Answer:
a
Explanation:
..................
Calculate the minimum area moment of inertia for a rectangular cross-section with side lengths 3 cm and 8 cm.
Lx = b h^3 /12
Ly= b^3 h/ 12
Where:
Lx= Area moment of inertia related to the x axis
Ly= Area moment of inertia related to the y axis
b= base = 3 cm
h= height = 8 cm
Replacing:
Lx = 3 (8)^3 / 12 = 3 (512) /12 = 128 cm^4
Ly= 3^3 * 8 / 12 = 27 (8) / 12 = 18 cm^4
Answer = 18 cm4
A 3 kg block is attached to a vertical spring. Initially, you exert a 50 N downwards force on the block, holding it in place, at rest. You let go. Find the instantaneous acceleration of the block immediately after you let go. What is the direction of the acceleration
As you were holding the block down and in place, the spring exerted an upward force that balanced the downward push by your hand and its own weight. So this restoring force has a magnitude of R such that
R - 50 N - (3 kg) g = 0 => R = 79.4 N
As soon as you remove your hand, the block has acceleration a such that, by Newton's second law,
R - (3 kg) g = (3 kg) a => a = (79.4 N - (3 kg) g) / (3 kg) ≈ 16.7 m/s^2
pointing upward.
The diagram below shows the velocity vectors for two cars that are moving
relative to each other.
45 m/s west
25 m/s east
Car 1
Car 2
From the frame of reference of car 1, what is the velocity of car 2?
OA. 20 m/s west
OB. 70 m/s east
C. 70 m/s west
D. 20 m/s east
The relative velocity of the car 2 with respect to car 1 is 70 m/s west. So, the correct option is C.
Velocity of the car 1 with respect to the ground, v₁ = 45 m/s west
Velocity of the car 2 with respect to the ground, v₂ = 25 m/s east
Let the direction towards the east be positive and the opposite direction towards the west be negative.
The expression for the relative velocity of the car 2 with respect to car 1 is given by,
v₂₁ = v₂ - v₁
v₂₁ = 25 - (-45)
Therefore, the relative velocity,
v₂₁ = 70 m/s west
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Attaching the image file here.
need help with these 4 ASAP. if u can answer even 1 of them that would be amazing! formula sheet provided
1. a) a mass of 5.0kg is at rest at the top of an inclined plane, 5.0m above the ground. if the inclined plane is frictionless, calculate the final velocity of the mass as it reaches the bottom of the inclined plane?
b) if the mass slides down the plane and encounters a spring with a force constant of 8000.00 N/m by how much will the spring be compressed when the mass comes to a stop?
c) if the mass is found to move down the inclined plane at a constant velocity what is the force of friction between the ramp and the mass? (the angle of incline is 30°)
2. a 12 m long ramp has its high end 4.0m above the low end. a crate of mass 10kg is given a sharp push so that it begins to slide down the ramp from the top of the ramp with an initial speed of 4.0m/s. the force of friction along the ramp is 20N. determine the speed of the crate at the bottom of the ramp.
3. a 7.0 kg crate is dropped from a window 8.0m above the ground. if the average air resistance on the crate is 20.0N, what is the crates kinetic energy and velocity upon reaching the ground?
4. a 0.25 kg puck moving with a constant speed across a 2.0m stretch of horizontal frictionless ice. how much work is done?
you may want to google the answers cuz that is a lot
A common method to measure thermal conductivity of a biomaterial is to insert a long metallic probe axially into the center of a long cylindrical container of the material. The probe is maintained at some uniform temperature Ti, and the outside of the container is maintained at a temperature To. Inside the metallic probe is an electrical heater for which the electrical power is measured. If the diameter of the probe maintained at 50°C is 1 cm, the outer diameter of the container maintained at 20°C is 4 cm, the length of the cylinder container is 60 cm, and the power input is 40.8 W, calculate the thermal conductivity of this material.
Answer:
The thermal conductivity of the biomaterial is approximately 1.571 watts per meter-Celsius.
Explanation:
Let suppose that thermal conduction is uniform and one-dimensional, the conduction heat transfer (\(\dot Q\)), measured in watts, in the hollow cylinder is:
\(\dot Q = \frac{2\cdot k\cdot L}{\ln \left(\frac{D_{o}}{D_{i}} \right)}\cdot (T_{i}-T_{o})\)
Where:
\(k\) - Thermal conductivity, measured in watts per meter-Celsius.
\(L\) - Length of the cylinder, measured in meters.
\(D_{i}\) - Inner diameter, measured in meters.
\(D_{o}\) - Outer diameter, measured in meters.
\(T_{i}\) - Temperature at inner surface, measured in Celsius.
\(T_{o}\) - Temperature at outer surface, measured in Celsius.
Now we clear the thermal conductivity in the equation:
\(k = \frac{\dot Q}{2\cdot L\cdot (T_{i}-T_{o})}\cdot \ln\left(\frac{D_{o}}{D_{i}} \right)\)
If we know that \(\dot Q = 40.8\,W\), \(L = 0.6\,m\), \(T_{i} = 50\,^{\circ}C\), \(T_{o} = 20\,^{\circ}C\), \(D_{i} = 0.01\,m\) and \(D_{o} = 0.04\,m\), the thermal conductivity of the biomaterial is:
\(k = \left[\frac{40.8\,W}{2\cdot (0.6\,m)\cdot (50\,^{\circ}C-20\,^{\circ}C)}\right]\cdot \ln \left(\frac{0.04\,m}{0.01\,m} \right)\)
\(k \approx 1.571\,\frac{W}{m\cdot ^{\circ}C}\)
The thermal conductivity of the biomaterial is approximately 1.571 watts per meter-Celsius.
24. A body A rests on a smooth horizontal table. Two bodies of mass 2 kg and 10 kg hanging freely, are attached to A by strings which pass over smooth pulleys at the edges of the table. The two strings are taut. When the system is released from rest, it accelerates at 2 m/s2 . Find the mass of A.
The two strings are taut. When the system is released from rest, it accelerates at 2 m/s2 then, Mass of A = 8m/5 kg.
Let the mass of the body A be ‘m’.
The two strings are taut so they exert a tension ‘T’ on body A.
Let ‘a’ be the acceleration produced in the system.
The free body diagram of body A is given below: mA + 2T = mA + ma = mA + m(2)mA + 10T = mA + ma = mA + m(2)
As the two strings are taut, we can say that tension in both strings is equal.
Therefore 2T = 10T or T = 5T As the body A is resting on a smooth horizontal table, there is no friction force acting on the body A.
The net force acting on body A is the force due to tension in the strings. ma = 2T – mg …(1)
As per the given problem, the system is released from rest.
Hence the initial velocity is zero.
Also, we are given that the system accelerates at 2 m/s2.
Therefore a = 2 m/s2 …(2)
From the equations (1) and (2), we get, m(2) = 2T – mg …(3)⇒ m(2) = 2×5m – mg⇒ 2m = 10m – g⇒ g = 8m/5
Thus, the mass of A is 8m/5 kg.
Answer: Mass of A = 8m/5 kg.
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ONLY ANSWER IF YOU KNOW 100% IF YOU DO A WRONG ANSWER I
What is a controlled procedure that tests the effect of one variable on another?
A. observation
B. experiment
C. model
D. argument
A controlled procedure that tests the effect of one variable on another is known as experiment.
What is experiment?An experiment is a set of actions and observations, performed in the context of solving a particular problem or question, to support or falsify a hypothesis or research concerning phenomena.
An experiment generally tests how one variable is affected by another.
Thus, a controlled procedure that tests the effect of one variable on another is known as experiment.
The correct answer is option B. "experiment"
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Answer:
The experiment.
In the human femur, bone tissue is strongest in resisting compressive force,
approximately half as strong in resisting tensile force, and only about one-
fifth
as strong in resisting shear force. If a tensile force of 8000 N is sufficient to
produce a fracture, how much compressive force will produce a fracture?
How much shear force will produce a fracture
The compressive force that would be enough to produce a fracture is 4000 N.
What is the force?We know that the femur is one of the most important bones that we have in the human body. In this case, we have been told that In the human femur, bone tissue is strongest in resisting compressive force, approximately half as strong in resisting tensile force, and only about one- fifth as strong in resisting shear force.
Then we know that;
Tensile force = 8000
The compressive force would be half of this magnitude as such;
Compressive force = 4000 N
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A 20 kg sled is pulled up a 10m tall hill. What work is done against gravity?
Answer:
1962
Explanation:
w = f × d
= 20×9.81 × 10
= 1962
The y component of a vector (in the xy plane) whose magnitude is 84.5 and whose x component is 68.4.
Given y component = 49.6,-49.6
What is the direction of this vector (angle it makes with the x axis)?
The direction of the vector, with a y component of 49.6 and -49.6, and an x component of 68.4, makes an angle of approximately 39.2 degrees with the positive x-axis.
To find the direction of the vector, we can use the inverse tangent (arctan) function. The arctan of the y component divided by the x component will give us the angle that the vector makes with the positive x-axis.
Calculate the angle using the arctan function:
For the y component of 49.6: arctan(49.6 / 68.4) = 38.7 degrees
For the y component of -49.6: arctan(-49.6 / 68.4) = -38.7 degrees
Since we have both positive and negative values for the y component, we need to consider the signs to determine the correct angle.
The positive y component of 49.6 corresponds to an angle of 38.7 degrees.
The negative y component of -49.6 corresponds to an angle of -38.7 degrees.
To determine the overall direction, we can take the average of the two angles: (38.7 + (-38.7)) / 2 = 0 degrees.
The direction of the vector, with a y component of 49.6 and -49.6, and an x component of 68.4, is approximately 0 degrees with the positive x-axis.
Therefore, the vector makes an angle of approximately 0 degrees with the x-axis.
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g A 4-foot spring is elongated167feet long after a mass weighing 16 pounds is attached to it. The medium throughwhich the mass moves offers a damping force numerically equivalent to92times the instantaneous velocity.(a) Find the equation of motion if the mass is initially released from the equilibrium position with a downwardvelocity of 2 ft/s. (Use the convention that displacements measured below the equilibrium position are positive).(b) What is the velocity of the mass whent
Answer: hello question b is incomplete attached below is the missing question
a) attached below
b) V = 0.336 ft/s
Explanation:
Elongation ( Xo) = 16/ 7 feet
mass attached to 4-foot spring = 16 pounds
medium has 9/2 times instanteous velocity
a) Find the equation of motion if the mass is initially released from the equilibrium position with a downward velocity of 2 ft/s
The motion is an underdamped motion because the value of β < Wo
Wo = 3.741 s^-1
attached below is a detailed solution of the question
2.
A ball player hits a home run, and the ball just clears the wall which is 22.0 m high.
The ball is hit at an angle of 37.0" with a velocity of 45.0 m/s. If the ball is hit from
a height of 0.750 m above the ground, (a) How long is the ball in the air until it
clears the wall while on its way down? (b) How far is the wall from home plate?
a.) the ball was going up at 0.95 seconds and coming down at 4.58 seconds
b.) the distance of the wall from the plate is 34 m
Given that a ball player hits a home run, and the ball just clears the wall which is 22.0 m high. The ball is hit at an angle of 37.0" with a velocity of 45.0 m/s.
Since the ball is hit at an angle of 37.0 degrees, We need to find the vertical and horizontal component of the velocity.
\(U_{y}\) = 45 Sin 37 = 27.08 m/s
\(U_{x}\) = 45 cos 37 = 35. 94 m/s
Let us first calculate the maximum height reached.
\(V^{2}\) = \(U_{y} ^{2}\) - 2gH
At maximum height, V = 0
0 = \(27.08^{2}\) - 2 x 9.8H
19.6H = 733.3
H = 733.33/19.6
H = 37.4 m
If the ball is hit from a height of 0.750 m above the ground,
(a) To calculate the time the ball stays in the air until it clears the wall while on its way down, we will use the formula below.
h = \(U_{y}\)t - 1/2g\(t^{2}\)
Substitute for all the parameters
22 - 0.75 = 27.08t - 0.5 x 9.8 x \(t^{2}\)
21.25 = 27.08t - 4.9\(t^{2}\)
4.9\(t^{2}\) - 27.08t + 21.25
We will use quadratic formula
a = 4.9
b = - 27.08
c = 21.25
t = \(\frac{-b+/- \sqrt{b^{2} - 4ac } }{2a}\)
t = \(\frac{27.08 +/- \sqrt{27.08^{2} - 4 * 4.9 * x^{2} 21.25 } }{2 * 4.9}\)
t = \(\frac{27.08 +/-\sqrt{733.3 - 416.5} }{9.8}\)
t = \(\frac{27.08 +/- 17.8}{9.8}\)
t = 44.88/9.8 or 9.28 / 9.8
t = 4.58 s or 0.95 s
This means that the ball was going up at 0.95 seconds and coming down at 4.58 seconds
(b) The distance of the wall from home plate will be the range which is
R = \(U_{x}\)t
R = 35.94 x 0.95
R = 34.143 m
Therefore, the distance of the wall from the plate is 34 m approximately
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i’ll give brainliest, help!!
An applied force of 10.0 N is applied to a 2.0 kg mass over a distance of 4.0 m. The
surface is horizontal and friction is negligible. Assume that te object is initially at rest.
(a)
(b)
Calculate the work done on the object.
Calculate the final speed of the object.
Answer:
A,W = 40 J
Explanation:
A, W =FS
W=10N(4m)
W = 40 J
(ASAP) would it be 125 m/s2 to calculate for her speeding up?
Answer:
\(0\:\mathrm{ m/s^2}\)
Explanation:
Recall the formula for acceleration:
\(\displaystyle\\a=\frac{v_f-v_i}{\Delta t}\), where \(v_f\) is final velocity, \(v_i\) is initial velocity, and \(\Delta t\) is elapsed time (change in velocity over this amount of time).
Let's look at our time vs velocity graph. At t=0 seconds, V=25 m/s. So her initial velocity is 25 m/s.
We want to find the acceleration during the first 5 seconds of motion. Well, looking at our graph, at t=5 seconds, isn't our velocity still 25 m/s? Therefore, final velocity is 25 m/s (for this period of 5 seconds).
We are only looking from t=0 seconds to t=5 seconds which is a total period of 5 seconds. Therefore, elapsed time is 5 seconds.
Substituting values in our formula, we have:
\(\displaystyle a=\frac{25-25}{5}=\frac{0}{5}=\boxed{0\:\mathrm{m/s^2}}\)
Alternative:
Without even worrying about plugging in numbers, let's think about what acceleration actually is! Acceleration is the change in velocity over a certain period of time. If we are not changing our velocity at all, we aren't accelerating! In the graph, we can see that we have a straight line from t=0 seconds to t=5 seconds, the interval we are worried about. This indicates that our velocity is staying the same! At t=0 seconds, we have a velocity of 25 m/s and that velocity stays the same until t=5 seconds. Even though we are moving, we haven't changed velocity, which means our average acceleration is zero!
A ball is projected at an angle of 53°. If the initial velocity is 48 meters/second, what is the vertical component of the velocity with which it was launched?
A.) 31 meters/second
B.) 38 meters/second
C.) 44 meters/second
D.) 55 meters/second
Answer:
The vertical component of the velocity can be found using the formula:
V₀y = V₀ * sin(θ)
where V₀ is the initial velocity, θ is the angle of projection, and V₀y is the vertical component of the velocity.
Substituting the given values, we have:
V₀y = 48 * sin(53°)
Using a calculator, we can evaluate sin(53°) to be approximately 0.799:
V₀y = 48 * 0.799
V₀y ≈ 38.352
Therefore, the vertical component of the velocity with which the ball was launched is approximately 38 meters/second, which corresponds to option B.
Answer:
B.) 38 meters/second
Explanation:
. A 5cm tall object is placed perpendicular to the principal axis of a convex lens of focal
length 10 cm. The distance of the object from the lens is 15 cm. Find the nature, position
and size of the image. Also find its magnification
The nature of the image formed by the convex lens is virtual, the position of the image is 30 cm away from the lens on the same side as the object, and the size of the image is twice the size of the object. The magnification is 2, meaning the image is magnified.
Given:
Object height (h) = 5 cm
Focal length of the convex lens (f) = 10 cm
Object distance (u) = 15 cm (positive since it's on the same side as the incident light)
To determine the nature, position, and size of the image, we can use the lens formula:
1/f = 1/v - 1/u
Substituting the given values:
1/10 = 1/v - 1/15
To simplify the equation, we find the common denominator:
3v - 2v = 2v/3
Simplifying further:
v = 30 cm
The image distance (v) is 30 cm. Since the image distance is positive, the image is formed on the opposite side of the lens from the object.
To find the magnification (M), we use the formula:
M = -v/u
Substituting the values:
M = -30 / 15 = -2
The magnification is -2, indicating that the image is inverted and twice the size of the object.
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with the aid of a diagrams, explainthe law of conservation of energy
According to laws of Conservation of energy , energy cannot be created nor be destroyed.
What are examples of energy?Energy conservation is to practice the less use of energy .
For example, Turning off the electrical appliances when they are not in use , walking should be first priority rather than driving are examples of energy conservation .
According to the law energy cannot be created nor be destroyed but it can be transfer from one form to another form of energy.
KE+PE=constant
Energy remains consvered when it is transferred from one form to another form .
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A unit used to measure force is a
a. newton.
b. newton-meter.
c. joule.
d. watt.
Answer:
Newton
Explanation:
Newton's are the standard unit of force.
The radius of the earth is 6.4 x 10^6 m . Calculate the angular and linear velocity of a participle at the earth's surface.
Answer: the angular velocity of a particle at the Earth's surface is approximately 7.27 x 10^-5 rad/s, and the linear velocity of the particle is approximately 464.1 m/s.
Explanation:
A gust of wind blows an apple from a tree. As the apple falls, the force of gravity on the apple is 9.13 N downward, and the force of the wind on the apple is 1.31 N to the right. What is the magnitude of the net external force on the apple? Answer in units of N.
Answer: See the diagram,from the vevtor addition we can say,net force acting on the apple is =
√
2.25
2
+
1.05
2
+
2
⋅
2.25
⋅
1.05
cos
90
=
2.48
N
And,this resultant force makes an angle of
tan
−
1
(
1.05
2.25
)
or,
25
degrees w.r.t vertical
How to find the angle in a projectile motion when displacement and heigh are the same
projectile's vertical speed v yv y v, start , y, drops prior to reaching its maximum height since its acceleration is in the opposite direction. Since the object's height is rising, the initial direction of the velocity is upward
.
At the projectile's highest point, vertical velocity zeroes out. After reaching the highest point, the vertical speed increases due of the same-direction acceleration. As the object's height lowers, the vertical velocity has a downward direction.
In projectile The beginning vertical velocity determines the maximum height. Increasing the launch angle raises the maximum height because steeper launch angles have a larger vertical velocity component.
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Water has a specific heat of 4.186 J/g°C. How much heat is required to increase 10.0 g of water from 25.0°C to 30.0°C?
Answer:
=0.2093J
Explanation:
Heat required(Q) = mcø
= (10/1000)*4.186*(30-25)
= 0.2093J
Answer:
its d
Explanation:
An electric charge of 1 microcoulomb is affected by an electric field of intensity 1000 newtons per coulomb at a point, then this point is a meter away from it by a distance
Answer:
r = 3 m
Explanation:
Given that,
Charge, \(q=1\ mu C\)
Electric field at a point, \(E=1000\ N/C\)
We need to find the distance away from it. We know that electric field at point r is given by :
\(E=\dfrac{kq}{r^2}\\\\r=\sqrt{\dfrac{kq}{E}} \\\\r=\sqrt{\dfrac{9\times 10^9\times 10^{-6}}{1000}} \\\\r=3\ m\)
So, the required distance is 3 m.
Shawn recently published his first novel. He sold 1,264 copies in the first month after it was published. Shawn's publisher predicts that the monthly number of copies sold will increase by a factor of I
- each month.
8
The exponential form is y = 1264 (1/8)ˣ.
Number of copies sold in first month = 1264
Factor by which the number of copies sold to be increased = 1/8
The exponential form can be written as,
y = abˣ
where a is the number of copies sold in first month, b is the factor by which the selling rate increased, x is the number of months and y is the total copies sold at the end of x months.
Therefore,
y = 1264 (1/8)ˣ
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Calculate the speed of the water jet emerging from the nozzle
Given:
• Height of ladder = 9.7 m
,• Diameter of water hose = 2.7 inch
,• Diameter of pump outlet = 3.47 inch
,• Guage of water pump = 236.49 kPa = 236.49 x 10³ Pa
,• Density of water = 1000 kg/m³
Let's find the speed of the water jet emerging from the nozzle.
To find the speed of the water jet, apply Bernoulli's equation.
We have:
\(\frac{1}{2}p(v^2_N-v^2_p)=p_{pumpguage}-pgh\)Rewrite the formula for VN:
\(\begin{gathered} v^2_N-v^2_p=(\frac{2}{p})p_{pumpguage}-2gh_{}_{} \\ \\ v^2_N-(\frac{A_N}{A_P})^2v^2_p=(\frac{2}{p})p_{pumpguage}-2gh \\ \\ v^2_N-(\frac{r^4_N}{r^4P_{}})^{}v^2_p=(\frac{2}{p})p_{pumpguage}-2gh \\ \\ \\ v_N=\sqrt[]{\frac{(\frac{2}{p})p_{pumpguage}-2gh}{1-(\frac{r^4_N}{r^4_p})}} \\ \end{gathered}\)• VN is the speed of the water jet from nozzle.
,• g is acceleration due to gravity = 9.8 m/s²
,• h is the height of ladder = 9.7 m
,• rN is the radius of nozzle = diameter of noozle/2 = 2.7/2 = 1.35 inches
,• rp is the radius of the pump outlet = diameter of pump outlet/2 = 3.47/2 = 1.735 inches
,• p is density of water = 1000 kg/m³
Input values into the formula and solve for vN:
\(\begin{gathered} v_N=\sqrt[]{\frac{(\frac{2}{1000})(236.49\times10^3)-2(9.8)(9.7)}{1-(\frac{1.35}{1.735})^4}} \\ \\ v_N=21.1315\text{ m/s} \end{gathered}\)Therefore, the speed of the water jet from the noozle is 21.1315 m/s
ANSWER:
21.1315 m/s
Select all correct answers....Covalent compounds
A cable stretches by an amount d as it supports a crate of mass M. The cable is cut in half. What is the mass of the load that can be supported by either half of the cable if the cable stretches by an amount d?
Answer:
2M
Explanation:
Let's say the initial length of the cable is L.
Now, if it is cut in two, it will have two lengths of L/2 each.
We are told that the cable stretches by d.
Now, young's modulus is;
E = stress/strain.
Stress = Force/Area = Mg/A
Strain = change in length/original length = d/(L/2)
Since g, A and L are constant, thus;
E α M/(L/2)
E α 2M/L
Thus, mass of the load that can be supported by either half of the cable if the cable stretches by an amount d is 2M
Answer:
The mass of the load that can be supported by either half of the cable = \(2M\)Explanation:
For the same stress as the full cable, the new cable would stretch by only \(\frac{d}{2}\).
Since it has stretched by d, the stress must now be two times the original. Therefore, the mass must also be two times the original which is \(2M\)
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*3.3B The potential at points in a plane is given by V = ax /(x² + y²) ³/² + b/(x²+y²)½ where x and y are the rectangular coordinates of a point, and a and b are constants. Find the components Ex, and Ey, of the electric intensity at any point.
To find the electric field components E(x) and E(y) at any point, we need to take the negative gradient of the potential function V.
Let's first find the partial derivative of potential function V with respect to x:
∂V/∂x = a(3x² - y²)/(x² + y²)^(5/2) - b x/(x² + y²)^(3/2)
Similarly, the partial derivative of V with respect to y is:
∂V/∂y = -2a xy/(x² + y²)^(5/2) - b y/(x² + y²)^(3/2)
Now, we can find the components of the electric field:
Ex = -∂V/∂x = -a(3x² - y²)/(x² + y²)^(5/2) + b x/(x² + y²)^(3/2)
Ey = -∂V/∂y = 2a xy/(x² + y²)^(5/2) + b y/(x² + y²)^(3/2)
Therefore, the components of the electric field at any point (x,y) are:
Ex = -a(3x² - y²)/(x² + y²)^(5/2) + b x/(x² + y²)^(3/2)
Ey = 2a xy/(x² + y²)^(5/2) + b y/(x² + y²)^(3/2)
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An object weighs 275 N when fully immersed in water and 325 N when fully immersed in oil of specific gravity 0.9. The volume of the object (m3) is most nearly:___________
Answer:
V = 0.05 m^3
Explanation:
The weight of the object of mass (m) and volume (V) in water is: 275 N and it is calculated using the buoyancy formula:
275 N = m g - V (dw) g
where (dw) is the density of water (1000 kg/m^3)
The weight of the object in oil is 325 N and it is calculated using the buoyancy formula as:
325 N = m g - V (do) g
where "do" is the density of oil which is: 0.9 x 1000 kg/m^3 = 900 kg/m^3
Therefore, subtracting term by term the two equations:
325 N -275 N = mg - mg - V (do) g + V (dw) g
50 N = V g (dw - do)
50 N = V g (1000 - 900)
50 N = V g (100)
solving for V (which will result in units of m^3):
V = 50 / 981 m^3
V = 0.05 m^3
A concave lens cannot produce a real image.
A. True
B. False
Answer:
B. False
A concave mirror and a converging lens will only produce a real image if the object is located beyond the focal point.
~Hoped this helped~
~Brainiliest?~