Answer:
The distance is [tex]y = 0.00162 \ m[/tex]
Explanation:
From the question we are told that
The wavelength is [tex]\lambda = 486 \ nm = 486 *10^{-9} \ m[/tex]
The slit spacing is [tex]d = 0.600 \ mm = 0.60 *10^{-3} \ m[/tex]
The distance of the screen is [tex]D = 2.0 \ m[/tex]
Generally the distance along the screen between adjacent bright fringes is mathematically represented as
[tex]y = \frac{\lambda * D}{d}[/tex]
substituting values
[tex]y = \frac{ 486 *10^{-9} * 2}{0.6*10^{-3}}[/tex]
[tex]y = 0.00162 \ m[/tex]
If two identical wires carrying a certain current in the same direction are placed parallel to each other, they will experience a force of repulsion.
a) true
b) false
Answer:
The answer is B. falseExplanation:
Current in the same direction
When current flow through to parallel conductors of a given length, when the current flows in the same direction
1. A force of attraction between the wires occurs and this tends to draw the wires inward
2. A magnetic field in the same direction is produced.
Current in opposite direction
when the current is in opposite direction
1. Force of repulsion between the two wires occurs, draws the wire outward
2. A magnetic field in opposite direction occurs
A charge of 15 is moving with velocity of 6.2 x17 which makes an angle of 48 degrees with respect to the magnetic field. If the force on the particle is 4838 N, find the magnitude of the magnetic field.
a. 06.0T.
b. 08.0T.
c. 07.0T.
d. 05.0 T.
Complete question:
A charge of 15C is moving with velocity of 6.2 x 10³ m/s which makes an angle of 48 degrees with respect to the magnetic field. If the force on the particle is 4838 N, find the magnitude of the magnetic field.
a. 0.06 T
b. 0.08 T
c. 0.07 T
d. 0.05 T
Answer:
The magnitude of the magnetic field is 0.07 T.
Explanation:
Given;
magnitude of the charge, q = 15C
velocity of the charge, v = 6.2 x 10³ m/s
angle between the charge and the magnetic field, θ = 48°
the force on the particle, F = 4838 N
The magnitude of the magnetic field can be calculated by applying Lorentz force formula;
F = qvBsinθ
where;
B is the magnitude of the magnetic field
B = F / vqsinθ
B = (4838) / (6.2 x 10³ x 15 x sin48)
B = 0.07 T
Therefore, the magnitude of the magnetic field is 0.07 T.
A solenoid is designed to produce a magnetic field of 3.50×10^−2 T at its center. It has a radius of 1.80 cm and a length of 46.0 cm , and the wire can carry a maximum current of 13.0 A.
Required:
a. What minimum number of turns per unit length must the solenoid have?
b. What total length of wire is required?
Answer:
a. 2143 turns/m
b. 111.5 m
Explanation:
a. The minimum number of turns per unit length (N/L) can be found using the following equation:
[tex] B = \frac{\mu_{0}NI}{L} [/tex]
[tex] \frac{N}{L} = \frac{B}{\mu_{0}I} = \frac{3.50 \cdot 10^{-2} T}{4\pi \cdot 10^{-7} Tm/A*13.0 A} = 2143 turns/m [/tex]
Hence, the minimum number of turns per unit length is 2143 turns/m.
b. The total length of wire is the following:
[tex] N = 2143 turns/m*L = 2143 turns/m*46.0 \cdot 10^{-2} m = 986 turns [/tex]
Since each turn has length 2πr of wire, the total length is:
[tex] L_{T} = N*2\pi r = 986 turn*2*\pi*1.80 \cdot 10^{-2} m = 111.5 m [/tex]
Therefore, the total length of wire required is 111.5 m.
I hope it helps you!
An archer practicing with an arrow bow shoots an arrow straight up two times. The first time the initial speed is vi and second
time he increases the initial sped to 4v. How would you compare the maximum height in the second trial to that in the first trial?
Answer:
The maximum height reached in the second trial is 16times the maximum height reached in the first trial.
Explanation:
The following data were obtained from the question:
First trial
Initial speed (u) = v
Final speed (v) = 0
Second trial
Initial speed (u) = 4v
Final speed (v) = 0
Next, we shall obtain the expression for the maximum height reached in each case.
This is illustrated below:
First trial:
Initial speed (u) = v
Final speed (v) = 0
Acceleration due to gravity (g) = 9.8 m/s²
Height (h₁) =.?
v² = u² – 2gh₁ (going against gravity)
0 = (v)² – 2 × 9.8 × h₁
0 = v² – 19.6 × h₁
Rearrange
19.6 × h₁ = v²
Divide both side by 19.6
h₁ = v²/19.6
Second trial
Initial speed (u) = 4v
Final speed (v) = 0
Acceleration due to gravity (g) = 9.8 m/s²
Height (h₂) =.?
v² = u² – 2gh₂ (going against gravity)
0 = (4v)² – 2 × 9.8 × h₂
0 = 16v² – 19.6 × h₂
Rearrange
19.6 × h₂ =16v²
Divide both side by 19.6
h₂ = 16v²/19.6
Now, we shall determine the ratio of the maximum height reached in the second trial to that of the first trial.
This is illustrated below:
Second trial:
h₂ = 16v²/19.6
First trial:
h₁ = v²/19.6
Second trial : First trial
h₂ : h₁
h₂ / h₁ = 16v²/19.6 ÷ v²/19.6
h₂ / h₁ = 16v²/19.6 × 19.6/v²
h₂ / h₁ = 16
h₂ = 16 × h₁
From the above illustrations, we can see that the maximum height reached in the second trial is 16times the maximum height reached in the first trial.
A plastic dowel has a Young's Modulus of 1.50 ✕ 1010 N/m2. Assume the dowel will break if more than 1.50 ✕ 108 N/m2 is exerted.
(a) What is the maximum force (in kN) that can be applied to the dowel assuming a diameter of 2.40 cm?
______Kn
(b) If a force of this magnitude is applied compressively, by how much (in mm) does the 26.0 cm long dowel shorten? (Enter the magnitude.)
mm
Answer:
a
[tex]F = 67867.2 \ N[/tex]
b
[tex]\Delta L = 2.6 \ mm[/tex]
Explanation:
From the question we are told that
The Young modulus is [tex]Y = 1.50 *10^{10} \ N/m^2[/tex]
The stress is [tex]\sigma = 1.50 *10^{8} \ N/m^2[/tex]
The diameter is [tex]d = 2.40 \ cm = 0.024 \ m[/tex]
The radius is mathematically represented as
[tex]r =\frac{d}{2} = \frac{0.024}{2} = 0.012 \ m[/tex]
The cross-sectional area is mathematically evaluated as
[tex]A = \pi r^2[/tex]
[tex]A = 3.142 * (0.012)^2[/tex]
[tex]A = 0.000452\ m^2[/tex]
Generally the stress is mathematically represented as
[tex]\sigma = \frac{F}{A}[/tex]
=> [tex]F = \sigma * A[/tex]
=> [tex]F = 1.50 *10^{8} * 0.000452[/tex]
=> [tex]F = 67867.2 \ N[/tex]
Considering part b
The length is given as [tex]L = 26.0 \ cm = 0.26 \ m[/tex]
Generally Young modulus is mathematically represented as
[tex]E = \frac{ \sigma}{ strain }[/tex]
Here strain is mathematically represented as
[tex]strain = \frac{ \Delta L }{L}[/tex]
So
[tex]E = \frac{ \sigma}{\frac{\Delta L }{L} }[/tex]
[tex]E = \frac{\sigma }{1} * \frac{ L}{\Delta L }[/tex]
=> [tex]\Delta L = \frac{\sigma * L }{E}[/tex]
substituting values
[tex]\Delta L = \frac{ 1.50*10^{8} * 0.26 }{ 1.50 *10^{10 }}[/tex]
[tex]\Delta L = 0.0026[/tex]
Converting to mm
[tex]\Delta L = 0.0026 *1000[/tex]
[tex]\Delta L = 2.6 \ mm[/tex]
What is temperature?
O A. The force exerted on an area
B. A measure of mass per unit volume
O C. The net energy transferred between two objects
OD. A measure of the movement of atoms or molecules within an
object
Answer:
The net energy transferred between two objects
Explanation:
The physical property of matter that expresses hot or cold is called temperature. It demonstrates the thermal energy. A thermometer is used to measure temperature. It defines the rate to which the chemical reaction occurs. It tells about the thermal radiation emitted from an object.
The correct option that defines temperature is option C.
Answer:
A measure of the movement of atoms or molecules within an object
Explanation:
Process of elimination
If one could transport a simple pendulum of constant length from the Earth's surface to the Moon's, where acceleration due to gravity is one-sixth (1/6) that on the Earth, by what factor would be the pendulum frequency be changed
Answer:
The frequency will change by a factor of 0.4
Explanation:
T = 2(pi)*sqrt(L/g)
Since g(moon) = (1/6)g(earth), the period would change by sqrt[1/(1/6)] = sqrt(6) ~ 2.5 times longer on the moon. Since the period & frequency are inverses, the frequency would be 1/2.5 or 0.4 times shorter on the moon.
With the same block-spring system from above, imagine doubling the displacement of the block to start the motion. By what factor would the following change?
A. Kinetic energy when passing through the equilibrium position.
B. Speed when passing through the equilibrium position.
Answer:
A) K / K₀ = 4 b) v / v₀ = 4
Explanation:
A) For this exercise we can use the conservation of mechanical energy
in the problem it indicates that the displacement was doubled (x = 2xo)
starting point. At the position of maximum displacement
Em₀ = Ke = ½ k (2x₀)²
final point. In the equilibrium position
[tex]Em_{f}[/tex] = K = ½ m v²
Em₀ = Em_{f}
½ k 4 x₀² = K
(½ K x₀²) = K₀
K = 4 K₀
K / K₀ = 4
B) the speed value
½ k 4 x₀² = ½ m v²
v = 4 (k / m) x₀
if we call
v₀ = k / m x₀
v = 4 v₀
v / v₀ = 4
A cylindrical container with a cross-sectional area of 66.2 cm2 holds a fluid of density 856 kg/m3 . At the bottom of the container the pressure is 119 kPa . Assume Pat = 101 kPa
A) What is the depth of the fuild?
B) Find the pressure at the bottom of the container after an additional 2.35×10−3 m3 of this fluid is added to the container. Assume that no fluid spills out of the container.
Answer:
A. h = 2.15 m
B. Pb' = 122 KPa
Explanation:
The computation is shown below:
a) Let us assume the depth be h
As we know that
[tex]Pb - Pat = d \times g \times h \\\\ ( 119 - 101) \times 10^3 = 856 \times 9.8 \times h[/tex]
After solving this,
h = 2.15 m
Therefore the depth of the fluid is 2.15 m
b)
Given that
height of the extra fluid is
[tex]h' = \frac{2.35 \times 10^{-3}}{ area} \\\\ h' = \frac{2.35 \times 10^{-3}} { 66.2 \times 10^{-4}}[/tex]
h' = 0.355 m
Now let us assume the pressure at the bottom is Pb'
so, the equation would be
[tex]Pb' - Pat = d \times g \times (h + h')\\\\Pb' = 856 \times 9.8 \times ( 2.15 + 0.355) + 101000[/tex]
Pb' = 122 KPa
(A) The depth of the fluid is 2.14 m.
(B) The new pressure at the bottom of container is 121972 Pa.
Given data:
The cross-sectional area of the container is, [tex]A =66.2 \;\rm cm^{2}=66.2 \times 10^{-4} \;\rm m^{2}[/tex].
The density of fluid is, [tex]\rho = 856 \;\rm kg/m^{3}[/tex].
The container pressure at bottom is, [tex]P=119 \;\rm kPa=119 \times 10^{3} \;\rm Pa[/tex].
The atmospheric pressure is, [tex]P_{at}=101 \;\rm kPa=101 \times 10^{3}\;\rm Pa[/tex].
(A)
The given problem is based on the net pressure on the container, which is equal to the difference between the pressure at the bottom and the atmospheric pressure. Then the expression is,
[tex]P_{net} = P-P_{at}\\\\\rho \times g \times h= P-P_{at}[/tex]
Here, h is the depth of fluid.
Solving as,
[tex]856\times 9.8 \times h= (119-101) \times 10^{3}\\\\h=\dfrac{ (119-101) \times 10^{3}}{856\times 9.8}\\\\h= 2.14 \;\rm m[/tex]
Thus, the depth of the fluid is 2.14 m.
(B)
For an additional volume of [tex]2.35 \times 10^{-3} \;\rm m^{3}[/tex] to the liquid, the new depth is,
[tex]V=A \times h'\\\\h'=\dfrac{2.35 \times 10^{-3}}{66.2 \times 10^{-4}}\\\\h'=0.36 \;\rm m[/tex]
Now, calculate the new pressure at the bottom of the container as,
[tex]P'-P_{at}= \rho \times g \times (h+h')\\\\\P'-(101 \times 10^{3})= 856 \times 9.8 \times (2.14+0.36)\\\\P'=121972 \;\rm Pa[/tex]
Thus, we can conclude that the new pressure at the bottom of container is 121972 Pa.
Learn more about the atmospheric pressure here:
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NASA is giving serious consideration to the concept of solar sailing. A solar sailcraft uses a large, low mass sail and the energy and momentum of sunlight for propulsion.
(a) Should the sail be absorbing or reflective? Why?
(b) How large a sail is necessary to propel a 10000kg
spacecraft against the gravitational force of the sun? Express your result in square kilometers.
(c) Explain why your answer to part (b) is independent of the distance from the sun.
The gravitational constant is G=6.67×10−11m3⋅s−2⋅kg−1.
The mass of the sun is Ms=1.99×1030kg.
Answer:
The complete question is
NASA is giving serious consideration to the concept of solar sailing. A solar sailcraft uses a large, low-mass sail and the energy and momentum of sunlight for propulsion. (a) Should the sail be absorbing or reflective? Why? (b) The total power output of the sun is 3.9 x 10^26 W. How large a sail is necessary to propel a 10,000-kg spacecraft against the gravitational force of the sun? Express your result in square kilometers. (c) Explain why your answer to part (b) is independent of the distance from the sun.
a) The sail should be reflective because, an incident electromagnetic wave, in this case, light wave, impacts twice the energy density on a reflective sail, and hence twice the force on a totally reflective sail as would be impacted on a sail that is totally absorbing.
For totally reflective, F = (2I/c)A ....1
for totally reflective, F = (I/c)A ....2
where I is the intensity of the light
c is the speed of light = 3 x 10^8 m/s
A is the area the sail
b) The intensity of the light from the sun = power/area
==> I = [tex]\frac{3.9*10^{26}}{4\pi r^{2} }[/tex]
where r is the distance from the sun and the sail
The Force from the sail from equation 1 is therefore
[tex]F[/tex] = [tex]\frac{2*3.9*10^{26}*A}{4\pi r^{2} *3*10^{8}}[/tex] = [tex]2.069*10^{17}\frac{A}{r^{2}}[/tex]
gravitational force between the sail and the sun [tex]F_{g}[/tex] = [tex]\frac{GMm}{r^{2}}[/tex]
where
G is the gravitational constant = 6.67 x 10^−11 m^3⋅s−2⋅kg−1.
m is the mass of the sail = 10000 kg
M is the mass of the sun = 1.99 x 10^30 kg.
==> [tex]F_{g}[/tex] = [tex]\frac{6.67*10^{-11}*1.99*10^{30}*10000}{r^{2}}[/tex] = [tex]\frac{1.33*10^{24}}{r^{2}}[/tex]
Equating the forces, we have
[tex]2.069*10^{17}\frac{A}{r^{2}}[/tex] = [tex]\frac{1.33*10^{24}}{r^{2}}[/tex]
the distance cancels out
A = (1.33 x 10^24)/(2.069 x 10^17) = 6428226.196 m^2
==> 6428.2 km^2
c) The force of the solar radiation is proportional to the intensity of the sun from the light, and the intensity is inversely proportional to the square of the distance from the source. Also, the force of gravitation is inversely proportional to the square of the distance, so they both cancel out.
When you slosh the water back and forth in a tub at just the right frequency, the water alternately rises and falls at each end, remaining relatively calm at the center. Suppose the frequency to produce such a standing wave in a 55m wide tub is 0.80 Hz.
Required:
What is the speed of the water wave?
Answer:
The speed of the water wave is [tex]v = 88 \ m/s[/tex]
Explanation:
From the question we are told that
The width of the tube is [tex]L = 55 \ m[/tex]
The fundamental frequency is [tex]f = 0.80 \ Hz[/tex]
Generally the fundamental frequency is mathematically represented as
[tex]f = \frac{v}{2 * L }[/tex]
=> [tex]v = f * 2 * L[/tex]
substituting values
[tex]v = 0.8 * 2 * 55[/tex]
[tex]v = 88 \ m/s[/tex]
The speed of the water wave will be 88 m/s.
Given information:
When you slosh the water back and forth in a tub at just the right frequency, the water alternately rises and falls at each end, remaining relatively calm at the center.
The frequency of the standing wave is [tex]f=0.8[/tex] Hz.
The width of the tub is [tex]w=55[/tex] m.
Let v be the speed of the standing wave.
The speed of the wave can be calculated as,
[tex]v=2wf\\v=2\times 55\times 0.8\\v=88\rm\; m/s[/tex]
Therefore, the speed of the water wave will be 88 m/s.
For more details, refer to the link:
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Does the moon light originate from the moon only
Answer:
No
Explanation:
Moon has no light of its own. It just shines because its surface reflects light from the sun and that's what we see.
:-)
Matter's resistance to a change in motion is called
Answer:
Inertia! I hope this helps!
Answer:
inertia
Explanation
Inertia.
PLEASE HELP FAST The object distance for a convex lens is 15.0 cm, and the image distance is 5.0 cm. The height of the object is 9.0 cm. What is the height of the image?
Answer:
The image height is 3.0 cm
Explanation:
Given;
object distance, [tex]d_o[/tex] = 15.0 cm
image distance, [tex]d_i[/tex] = 5.0 cm
height of the object, [tex]h_o[/tex] = 9.0 cm
height of the image, [tex]h_i[/tex] = ?
Apply lens equation;
[tex]\frac{h_i}{h_o} = -\frac{d_i}{d_o}\\\\ h_i = h_o(-\frac{d_i}{d_o})\\\\h_i = -9(\frac{5}{15} )\\\\h_i = -3 \ cm[/tex]
Therefore, the image height is 3.0 cm. The negative values for image height indicate that the image is an inverted image.
A circular loop of wire has radius of 9.50 cmcm. A sinusoidal electromagnetic plane wave traveling in air passes through the loop, with the direction of the magnetic field of the wave perpendicular to the plane of the loop. The intensity of the wave at the location of the loop is 0.0215 W/m2W/m2, and the wavelength of the wave is 6.90 mm.What is the maximum emf induced in the loop?
Express your answer with the appropriate units.
Answer:
The induced emf is [tex]\epsilon = 0.1041 \ V[/tex]
Explanation:
From the question we are told that
The radius of the circular loop is [tex]r = 9.50 \ cm = 0.095 \ m[/tex]
The intensity of the wave is [tex]I = 0.0215 \ W/m^2[/tex]
The wavelength is [tex]\lambda = 6.90\ m[/tex]
Generally the intensity is mathematically represented as
[tex]I = \frac{ c * B^2 }{ 2 * \mu_o }[/tex]
Here [tex]\mu_o[/tex] is the permeability of free space with value
[tex]\mu_o = 4 \pi *10^{-7} N/A^2[/tex]
B is the magnetic field which can be mathematically represented from the equation as
[tex]B = \sqrt{ \frac{ 2 * \mu_o * I }{ c} }[/tex]
substituting values
[tex]B = \sqrt{ \frac{ 2 * 4\pi *10^{-7} * 0.0215 }{ 3.0*10^{8}} }[/tex]
[tex]B = 1.342 *10^{-8} \ T[/tex]
The area is mathematically represented as
[tex]A = \pi r^2[/tex]
substituting values
[tex]A = 3.142 * (0.095)^2[/tex]
[tex]A = 0.0284[/tex]
The angular velocity is mathematically represented as
[tex]w = 2 * \pi * \frac{c}{\lambda }[/tex]
substituting values
[tex]w = 2 * 3.142 * \frac{3.0*10^{8}}{ 6.90 }[/tex]
[tex]w = 2.732 *10^{8} rad \ s^{-1}[/tex]
Generally the induced emf is mathematically represented as
[tex]\epsilon = N * B * A * w * sin (wt )[/tex]
At maximum induced emf [tex]sin (wt) = 1[/tex]
So
[tex]\epsilon = N * B * A * w[/tex]
substituting values
[tex]\epsilon = 1 * 1.342 *10^{-8} * 0.0284 *2.732 *10^{8}[/tex]
[tex]\epsilon = 0.1041 \ V[/tex]
The block moves up an incline with constant speed. What is the total work WtotalWtotalW_total done on the block by all forces as the block moves a distance LLL
Answer:
External force W₁ = F L
Friction force W₂ = - fr L
weight component W₃ = - mg sin θ L
Y Axis Force W=0
Explanation:
When the block rises up the plane with constant velocity, it implies that the sum of the forces is zero.
For these exercises it is indicated to create a reference system with the x axis parallel to the plane and the y axis perpendicular
let's write the equations of translational equilibrium in given exercise
X axis
F - fr -Wₓ = 0
F = fr + Wₓ
the components of the weight can be found using trigonometry
Wₓ = W sin θ
[tex]W_{y}[/tex] = W cos θ
let's look for the work of these three forces
W = F x cos θ
External force
W₁ = F L
since the displacement and the force have the same direction
Friction force
W₂ = - fr L
since the friction force is in the opposite direction to the displacement
For the weight component
W₃ = - mg sin θ L
because the weight component is contrary to displacement
Y Axis
N- Wy = 0
in this case the forces are perpendicular to the displacement, the angle is 90º and the cosine 90 = 0
therefore work is worth zero
The linear density rho in a rod 3 m long is 8/ x + 1 kg/m, where x is measured in meters from one end of the rod. Find the average density rhoave of the rod.
Answer:
The average density of the rod is 1.605 kg/m.
Explanation:
The average density of the rod is given by:
[tex] \rho = \frac{m}{l} [/tex]
To find the average density we need to integrate the linear density from x₁ = 0 to x₂ = 3, as follows:
[tex] \int_{0}^{3} \frac{8}{3(x + 1)}dx [/tex]
[tex] \rho = \frac{8}{3} \int_{0}^{3} \frac{1}{(x + 1)}dx [/tex] (1)
Using u = x+1 → du = dx → u₁= x₁+1 = 0+1 = 1 and u₂ = x₂+1 = 3+1 = 4
By entering the values above into (1), we have:
[tex] \rho = \frac{8}{3} \int_{0}^{3} \frac{1}{u}du [/tex]
[tex]\rho = \frac{8}{3}*log(u)|_{1}^{4} = \frac{8}{3}[log(4) - log(1)] = 1.605 kg/m[/tex]
Therefore, the average density of the rod is 1.605 kg/m.
I hope it helps you!
The average density of the rod is [tex]1.605 \;\rm kg/m^{3}[/tex].
Given data:
The length of rod is, L = 3 m.
The linear density of rod is, [tex]\rho=\dfrac{8}{x+1} \;\rm kg/m[/tex].
To find the average density we need to integrate the linear density from x₁ = 0 to x₂ = 3, The expression for the average density is given as,
[tex]\rho' = \int\limits^3_0 { \rho} \, dx\\\\\\\rho' = \int\limits^3_0 { \dfrac{m}{L}} \, dx\\\\\\\rho' = \int\limits^3_0 {\dfrac{8}{3(x+1)}} \, dx[/tex]............................................................(1)
Using u = x+1
du = dx
u₁= x₁+1 = 0+1 = 1
and
u₂ = x₂+1 = 3+1 = 4
By entering the values above into (1), we have:
[tex]\rho' =\dfrac{8}{3} \int\limits^3_0 {\dfrac{1}{u}} \, du\\\\\\\rho' =\dfrac{8}{3} \times [log(u)]^{4}_{1}\\\\\\\rho' =\dfrac{8}{3} \times [log(4)-log(1)]\\\\\\\rho' =1.605 \;\rm kg/m^{3}[/tex]
Thus, we can conclude that the average density of the rod is [tex]1.605 \;\rm kg/m^{3}[/tex].
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CAN SOMEONE HELP ME PLEASE ITS INTEGRATED SCIENCE AND I AM STUCK
Answer:
[tex]\huge \boxed{\mathrm{Option \ D}}[/tex]
Explanation:
Two forces are acting on the object.
Subtracting 2 N from both forces.
2 N → Object ← 5 N
- 2 N - 2N
0 N → Object ← 3 N
The force 3 N is pushing the object to the left side.
The mass of the object is 10 kg.
Applying formula for acceleration (Newton’s Second Law of Motion).
a = F/m
a = 3/10
a = 0.3
The advantage of a hydraulic lever is A : it transforms a small force acting over a large distance into a large force acting over a small distance. B : it transforms a small force acting over a small distance into a large force acting over a large distance. C : it allows you to exert a larger force with less work. D : it transforms a large force acting over a large distance into a small force acting over a small distance. E : it transforms a large force acting over a small distance into a small force acting over a large distance.
Answer:
A) it transforms a small force acting over a large distance into a large force acting over a small distance.
Explanation:
The hydraulic lever works based on Pascal's law of transmission of pressure through a fluid. In the hydraulic lever, the pressure transmitted is the same.
Pressure transmitted P = F/A
where F is the force applied
and A is the area over which the force is applied.
This pressure can be manipulated on the input end as a small force applied over a small area, and then be transmitted to the output end as a large force over a large area.
F/A = f/a
where the left side of the equation is for the output, and the right side is for the input.
The volume of the displaced fluid will be the same on both ends of the hydraulic lever. Since we know that
volume V = (area A) x (distance d)
this means that the the piston on the input smaller area of the hydraulic lever will travel a greater distance, while the piston on the larger output area of the lever will travel a small distance.
From all these, we can see that the advantage of a hydraulic lever is that it transforms a small force acting over a large distance into a large force acting over a small distance.
A wooden ice box has a total area of 1.50 m2 amd walls with an average thickness of 2.0 cm. The box contains ice at 0.0 oC. The inside of the box is kept cold by melting ice. How much ice melts in one day if the ice box is kept in the shade of tree at 29 oC. (Assume the thermal conductivity of wood is 0.16 kJ/s m oC
Answer:
m = 9.1 x 10⁶ kg
Explanation:
First, we need to find the rate of heat transfer through the box to the ice. For this purpose, we use Fourier's Law of Heat Conduction:
Q = KA ΔT/L
where,
Q = Rate Of Heat Transfer = ?
K = Thermal Conductivity = 0.16 KW/m.°C = 160 W/m.°C
A = Area = 1.5 m²
ΔT = Difference in Temperature = 29°C - 0°C = 29°C
L = Thickness of wall = 2 cm = 0.002 m
Therefore,
Q = (160 W/m °C)(1.5 m²)(29°C)/(0.002 m)
Q = 3.48 x 10⁶ W
Now, we find the amount of heat transferred in one day to the ice:
q = Qt
where,
q = amount of heat = ?
t = time = (1 day)(24 h/1 day)(3600 s/1 h) = 86400 s
Therefore,
q = (3.48 x 10⁶ W)(8.64 x 10⁴ s)
q = 3 x 10¹¹ J
Now, for mass of ice melted in a day:
q = m H
m = q/H
where,
m = mass of ice melted in a day = ?
H = latent heat of fusion of ice = 3.3 x 10⁵ J/kg
Therefore,
m = (3 x 10¹¹ J)/(3.3 x 10⁵ J/kg)
m = 9.1 x 10⁶ kg
"Determine the magnitude of the net force of gravity acting on the Moon during an eclipse when it is directly between Earth and the Sun."
Answer:
Net force = 2.3686 × 10^(20) N
Explanation:
To solve this, we have to find the force of the earth acting on the moon and the force of the sun acting on the moon and find the difference.
Now, from standards;
Mass of earth;M_e = 5.98 × 10^(24) kg
Mass of moon;M_m = 7.36 × 10^(22) kg
Mass of sun;M_s = 1.99 × 10^(30) kg
Distance between the sun and earth;d_se = 1.5 × 10^(11) m
Distance between moon and earth;d_em = 3.84 × 10^(8) m
Distance between sun and moon;d_sm = (1.5 × 10^(11)) - (3.84 × 10^(8)) = 1496.96 × 10^(8) m
Gravitational constant;G = 6.67 × 10^(-11) Nm²/kg²
Now formula for gravitational force between the earth and the moon is;
F_em = (G × M_e × M_m)/(d_em)²
Plugging in relevant values, we have;
F_em = (6.67 × 10^(-11) × 5.98 × 10^(24) × 7.36 × 10^(22))/(3.84 × 10^(8))²
F_em = 1.9909 × 10^(20) N
Similarly, formula for gravitational force between the sun and moon is;
F_sm = (G × M_s × M_m)/(d_sm)²
Plugging in relevant values, we have;
F_se = (6.67 × 10^(-11) × 1.99 × 10^(30) ×
7.36 × 10^(22))/(1496.96 × 10^(8))²
F_se = 4.3595 × 10^(20) N
Thus, net force = F_se - F_em
Net force = (4.3595 × 10^(20) N) - (1.9909 × 10^(20) N) = 2.3686 × 10^(20) N
The magnetic field of a plane-polarized electromagnetic wave moving in the z-direction is given by in SI units. What is the frequency of the wave
Complete Question
The magnetic field of a plane-polarized electromagnetic wave moving in the z-direction is given by
[tex]B=1.2* 10^{-6} sin [2\pi[(\frac{z}{240} ) - ( \frac{t * 10^7}{8} ) ] ][/tex] in SI units.
Answer:
The value is [tex]f = 1.98918*10^{5}\ Hz[/tex]
Explanation:
From the question we are told that
The magnetic field is [tex]B=1.2* 10^{-6} sin [2\pi[(\frac{z}{240} ) - ( \frac{t * 10^7}{8} ) ] ][/tex]
This above equation can be modeled as
[tex]B=1.2* 10^{-6} sin [2\pi[(\frac{z}{240} ) - ( \frac{t * 10^7}{8} ) ] ] \equiv A sin ( kz -wt )[/tex]
So
[tex]w = \frac{10^7}{8}[/tex]
Generally the frequency is mathematically represented as
[tex]f = \frac{w}{2 \pi}[/tex]
=> [tex]f = \frac{ \frac{10^7}{8} }{2 \pi}[/tex]
=> [tex]f = 1.98918*10^{5}\ Hz[/tex]
A laboratory electromagnet produces a magnetic field of magnitude 1.38 T. A proton moves through this field with a speed of 5.86 times 10^6 m/s.
a. Find the magnitude of the maximum magnetic force that could be exerted on the proton.
b. What is the magnitude of the maximum acceleration of the proton?
c. Would the field exert the same magnetic force on an electron moving through the field with the same speed? (Assume that the electron is moving in the direction as the proton.)
1. Yes
2. No
.Answer;
Using Fmax=qVB
F=(1.6*10^-19 C)(5.860*10^6 m/s)(1.38 T)
ANS=1.29*10^-12 N
2. Using Amax=Fmax/ m
Amax =(1.29*10^-12 N) / (1.67*10^-27 kg)
ANS=1.93*10^15 m/s^2*
3. No, the acceleration wouldn't be the same. Since The magnitude of the electron is equal to that of the proton, but the direction would be in the opposite direction and also Since an electron has a smaller mass than a proton
A stereo speaker produces a pure "G" tone, with a frequency of 392 Hz. What is the period T of the sound wave produced by the speaker?
Answer:
The period is [tex]T = 0.00255 \ s[/tex]
Explanation:
From the question we are told that
The frequency is [tex]f = 392 \ Hz[/tex]
Generally the period is mathematically represented as
[tex]T = \frac{1}{f}[/tex]
=> [tex]T = \frac{1}{ 392}[/tex]
=> [tex]T = 0.00255 \ s[/tex]
An oil film (n = 1.48) of thickness 290 nm floating on water is illuminated with white light at normal incidence. What is the wavelength of the dominant color in the reflected light? A. Blue (470 nm) B. Green (541 nm) C. Violet (404 nm) D. Yellow (572 nm)
Answer:
The correct option is D
Explanation:
From the question we are told that
The refractive index of oil film is [tex]k = 1.48[/tex]
The thickness is [tex]t = 290 \ nm = 290*10^{-9} \ m[/tex]
Generally for constructive interference
[tex]2t = [m + \frac{1}{2} ]* \frac{\lambda}{k}[/tex]
For reflection of a bright fringe m = 1
=> [tex]2 * (290*10^{-9}) = [1 + \frac{1}{2} ]* \frac{\lambda}{1.48}[/tex]
=> [tex]\lambda = 5.723 *10^{-7} \ m[/tex]
This wavelength fall in the range of a yellow light
How much heat is required to convert 5.0 kg of ice from a temperature of - 20 0C to water at a temperature of 205 0F
Answer:
Explanation:
To convert from °C to °F , the formula is
( F-32 ) / 9 = C / 5
F is reading fahrenheit scale and C is in centigrade scale .
F = 205 , C = ?
(205 - 32) / 9 = C / 5
C = 96°C approx .
Let us calculate the heat required .
Total heat required = heat required to heat up the ice at - 20 °C to 0°C + heat required to melt the ice + heat required to heat up the water at 0°C to
96°C.
= 5 x 2.04 x (20-0) + 5 x 336 + 5 x ( 96-0 ) x 4.2 kJ .
= 204 + 1680 + 2016
= 3900 kJ .
A string is stretched and fixed at both ends, 200 cm apart. If the density of the string is 0.015 g/cm, and its tension is 600 N, what is the wavelength (in cm) of the first harmonic?
Answer:
200cm
Explanation:
Answer:
100cm
Explanation:
Using
F= ( N/2L)(√T/u)
F1 will now be (0.5*2)( √600/0.015)
=> L( wavelength)= 200/2cm = 100cm
We can reasonably model a 75 W incandescent light bulb as a sphere 6.0 cm in diameter. Typically only about 5% of the energy goes to visible light; the rest goes largely to non-visible infrared radiation. (a) What is the visible light intensity at the surface of the bulb
Answer:
Visible light intensity at the surface of the bulb (I) = 331 W/m²
Explanation:
Given:
Energy = 75 W
Radius = 6 /2 = 3 cm = 3 × 10⁻² m
Energy goes to visible light = 5% = 0.05
Find:
Visible light intensity at the surface of the bulb (I)
Computation:
Visible light intensity at the surface of the bulb (I) = P / 4A
Visible light intensity at the surface of the bulb (I) = (0.05)(75) / 4π(3 × 10⁻²)²
Visible light intensity at the surface of the bulb (I) = 3.75 / 4π(9 × 10⁻⁴)
Visible light intensity at the surface of the bulb (I) = 331 W/m²
A particle undergoes damped harmonic motion. The spring constant is 100 N/m, the damping constant is 8.0 x 10-3 kg.m/s, and the mass is 0.050 kg. If the particle starts at its maximum displacement, x = 1.5 m, at time t = 0. What is the amplitude of the motion at t = 5.0 s?
Answer:
The amplitude [tex]A(5) = 1 \ m[/tex]
Explanation:
From the question we are told that
The spring constant is [tex]k = 100 \ N/m[/tex]
The damping constant is [tex]b = 8.0 *10^{-3} \ kg \cdot m/s[/tex]
The mass is [tex]m = 0.050 \ kg[/tex]
The maximum displacement is [tex]A_o = 1.5 \ m \ at t = 0[/tex]
The time considered is t = 5.0 s
Generally the displacement(Amplitude) of damped harmonic motion is mathematically represented as
[tex]A(t) = A_o * e ^{ - \frac{b * t}{2 * m} }[/tex]
substituting values
[tex]A(5) = 1.5 * e ^{ - \frac{ 8.0 *10^{-3} * 5}{2 * 0.050} }[/tex]
[tex]A(5) = 1 \ m[/tex]
A rectangular coil having N turns and measuring 15 cm by 25 cm is rotating in a uniform 1.6-T magnetic field with a frequency of 75 Hz. The rotation axis is perpendicular to the direction of the field. If the coil develops a sinusoidal emf of maximum value 56.9 V, what is the value of N?
A) 2
B) 4
C) 6
D) 8
E) 10
Answer:
A) 2
Explanation:
Given;
magnetic field of the coil, B = 1.6 T
frequency of the coil, f = 75 Hz
maximum emf developed in the coil, E = 56.9 V
area of the coil, A = 0.15 m x 0.25 m = 0.0375 m²
The maximum emf in the coil is given by;
E = NBAω
Where;
N is the number of turns
ω is the angular velocity = 2πf = 2 x 3.142 x 75 = 471.3 rad/s
N = E / BAω
N = 56.9 / (1.6 x 0.0375 x 471.3)
N = 2 turns
Therefore, the value of N is 2
A) 2
