Answer:
frequency is the correct answer for the question if the answer is correct plz mark me as brainliest.
Imagine an alternate universe where all of the quantum number rules were identical to ours except m_{s} had three allowed values (rather than two as it does in our universe). If this were the case, and the Pauli Exclusion Principle still applies, how many electrons would be allowed in each orbital
Answer:
so in a given orbital there can be 3 electrons.
Explanation:
The Pauli exclusion principle states that all the quantum numbers of an electron cannot be equal, if the spatial part of the wave function is the same, the spin part of the wave function determines how many electrons fit in each orbital.
In the case of having two values, two electrons change. In the case of three allowed values, one electron fits for each value, so in a given orbital there can be 3 electrons.
What is the voltage drop across an alarm clock that is connected to a circuit with a current of 1.10A and a resistance of 90Ω?
V = 99 volts
Explanation:
The voltage drop can be calculated using Ohm's law:
V = IR
= (1.10 A)(90 Ω)
= 99 volts
g A circular swimming pool has a diameter of 10 m, the sides are 4 m high, and the depth of the water is 3 m. How much work (in Joules) is required to pump all of the water over the side
Answer:
[tex]W=2.69*10^{6}J[/tex]
Explanation:
Generally the equation for momentum is mathematically given by
Diameter [tex]d=10m[/tex]
Radius [tex]r =\frac{d}{2}=>5m[/tex]
Height [tex]h=4m[/tex]
Depth [tex]d=3m[/tex]
Generally the equation for Area of the circle is mathematically given by
[tex]A=\pi r^2[/tex]
[tex]A=\pi 5^2[/tex]
[tex]A=78.5m^2[/tex]
Generally the equation for Work done is mathematically given by
[tex]d_w=d_f*x[/tex]
Where
[tex]d_f=(\rho*d_v)*g[/tex]
Therefore
[tex]d_w=((\rho*d_v)*g)*x[/tex]
[tex]d_w=((1000*78.5)*9.8)*x dx[/tex]
[tex]d_w=769300x dx[/tex]
[tex]W=\int dw[/tex]
[tex]\int dw=769300 \int^4_{3}x dx[/tex]
[tex]W=(769300\2)*7[/tex]
[tex]W=2.69*10^{6}J[/tex]
A 50-ohm resistor in a 0.5 Amp circuit produces how much voltage drop?
25 volts
Explanation:
Use Ohm's law to find the potential drop:
V = IR
= (0.5 A)(50 ohms)
= 25 volts
An astronaut weighing 248 lbs on Earth is on a mission to the Moon and Mars. (a) What would he weigh in newtons when he is on the Moon
Answer:
[tex]W_m=183.495N[/tex]
Explanation:
From the question we are told that:
Weight [tex]W=248Ibs[/tex]
Mass of Weight [tex]m= 248*0.453[/tex]
[tex]m=112.344kg[/tex]
Generally the acceleration due to gravity on the Moon is one-sixth that on Earth.
Therefore
The equation for Weight on Moon is given as
[tex]W_m=m*g/6[/tex]
[tex]W_m=122.344*\frac{9.8}{6}[/tex]
[tex]W_m=183.495N[/tex]
Answer:
F ’= 40.9 lb
Explanation:
The weight of a body is the attraction of the Earth on the body
F = [tex]G \frac{m M_e}{R_e^2}[/tex]
F = mg
m = F / g
m = 248/32
m = 7.75 slug
The weight of the body on the moon is the attraction of the body for the satellite
F ’= [tex]G \frac{mM}{R^2}[/tex]
from the tables the mass and radius of the moon are M = 7.34 10²² kg and R = 1.74 10⁶ m
let's reduce the mass to the SI system
m = 7.75 slug (14.59 kg / 1 slug) = 113 kg
F ’= 6.67 10⁻¹¹ 113 7.34 10²² / (1.74 10⁶) ²
F ’= 1.82 10² N
F ’= 1.82 10² N (0.2248 lb / 1 N)
F ’= 40.9 lb
convert 37 microgram to Gigagram
Answer:
3.7e-14
Explanation:
mark me brainliest!!
Ftension = 120 N
10 kg
Fg
What is the weight (not mass) of the box?
O 10 kg
0 19 N
98 N
O 98 kg
PLZ HELP
Answer:
98N
Explanation:
El peso se mide en kg y la fuerza no afecta
A proton moves across a magnetic field and feels a force. If an electron were to
move at the same speed in the same direction across the same magnetic field, the
force
O 1) would be smaller in the opposite direction
O2) would be the same magnitude in the same direction
3) would be the same magnitude in the opposite direction
4) would be larger in the opposite direction
5) would be smaller in the same direction
potang ina mooooooo bubu hayop kaaaaapestrng yawa
Explanation:
peste kakkkkaaaaaaaa bubu pesteng yawaaaayaka kaayu kaaaaaaa
Allison pulls a sled up a hill, which has an incline of 20 degrees to the horizontal. Of the sled has a mass of 20 kg, and the coefficient of kinetic friction between the sled and the ground is 0.1, what is the minimum force Allison must apply to the sled to keep it moving forward at a constant speed
Answer:
[tex]F=48.62N[/tex]
Explanation:
From the question we are told that:
Angle [tex]\theta= 20[/tex]
Mass [tex]m= 20kg[/tex]
Coefficient of kinetic friction [tex]\mu=0.1[/tex]
Generally the equation for Force Required to jeep sled moving is mathematically given by
[tex]F= mg sin \theta - \mu N[/tex]
Where N is normal force
[tex]F_N=Wcos\theta[/tex]
[tex]F_N=20*9.8*cos 20[/tex]
[tex]F_N=184.18N[/tex]
Therefore
[tex]F= (20*9.8) sin 20 - (0.1) (184.18)[/tex]
[tex]F=48.62N[/tex]
A solid piece of clear transparent materialhas an index of refraction of 1.61. If you place itinto a clear transparentsolution and it seems to disappear, approximately what is the index of refraction of the solution
Answer: The index of refraction of a solution is 1.61
Explanation:
Refractive index or index of refraction is defined as the measure of bending of light when passed through a medium. It is simply the ratio of the speed of light in a vacuum and the speed of light in the phase.
We are given:
Index of refraction of a material = 1.61
When the material is put in a solution, it disappears. This means that there is no bending of light to distinguish between the two phases (material and solution). Thus, the refractive index of the solution is equal to the refractive index of the material.
Index of refraction of a solution = Index of refraction of a material = 1.61
Hence, the index of refraction of a solution is 1.61
1. What did you observe about the magnitudes of the forces on the two charges? Were they the same or different? Does your answer depend on whether the charges were of the same magnitude or different? How does this relate to Newton’s 3rd law?
Answer:
Following are the solution to the given question:
Explanation:
Its strength from both charges is equivalent or identical. The power is equal. And it is passed down
[tex]F=\frac{kq_1q_2}{r^2}[/tex]
Therefore, the extent doesn't rely on the fact that charges are the same or different. Newton's third law complies with Electrostatic Charges due to a couple of charges. They are similar in magnitude, and they're in the other way.
[tex]|F_{12}| = |F_{21}|[/tex]
An object travels down a ramp at a constant acceleration. The object experiences a force of friction and a gravitational force. Which of the following could be true about the motion of the object?
a. The force of friction acts in the same direction as the object travels.
b. The force of friction between the surface and the object is less than the component of the gravitational force that is parallel to the ramp.
c. If the object increased in mass, the object's acceleration would change.
d. If the object increased in mass, the normal force exerted on the object would remain the same.
Answer:
Option (b) is correct.
Explanation:
An object travels down a ramp at a constant acceleration.
Friction force and the gravitational force is there.
(a). It is false. As the friction acts in the opposite direction of motion of the object.
(b). It is true. As the object moves down the ramp so the force of friction is less the component of gravitational force along the ramp.
(c). It is false. the acceleration does not depends on the mass.
a = g sin A - u g cos A
where, A is the angle of inclination and u is the coefficient of friction.
(d). It is false. The normal force is given by
N = m g cos A
so it depends on the mass.
With the frequency set at the mid-point of the slider and the amplitude set at the mid-point of the slider, approximately how many grid marks is the wavelength of the wave (use the pause button and step button as you need to in order to get a good measure, and round to the nearest whole grid mark)?
If the amplitude is increased the wavelength:___________.
a. decreases
b. stays the same
c. increases
Answer:
correct answer is b
Explanation:
The frequency of a wave depends on the properties of medium density and the elasticity properties change the amplitude depends on the energy carried by the wave, that is, the amplitude is proportional to the height of the wave (oscillation).
Consequently the amplitude of independent of the frequency because it depends on different factors.
Therefore when changing the amplitude the wavelength remains constant
the correct answer is b
which wave characteristic is the same for all electromagnetic waves traveling through a vacuum?
Answer:
All waves travel at the same speed c.
How much power is required to light a lightbulb at 100V of voltage when the lightbulb has a resistance of 500 Ohms?
Answer:
20 Watt
Explanation:
Applying
P = V²/R................... Equation 1
Where V = Voltage, P = Power, R = resistance
From the question,
Given: V = 100V, R = 500 ohms
Substitute these values into equation 1
P = (100²)/500
P = 10000/500
P = 20 Watt.
Hence the power required to light the bulb is 20 W
in a Mercury thermometer the level of Mercury Rises when its bulb comes in contact with a hot object what is the reason for this rise in the level of Mercury
Answer:
because thermometric liquid readily expands on heating or contracts on cooling even for a small difference in the temperature of the body.
Find the acceleration due to gravity on the surface of a planet with a mass of 3.5 * 10^24 kg and an average radius of 4.5 * 10^6 m.
A truck is driving over a scale at a weight station. When the front wheels drive over the scale, the scale reads 5800 N. When the rear wheels drive over the scale, it reads 6500 N. The distance between the front and rear wheels is 3.20 m Determine the distance between the front wheels and the truck's center of gravity.
Answer:
[tex]x_2=1.60m[/tex]
Explanation:
From the Question We are told that
Initial Force [tex]F_1=5800N[/tex]
Final Force [tex]F_2=6500N[/tex]
Distance between the front and rear wheels \triangle x=3.20 m
Since
[tex]\triangle x=3.20 m[/tex]
Therefore
[tex]x_1+x_2=3.20[/tex]
[tex]x_1=3.20-x_2[/tex]
Generally the equation for The center of mass is at x_2 is mathematically
given by
[tex]x_2 =\frac{(F_1x_1+F_2x_2)}{(F_1+F_2)}[/tex]
[tex]x_2=3.20F_1-\frac{x_2F_1+F_2x_2}{(F_1+F_2)}[/tex]
[tex]2*F_1*x_2 =3.20F_1[/tex]
[tex]x_2=1.60m[/tex]
Center of gravity of a body is the sum of its moments divided by the overall weight of the object. The distance between the front wheels and the truck's center of gravity is 1.6 meters.
Given-
Scale reading value when the front wheels drive over the scale [tex]m_{1}[/tex] is 5800 N.
Scale reading value when the rear wheels drive over the scale [tex]m_{2}[/tex] is 6500 N
Distance between the front and rear wheel [tex]\bigtriangleup x[/tex] is 3.20 meters.
Let, the distance between the front wheels and the truck's center of gravity is [tex]x_{2}[/tex].
Since sum of the distance between front wheel to truck's center of gravity [tex]x_{1}[/tex], and rear wheel to truck's center of gravity [tex]x_{2}[/tex], is equal to the distance between the front and rear wheel [tex]\bigtriangleup x[/tex]. Therefore,
[tex]\bigtriangleup x=x_{1} +x_{2}[/tex]
[tex]3.20=x_{1} +x_{2}[/tex]
[tex]x_{1} =3.20-x_{2}[/tex]
For the distance between the front wheels and the truck's center of gravity is the formula of center of gravity can be written as,
[tex]x_{2} =\dfrac{m_{1}x_{1}+m_{2} x_{2} }{m_{1} +m_{2} }[/tex]
[tex]x_{2} =\dfrac{5800\times (3.20- x_{2})+6500\times x_{2} }{5800 +6500 }[/tex]
[tex]1230 x_{2} ={18560-5800 x_{2}+6500 x_{2} }[/tex]
[tex]x_{2}= 1.6[/tex]
Hence, the distance between the front wheels and the truck's center of gravity is 1.6 meters.
For more about the center of gravity, follow the link below-
https://brainly.com/question/20662119
150 0.0000
2. Use the distance from the previous problem to calculate how long it takes for light to travel from the
Sun to Earth.
300,000,000X 15000,000,
Answer:
t = 5 10² s
Explanation:
Light is an electromagnetic wave that travels at a constant speed of 3 10⁸ m/s in vacuum.
The distance from the Sun to the Earth is 150 10⁶ km
Let's reduce the distance to the SI system
x = 150 10⁶ km (10³ m / 1 km) = 150 10⁹ m
Since the velocity is constant, we can use the uniform motion relations
v = x / t
t = x / v
let's calculate
t = 150 10⁹ / 3 10⁸
t = 5 10² s
An ideal spring is used to stop blocks as they slide along a table without friction. A 0.80 kg block traveling at a speed of 2.2 m/s can be stopped over a distance of 0.11 m once it makes contact with the spring.
A rectangular block on a level surface moves at velocity v to the right towards a spring that rests on the surface and is attached to a fixed mount on the right.
What distance would a 1.40 kg block travel after making contact with the spring if the block is traveling at a speed of 3.0 m/s before it makes contact with the spring?
Use the work-energy theorem: the total work done on the 0.80 kg block by the spring to make it come to a stop is equal to the change in the block's kinetic energy.
If we take the block's initial direction of motion to be positive, then the spring does negative work on the block, and
-1/2 k (0.11 m)² = 0 - 1/2 (0.80 kg) (2.2 m/s)²
Solve for the spring constant k :
k = (1/2 (0.80 kg) (2.2 m/s)²) / (1/2 (0.11 m)²) = 320 N/m
We can use the same equation as above to find the distance the 1.40 kg block would travel as it is slowed down by the same spring:
-1/2 (320 N/m) x ² = 0 - 1/2 (1.40 kg) (3.0 m/s)²
Solve for the displacement x :
x = √((1/2 (1.40 kg) (3.0 m/s)²) / (1/2 (320 N/m))) ≈ 0.20 m
Which statement describe. good hypothesis ? Check all that apply
How do we get heat on Earth? Does thermal energy travel directly from the sun?
A concert loudspeaker suspended high off the ground emits 26.0 W of sound power. A small microphone with a 1.00 cm2 area is 53.0 m from the speaker. Part A What is the sound intensity at the position of the microphone
Answer:
the sound intensity at the position of the microphone is 7.4 × 10⁻⁴ W/m²
Explanation:
Given the data in the question;
Sound power P = 26.0 W
Area of microphone A = 1.00 cm²
Radius r = 53.0 m
sound intensity at the position of the microphone = ?
Now, intensity at the position of the microphone can be determined using the following expression;
[tex]I[/tex] = P / 4πr²
We substitute
[tex]I[/tex] = 26.0 / ( 4 × π × (53.0 )² )
[tex]I[/tex] = 26.0 / ( 4 × π × 2809 )
[tex]I[/tex] = 26.0 / 35298.935
[tex]I[/tex] = 26.0 / ( 4 × π × (53.0 )² )
[tex]I[/tex] = 0.000736566
[tex]I[/tex] = 7.4 × 10⁻⁴ W/m²
Therefore, the sound intensity at the position of the microphone is 7.4 × 10⁻⁴ W/m²
acceleration of a refrigerator 30s after a person begins pushing it at a force of 400 N
Question: Find acceleration of a refrigerator 30s after a person begins pushing it at a force of 400 N, If the mass of the refrigerator is 10 kg.
Answer:
40 m/s²
Explanation:
Applying,
F = ma................Equation 1
Where F = Force applied to the refrigerator, m = mass of the refrigerator, a = acceleration of the refrigerator.
make a the subject of the equation
a = F/m............ Equation 2
From the question,
Given: F = 400 N, m = 10 kg
Substitute these values into equation 2
a = 400/10
a = 40 m/s²
A thermodynamic system consists of an ideal gas at a volume of 3.50 L and initial pressure of 6.2 × 104 Pa. As the volume is held constant, the pressure is increased to 8.2 × 104 Pa. What work is involved in this process?
Answer:
0 J
Explanation:
Since work done W = PΔV where P = pressure and ΔV = change in volume.
Since the volume is constant, ΔV = 0
So, Work done, W = PΔV = P × 0 = 0 J
So, the work done is 0 J.
Question 24 of 33 Which of the following is an example of uniform circular motion? A. A car speeding up as it goes around a curve O B. A car slowing down as it goes around a curve 2 C. A car maintaining constant speed as it goes around a curve D. A car traveling along a straight road
Answer:
Uniform Circular Motion is the Movement or Rotation of an Object along a circular Path at constant speed.
OPTION C IS YOUR ANSWER!.
You need to produce a set of cylindrical copper wire 3.5 m long that will have a
resistance of 0.125 Ω each. What will be the mass of each of these wires?
(ρ = 1.72X10-8 Ωm, density of copper = 8.9X103 kg/m3)
Solution :
We know, resistance is given by :
[tex]R = \dfrac{\rho l}{A}[/tex]
[tex]A = \dfrac{\rho l }{R}\\\\A = \dfrac{1.72\times 10^{-8} \times 3.5 }{0.125}\\\\A = 4.816 \times 10^{-7} \ m^2[/tex]
Now, we know mass of wire is given by :
[tex]Mass = Density \times Volume\\\\\M = 8.9 \times 10^3 \times 4.816 \times 10^{-7} \times 3.5 kg\\\\M = 0.01500\ kg\\\\M = 15.00\ gram[/tex]
Hence, this is the required solution.
Given:
Length of wire, l = 3.5 mResistance, R = 0.125 ΩThe resistance will be:
→ [tex]R = \frac{\rho l}{A}[/tex]
or,
→ [tex]A = \frac{\rho l}{R}[/tex]
By substituting the values, we get
[tex]= \frac{1.72\times 10^{-8}\times 3.5}{0.125}[/tex]
[tex]= 4.816\times 10^{-7} \ m^2[/tex]
hence,
The mass will be:
→ [tex]Mass = Density\times Volume[/tex]
[tex]= 8.9\times 10^3\times 4.816\times 10^{-7}\times 3.5[/tex]
[tex]= 0.01500 \ kg[/tex]
[tex]= 15.00 \ g[/tex]
Thus the above answer is right.
Learn more about mass here:
https://brainly.com/question/17108656
A 45.00 kg person in a 43.00 kg cart is coasting with a speed of 19 m/s before it goes up a hill.Assuming there is no friction, what is the maximum vertical height the person in the cart can reach?
Answer:
the maximum vertical height the person in the cart can reach is 18.42 m
Explanation:
Given;
mass of the person, m₁ = 45 kg
mass of the cart, m₂ = 43 kg
velocity of the system, v = 19 m/s
let the maximum vertical height reached = h
Apply the principle of conservation mechanical energy;
[tex]P.E = K.E\\\\mgh_{max} = \frac{1}{2} mv^2_{max}\\\\gh_{max} = \frac{1}{2} v^2_{max}\\\\h_{max} = \frac{v_{max}^2}{2g} \\\\h_{max} = \frac{19^2}{2\times 9.8} \\\\h_{max} = 18.42 \ m[/tex]
Therefore, the maximum vertical height the person in the cart can reach is 18.42 m
A gymnast is swinging on a high bar. The distance between his waist and the bar is 1.06 m, as the drawing shows. At the top of the swing his speed is momentarily zero. Ignoring friction and treating the gymnast as if all of his mass is located at his waist, find his speed at the bottom of the swing.
Answer: The speed of gymnast at the bottom of the swing is 6.44 m/s.
Explanation:
Given: Distance = 1.06 m
According to the law of conservation of energy, the speed is calculated as follows.
[tex]mgh = - mgh + \frac{1}{2}mv^{2}\\gh = - gh + \frac{1}{2}v^{2}\\v = \sqrt{4gh}[/tex]
where,
g = acceleration due to gravity = 9.8 [tex]m/s^{2}[/tex]
h = distance
v = speed
Substitute the values into above formula as follows.
[tex]v = \sqrt{4gh}\\= \sqrt{4 \times 9.8 m/s^{2} \times 1.06}\\= 6.44 m/s[/tex]
Thus, we can conclude that speed of gymnast at the bottom of the swing is 6.44 m/s.
Si un electrón recorre el acelerador lineal de Stanford de 2 millas de longitud a 99% de la velocidad de la luz, ¿Cuál es la longitud del acelerador según el electrón?
