes at 0 ∘C?
Answer:
Explanation:
All
Two point particles, with masses m and 2 m, are separated by some distance Lin an uniform gravitational field. The center of gravity of this two particle system is:________
А. located L/2 distance away from either mass and on the line joining the two particles.
B. located 1/3 distance away from 2 m and on the line joining the two particles.
C. located L/4 distance away from 2 m and on the line joining the two particles.
D. located L/3 distance away from mand on the line joining the two particles.
E. located L/4 distance away from m and on the line joining the two particles.
F. located where the mass 2 m is located.
The center of gravity of this two particle system is located 1/3 distance away from 2 m and on the line joining the two particles.
The given parameters:
Mass of the first particle, = mMass of the second particle, = 2mDistance between the two particles, = LThe center of gravity of the two particles when first particle is fixed is calculated as;
[tex]C_G = \frac{m(0 ) \ +2m(L) }{m+ 2m} \\\\C_G = \frac{2mL}{3m} \\\\C_G = \frac{2L}{3}[/tex]
The center of gravity of the two particles when second particle is fixed is calculated as;
[tex]C_G = \frac{m(L) \ + \ 2m(0)}{m + 2m} \\\\C_G = \frac{mL}{3m} \\\\C_G = \frac{L}{3}[/tex]
Thus, the center of gravity of this two particle system is located 1/3 distance away from 2 m and on the line joining the two particles.
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10. On Earth, where is hydrogen not found?
A. Natural gas well
B. Water
C. Atmosphere
o D. Minę
Answer:
i think its D or B
Explanation:
i just think
On Earth,in atmosphere the hydrogen is found in the less quantity. Option C is correct.
What is hydrogen?The chemical element hydrogen has the atomic number 1 and the symbol H. The smallest element is hydrogen.
Hydrogen is found in large numbers on Earth in combination with other elements, such as water and hydrocarbons, yet it is only 0.00005 percent present in Earth's atmosphere.
On Earth,in atmosphere the hydrogen is found in the less quantity.
Option C is correct.
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What are the customary units for real power? volt-amperes reactive (VAR) volt-amperes (VA) watts (W)
Answer:
watts is for real power
volt amperes reactive (VA) for reactive power
volt amperes (VA) for apparent power
HELP ASAPPPPPPPPPPPPPP
Answer:
F=15N
Explanation:
F=m.a
m=1500g ÷1,000 = 1.5kg
a= 10m/s/s
F =1.5 × 10
F = 15 Newton
What’s Newton’s second law? Explain and mention some examples in daily life
Answer:
Newton's second law states that .
The rate of change of linear momentum is directly proportional to the force applied.Formulically
F=maF=Force
m=mass
a=acceleration
The best example is hitting a tennis ball.
Define critical angle
Answer:
Critical angle is a angle of incidence, for which refraction is 90 degree, this is a least angle of incidence at which total reflection takes place, please mark me Brainliest
50 points please help ASAP! !
What is the average velocity of an object that moves from 6 meter to 2 meter relative to origin in 2 second?
[tex]\\ \rm\rightarrowtail Avg\: velocity=\dfrac{Total\: Displacement }{Total\: time}[/tex]
[tex]\\ \rm\rightarrowtail Avg\: velocity=\dfrac{4}{2}=2m/s[/tex]
A 27 kg chair initially at rest on a horizontal
floor requires a 173 N horizontal force to set
it in motion. Once the chair is in motion, a
148 N horizontal force keeps it moving at a
constant velocity.
The acceleration of gravity is 9.81 m/s.
a) What is the coefficient of static friction
between the chair and the floor?
Answer:
4653
4565445655687568677667876
A fat wire, radius a, carries a constant current I, uniformly distributed over its cross section. A narrow gap in the wire, of width w << a, forms a parallel-plate capacitor. Find the magnetic field in the gap, at a distance s < a from the axis.
The magnetic field in the gap is [tex]\mathbf{B = \dfrac{\mu_oI s}{2 \pi a^2}}[/tex]
In a circuit, a magnetic flux is circulated or followed via a confined environment or passage called a magnetic field gap. The narrow air gap is a non-magnetic component of a magnetic circuit that is normally connected to the remainder of the circuit magnetically in series. This enables a significant amount of magnetic flux to pass via the gap.
The magnetic field in the gap at a distance s < a can be computed by using the formula:
[tex]\mathbf{ \oint Bdl = \mu_oI_{enclosed}}[/tex]
where;
Magnetic flux density = Bdistance = d[tex]\mathbf{B( 2 \pi d) = \mu _o \oint _s J_d da }[/tex]
where;
[tex]\mathbf{J_d}[/tex] = drift current density[tex]\mathbf{B( 2 \pi d) = \mu _o J_d \oint _sda }[/tex]
[tex]\mathbf{B( 2 \pi d) = \mu _o J_d (\pi d^2) }[/tex]
Making the magnetic flux density the subject, we have:
[tex]\mathbf{B =\dfrac{ \mu _o J_d (\pi d^2) }{( 2 \pi d)}}[/tex]
[tex]\mathbf{B =\dfrac{ \mu _o J_dd}{ 2 }}[/tex]
Recall that, the drift current density [tex]\mathbf{J_d = \dfrac{I}{\pi a^2}}[/tex]
[tex]\mathbf{B = \dfrac{\mu_o d}{2}(\dfrac{I}{\pi a^2})}[/tex]
Recall that distance in question is said to be (s);
∴
[tex]\mathbf{B = \dfrac{\mu_oI s}{2 \pi a^2}}[/tex]
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What are the two factors that affect the frictional force between objects
In an oscillating LC circuit, when 86.6% of the total energy is stored in the inductor's magnetic field, (a) what multiple of the maximum charge is on the capacitor and (b) what multiple of the maximum current is in the inductor
We have that for the Question "(a) what multiple of the maximum charge is on the capacitor and (b) what multiple of the maximum current is in the inductor"
Answer:
a) maximum charge = [tex]0.366Q_{max[/tex]b) maximum current = [tex]0.931I_{max}[/tex]
From the question we are told
In an oscillating LC circuit, when 86.6% of the total energy is stored in the inductor's magnetic field
A) When 86.6\% energy is stored in inductor
[tex]\%[/tex]of energy stored in electric field = [tex]1 - 0.866 = 13.4\%[/tex]
[tex]\frac{V_E}{V} = \frac{\frac{q^2}{2c}}{\frac{Q^2}{2c}} = 0.134\\\\\frac{q}{Q} = \sqrt0.134\\\\\frac{q}{Q} = 0.366\\\\q = 0.366Q_{max[/tex]
B)
[tex]\frac{V_B}{V} = \frac{\frac{Li^2}{2}}{\frac{LI^2}{2}} = 0.866\\\\\frac{i}{I} = \sqrt0.866\\\\\frac{i}{I} = 0.931\\\\i = 0.931I_{max[/tex]
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Why did humans first develop culture?
A. It helped humans distinguish themselves from wild animals.
B. It helped humans become the dominant species on Earth.
C. It helped the members of a society survive in their environment.
D. It helped individuals survive without the support of a society.
SUBMIT
Answer: C.
Correct me if I'm wrong.
An applied force accelerates a 4.00 kg block to an initial velocity of 11 m/s across a rough horizontal surface, in the positive x direction. As the block reaches 11 m/s, the applied force is removed. The block then slows to 1.5 m/s at a distance of 4.00 m beyond where the applied force was removed. Determine the magnitude and the direction of the non-conservative force acting on the box as it slides.
The only force opposing the block's sliding as it slows down is friction with magnitude f . By Newton's second law, the net force in this direction is
∑ F = -f = ma = (4.00 kg) a
Assuming constant acceleration a , the acceleration applied by friction is such that
(1.5 m/s)² - (11 m/s)² = 2a (4.00 m)
Solve for the acceleration :
a = ((1.5 m/s)² - (11 m/s)²) / (8.00 m) ≈ -14.8 m/s²
Then the frictional force exerted a magnitude of
-f = (4.00 kg) (-14.8 m/s²)
f ≈ 59.4 N
and was directed opposite the block's motion.
A pumpkin is launched in the air and travels at a horizontal velocity of 25 meters per second for 5 seconds. How far does it travel horizontally?
Answer:
30.3 meters, 172 degrees
Explanation:
To insure the most accurate solution, this problem is best solved using a calculator and trigonometric principles. The first step is to determine the sum of all the horizontal (east-west) displacements and the sum of all the vertical (north-south) displacements.
Horizontal: 2.0 meters, West + 31.0 meters, West + 3.0 meters, East = 30.0 meters, West
Vertical: 12.0 meters, North + 8.0 meters, South = 4.0 meters, North
The series of five displacements is equivalent to two displacements of 30 meters, West and 4 meters, North. The resultant of these two displacements can be found using the Pythagorean theorem (for the magnitude) and the tangent function (for the direction). A non-scaled sketch is useful for visualizing the situation.
Applying the Pythagorean theorem leads to the magnitude of the resultant (R).
R2 = (30.0 m)2 + (4.0 m)2 = 916 m2
R = Sqrt(916 m2)
R = 30.3 meters
The angle theta in the diagram above can be found using the tangent function.
tangent(theta) = opposite/adjacent = (4.0 m) / (30.0 m)
tangent(theta) = 0.1333
theta = invtan(0.1333)
theta = 7.59 degrees
This angle theta is the angle between west and the resultant. Directions of vectors are expressed as the counterclockwise angle of rotation relative to east. So the direction is 7.59 degrees short of 180 degrees. That is, the direction is ~172 degrees.
The axis of the earth is
Tilted about 23.5 degrees
Vertical
Vertical
Answer:
The axis of rotation of the Earth is tilted at an angle of 23.5 degrees away from vertical, perpendicular to the plane of our planet's orbit around the sun. The tilt of the Earth's axis is important, in that it governs the warming strength of the sun's energy.
Explanation:
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At what speed must the electron revolve round the nucleus of
the hydrogen in its ground state in order that it may not be pulled into the
nucleus by electrostatic attraction
Explanation:
I think this is it, give it a try
Suppose 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
Please solve this This is my exam question please be fast
Answer:
OK sure why not!!!!!!!!!!!!
At a construction site, a 68.0 kg bucket of concrete hangs from a light (but strong) cable that passes over a light friction-free pulley and is connected to an 85.0 kg box on a horizontal roof (see the figure (Figure 1)). The cable pulls horizontally on the box, and a 46.0 kg bag of gravel rests on top of the box. The coefficients of friction between the box and roof are shown. The system is not moving.
Following are the calculation to the friction force:
Please see the diagram below for an illustration of the forces acting on objects.Therefore, m denotes the mass of a gravel bag, and M denotes the overall mass of the box.Whenever the system is stationary, the friction force on the box equals the tension in the string.The substring tension force is provided as follows:
[tex]\to T-m_{1}g=0 \\\\\to T = m_{1}g \\\\[/tex]
[tex]= (68\ kg) (9.81 \ \frac{m}{s^2} ) (\frac{ 1\ N}{ 1\ kg \cdot \frac{m}{s^2}}) \\\\ = (68 ) (9.81 ) ( 1\ N) \\\\= 667.08\ N \\\\[/tex]
Therefore, the friction force on the box is "667 N".
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