Answer:
"the minimum frequency of radiation that will produce a photoelectric effect."
Explanation:
That answer was derived from gogle cuz my explanations was harder to explain but good luck
Suppose you exert a force of 314 N tangential to a grindstone (a solid disk) with a radius of 0.281 m and a mass of 84.2 kg What is the resulting angular acceleration of the grindstone assuming negligible opposing friction
Answer:
The angular acceleration is 26.6 rad/s^2.
Explanation:
Force, F = 314 N
radius, r = 0.281 m
mass, m = 84.2 kg
The grindstone is a disc.
The torque is given by
torque = force x radius
Torque = 314 x 0.281 = 88.234 Nm
The torque is given by
Torque = Moment of inertia x angular acceleration
[tex]88.234 = 0.5 mr^2 \alpha \\\\88.234 = 0.5\times 84.2\times 0.281\times 0.281\times \alpha \\\\\alpha = 26.6 rad/s^2[/tex]
A spherical, concave shaving mirror has a radius of curvature of 0.983 m. What is the magnification of a person's face when it is 0.155 m from the vertex of the mirror (answer sign and magnitude)
Answer:
Magnification = 1
Explanation:
given data
radius of curvature r = - 0.983 m
image distance u = - 0.155
solution
we get here first focal length that is
Focal length, f = R/2 ...................1
f = -0.4915 m
we use here formula that is
[tex]\frac{1}{v} + \frac{1}{u} + \frac{1}{f}[/tex] .................2
put here value and we get
[tex]\frac{1}{v} = \frac{1}{0.155} - \frac{1}{4915}[/tex]
v = 0.155 mso
Magnification will be here as
m = [tex]- \frac{v}{u}[/tex]
m = [tex]\frac{0.155}{0.155}[/tex]
m = 1Answer:
The magnification is 1.5.
Explanation:
radius of curvature, R = - 0.983 m
distance of object, u = - 0.155 m
Let the distance of image is v.
focal length, f = R/2 = - 0.492 m
Use the mirror equation
[tex]\frac{1}{f}=\frac{1}{v}+\frac {1}{u}\\\\\frac{-1}{0.492}=\frac{1}{v}-\frac{1}{0.155}\\\\\frac{1}{v}=\frac{1}{0.155}-\frac{1}{0.492}\\\\\frac{1}{v}=\frac{0.492-0.155}{0.155\times 0.492}\\\\\frac{1}{v}=\frac{0.337}{0.07626}\\ \\v = 0.226 m[/tex]
The magnification is given by
m = - v/u
m = 0.226/0.155
m = 1.5
Imagine you’re driving along a road and you approach a bridge. You notice a sign that reads, “Bridge freezes before road.” Why do bridges become covered with ice before roads do? Research this question and respond in depth, writing a full paragraph. Be sure to include examples. At the end of your response, provide at least two authoritative sources that you used in your research.
Answer:
During wet and freezing temperatures, ice is able to form at a faster pace on bridges because freezing winds blow from above and below and both sides of the bridge, causing heat to quickly escape. The road freezes slower because it is merely losing heat through its surface.Sources:
-- https://intblog.onspot.com/en-us/why-do-bridges-become-icy-before-roads
and
-- https://www.accuweather.com/en/accuweather-ready/why-bridges-freeze-before-roads/687262
I hope this helps you! ^^
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.
Accommodation of the eye refers to its ability to __________. see on both the brightest days and in the dimmest light see both in air and while under water move in the eye socket to look in different directions focus on both nearby and distant objects
Answer:
to adjust from distant to the near objects
Explanation:
The process of accommodation is achieved by changing in the shape and position of the eye ball. Just like adjusting the lens of the camera.Answer:
The ability of eye lens to change the focal length of eye lens is called accommodation power of eye.
Explanation:
The human eye is the optical instrument which works on the refraction of light.
The ability of eye lens to change its focal length is called accommodation power of eye.
The focal length of eye lens is changed by the action of ciliary muscles.
When the ciliary muscles are relaxed then the thickness of lens is more and thus the focal length is small. When the ciliary muscles is stretched, the lens is thin and then the focal length is large.
Two concentric current loops lie in the same plane. The smaller loop has a radius of 3.0 cm and a current of 12 A. The bigger loop has a current of 20 A. The magnetic field at the center of the loops is found to be zero.
Required:
What is the radius of the bigger loop?
Answer:
the radius of the bigger loop is 5 cm.
Explanation:
Given;
current in the smaller loop, I₁ = 12 A
current in the larger loop, I₂ = 20 A
radius of the smaller loop, r₁ = 3 cm
let the radius of the larger loop, = r₂
Apply Biot-Savart's law to determine the magnetic field at the center of the circular loops.
[tex]B= \frac{\mu_0 I}{2r}[/tex]
The magnetic field at the center of the smaller loop;
[tex]B_1 = \frac{\mu_0 I_1}{2 r_1}[/tex]
The magnetic field at the center of the bigger loop;
[tex]B_2 = \frac{\mu_0 I_2}{2 r_2}[/tex]
If the magnetic field at the center is zero, then B₁ = B₂
[tex]B_1 = B_2 = \frac{\mu_0 I_1}{2 r_1} = \frac{\mu_0 I_2}{2 r_2} \\\\\frac{I_1}{ r_1} = \frac{ I_2}{r_2} \\\\r_2 = \frac{I_2 r_1}{ I_1} = \frac{(20 \ A) \times (3.0 \ cm)}{12 \ A} = 5 \ cm[/tex]
Therefore, the radius of the bigger loop is 5 cm.
Trình bày những hiểu biết của em về đại lượng vận tốc dài, vận tốc góc(định nghĩa, công thức, ý nghĩa, đơn vị, loại đại lượng).
What is the percentage of the population that wanted both the swimming pool and the soccer complex? Use your knowledge
of the addition rule and the Venn diagram to answer.
Answer:
The percentage of people who wanted both the swimming pool and the soccer complex is 0.6 + 0.6 – 0.95 = 0.25. This can also be seen in the Venn diagram.
Explanation:
Edmentum
In these formulas, it is useful to understand which variables are parameters that specify the nature of the wave. The variables E0E0E_0 and B0B0B_0 are the __________ of the electric and magnetic fields. Choose the best answer to fill in the blank.
why did Rita's hands get hot when she rubbed them ?
Answer:
due to production of heat through friction
Explanation:
because of the friction produce between her hands
For a spring-mass oscillator if you double the mass but keep the stiffness the same, by what numerical factor does the pena original period was and the new period is DT, what is b7 It is useful to write out the expression for the period and ask yours you doubled the mass.
b = _____
If, instead, you double the spring stiffness but keep the mass the same, what is the factor b?
b = _____
If, instead, you double the mass and also double the spring stiffness, what is the factor b?
b = _____
If, instead, you double the amplitude (keeping the original mass and spring stiffness), what is the factor b?
b = _____
Answer:
ygguguguhhihhihijijijjojojinjbgy
Two masses of 3 kg and 5 kg are connected by a light string that passes over a smooth polley as shown in the Figure.
QL
Determine:
i. the tension in the string,
ii. the acceleration of each mass, and
iii. the distance each mass moves in the first second of motion if they start from rest
i. [tex]T = 36.8\:\text{N}[/tex]
ii. [tex]a = 2.45\:\text{m/s}^2[/tex]
iii. [tex]x = 1.23\:\text{m}[/tex]
Explanation:
Let's write Newton's 2nd law for each object. We will use the sign convention assigned for each as indicated in the figure. Let T be the tension on the string and assume that the string is inextensible so that the two tensions on the strings are equal. Also, let a be the acceleration of the two masses. And [tex]m_1 = 3\:\text{kg}[/tex] and [tex]m_2 = 5\:\text{kg}[/tex]
Forces acting on m1:
[tex]T - m_1g = m_1a\:\:\:\:\:\:\:(1)[/tex]
Forces acting on m2:
[tex]m_2g - T = m_2a\:\:\:\:\:\:\:(2)[/tex]
Combining Eqn(1) and Eqn(2) together, the tensions will cancel out, giving us
[tex]m_2g - m_1g = m_2a + m_1a[/tex]
or
[tex](m_2 - m_1)g = (m2 + m_1)a[/tex]
Solving for a,
[tex]a = \left(\dfrac{m_2 - m_1}{m_2 + m_1}\right)g[/tex]
[tex]\:\:\:\:= \left(\dfrac{5\:\text{kg} - 3\:\text{kg}}{5\:\text{kg} + 3\:\text{kg}}\right)(9.8\:\text{m/s}^2)[/tex]
[tex]\:\:\:\:= 2.45\:\text{m/s}^2[/tex]
We can solve for the tension by using this value of acceleration on either Eqn(1) or Eqn(2). Let's use Eqn(1).
[tex]T - (3\:\text{kg})(9.8\:\text{m/s}^2) = (3\:\text{kg})(2.45\:\text{m/s}^2)[/tex]
[tex]T = (3\:\text{kg})(9.8\:\text{m/s}^2) + (3\:\text{kg})(2.45\:\text{m/s}^2)[/tex]
[tex]\:\:\:\:= 29.4\:\text{m/s}^2 + 7.35\:\text{m/s}^2 = 36.8\:\text{N}[/tex]
Assuming that the two objects start from rest, the distance that they travel after one second is given by
[tex]x = \frac{1}{2}at^2 = \frac{1}{2}(2.45\:\text{m/s}^2)(1\:\text{s})^2 = 1.23\:\text{m}[/tex]
Pascal's principle says: a A change in pressure at one point in an incompressible fluid is felt at every other point in the fluid. b The buoyant force equals the weight of the displaced fluid. c Matter must be conserved in a flowing, ideal fluid. d Energy is conserved in a flowing, ideal fluid. e A small input force causes a large output force.
Answer:
A change in pressure at one point in an incompressible fluid is felt at every other point in the fluid.
Explanation:
Pascal's principle states that ''pressure applied to an enclosed fluid will be transmitted without a change in magnitude to every point of the fluid and to the walls of the container.''(Science direct).
The implication of this law is; that a change in pressure at one point in an incompressible fluid is felt at every other point in the fluid. Hence the correct answer chosen above.
The Pascal's principle is applied in hydraulic jacks and automobile brakes.
how can you convert galvanometer into ammeter?
Answer:
A galvanometer is converted into an ammeter by connecting a low resistance in parallel with the galvanometer.
Explanation:
This low resistance is called shunt resistance S. The scale is now calibrated in ampere and the range of the ammeter depends on the values of the shunt resistance.
Use the pressure meter to read the pressure in Fluid A at the bottom of the tank. Do not move the pressure meter. Switch to Fluid B and read the pressure in fluid B. Based on the two readings, compare the density of fluid B to the density of fluid A. Which statement is correct?
Answer:
[tex]P_b = \frac{\rho_b}{\rho_a} \ P_a[/tex]
Explanation:
The pressure at a depth of a fluid is
P = ρ g y
where ρ is the density of the fluid, y the depth of the gauge measured from the surface of the fluid.
In this case the pressure for fluid A is
Pa = ρₐ g y
the pressure for fluid B is
P_b = ρ_b g y
depth y not changes as the gauge is stationary
if we look for the relationship between these pressures
[tex]\frac{P_a}{P_b} = \frac{ \rho_a}{\rho_b}[/tex]
[tex]P_b = \frac{\rho_b}{\rho_a} \ P_a[/tex]
therefore we see that the pressure measured for fluid B is different from the pressure of fluid A
if ρₐ < ρ_b B the pressure P_b is greater than the initial reading
ρₐ> ρ_b the pressure in B decreases with respect to the reading in liquid A
It was recorded that the temperature of a body was 320 degree F determine the value of the temperature in kelvin
Answer:
433.15K
Explanation:
(320°F − 32) × 5/9 + 273.15 = 433.15K
Three wires are connected at a branch point. One wire carries a positive current of 18 A into the branch point, and a second wire carries a positive current of 7 A away from the branch point. Find the current carried by the third wire into the branch point.
Answer:
The current in third branch is 11 A.
Explanation:
incoming current in one branch = 18 A
outgoing current in the other branch = 7 A
let the current in the third branch is i.
According to the Kirchoff's fist law in electricity
incoming current = out going current
18 = 7 + i
i = 11 A
The current in third branch is 11 A.
What are the examples of pulley? Plz tell the answer as fast as possible plz.
Answer:
elevators
Theatre system
construction pulley
lifts
Answer:
elevator,cargo lift system
Condensation is the process of ____________________.
a. planetesimals accumulating to form protoplanets.
b. planets gaining atmospheres from the collisions of comets.
c. clumps of matter adding material a small bit at a time.
d. clumps of matter sticking to other clumps.
e. clouds formed from volcanic eruptions.
A 100 kg man is one fourth of the way up a 4.0 m ladder that is resting against a smooth, frictionless wall. The ladder has mass 25 kg and makes an angle of 56 degrees with the ground. What is the magnitude of the force of the wall on the ladder at the point of contact, if this force acts perpendicular to the wall and points away from the wall
Answer:
[tex]N_f=248N[/tex]
Explanation:
From the question we are told that:
Mass [tex]m=100kg[/tex]
Ladder Length [tex]l=4.0m[/tex]
Mass of Ladder [tex]m_l=25kg[/tex]
Angle [tex]\theta=56 \textdegree[/tex]
Generally the equation for Co planar forces is mathematically given by
[tex]mgcos \theta *2+Mgcos\theta*1 -N_fsin \theta*4=0[/tex]
Therefore
[tex]25*9.81cos 56 *2+100*9.81cos56*1 -N_fsin 56*4=0[/tex]
[tex]N_f=248N[/tex]
A roller coaster has a mass of 1200.0kg. The coaster is going 22.0 m/s at the bottom
of the third loop-the-loop that is 2.5m above the ground. Determine the height of
the first hill that is required, assuming the cart is stationary at the top of the first hill
before it falls.
Answer:
h = 27.17 m
Explanation:
First, we will calculate the total mechanical energy of the system at the bottom point of the third loop:
Mechanical Energy = Kinetic Energy + Potential Energy
[tex]E = \frac{1}{2}mv^2 + mgh[/tex]
where,
E = Total Mechanical Energy = ?
m = mass of the roller coaster = 1200 kg
v = velocity of the roller coaster = 22 m/s
g = acceleration due to gravity = 9.81 m/s²
h = height of roller coaster = 2.5 m
Therefore,
[tex]E = \frac{1}{2}(1200\ kg)(22\ m/s)^2+(1200\ kg)(9.81\ m/s^2)(2.5\ m)\\\\E = 290400 J +29430\ J\\\\E = 319830\ J = 319.83\ KJ[/tex]
Now, the total mechanical energy at the top position of the first hill must also be the same:
[tex]E = \frac{1}{2}mv^2 + mgh[/tex]
where,
v = 0 m/s
h = ?
Therefore,
[tex]319830\ J = \frac{1}{2}(1200\ kg)(0\ m/s)^2+(1200\ kg)(9.81\ m/s^2)(h)\\\\h = \frac{319830\ J}{11772\ N}\\\\[/tex]
h = 27.17 m
A planet of mass m = 4.25 x 1024 kg orbits a star of mass M = 6.75 x 1029 kg in a circular path. The radius of the orbits R = 8.85 x 107 km. What is the orbital period Tplanet of the planet in Earth days?
285.3 days
Explanation:
The centripetal force [tex]F_c[/tex] experienced by the planet is the same as the gravitational force [tex]F_G[/tex] so we can write
[tex]F_c = F_G[/tex]
or
[tex]m\dfrac{v^2}{R} = G\dfrac{mM}{R^2}[/tex]
where M is the mass of the star and R is the orbital radius around the star. We know that
[tex]v = \dfrac{C}{T} = \dfrac{2\pi R}{T}[/tex]
where C is the orbital circumference and T is orbital period. We can then write
[tex]\dfrac{4\pi^2R}{T^2} = G\dfrac{M}{R^2}[/tex]
Isolating [tex]T^2[/tex], we get
[tex]T^2 = \dfrac{4\pi^2R^3}{GM}[/tex]
Taking the square root of the expression above, we get
[tex]T = 2\pi \sqrt{\dfrac{R^3}{GM}}[/tex]
which turns out to be [tex]T = 2.47×10^7\:\text{s}[/tex]. We can convert this into earth days as
[tex]T = 2.47×10^7\:\text{s}×\dfrac{1\:\text{hr}}{3600\:\text{s}}×\dfrac{1\:\text{day}}{24\:\text{hr}}[/tex]
[tex]\:\:\:\:\:= 285.3\:\text{days}[/tex]
Michelson and Morley concluded from the results of their experiment that Group of answer choices the experiment was successful in not detecting a shift in the interference pattern. the experiment was a failure since they detected a shift in the interference pattern. the experiment was a failure since there was no detectable shift in the interference pattern. the experiment was successful in detecting a shift in the interference pattern.
Answer:
The results of the experiment indicated a shift consistent with zero, and certainly less than a twentieth of the shift expected if the Earth's velocity in orbit around the sun was the same as its velocity through the ether.
Explanation:
If a body travels 6km in 30 minutes in a fixed direction, calculate it's velocity.
Plz show me the process too.
We know
[tex]\boxed{\large{\sf Velocity=\dfrac{Distance}{Time}}}[/tex]
[tex]\\ \Large\sf\longmapsto Velocity=\dfrac{6}{\dfrac{1}{2}}[/tex]
[tex]\\ \Large\sf\longmapsto Velocity=6\times 2[/tex]
[tex]\\ \Large\sf\longmapsto Velocity=12km/h[/tex]
Is this the right answer??
We should keep km and min in smallest SI unit
why is it necessary to have end correction in the organ pipe?
Answer:
The vibrating length of the air column is greater than the actual length of the organ pipe
find the rate of energy radiated by a man by assuming the surface area of his body 1.7m²and emissivity of his body 0.4
The rate of energy radiated by the man is 3.86 x [tex]10^{-8}[/tex] J/s. [tex]m^{2}[/tex].
The amount of energy radiated by an object majorly depends on the area of its surface and its temperature. The is well explained in the Stefan-Boltzmann's law which states that:
Q(t) = Aeσ[tex]T^{4}[/tex]
where: Q is the quantity of heat radiated, A is the surface area of the object, e is the emmisivity of the object, σ is the Stefan-Boltzmann constant and T is the temperature of the object.
To determine the rate of energy radiated by the man in the given question;
[tex]\frac{Q(t)}{T^{4} }[/tex] = Aeσ
But A = 1.7 m², e = 0.4 and σ = 5.67 x [tex]10^{-8}[/tex] J/s.
So that;
[tex]\frac{Q(t)}{T^{4} }[/tex] = 1.7 * 0.4 * 5.67 x [tex]10^{-8}[/tex]
= 3.8556 x [tex]10^{-8}[/tex]
= 3.86 x [tex]10^{-8}[/tex] J/s. [tex]m^{2}[/tex]
Thus, the rate of energy radiated by the man is 3.86 x [tex]10^{-8}[/tex] J/s. [tex]m^{2}[/tex].
Learn more on energy radiation of objects by visiting: https://brainly.com/question/12550129
Let A^=6i^+4j^_2k^ and B= 2i^_2j^+3k^. find the sum and difference of A and B
Explanation:
Let [tex]\textbf{A} = 6\hat{\textbf{i}} + 4\hat{\textbf{j}} - 2\hat{\textbf{k}}[/tex] and [tex]\textbf{B} = 2\hat{\textbf{i}} - 2\hat{\textbf{j}} + 3\hat{\textbf{k}}[/tex]
The sum of the two vectors is
[tex]\textbf{A + B} = (6 + 2)\hat{\textbf{i}} + (4 - 2)\hat{\textbf{j}} + (-2 + 3)\hat{\textbf{k}}[/tex]
[tex] = 8\hat{\textbf{i}} + 2\hat{\textbf{j}} + \hat{\textbf{k}}[/tex]
The difference between the two vectors can be written as
[tex]\textbf{A - B} = (6 - 2)\hat{\textbf{i}} + (4 - (-2))\hat{\textbf{j}} + (-2 - 3)\hat{\textbf{k}}[/tex]
[tex]= 4\hat{\textbf{i}} + 6\hat{\textbf{j}} - 5\hat{\textbf{k}}[/tex]
In many cartoon shows, a character runs of a cliff, realizes his predicament and lets out a scream. He continues to scream as he falls. If the physical situation is portrayed correctly, from the vantage point of an observer at the foot of the cliff, the pitch of the scream should be Group of answer choices
Answer:
Increasing until terminal velocity is reached
Explanation:
Provided the scream is a constant pitch at the source, Doppler effect will make the pitch increase as the velocity of the source towards the listener increases.
if 6000j of energy is supplid to a machine to lift a load of 300N through a vvertical height of 1M calculatework out put
Answer:
300J
Explanation:
Work done = Force x the distance travelled in the direction of the force
=300 x 1
=300J
calculate the length of wire.
Answer:
L = 169.5 m
Explanation:
Using Ohm's Law:
V = IR
where,
V = Voltage = 1.5 V
I = Current = 10 mA = 0.01 A
R = Resistance = ?
Therefore,
1.5 V = (0.01 A)R
R = 150 Ω
But the resistance of a wire is given by the following formula:
[tex]R = \frac{\rho L}{A}[/tex]
where,
ρ = resistivity = 1 x 10⁻⁶ Ω.m
L = length of wire = ?
A = cross-sectional area of wire = πr² = π(0.6 mm)² = π(0.6 x 10⁻³ m)²
A = 1.13 x 10⁻⁶ m²
Therefore,
[tex]150\ \Omega = \frac{(1\ x\ 10^{-6}\ \Omega .m)L}{1.13\ x\ 10^{-6}\ m^2}\\\\L = \frac{150\ \Omega(1.13\ x\ 10^{-6}\ m^2)}{1\ x\ 10^{-6}\ \Omega .m}\\\\[/tex]
L = 169.5 m
