Answer:
The correct answer is "[tex]21.344^{\circ}[/tex]" and "[tex]39.56^{\circ}[/tex]".
Explanation:
According to the question,
Slit width,
[tex]d=\frac{1}{910 \ lines/mm}[/tex]
[tex]=\frac{1}{910\times 10^3}[/tex]
[tex]=1.099\times 10^{-6} \ m[/tex]
The condition far first order maxima will be:
⇒ [tex]d Sin \theta = 1 \lambda[/tex]
Now,
⇒ [tex]\Theta_{min} = Sin^{-1} (\frac{\lambda}{d} )[/tex]
[tex]=Sin^{-1} (\frac{400\times 10^{-9}}{1.099\times 10^{-6}} )[/tex]
[tex]=21.344^{\circ}[/tex]
⇒ [tex]\Theta_{max} = Sin^{-1} (\frac{\lambda}{d} )[/tex]
[tex]=Sin^{-1} (\frac{700\times 10^{-9}}{1.099\times 10^{-6}} )[/tex]
[tex]=39.56^{\circ}[/tex]
A man standing in an elevator holds a spring scale with a load of 5 kg suspended from it. What would be the reading of the scale, if the elevator is accelerating downward with an acceleration 3.8 m/s?.
Answer:
3.1 kg
Explanation:
Applying,
R = m(g-a)..................... Equation 1
Where R = weight of the scale when the elevator is coming down, a = acceleration of the elevator, g = acceleration due to gravith.
From the question,
Given: m = 5 kg, a = 3.8 m/s²
Constant: g = 9.8 m/s²
Substitute these values into equation 1
R = 5(9.8-3.8)
R = 5(6)
R = 30 N
Hence the spring scale is
m' = R/g
m' = 30/9.8
m' = 3.1 kg
A cylindrical swimming pool has a radius 2m and depth 1.3m .it is completely filled with salt water of specific gravity 1.03.The atmospheric preassure is 1.013 x 10^5 Pa.
a.calculate the density of salt water.
Answer:
the density of the salt water is 1030 kg/m³
Explanation:
Given;
radius of the cylindrical pool, r = 2 m
depth of the pool, h = 1.3 m
specific gravity of the salt water, γ = 1.03
The atmospheric pressure, P₀ = 1.013 x 10⁵ Pa
Density of fresh water, [tex]\rho _w[/tex] = 1000 kg/m³
The density of the salt water is calculated as;
[tex]Specific \ gravity \ of \ salt\ water \ (\gamma _s_w) = \frac{density \ of \ salt \ water \ (\rho_{sw})}{density \ of \ fresh \ water \ (\rho_{w})} \\\\1.03 = \frac{\rho_{sw}}{1000 \ kg/m^3}\\\\\rho_{sw} = 1.03 \times 1000 \ kg/m^3\\\\\rho_{sw} = 1030 \ kg/m^3[/tex]
Therefore, the density of the salt water is 1030 kg/m³
g Light that is incident upon the eye is refracted several times before it reaches the retina. As light passes through the eye, at which boundary does most of the overall refraction occur?
Answer
Explanation
:giác mạc
5. Steve is driving in his car to take care of some errands. The first errand has him driving to a location 2 km East and 6 km North of his starting location. Once he completes that errand, he drives to the second one which is 4 km East and 2 km South of the first errand. What is the magnitude of the vector that describes how far the car has traveled from its starting point, rounded to the nearest km?
Answer:
gshshs
Explanation:
hshsksksksbsbbshd
Chameleons catch insects with their tongues, which they can rapidly extend to great lengths. In a typical strike, the chameleon's tongue accelerates at a remarkable 220 m/s^2 for 20 msms, then travels at constant speed for another 30 ms.
Required:
During this total time of 50 ms, 1/20 of a second, how far does the tongue reach?
Solution :
We know,
Distance,
[tex]$S=ut+\frac{1}{2}at^2$[/tex]
[tex]$S=ut+0.5(a)(t)^2$[/tex]
For the first 20 ms,
[tex]$S=0+0.5(220)(0.020)^2$[/tex]
S = 0.044 m
In the remaining 30 ms, it has constant velocity.
[tex]$v=u+at$[/tex]
[tex]$v=0+(220)(0.020)[/tex]
v = 4.4 m/s
Therefore,
[tex]$S=ut+0.5(a)(t)^2$[/tex]
[tex]$S'=4.4 \times 0.030[/tex]
S' = 0.132 m
So, the required distance is = S + S'
= 0.044 + 0.132
= 0.176 m
Therefore, the tongue can reach = 0.176 m or 17.6 cm
Answer:
The total distance is 0.176 m.
Explanation:
For t = 0 s to t = 20 ms
initial velocity, u = 0
acceleration, a = 220 m/s^2
time, t = 20 ms
Let the final speed is v.
Use first equation of motion
v = u + at
v = 0 + 220 x 0.02 = 4.4 m/s
Let the distance is s.
Use second equation of motion
[tex]s = u t + 0.5 at^2\\\\s = 0 + 0.5 \times 220 \times 0.02\times 0.02\\\\s = 0.044 m[/tex]
Now the distance is
s' = v x t
s' = 4.4 x 0.03 = 0.132 m
The total distance is
S = s + s' = 0.044 + 0.132 = 0.176 m
A 700N marine in basic training climbs a 10m vertical rope at constant speed in 8sec. what is power put
Answer:
875 Watts
Explanation:
P = W/t = mgh/t = 700(10)/8 = 875 Watts
. A ball of mass 0.50 kg is rolling across a table top with a speed of 5.0 m/s. When the ball reaches the edge of the table, it rolls down an incline onto the floor 1.0 meter below (without bouncing). What is the speed of the ball when it reaches the floor?
Answer:
4
Explanation:
Choose the appropriate explanation how such a low value is possible given Saturn's large mass - 100 times that of Earth.
a. This low value is possible because the magnetic field of Saturn is so strong.
b. This low value is possible because the magnetic field of Saturn is so weak.
c. This low value is possible because the density of Saturn is so high.
d. This low value is possible because the density of Saturn is so low.
Answer:
Explanation:
That is an amazing fact.
The minus sign is what you have to pay attention to. The earth has a mass of 100 times that of Saturn. As someone on here once noted, Saturn has such a low density that it would float in water.
The answer is D
Choose one. 5 points
Use the equation from week 3:
frequency =
wavespeed
wavelength
and the wavelength you found in #3 to calculate the frequency of this photon (remember the speed of
light is 3E8 m/s);
7.6E14 Hz
6.0E14 Hz
4,6E14 Hz
The frequency is 4,6E14 Hz.
What is the frequency?
Frequency is the fee at which modern changes direction in step with 2nd. it's far measured in hertz (Hz), a worldwide unit of degree wherein 1 hertz is identical to 1 cycle in line with 2d. Hertz (Hz) = One hertz is the same as 1 cycle in step with the second. Cycle = One entire wave of alternating present-day voltage.
Frequency describes the number of waves that pass a hard and fast place in a given quantity of time. So if the time it takes for a wave to skip is half of 2d, the frequency is 2 per 2nd. If it takes 1/one hundred of an hour, the frequency is a hundred in step with hour.
Learn more about frequency here:-https://brainly.com/question/254161
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Derive the dimension of coefficient of linear expansivity
Answer:
The SI unit of coefficient of linear expansion can be expressed as °C-1 or °K-1. ... The dimension of coefficient of linear expansion will be M0L0T0K−1.
Light of a given wavelength is used to illuminate the surface of a metal, however, no photoelectrons are emitted. In order to cause electrons to be ejected from the surface of this metal you should: ___________
a. use light of the same wavelength but increase its intensity.
b. use light of a shorter wavelength.
c. use light of the same wavelength but decrease its intensity.
d. use light of a longer wavelength.
Answer:
use light of the same wavelength but decrease it's intensity
potential diffetence
Answer:
6v
Explanation:
V=IR
V= 2* 3
V= 6 volts
Q)what are convex mirrors?
Answer:
A curved mirror is a mirror with a curved reflecting surface. The surface may be either convex or concave. Most curved mirrors have surfaces that are shaped like part of a sphere, but other shapes are sometimes used in optical devices.
A convex mirror (or lens) is one constructed so that it is thicker in the middle than it is at the edge.
A cylinder is given a push and then rolls up an inclined plane. If the origin is the starting point, sketch the position, velocity, and acceleration of the cylinder vs. time as it goes up and then down the plane.
Cuando el pistón tiene un volumen de 2x10^-4 m^3, el gas en el pistón está a una presión de 150 kPa. El área del pistón es 0.00133 m^2. Calcular la fuerza que el gas ejerce sobre el embolo del pistón.
Answer:
F = 1.128 10⁸ Pa
Explanation:
Pressure is defined by
P = F / A
If the gas is ideal for equal force eds on all the walls, so on the piston area we have
F = P A
We reduce the pressure to the SI system
P = 150 kpa (1000 Pa / 1kPa = 150 103 Pa
we calculate
F = 150 10³ / 0.00133
F = 1.128 10⁸ Pa
A 1.5kg block slides along a frictionless surface at 1.3m/s . A second block, sliding at a faster 4.3m/s , collides with the first from behind and sticks to it. The final velocity of the combined blocks is 2.0m/s . What was the mass of the second block?
Answer:
The mass of the second block=0.457 kg
Explanation:
We are given that
m1=1.5 kg
v1=1.3m/s
v2=4.3 m/s
V=2.0 m/s
We have to find the mass of the second block.
[tex]m_1v_1+m_2v_2=(m_1+m_2)V[/tex]
Let m2=m
Substitute the values
[tex]1.5(1.3)+m(4.3)=(1.5+m)(2)[/tex]
[tex]1.95+4.3m=3+2m[/tex]
[tex]4.3m-2m=3-1.95[/tex]
[tex]2.3m=1.05[/tex]
[tex]m=\frac{1.05}{2.3}[/tex]
[tex]m=0.457 kg[/tex]
Hence, the mass of the second block=0.457 kg
What is the volume of a metal block 3cm long by 2cm wide by 4cm high? What would be the volume of a block twice as long, wide, and high?
Answer:
Volume of a metal block = 24 cm^3
Volume of a block twice as long, wide and high = 192 cm^3
Explanation:
Volume of a block is measured in l*w*h and in the first block, the sides are 3, 2 and 4 and 3*2*4 = 24
Second block, just double each of the lengths to get 6*4*8 = 192
A parallel-plate capacitor consists of two plates, each with an area of 29 cm2cm2 separated by 3.0 mmmm. The charge on the capacitor is 7.8 nCnC . A proton is released from rest next to the positive plate. Part A How long does it take for the proton to reach the negative plate
Answer:
t = 2.09 10⁻³ s
Explanation:
We must solve this problem in parts, first we look for the acceleration of the electron and then the time to travel the distance
let's start with Newton's second law
∑ F = m a
the force is electric
F = q E
we substitute
q E = m a
a = [tex]\frac{q}{m} \ E[/tex]
a = [tex]\frac{1.6 \ 10^{-19}}{ 9.1 \ 10^{-31} } \ 7.8 \ 10^{-9}[/tex]
a = 1.37 10³ m / s²
now we can use kinematics
x = v₀ t + ½ a t²
indicate that rest starts v₀ = 0
x = 0 + ½ a t²
t = [tex]\sqrt{\frac{2x}{a} }[/tex]
t = [tex]\sqrt{\frac {2 \ 3 \ 10^{-3}}{ 1.37 \ 10^3} }[/tex]
t = 2.09 10⁻³ s
The cart travels the track again and now experiences a constant tangential acceleration from point A to point C. The speeds of the cart are 11.0 ft/s at point A and 18.0 ft/s at point C. The cart takes 5.00 s to go from point A to point C, and the cart takes 1.30 s to go from point B to point C. What is the cart's speed at point B
Answer:
The speed at B is 16.18 ft/s .
Explanation:
Speed at A, u = 11 ft/s
Speed at C, v' = 18 ft/s
Time from A to C = 5 s
Time from B to C = 1.3 s
Let the speed of car at B is v.
Let the acceleration is a.
From A to B
Use first equation of motion
v = u + a t
18 = 11 + a x 5
a = 1.4 ft/s^2
Let the time from A to B is t' .
t' = 5 - 1.3 = 3.7 s
Use first equation of motion from A to B
v = 11 + 1.4 x 3.7 = 16.18 ft/s
If a proton and electron both move through the same displacement in an electric field, is the change in potential energy associated with the proton equal in magnitude and opposite in sign to the change in potential energy associated with the electron?
a. The magnitude of the change is smaller for the proton.
b. The magnitude of the change is larger for the proton.
c. The signs Of the two changes in potential energy are opposite.
d. They are equal in magnitude.
e. The signs of the two changes in potential energy are the same.
Answer: They are equal in magnitude.
- The signs of the two changes in potential energy are opposite
Explanation:
When the proton and electron both move through the same displacement in an electric field, the change in potential energy that is associated with the proton is equal in magnitude.
Also, it should be noted that the signs of the two changes in potential energy are opposite.
A space ship has four thrusters positioned on the top and bottom, and left and right as shown below. The thrusters can be operated independently or together to help the ship navigate in all directions.
Initially, the Space Probe is floating towards the East, as shown below, with a velocity, v. The pilot then turns on thruster #2.
Select one:
a.
Space ship will have a velocity to the West and will be speeding up.
b.
Space ship will have a velocity to the East and will be speeding up.
c.
Space ship will have a velocity to the East and will be slowing down.
d.
Space ship will have a velocity to the West and will be slowing down.
e.
Ship experiences no change in motion.
Answer:
The correct answer is - c. Spaceship will have a velocity to the East and will be slowing down.
Explanation:
In this case, if turned on thruster #2 then it will exert force on the west side as thruster 2 is on the east side and it can be understood by Newton's third law that says each action has the same but opposite reaction.
As the spaceship engine applies force on the east side then according to the law the exhauster gas applies on towards west direction. It will try to decrease the velocity of the spaceship however, the direction of floating still be east side initally.
A large metal sphere has three times the diameter of a smaller sphere and carries three times the charge. Both spheres are isolated, so their surface charge densities are uniform. Compare (a) the potentials (relative to infinity) and (b) the electric field strengths at their surfaces.
Answer:
A. Equals to that of the smaller sphere
B. 3 times less than that of the smaller sphere
Explanation:
(a) Equals to that of the smaller sphere
The potential of an isolated metal sphere, with charge Q and radius R, is kQ=R, so a sphere with charge 3Q and radius 3R has the same potential
b) 3 times less than that of the smaller sphere
However, the electric field at the surface of the smaller sphere is ?=? 0 = kQ=R2 , so tripling Q and R reduces the surface field by a factor of 1/3
For example, we can take Water
In (A) Water has same mass and great volume
In (B) Water has same mass and lower volume
Will there be any change in its density then?
Answer:
yes there will be change in its density
In the diagram, the crest of the wave is show by:
A
B
C
D
Answer:
D.
Explanation:
The crest of a wave refers to the highest point of a wave. This is illustrated by D.
An electron is released from rest at a distance of 9.00 cm from a fixed proton. How fast will the electron be moving when it is 3.00 cm from the proton
Answer:
the speed of the electron at the given position is 106.2 m/s
Explanation:
Given;
initial position of the electron, r = 9 cm = 0.09 m
final position of the electron, r₂ = 3 cm = 0.03 m
let the speed of the electron at the given position = v
The initial potential energy of the electron is calculated as;
[tex]U_i = Fr = \frac{kq^2}{r^2} \times r = \frac{kq^2}{r} \\\\U_i = \frac{(9\times 10^9)(1.602\times 10^{-19})^2}{0.09} \\\\U_i = 2.566 \times 10^{-27} \ J[/tex]
When the electron is 3 cm from the proton, the final potential energy of the electron is calculated as;
[tex]U_f = \frac{kq^2}{r_2} \\\\U_f = [\frac{(9\times 10^9)\times (1.602 \times 10^{-19})^2}{0.03} ]\\\\U_f = 7.669 \times 10^{-27} \ J \\\\\Delta U = U_f -U_i\\\\\Delta U = (7.699\times 10^{-27} \ J ) - (2.566 \times 10^{-27} \ J)\\\\\Delta U = 5.133 \times 10^{-27} \ J[/tex]
Apply the principle of conservation of energy;
ΔK.E = ΔU
[tex]K.E_f -K.E_i = \Delta U\\\\initial \ velocity \ of \ the \ electron = 0\\\\K.E_f - 0 = \Delta U\\\\K.E_f = \Delta U\\\\\frac{1}{2} mv^2 = \Delta U\\\\where;\\\\m \ is \ the \ mass \ of\ the \ electron = 9.1 1 \times 10^{-31} \ kg\\\\v^2 = \frac{ 2 \Delta U}{m} \\\\v = \sqrt{\frac{ 2 \Delta U}{m}} \\\\v = \sqrt{\frac{ 2 (5.133\times 10^{-27})}{9.11\times 10^{-31}}}\\\\v = \sqrt{11268.935} \\\\v = 106.2 \ m/s[/tex]
Therefore, the speed of the electron at the given position is 106.2 m/s
A object of mass 3.00 kg is subject to a force Fx that varies with position as in the figure below. A coordinate plane has a horizontal axis labeled x (m) and a vertical axis labeled Fx (N). There are three line segments. The first segment runs from the origin to (4,3). The second segment runs from (4,3) to (11,3). The third segment runs from (11,3) to (17,0). (a) Find the work done by the force on the object as it moves from x = 0 to x = 4.00 m. J (b) Find the work done by the force on the object as it moves from x = 4.00 m to x = 11.0 m. J (c) Find the work done by the force on the object as it moves from x = 11.0 m to x = 17.0 m. J (d) If the object has a speed of 0.450 m/s at x = 0, find its speed at x = 4.00 m and its speed at x = 17.0 m.
Answer:
Explanation:
An impulse results in a change of momentum.
The impulse is the product of a force and a distance. This will be represented by the area under the curve
a) W = ½(4.00)(3.00) = 6.00 J
b) W = (11.0 - 4.00)(3.00) = 21.0 J
c) W = ½(17.0 - 11.0)(3.00) = 9.00 J
d) ASSUMING the speed at x = 0 is in the direction of applied force
½(3.00)(v₄²) = ½(3.00)(0.450²) + 6.00
v₄ = 2.05 m/s
½(3.00)(v₁₇²) = ½(3.00)(0.450²) + 6.00 + 21.0 + 9.00
v₁₇ = 4.92 m/s
If the initial speed is NOT in the direction of applied force, the final speed will be slightly less in both cases.
Do you believe in ghost
Answer:
well its about our thinking but i do believe in ghost a little
One charge is fixed q1 = 5 µC at the origin in a coordinate system, a second charge q2 = -3.2 µC the other is at a distance of x = 90 m from the origin.
What is the potential energy of this pair of charges?
Answer:
5.4uC
Explanation:
Rays of light coming from the sun (a very distant object) are near and parallel to the principal axis of a concave mirror. After reflecting from the mirror, where will the rays cross each other at a single point?
The rays __________
a. will not cross each other after reflecting from a concave mirror.
b. will cross at the center of curvature.
c. will cross at the point where the principal axis intersects the mirror.
d. will cross at the focal point. will cross at a point beyond the center of curvature.
A concave mirror is an example of curved mirrors. So that the appropriate answer to the given question is option D. The rays will cross at the focal point.
A concave mirror is a type of mirror in which its inner part is the reflecting surface, while its outer part is the back of the mirror. This mirror reflects all parallel rays close to the principal axis to a point of convergence. It can also be referred to as the converging mirror.
In this type of mirror, all rays of light parallel to the principal axis of the mirror after reflection will cross at the focal point.
Therefore, the required answer to the given question is option D. i.e The rays will cross at the focal point.
For reference: https://brainly.com/question/20380620
Suppose a 60-turn coil lies in the plane of the page in a uniform magnetic field that is directed out of the page. The coil originally has an area of 0.325 m2. It is stretched to have no area in 0.100 s. What is the magnitude (in V) and direction (as seen from above) of the average induced emf if the uniform magnetic field has a strength of 1.60 T
Answer:
emf = 312 V
Explanation:
In this exercise the electromotive force is asked, for which we must use Faraday's law
emf = [tex]- N \frac{d \Phi }{dt}[/tex]- N dfi / dt
Ф = B. A = B A cos θ
bold type indicates vectors.
They indicate that the magnetic field is constant, the angle between the normal to the area and the magnetic field is parallel by local cosine values 1
It also indicates that the area is reduced from a₀ = 0.325 me² to a_f = 0 in a time interval of ΔT = 0.100 s, suppose that this reduction is linear
emf = -N B [tex]\frac{dA}{dT}[/tex]
emf = - N B (A_f - A₀) / Dt
we calculate
emf = - 60 1.60 (0 - 0.325) /0.100
emf = 312 V
The direction of this voltage is exiting the page
