The cutoff frequency for magnesium is 8.93 x 10¹⁴ Hz.
What is cutoff frequency?The work function is related to the frequency as
W0 = h x fo
where, fo = cutoff frequency and h is the Planck's constant
Given is the work function for magnesium is 3.70 eV.
fo = 3.7 x 1.6 x 10⁻¹⁹ / 6.626 x 10⁻³⁴
fo = 8.93 x 10¹⁴ Hz.
Thus, the cut off frequency is 8.93 x 10¹⁴ Hz.
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'The wave' is a particular type of pulse that can propagate through a large crowd gathered at a sports arena to watch a soccer, hockey, or CFL game. The elemets of the medium are the spectators, with zero position corresponding to their being seated and maximum position corresponding to their standing and raising their arms. When a large fraction of the spectators participate in the wave motion, a somewhat stable pulse shape can develop. The wave speed depends on people's reaction time, which is typically on the order of 0,1s. Estimate the order of magnitude, in minutes, of the time required for such a pulse to make one circuit around BC Place Stadium in Vancouver.
State all the assumptions that you've made.
Information about BC Place: dimensions are approximately 100 m X 85 m.
The total time required is 1 minute and 0.3 seconds.
Assumption: The distance between the people is 1 m and the stadium is a circle with a radius of 100m.
Here, the time taken by the person is 0.1 seconds.
Total distance covered by the wave = Circumference of the circle
So, total distance = 2 πr = 2 × 3.14 × 100 = 618 m
As the distance between each of the people is 1 m.
So, the number of personal interactions is 618.
Time taken by each person is 0.1 seconds.
So, total time, t = 0.1 × 618 = 61.8 seconds.
So, the order of the magnitude of the time required is 1 minute and 0.3 seconds for such a pulse to make one circuit around BC Place Stadium in Vancouver.
Hence, the total time required is 1 minute and 0.3 seconds.
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A baseball is thrown at angle of 65 degrees with an initial velocity of 45 m/s. How long will the baseball remain in the air?
Assume yi = yf =0 m
G= -9.81 m/s^2
A) 4.15 seconds
B) 8.31 seconds
C) 9.37 seconds
D) 1.44 seconds
The time for which the baseball remains in the air will be 9.37 sec. Option C is correct.
What is the maximum height achieved in projectile motion?It is the height achieved by the body when a body is thrown at the same angle and the body is attaining the projectile motion.
From Newton's first equation of motion;
[tex]\rm v_y= u_y+gt \\\\ 0 = usin \theta - gt \\\\ t = \frac{u sin\theta }{g}\\\\t = 4.68 \ sec[/tex]
The time the baseball remains in the air will be twice the time he travels for one side motion;
T' = 2t
T'=2 × 4.68 sec
T'= 9.37 sec
Hence for 9.37 sec, the baseball willl remains in the air. Option C is correct.
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The following table contains financial information for Trumpter Inc. before closing entries:
Cash
Supplies
Prepaid Rent
Salaries Expense
Equipment
Service Revenue
Miscellaneous Expenses
Dividends
Accounts Payable
Common Stock
Retained Earnings
$13,100
6,000
3,800
5,000
65,900
28,000
20,100
4,000
3,100
66,100
20,700
What is the amount of Trumpter's total assets?
The Total assets based on the given table is $88,800.
What are assets?An asset refers to a resource owned by an individual or company which has tangible value and whose value is expected to rise in the future.
The Total assets based on the table = Cash + Supplies + Prepaid rent + Equipment
Total assets = 13,100 + 6,000 + 3,800 + 65,900
Total assets = $88,800
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The value of acceleration due to gravity (g) on a point 10,000 kilometers above sea level is about 1.49 meters/second2. How much will an object, which weighs 98 newtons on the surface of Earth, weigh on this point? The value of acceleration due to gravity on Earth is 9.8 meters/second2.
An object, which weighs 98 N on the surface of Earth, weigh on this point is 2.23 x 10²⁴ kg.
What is gravity?The force of attraction felt by a person at the center of a planet or Earth is called as the gravity.
Given, the Earth has the acceleration due to gravity, g = 9.81 m/s².
Force of gravity W = mass x acceleration due to gravity
98N = m x 9.8m/s²
m = 10 kg
The value of acceleration due to gravity (g) on a point 10,000 km above sea level is about 1.49 m/s².
The acceleration due to gravity and mass is related as
g = GM/R²
where G = gravitational constant = 6.67 x 10⁻¹¹ N.m²/kg² and R is the distance between two masses.
Substituting the values, we get
1.49 m/s² = 6.67 x 10⁻¹¹ N.m²/kg² x M/ (10000 x 10³ m)²
M = 2.23 x 10²⁴ kg
Therefore, an object will weigh on this point is 2.23 x 10²⁴ kg
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Two small plastic spheres are given positive electric charges. When they are 15 cm apart the repulsive force between them has magnitude 0.22 N. What is the charge on each sphere?
The charge on each sphere is 742 n
What is Charge?In physics, Charge, also known as electric charge is a electrical charge, or electrostatic charge and symbolized q, is a characteristic of a unit of matter that expresses the extent to which it has more or fewer electrons than protons.
According to the question,
Distance = 15cm or 0.15m
Force = 0.22N
[tex]K=9*10^{9} kgm^2/c^2[/tex] ( this is a constant value of K)
By using the formula,
F = [tex]\frac{kQq}{d^{2} }[/tex]
0.220 N = 8.99
N·m²/C² * Q² / (0.15m)²
Q = 7.42e-7
C = 742 n
C is the charge of each sphere .
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As a warehouse worker pushes a crate across a concrete floor, the force he
applies is not perfectly horizontal, as shown in the image below. If the
coefficient of kinetic friction between the crate and concrete floor is 0.5, what
is the net force on the crate?
A. 136 N
B. 99 N
C. 73 N
D. 112 N
The net force on the crate will be 99 N.Option B is correct.
What is the friction force?It is a type of opposition force acting on the surface of the body that tries to oppose the motion of the body. its unit is Newton (N). it is defined as the product of the coefficient of friction and normal reaction.
On resolving the given force and acceleration.Mathematically in the different components and balancing the equation gets.Components in the x-direction.
When all the forces get resolved the y-direction of forces are;
300 sin 10° +N = 445
N = 445-295.4
N = 392.9 N
The normal force is 392.9 N.
The net force on the crate is found by resolving the force in the x-direction;
F = 300 cos 10° - μN
F= 295.44-196.45
F= 98.99
F=99
Hence, the net force on the crate will be 99 N.Option B is correct.
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Select. The star cycle that is accurate
One stellar-mass star, red giants, white dwarfs, or planetary nebulae will make up the correct star cycle. Option C is correct.
What is the solar system?The satellites of the planet, countless comets, asteroids, and meteoroids, as well as the interplanetary medium, make up the solar system.
The complete question is;
"Choose the correct star life cycle.
A. Supernova, star as well as red giant, the nebula is incorrect.
B. Nebula, white dwarf, planetary nebula That is incorrect.
C. Planetary nebula, red giant, white dwarf, and star with a single stellar mass That is true!
D. Nebula, a star of one stellar mass, is not correct. It is correct if the star had four or more stellar masses."
Star of one stellar mass, red giant, white dwarf, and planetary nebula make up the proper life cycle.
Hence option C is correct.
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The wavelength of the wave pictured below is... Select one: a. 1 cm b. 1.5 cm c. 2 cm d. 3 cm
Answer:
D. 3 cm
Wavelength is the distance between two crests or two troughs.
In the given picture, the distance between two adjacent crests is 3 cm.
Hence, the wavelength is 3 cm.
please help me in this questions
Answer:
Rainy daywindy daysunny daycloudy daysorry I don't know the answer of question 8.sunglassesumbrella:-(:-):-):-(day ☀️night day ☀️night Day ☀️Day ☀️night nightExplanation:
Hope I give all correct answer please mark as brainlest answer
It is important that a frame of reference
Answer:
enable us to take in a wide variety of information, and process it based on our past experience and values.
Explanation:
Which refers to a material that causes a wave to bounce off of it?
A. absorber
B. reflector
C. conductor
D. insulator
Answer:
the answer to the question is a reflector
A contestant in a winter games event pushes a 54.0 kg block of ice across a frozen lake as shown in Figure 4.29(a). The coefficient of static friction is 0.1 and the coefficient of kinetic friction is 0.03.
a) Calculate the minimum force F he must exert to get the block moving.
N
(b) What is its acceleration once it starts to move, if that force is maintained?
m/s2
Answer:
54.0 x0.1=5.4 x0.03=0.162
kinetic force
If a contestant in a winter games event pushes a 54.0 kg block of ice across a frozen lake. The coefficient of static friction is 0.1 and the coefficient of kinetic friction is 0.03, then the minimum force F he must exert to get the block moving would be 5.2974 Nwrons.
What is friction?
Friction is a type of force that resists or prevents the relative motion of two physical objects when their surfaces come in contact.
As given in the problem, If a contestant in a winter games event pushes a 54.0 kg block of ice across a frozen lake. The coefficient of static friction is 0.1 and the coefficient of kinetic friction is 0.03,
The force required to get the block moving = μMg
= 0.01×54×9.81
= 5.2974 Newtons
Thus, the minimum force required to move the block would be 5.2974 Newtons.
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A hydrogen bond forms by the electrostatic interaction of opposite charges in two molecules. If
the bond length is 2x10 -10 m and the magnitude of the charges involved is
approximately 1.60x10 -20C, what is the force between the molecules involved in the bond?
The force between the molecules involved in the bond is 6. 426 *10^-11 Newton
How to determine the force
Using the formula:
F = K[q1 x q2]/D^2
where K is coulombs constant =9 *10 ^9 Nm^2/C^2.
q1 and q2 = charges = 1.60x10 -20C
d = distance between the charges = 2x10 -10 m
Substitute the values into the formula
F = [tex]9 * 10^9\frac{ 1.60*10^ -20 * 1.60 *10^ -20}{2x10^ -10^{2} }[/tex]
F = [tex]9 *10^9\frac{2. 856* 10^-40}{4* 10^-20}[/tex]
F = [tex]9* 10^9 * 7. 14* 10^-21[/tex]
F = [tex]6. 426 * 10^-11[/tex] Newton
Thus, the force between the molecules involved in the bond is 6. 426 *10^-11 Newton
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A teacher wants to demonstrate that the radioactive source emits alpha beta and gamma radiation. Describe a method the teacher could use (IMPORTANT REPLAY ASAP) I will give Brainly
By using an electric field, it is feasible to differentiate between these different forms of radiation.
What is a radioactive source?A source that emits radiation like gamma, beta, and alpha rays is said to be radioactive. Using an electric field, we can discriminate between these different forms of radiation.
The field does not deflate the gamma rays, but it does deflate the alpha and beta rays, with the alpha being deflated to the field's negative portion and the beta to its positive part.
Hence, by using an electric field, it is feasible to differentiate between these different forms of radiation.
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Which has more mass - a 5-kg bag of feathers or a 5-kg
cannon ball?
Answer:equal mass
Explanation:the mass of both is 5kg
. Two planets, both of mass m, are separated by a distance d. Their relative velocity is negligible, and there is an inertial frame in which both planets are essentially at rest. The gravitational potential u(r) at the position r in the presence of the two planets, located at R1 and R2, is given as u(r) = − Gm R1 −r − Gm R2 −r . This problem takes place far out in space and there are no other massive objects in the vicinity of the two planets. (a) Draw a graph of u as a function of position (r) along the line between the two planets. (b) There are space stations Alpha and Beta located on the line between the planets. Both space stations are at rest with respect to the planets. Alpha is at distance d 4 from planet 1 and Beta is at distance d 3 from planet 2. A projectile of mass m is fired from station Alpha, with its velocity v pointing directly at planet 2. What is the minimum speed v which will permit the projectile to reach station Beta?
(a) A graph of u as a function of position (r) along the line between the two planets is attached below.
(b) The minimum speed v which will permit the projectile to reach station Beta is √ [Gm/3d]
What is gravitational potential energy?If an object is lifted, work is done against gravitational force. The object gains energy.
Given are two planets, both of mass m, are separated by a distance d. Their relative velocity is negligible, and there is an inertial frame in which both planets are essentially at rest. The gravitational potential u(r) at the position r in the presence of the two planets, located at R1 and R2, is given as
u(r) = − (Gm /R1 −r) − (Gm / R2 −r) .
This problem takes place far out in space and there are no other massive objects in the vicinity of the two planets
(a) A graph of u (r) versus position (r) along the line between the two planets is attached in answer.
(b) There are space stations Alpha and Beta located on the line between the planets. Both space stations are at rest with respect to the planets. Alpha is at distance d 4 from planet 1 and Beta is at distance d 3 from planet 2. A projectile of mass m is fired from station Alpha, with its velocity v pointing directly at planet 2.
The range of the projectile is given by R = v²sin2θ / g
g = gravitational acceleration of Earth
If g = g(p) for planet , range R = v²sin2θ / g(p)..................(1)
The gravitational force of attraction = weight force
Gm² /d² = m g(p)
g(p) = Gm/d².........................(2)
For R = d/3, from equation (1), we have
d/3 = v²sin2θ / g(p)
Plug the expression for g(p) , we get
v = √ [Gm/3dsin2θ ]
For velocity to be minimum, sin2θ =1
So, the minimum velocity will be
v = √ [Gm/3d]
Thus, the minimum speed v which will permit the projectile to reach station Beta is √ [Gm/3d]
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Can i use a 1000mAh battery instead of 600mAh Battery?
please tell me quickly
To deliver more power or voltage in the circuit use a battery with more current rating which is 1000 mAh.
Power generated in a circuitThe power generated in a circuit is directly proportional to the current flowing in the circuit.
P = IV
where;
I is currentV is voltageVoltage in the circuit also increases with increase in the current flowing in the circuit.
Thus, to deliver more power or voltage in the circuit use a battery with more current rating which is 1000 mAh.
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A comet goes around the sun in an elliptical orbit. At its farthest point, 455 million miles from the sun, it is traveling with a speed of 15,135 mi/h. How fast is it traveling at its closest approach to the sun, at a distance of 116 million miles?
Answer:
A 2kg block is moved a long a level floor by a horizontal force of 20N if the coefficient of sliding friction is 0.4
what is the acceleration of the blove
Please help me!
If ball C is 3 times the volume of ball D and ball D has 1/3 the mass of ball C, which has the greater density?
A. Ball C
B. Ball D
C. The Densities are equal
C. The Densities are equal.
What is density?
Density is mass per unit volume or mass of a unit volume of a material substance.
If m1, V1 and D1 = mass, volume and density respectively of ball C
m2, V2 and D2 = mass, volume and density respectively of ball D
According to the Question ,
[tex]V_{1} = 3V_{2} , m_{2} = \frac{1}{3} (m_{1} ) \\ \\= m_{1} = 3m_{2}[/tex]
Therefore,
[tex]\frac{D_{1} }{D_{2} } = (\frac{m_{1} }{V_{1} } )* (\frac{m_{2} }{V_{2} } )\\ \\= (\frac{3m_{2} }{3V_{2} })*(\frac{V_{2} }{m_{2} }) \\\\= 1[/tex]
Hence, D1 = D2
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A uniform electric field exist between two parallel plate seperated by 1.2 cm. The intensity of the field is 23 KN/C. What is the potential difference between the plates
Explanation:
this is the answer 276 volts
A sled is moving down a steep hill. The mass of the sled is 50 kg and the net force acting on it is 20 n. What must be done to find the acceleration of the sled? check all that apply.
The acceleration of this sled can be calculated by doing: option A, B, E and F.
How to calculate the acceleration?The acceleration of this sled can be calculated by using Newton's Second law of motion. Mathematically, the acceleration of an object is given by this formula:
Net force = Mass × acceleration
Deductively, the acceleration of this sled can be calculated by doing the following:
The force of gravity must be broken into its parallel and perpendicular components.Acceleration can be found by dividing the net force by mass.Trigonometry can be used to solve for the magnitude of the force components.Net force must be found before acceleration can be found.Read more on acceleration here: brainly.com/question/14344386
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Complete Question:
A sled is moving down a steep hill. The mass of the sled is 50 kg and the net force acting on it is 20 n. What must be done to find the acceleration of the sled? check all that apply.
A. The force of gravity must be broken into its parallel and perpendicular components.
B. Acceleration can be found by dividing the net force by mass.
C. Acceleration can be found by multiplying the net force by mass.
D. The magnitude of the force components must be multiplied by gravity.
E. Trigonometry can be used to solve for the magnitude of the force components.
F. Net force must be found before acceleration can be found.
Answer:
To find the acceleration, you do 20N/50kg = 0.4 m/s^2
Explanation:
example: a = F/m = 10/2 = 5 m/s2
a 2kg aluminum block and a6kg copper block are connected by alight string over a frictionless pulley and fixed steel block of angle 30 degree .if the coefficient of friction on the surface is 0.2,find the acceleration of the two block and the tension in the string?
(a) The acceleration of the two block is determined as 0.93 m/s².
(b) The tension in the string is 5.78 N.
Net force on the aluminum block
The net force on the aluminum block is calculated as follows;
[tex]T - \mu_a m_a g = m_a a \ --- \ (1)[/tex]
Net force on the copper blockThe net force on the copper block is calculated as follows;
[tex]m_cg sin(30) - T -\mu _cm_c gcos(30) = m_c a --- (2)[/tex]
where;
T is tension in the stringma is mass of aluminummc is mass of copperg is acceleration due to gravitySolve for T using (1) and (2)
[tex]a = g(\frac{m_c sin30\ - \ \mu _c m_c cos30 \ - \mu_ a m_a}{ma_a + m_c} )\\\\a = 9.8(\frac{6 sin30\ - \ 0.2 (6) cos30 \ - 0.2 (6)}{2 +6} )\\\\a = 0.93 \ m/s^2[/tex]
Tension in the stringFrom equation (1);
T = μm_ag + m_aa
T = 0.2(2)(9.8) + 2(0.93)
T = 5.78 N
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What quantity is represented by a unit called a newton (N)?
A. Inertia
B. Acceleration
C. Force
D. Mass
[tex]\textsf {C. Force}[/tex]
[tex]\textsf {The quantity which is represented by a unit called Newton (N)}\\\textsf {is Force.}[/tex]
Please I need help fast. Assignment is due soon.
1) A ball is thrown from the top of a building with an initial speed of 8.0 m/s at an angle of 35° above the horizontal. The building is 18 m tall.
a) How long is the ball in the air?
b) How far from the building does the ball land?
c) What is its impact speed?
2) A 65-kg person driving a car hits the gas, accelerating the car at a rate of 3.9 m/s^2. Find the magnitude and direction of the force exerted by the seat on the person's body. Remember to include both the horizontal and the vertical components.
(a) The time spent in the air by the ball is 1.5 seconds.
(b) The horizontal distance of the ball is 9.83 m.
(c) The impact speed of the ball is 20.37 m/s.
(2) The magnitude of the force is 253.5 N in horizontal direction.
Time of motion of the ball
The time of motion of the ball is calculated as follows;
h = vt + ¹/₂gt²
18 = 8sin(35)t + (0.5)(9.8)t²
18 = 4.589t + 4.9t²
4.9t² + 4.589t - 18 = 0
solve the quadratic equation using formula method,
t = 1.5 s
Horizontal distance of the projectileX = vcosθ(t)
X = 8cos(35) x 1.5
X = 9.83 m
Impact speed of the projectilevyf = vyi + gt
vyf = 8sin(35) + 9.8(1.5)
vyf = 19.289 m/s
vxf = vxi
vxf = 8 x cos(35)
vxf = 6.55 m/s
Resultant speed = √(19.289² + 6.55²) = 20.37 m/s
Magnitude of the force exerted by the seat on the personF = ma
F = 65 x 3.9
F = 253.5 N
Since the motion is horizontal, the angle of the motion is zero.
Fx = 253.5cos(0) = 253.5 N
Fy = 253.5sin(0) = 0 N
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Using what you learned above, try to predict the missing entries in the table below. Use the simulation to check your answer and type in the correct answer.
there is no pic or whatever to show
A 545 N sled is pulled a distance of 385 m. The task is done by pulling on a rope with a force of 1225 N at an angle of 19° with the horizontal. How much work is done in pulling the sled? What is the acceleration in the x direction, assuming that friction is negligible? Assuming the sled started at rest, how long does it take to pull the sled 185 m?
The work done in pulling the sled is 445,930.2 Joules.
The acceleration in the x direction, assuming that friction is negligible, is 20.85 m/s².
Time taken to pull the sled 185 m is 4.21s.
What is work done?Work done is equal to product of force applied and distance moved.
Given is a 545 N sled is pulled a distance of 385 m. The task is done by pulling on a rope with a force of 1225 N at an angle of 19° with the horizontal.
Work = Force x Distance x cos(angle)
W= 1225 x 385 x cos 19°
W = 445,930.2 Joules
Thus, the work done in pulling the sled is 445,930.2 Joules
From the Newton's second law of motion, we have'
F = ma
acceleration, a = 1225cos19° / ( 545 /9.81)
a = 20.85 m/s²
Thus, the acceleration in the x direction is 20.85 m/s²
Using the second equation of motion, we get
s = ut+ 1/2 at²
Substitute the values, we have
185 m = 0 +1/2 x 20.85 x t²
t = 4.21 s
Thus, the time taken to pull the sled 185 is 4.21s.
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Lucia wants to change the motion map shown so that it’s shows uniform circular motion. What change should Lucia make ?
Answer:
The last one - each vector pointing towards the center of the circle must be the same length for uniform circular motion
A snowboarder on a slope starts from rest and reaches a speed of 4.2 m/s after 7.3 s
The car's acceleration will be 0.575 m/s².The unit of acceleration is m/sec².
What is acceleration?The rate of velocity change concerning time is known as acceleration.
Given data;
Initial velocity, u= 0 m/s
Final velocity, v= 4.2 m/s
Time elapsed, t = 7.3 seconds.
To find ;
Acceleration, a
The acceleration when the change in velocity is observed by the formula as:
a= (v-u)/(t)
Substitute the given values:
a= (4.2-0)/(7.3)
a=(4.2)/(7.3)
a= 0.575 m/s²
Hence, the car's acceleration will be 0.575 m/s².
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A uniform plate of height 1.870 m is cut in the form of a parabolic section.
The lower boundary of the plate is defined by: y = 0.500[tex]x^{2}[/tex]. Find the distance from the rounded tip of the plate to the center of mass.
The distance from the rounded tip of the plate to the center of mass is 1.87 m.
What is center of mass?The center of mass is a point inside or outside the mass where all of the mass is concentrated.
The y-coordinate of the centroid is given by the ratio of two definite integrals;
Yc = ∫ydm/∫dm,
where dm is a density function
For the uniform plate, δ does not change with position in the plate.
Yc = ∫yδdA/∫δdA
Yc = ∫ydA/∫dA.
dA is a horizontal slice of the plate with dimensions xdy.
Solving the parabola for x,
y = 0.5x²
x = ± √(y/0.50), where the negative value corresponds to the left half of the parabola and the positive to the right half.
dA = (√(y/0.50)
= √(y/0.50))dy
= 2(√(y/0.50))dy
The limits of integration are from zero to 1.870, the top of the plate.
∫ydA = ∫2y√(y/0.50)dy = 7.232 m³
∫dA = ∫2√(y/0.50)dy = 3.868 m²
∫ydA/∫dA = 7.232 m³/3.868 m²
∫ydA/∫dA = 1.869700 m
∫ydA/∫dA = 1.87 m
Thus, the distance from the rounded tip of the plate to the center of mass is 1.87 m.
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Need help with this question
Answer:
1. 92, 138, 92
2. 7, 7, 7
3. 17, 18, 17
4. 29, 34, 29
Explanation:
the number on the bottom is always going to be the atomic number, or the number of protons.
the number on the top is going to be the atomic mass number, which is the sum of the number of protons and neutrons.
there is always going to be the same number of electrons as protons, because that was the overall charge of each atom is 0.
for example, for the first one:
there are 92 protons
there are 230 - 92 = 138 neutrons
there are 92 electrons
