Given:
We have changed our speed limit signs to metric.
To find:
(a) 30 mi/h in km/h
(b) 75 mi/h in km/h
Explanation:
We know,
[tex]1\text{ mi=1.61 km}[/tex](a)
So,
[tex]\begin{gathered} 30\text{ mi/h} \\ =30\times1.61\text{ km/h} \\ =48.3\text{ km/h} \end{gathered}[/tex]Hence,
[tex]30\text{ m/h=48.3 km/h}[/tex]b)
Now,
[tex]\begin{gathered} 75\text{ mi/h} \\ =75\times1.61\text{ km/h} \\ =120.75\text{ km/h} \end{gathered}[/tex]Hence,
[tex]75\text{ mi/h=120.75 km/h}[/tex]The speed of a moving bullet can be determined by allowing the bullet to pass throughtwo rotating paper disks mounted a distance75 cm apart on the same axle. From theangular displacement 33.9◦of the two bullet holes in the disks and the rotational speed1209 rev/min of the disks, we can determinethe speed of the bullet.33.9◦v1209 rev/min75 cmconroy (klc4842) – Homework 7, rotation 21-22 – dowd – (TorresLPHY1 3) 2What is the speed of the b
Answer:
160.49 m /s
Explanation:
We know that the angular speed of the disks is 1209 rev/min.
[tex]\omega=1209\text{rev}/\min [/tex]Let us convert this into degrees / s.
Now
1 rev = 360 degreees and 1 min = 60 s; therefore,
[tex]\omega=\frac{1209\cdot360^o}{60s}[/tex][tex]\boxed{\omega=7254^o/s\text{.}}[/tex]Now, we know that the angular separation of the holes in the two disks is 33.9 degrees. How long did it take the paper disk to rotate by this amount? The answer is
[tex]33.9=(7254^o/s)t[/tex]where t is the time needed.
Solving for t gives
[tex]t=\frac{33.9}{7254}s[/tex][tex]\boxed{t=4.67\times10^{-3}s}[/tex]This means 4.67 * 10^-3 seconds passed before the bullet hit the second disk after hitting the first.
Now, we know that in those 4.67 * 10^-3 seconds, the bullet certainly travelled 75 cm = 0.75 m; therefore, its speed must be given by
[tex]\begin{gathered} d=vt \\ \Rightarrow0.75=v(4.67\cdot10^{-3}s) \end{gathered}[/tex]dividing both sides by 4.67 * 10^-3 s gives
[tex]v=\frac{0.75m}{4.67\cdot10^{-3}s}[/tex][tex]v=160.49m/s[/tex]rounded to the nearest hundredth.
Hence, the velocity of the bullet (rounded to the nearest hundredth) is 160.49 m/s.
Which factor affects the angle of sunlight on Earth? The distance between Earth and the sun Earth's tilt from its axis The path of Earth's orbit Earth's speed of rotation
2 If we want to test the strength of electrical forces between a pair of charged objects, which of the following questions should we ask? A B How much mass does each object have? D Are the objects made up of the same material? C Does each object sink or float in water? What happens when the objects are moved further apart?
Electric force is the attracting or repulsive interaction between any two charged things. Similar to any force, Newton's laws of motion describe how it affects the target body and how it does so. One of the many forces that affect objects is the electric force.
The electric force F, or Coulomb force, exerted per unit positive electric charge q at that place can be used to determine an electric field's strength, or E = F/q.
The amount of charge on the source charge (Q) and the distances from the source charge (d) both affect how strong the electric field is.
To test the strength of electrical forces we should ask the following questions:
What is the mass of each object?Is the substance the objects are made of the same?The volt per meter (V/m or Vm-1) is the common unit. A potential difference of 1 V between sites separated by 1 meter is represented by a field strength of 1 V/m. Electric field intensity is another name for electric field strength. An electric field is created by any electrically charged item.
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is h20 an element, atom, compound, or molecule
A 7.00 ✕ 105 kg subway train is brought to a stop from a speed of 0.500 m/s in 0.800 m by a large spring bumper at the end of its track. What is the force constant k of the spring? N/m
Given:
The mass of the subway train is: m = 7 × 10⁵ kg
The speed of the train is: v = 0.5 m/s
The train stops by covering 0.8 m of distance by the spring bumper. Thus, x = 0.8 m
To find:
The force constant of the spring.
Explanation:
The moving train is stopped by the spring bumper this means that the kinetic energy of the moving is stored as the elastic potential energy of the spring. Thus,
Kinetic Energy of Train = Elastic Potential Energy of Spring
The kinetic energy of the train is given as:
[tex]KE=\frac{1}{2}mv^2[/tex]The elastic potential energy of the train is given as:
[tex]PE=\frac{1}{2}kx^2[/tex]Here, k is the spring constant or force constant.
Equating KE and PE, we get:
[tex]\begin{gathered} \frac{1}{2}mv^2=\frac{1}{2}kx^2 \\ \\ mv^2=kx^2 \\ \\ k=\frac{mv^2}{x^2} \end{gathered}[/tex]Substituting the values in the above equation, we get:
[tex]\begin{gathered} k=\frac{7\times10^5\text{ kg}\times(0.5\text{ m/s\rparen}^2}{(0.8\text{ m\rparen}^2} \\ \\ k=\frac{7\times10^5\text{ kg}\times0.25\text{ m}^2\text{/s}^2}{0.64\text{ m}^2} \\ \\ k=273437.5\text{ N/m} \end{gathered}[/tex]Final answer:
Thus, the force constant of the spring is 273437.5 N/m.
31. When water boils (or vaporizes)...Select one:a. water molecules become more free to move around.b. water molecules become less free to move around.c. water molecules are not affected.d. water molecules stay the same, but their temperature changes.
a. water molecules become more free to move around.
gases have a lower density and the molecules are free to move more and faster
In the absence of air resistance, a projectile launched at an angle of 33° above the horizontal will have the same range as a projectile launched at which of the following angles?a.)57°b.)38° c.)45°
ANSWER
[tex]a)\text{ }57\degree[/tex]EXPLANATION
We want to find the angle at which the range for the projectiles will be the same.
The range for a projectile is given by:
[tex]R=\frac{u^2\sin2\theta}{g}[/tex]where u = initial velocity
θ = angle of the projectile
g = acceleration due to gravity
For a projectile to have the same range as one with an angle of 33° (given that other values are constant), the value of sin(2θ) must be equal for both projectiles.
Let us find the value of sin(2(33°)):
[tex]\sin(2*33)=\sin66=0.9135[/tex]Let us find the same for the angles in the options:
[tex]\begin{gathered} \sin(2*57)=\sin114=0.9135 \\ \\ \sin(2*38)=\sin76=0.9703 \\ \\ \sin(2*45)=\sin90=1 \end{gathered}[/tex]As we can see, the angle that will result in the same range as 33° is 57°.
The answer is option a.
A circuit consists of one resistor connected to a battery. How will adding another resistor in series to the first resistor change the current of the circuit? Briefly justify your answer. (2 Points)
When two resistors are connected in series, their action is equivalent to a single resistor with an equivalent resistance which is the sum of the individual resistances:
[tex]R_{eq}=R_1+R_2[/tex]This means that the equivalent resistance must always be greater than any of the resistances that the series association is made of.
Since the current I, the resistance R and the voltage V are related through the following equation:
[tex]I=\frac{V}{R}[/tex]If we increase the value of R, then the value of I will decrease.
Therefore, adding another resistor in series will decrease the current through the circuit.
Find the answer to the problem
Answer:
1,51 m/s²
Explanation:
Since both forces F₁ and F₂ act in the same horizontal direction(East) with no vector components in any other direction the net force acting on the object is
[tex]F_{NET} = F_1 + F_2 = 10 + 6 = 16N[/tex] acting due east
The mass of the object is 10.6kg
The equation relating Force, mass and acceleration is
[tex]F = m \cdot a[/tex]
where
F = force in Newtons
m = mass in kg
a = acceleration in m/s²
Plugging in known values into the equation for Force gives us:
16 N = 10.6 kg x a m/s²
Therefore a = 16N/10.6 = 1,51 m/s²
GUYS THIS IS URGENTTTTTT CAN YOU PLS ANSWER???
It takes a machine 90 seconds to do 54,000 J of work moving a conveyor belt. How much power is used to move the conveyor belt?
600 watts
540 watts
1.6 x 10-3 watts
4.8 x 106 watts
Answer:
600 watts is used to move the conveyor belt
Kate is at the playground and goes down a slide
that makes an angle with the horizontal of 35°. If
the coefficient of kinetic friction between Kate and
the slide is 0.25 and Kate’s mass is 41 kg, what is
the kinetic friction force on her?
The Kinetic friction force on her 84.05 N.
Kinetic friction is defined as a force that acts between shifting surfaces. A body moving at the surface reports pressure inside the contrary path of its motion. The significance of the pressure will rely on the coefficient of kinetic friction between the 2 materials.
In static friction, the frictional force resists pressure this is applied to an item, and the object stays at relaxation until the force of static friction is conquered. In kinetic friction, the frictional force resists the motion of an item.
Kinetic friction is produced whilst brakes are applied to tires, when an item like a field slides throughout the floor, or whilst sandpaper is rubbed throughout a surface.
calculation:-
μ = 0.25
mass = 41 kg
θ = 35°
N = 410 cos 35° N
= 410 × 0.82 N
= 336.2 N
F = μkN
= 0.25 × 336.2 N
Kinetic friction force = 84.05 N
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please solve exponential function10^2a+3=10^a-4
Given the exponential function:
[tex]10^{2a+3}=10^{a-4}[/tex]Let's evaluate the given exponential function.
Since the base of both sides are the same, the expressions are equal if the exponents are equal.
Remove the bases and equate the exponents:
[tex]2a+3=a-4[/tex]Subtract 3 from both sides:
[tex]\begin{gathered} 2a+3-3=a-4-3 \\ \\ 2a=a-7 \end{gathered}[/tex]Subtract a from both sides:
[tex]\begin{gathered} 2a-a=a-a-7 \\ \\ a=-7 \end{gathered}[/tex]ANSWER:
a = -7
A vector starts at point (-3, 4) and ends at point (6,-3). What is the magnitude of the vector? AnswerTo two decimal places
Given data
*A vector starts with a point is P = (-3, 4)
*A vector ends with a point is Q = (6, -3)
The formula for the magnitude of the vector PQ is given as
[tex]\lvert\vec{PQ}\rvert=\sqrt[]{(x_2-x_1)^2+(y_2-y_1)^2}[/tex]Substitute the known values in the above expression as
[tex]\begin{gathered} \lvert\vec{PQ}\rvert=\sqrt[]{(6-(-3))^2+(-3-4)^2} \\ =\sqrt[]{(9)^2+(-7)^2} \\ =\sqrt[]{81+49} \\ =\sqrt[]{130} \\ =11.40 \end{gathered}[/tex]Which of the following should aspiring technicians expect to accomplish before obtaining a license?
Aspire Licensing and Administrative Services offers a one-stop shop for all licensing services applied under the License Act 2003.
What do we accomplish before obtaining a license?There are two types of licenses for Lexia Aspire i.e. participant licenses and leader licenses. Each license is valid for one year. Like traditional teacher licensure programs, Aspire puts teaching careers within reach through high quality. No other endorsements can be added to the teaching license. The teacher's license is renewable. Instructional Leadership License Aspiring This 1- or 2-year certificate allows you to teach while you complete a certification program. This license can only be issued once you are hired to teach the training. Professional licensure protects people by enforcing standards that restrict practice to qualified persons who have met specific qualifications in education, work, and experience.
So we can conclude that for gaining a license, Aspire Licensing and Administrative Services are the best options.
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20) At some point along the direct path from the center of the Earth to the center of the Moon, the gravitational force of attraction on a spacecraft from the Moon becomes greater than the force from the Earth. (a) How far from the center of the Earth does this occur? (b) At this location, how far is the spacecraft from the surface of the Moon? How far is it from the surface of the Earth?
At some point along the direct path from the center of the Earth to the center of the Moon, the gravitational force of attraction on a spacecraft from the Moon becomes greater than the force from the Earth.
(a) How far from the center of the Earth does this occur?
The distance from the center of the Earth to the point at which the gravitational force of attraction from the Moon is greater than the force from the Earth is given by
[tex]\frac{GM}{x^2}=\frac{Gm}{(L-x)^2}[/tex]Where M is the mass of the Earth.
m is the mass of the Moon.
G is the gravitational constant.
L is the distance between the Earth and the Moon.
x is the distance that we need to find out.
Let us simplify the above equation,
[tex]\begin{gathered} \frac{GM}{x^2}=\frac{Gm}{(L-x)^2} \\ \frac{M}{x^2}=\frac{m}{(L-x)^2} \\ \frac{M}{m}=\frac{x^2}{(L-x)^2} \\ \frac{M}{m}=(\frac{x}{L-x^{}})^2 \end{gathered}[/tex]The ratio of Earth's mass to the Moon's mass is (M/m = 81)
[tex]\begin{gathered} 81=(\frac{x}{L-x^{}})^2 \\ \sqrt{81}=\sqrt{(\frac{x}{L-x^{}})^2} \\ 9=\frac{x}{L-x} \\ 9L-9x=x \\ 10x=9L \\ x=\frac{9L}{10} \end{gathered}[/tex]The distance between the Earth and the Moon is L = 384400 km
[tex]\begin{gathered} x=\frac{9\cdot384400}{10} \\ x=345960\; km \end{gathered}[/tex]Therefore, the gravitational force of attraction on a spacecraft from the Moon becomes greater than the force from the Earth at a distance of 345960 km from the center of the Earth does this occur.
(b) At this location, how far is the spacecraft from the surface of the Moon?
The distance between the spacecraft from the surface of the Moon is given by
[tex]d_{moon}=L-x-r[/tex]Where r is the radius of the Moon that is 1740 km
Substituting all the known values into the above equation, we get
[tex]\begin{gathered} d_{moon}=L-x-r \\ d_{moon}=384400-345960-1740 \\ d_{moon}=36700\; km \end{gathered}[/tex]Therefore, the distance between the spacecraft from the surface of the Moon is 36700 km
How far is it from the surface of the Earth?
The distance between the spacecraft from the surface of the Earth is given by
[tex]d_{earth}=x-R[/tex]Where R is the radius of the Earth that is 6370 km
[tex]\begin{gathered} d_{earth}=x-R \\ d_{earth}=345960-6370 \\ d_{earth}=339590\; km \end{gathered}[/tex]Therefore, the distance between the spacecraft from the surface of the Earth is 339590 km
How does the USDA define a concentrated animal feeding operation (CAFO)?
A. A CAFO is defined as an AFO with over 1,000 animals of any size kept for over 45 days a year.
B. A CAFO is defined as an AFO with over 1,000 animal units kept for the majority of a year.
C. A CAFO is defined as an AFO with over 1,000 animals of any size kept for the majority of a year.
D. A CAFO is defined as an AFO with over 1,000 animal units kept for over 45 days a year.
please help!!!
A concentrated animal feeding operation (CAFO) is defined as an AFO with over 1,000 animal units kept for over 45 days a year. Therefore, option (D) is correct.
What is a concentrated animal feeding operation?A concentrated animal feeding operation (CAFO) can be defined by the USDA as an intensive animal feeding operation (AFO) in which over 1,000 animal units are confined for over 45 days a year in animal husbandry.
The Environmental Protection Agency (EPA) has focused on regulating CAFOs because they produce millions of tons of manure every year. When improperly managed can pose risks to the environment and public health.
CAFO operators have developed agricultural wastewater treatment plans in order to manage their waste. The most common facility utilized in these plans, the anaerobic lagoon, has contributed to environmental and health problems attributed to the CAFO.
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12) A bus makes three displacements in the following order:
1) 38 mi, 48° east of north
41.42² +67.44²/
2) 79 mi, 46 west of north; and
3) 55 mi, south-east
a. Find the X and Y components of displacement D₁.
b. Find the X and Y components of displacement D2.
Find the X and Y components of displacement D3.
d. Find the magnitude and direction of the resultant vector.
(a) The X and Y components of vector D₁ are 28.24i and 25.43j respectively.
(b) The X and Y components of D₂ are - 56.83i and 57.88j respectively.
(c) The X and Y components of D₃ are 38.89i and -38.89j respectively.
(d) The magnitude and the direction of the resultant vector are 45.6 miles and 13.1° east of north respectively.
a) The X and Y parts of D₁ are as follows:
The X component of D₁ is:
D₁ₓ = 38 mi × sin(48°) = 38 mi × cos(90° - 48°) ≈ 28.24 mi
In vectors, [tex]\overrightarrow{D_{1x}}}[/tex] ≈ 28.24·i
The Y component is:
[tex]D_y[/tex] = 38 mi × cos(48°) = 38 mi × sin(90° - 48°) ≈ 25.43 mi
In vectors, [tex]\overrightarrow{D_{1y}}}[/tex] ≈ 25.43·j
b)The following is a list of the X and Y components of D₂:
D₂ₓ = 79 mi × sin(46°) = 79 mi × cos(90° - 46°) ≈ 56.83 mi
[tex]\overrightarrow{D_{2x}}}[/tex] ≈ -56.83·i
The Y component is:
[tex]D_{2y}[/tex] = 79 mi × cos(46°) = 79 mi × sin(90° - 48°) ≈ 57.88 mi
[tex]\overrightarrow{D_{2y}}}[/tex] ≈ 57.88·j
c) Given that D₃ points southwest, which is south 45° west, the X and Y components of D₃ are as follows:
D₃ₓ = 55 mi × sin(45°) = 55 mi × cos(90° - 45°) ≈ 38.89 mi
[tex]\overrightarrow{D_{3x}}}[/tex] ≈ 38.89·i
The Y component will be:
[tex]D_{3y}[/tex] = 55 mi × cos(45°) = 55 mi × sin(90° - 45°) ≈ 38.89 mi
[tex]\overrightarrow{D_{3y}}[/tex] ≈ -38.89·j
d) So, the resultant vector will be:
(28.24-56.83+38.89)·i + (25.43+57.88-38.89)·j = 10.3·i + 44.42·j
The resultant vector, [tex]\overrightarrow{D}}[/tex] ≈ 10.3·i + 44.42·j
The magnitude of the resultant vector,[tex]|\overrightarrow{D}|=\sqrt{10.3^2 + 44.42^2}[/tex] = 45.6
The resultant vector is approximately 45.6 miles
So, the angle θ is:
θ = tan⁻¹( 44.42 / 10.3 ) = 76.9°
The direction of the resultant is 76.9° north of east which is (90° - 76.9°) ≈ 13.1° east of north
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What amount of force is required to accelerate a 1,264 kg car at a rate of 13 m/s²?
(Round you answer to the nearest whole number. Do not use units or decimals in your answer.)
16432 N force is required to accelerate a 1,264 kg car at a rate of 13 m/s²
What is force?Force is an external agent that may change the condition of rest or motion of a body. It has a magnitude as well as a direction. The direction of the force is the place where force is applied, and the application of force is the location where force is applied. Newton (N) is the SI unit of force.
Given that,
Mass of the car (m) = 1264 kg
Acceleration (a) = 13 m/s²
Now calculate how much force (F) is required,
F = m × a
F = 1264 × 13 kg m/s²
F = 16432 N
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George is trying to move a heavy bookshelf across the classroom. He pushes on the bookshelf with a force of 57 N to the right Friction is resisting his movement with a force of78 N. The bookshelf does not move. A friend comes to help George, The friend pushes the bookshelf in the same direction as George with a force of 25 N. What will happen tothe bookshelf?The bookshelf will not move,0000The bookshelf will move to the left with a force of 25 NThe bookshelf will move to the right with a force of 82 NThe bookshelf will move to the right with a force of 4 N.
Given data:
* The force applied on the bookshelf towards the direction of motion by George is,
[tex]F_1=57\text{ N}[/tex]* The force applied on the bookshelf towards the direction of motion by the friend of George is,
[tex]F_2=25\text{ N}[/tex]* The force of friction opposite to the direction of motion is,
[tex]F=78\text{ N}[/tex]Solution:
Thus, the net force acting on the bookshelf is,
[tex]F_{net}=F_1+F_2-F[/tex]Here, the negative sign before the friction force indicates the direction of friction force is oppoiste to the direction of motion,
Substituting the known values,
[tex]\begin{gathered} F_{net}=57+25-78 \\ F_{net}=4\text{ N} \end{gathered}[/tex]Here, the positive value of net force indicates the net force is acting in the direction of motion of the bookshelf.
Thus, the bookshelf will move to the right with a force of 4 N.
Hence, 4th option is the correct answer.
Is more mass weaker gravity or stronger gravity
According to the law of universal gravitation:
gravitational force is directly proportional to the mass of the object as shown in the equation below
[tex]F=\frac{Gm_1m_2}{r^2}[/tex]Since the force of gravity varies directly with the mass, as the mass incr
A 0.6 kg object moves at a constant velocity of 2 m/s, traveling a total distance of 30 m. Using the work-energy theorem, what is the net work on the object? (1 point)
A. 2.4 J
B. 0 J
C. 1.2 J
D. 36 J
The net work on an object with a mass of 0.6 kg when moving with a constant velocity through a total distance of 30 m is 0 J
What is work?Work can be defined as the product of force and distance.
To calculate the net work on the obeject, we use the formula below.
Formula:
W = (ma)d......... Equation 1Where:
W = net work on the objectm = Mass of the objecta = Acceleration of the objectH = Distance traveled by the objectFrom the question,
Given:
m = 0.6 kga = 0 m/s² ( Constant velocity)d = 30 mSubstitute the values into equation 1
W = 0.6×0×30W = 0 JHence, the right option is B. 0 J.
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Question 2 of 10Voltage:O A. makes the current flow in a circuit.O B. slows down the electrons in a circuit.aC. opposes energy flow in the circuit.O D. is measured in ohms.SIRA
When voltage is supplied to a circuit, potential difference is generated in the circuit.
The potential difference causes the current to flow in the circuit.
Thus, the voltage makes the current flow in a circuit.
A pendulum is constructed by attaching a small metal ball to one end of a string of length =1.20 m that hangs from the ceiling, as shown in the figure. The ball is released when it is raised high enough for the string to make an angle of =20.0∘ with the vertical.
With what speed is the ball moving at the bottom of its swing?
The speed of the ball when released from the top will be 2.97m/s at the bottom of the swing.
The length L of the string of ball is 1.2 meter.
The angle made by the string while going up is 20°.
We know,
According to the law of conservation of energy,
Total energy remains constant at every point of the motion, because energy can neither be created nor be destroyed,
So,
Total energy at the top point = Total energy at the bottom point
MgH = 1/2Mv²
Where,
M is the mass of the ball,
g is the acceleration due to gravity,
v is the speed at the bottom point,
H is the height till it is raised.
We can find H,
Total length minus vertical component of the length,
H = L - LcosA
H = 1.2(1-cos20°)
H = 1.2(1-0.93)
H = 1.2(0.37)
H = 0.444 meters.
Now, putting all the values, to find the value of v,
MgH = 1/2Mv²
gH = 1/2v²
v = √2gH
v = √2x10x0.444
v = 2.97m/s
Speed of the ball at the bottom is 2.9m/s.
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diameter is 0.9 mm what is area in meter square
Given,
diameter of the circle, d=0.9 mm=0.9×10⁻³ m
The radius is given by,
[tex]r=\frac{d}{2}[/tex]Therefore the radius of the circle is,
[tex]r=\frac{0.9\times10^{-3}^{}}{2}=0.45\times10^{-3}\text{ m}[/tex]The area of the circle is given by,
[tex]A=\pi r^2[/tex]On substituting the know values,
[tex]A=3.14\times(0.45\times10^{-3})^2=6.36\times10^{-7}m^2[/tex]Therefore the area of the circle is 6.36×10⁻⁷ m ²
An electromagnetic wave has a wavelength of 6.11 ✕ 10−5 m. What is the frequency of the wave (in Hz)? answer in:___ Hz In what region of the electromagnetic spectrum is this radiation? a.)Infrared b.)Visible c.)Ultraviolet d.)X-rays
We are given the wavelength of an electromagnetic wave which is 6.11×10-5 m.
We are asked to determine its frequency.
Recall that the relationship between wavelength and frequency is given by
[tex]f=\frac{v}{\lambda}[/tex]Where λ is the wavelength, f is the frequency, and v is the speed of light.
The speed of light is 3.00×10^8 m/s
[tex]f=\frac{3.00\times10^8}{6.11\times10^{-5}}=4.91\times10^{12}\;Hz[/tex]Therefore, the frequency of the wave is 4.91×10^12 Hz.
We are also asked in which region of the electromagnetic spectrum is this radiation.
The above frequency range (4.91×10^12 Hz) is in the infrared region of the electromagnetic spectrum.
Therefore, option a.) Infrared is the correct answer.
Question #5: On the moon, what would be the force of gravity acting on an object that has a mass of 7 kg?
Explanation
the acceleration due to gravity formula is given by:
[tex]\begin{gathered} g=\frac{GM}{R^2}^{} \\ \text{where} \\ G\text{ is the universal gravitational constatn G=6.674 }\cdot10^{-11}\frac{m^3}{\operatorname{kg}\cdot s^2} \\ R\text{ is the radius of the massive body ( in meters)} \\ g\text{ is the acceleration due to gravity} \\ M\text{ is the mass of the massive body ( kg)} \end{gathered}[/tex]Step 1
Find g(acceleration due to gravity)
Let
[tex]\begin{gathered} M=\text{ mass of the moon=7.35}\cdot10^{22}\operatorname{kg} \\ G=\text{=6.674 }\cdot10^{-11}\frac{m^3}{\operatorname{kg}\cdot s^2} \\ R=\text{radius of the moon=}1.74\cdot10^6m \end{gathered}[/tex]now, replace in the formula
[tex]\begin{gathered} g=\frac{GM}{R^2}^{} \\ g=\frac{\text{6.674 }\cdot10^{-11}\frac{m^3}{\operatorname{kg}\cdot s^2}\cdot\text{7.35}\cdot10^{22}\operatorname{kg}}{(1.74\cdot10^6m)^2} \\ g=\frac{4.90539\cdot10^{12}}{(3.0276\cdot10^{12})} \\ g=1.62\frac{m}{s^2} \end{gathered}[/tex]Step 2
to find the force , use the formula
[tex]\begin{gathered} F=\text{mg} \\ \text{where m is the mass of the object} \\ g\text{ is the acceleration due to gravity} \end{gathered}[/tex]replace
[tex]\begin{gathered} F=7\operatorname{kg}\cdot1.625\text{ }\frac{m}{s^2} \\ F=11.375\text{ Newtons} \end{gathered}[/tex]therefore, the force would be
11.375 N
A ceramic container used for melting metals (called a crucible) contains 1.30 kg of a molten metal. The liquid metal cools until it reaches its melting point of 1,749°C. A scientist then measures that 2.99 ✕ 10^4 J of heat is transferred out of the metal before it completely solidifies. What is the latent heat of fusion of this metal, in J/g? (Be careful with units!)
Answer:
23 J/g
Explanation:
The heat transferred in a phase change like this case is calculated as:
[tex]Q=mH_f[/tex]Where m is the mass and Hf is the latent heat of fusion. We can solve the equation for Hf
[tex]H_f=\frac{Q}{m}[/tex]Now, replacing Q = 2.99 x 10⁴ J and the mass = 1300 g, we get:
[tex]H_f=\frac{2.99\times10^4J}{1300\text{ g}}=23\text{ J/g}[/tex]We use 1300 g instead of 1.30 kg because we want the units in J/g.
Therefore, the latent hear of fusion of this metal is 23 J/g
Annular and penumbral refer, respectively, to which phenomenon?
O A.
OB.
O C.
O D.
spring and autumn tides
solstice and equinoxes
solar and lunar eclipses
elliptical and round orbits
Annular and penumbra are terms that are associated with Solar and lunar eclipses. Option C is the correct answer
What is an Eclipse ?An eclipse is a result of a shadow cast by one heavenly body on another. There are eclipse of the sun or solar eclipse and eclipse of the moon or lunar eclipse.
Annular and penumbra actually refer, respectively, to the phenomenon known as solar and lunar eclipses.
Penumbra is a region of outer light or grey area of a shadow cast by an object or heavenly body.
While Annular can be formed under condition where the extreme rays at the moon's edge intersect before reaching the earth.
Therefore, Annular and penumbra refer to the solar and lunar eclipses respectively.
Learn more about Eclipse here: https://brainly.com/question/1077760
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A copper block of mass 0.75 g is removed from a furnace and quickly transferred into a glass beaker of mass 300.0 g containing 200.0 g of
water. The temperature of the water rises from 12 0 °C to 27 0 °C. What was the temperature of the furnace?
The furnace has a temperature of 585 degrees Celsius.
Temperature is an expression denoting hotness or coolness on any scale, including Fahrenheit and Celsius. According to temperature, heat energy will naturally move from a hotter (body with a higher temperature) to a colder (body with a lower temperature) (one at a lower temperature).
Agriculture, food production, and medical care—for both humans and animals—all rely largely on
A spring pushes a block with 120 N of force for 0.250 m. Solve for the energy given by the spring. (ans: 30 j)
Answer:
Energy = 30J
Explanation:
Energy = Force x Distance
Energy = 120 x 0.250m
Energy = 30J
