(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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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.
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
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
A lever is being used to try to lift a heavy object, but it is not working. Which of these actions would increase the output force the most?
The given problem can be exemplified in the following diagram:
Where:
[tex]\begin{gathered} F_i=\text{ input force} \\ d_i=\text{ input displacement} \\ d_0=\text{ output displacement} \\ F_o=\text{ output force} \end{gathered}[/tex]Now, we add the torques in the lever and we get:
[tex]\Sigma M=F_0d_0-F_id_i[/tex]If we consider the system in equilibrium then the sum of torque is equal to zero, therefore:
[tex]F_0d_0-F_id_i=0[/tex]Now we solve for the output force, first by adding "Fidi" to both sides:
[tex]F_0d_0=F_id_i[/tex]Now we divide both sides by the output distance:
[tex]F_0=\frac{F_id_i}{d_0}[/tex]Therefore, we get an expression for the output force. In this expression, we see that in order to increase the output force we need to decrease the output distance and increase the input force.
A team of horses is pulling a 910 kg wagon with a force of 4500 N. Assuming there are no other forces on the wagon, what is the wagon's acceleration?
What is the half life of an isotope that has a decay constant of 0.00567?
We are asked to determine the half-life of an isotope. To do that we will use the following formula:
[tex]T=\frac{ln(2)}{\lambda}[/tex]Where:
[tex]\begin{gathered} T=\text{ half-life} \\ \lambda=\text{ decay constant} \end{gathered}[/tex]Now, we plug in the values:
[tex]T=\frac{ln(2)}{0.00567}[/tex]Solving the operations:
[tex]T=122.25[/tex]Therefore, the half-life is 122.25
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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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.
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.
A car applies a force of 36.8 Newtons for a 668-meter loop; what was the work done? (1 point)
O 18.2 J
O 705 J
O OJ
O 24,600 J
Answer:OJ
Explanation: it ok you can trust me (-_-)
Answer:
(Question) What is the mathematical relationship between force and work?
(Answer) W=F×d
(Question) If W=F×d, which of the following equations shows work being calculated using the correct units?
(Answer) 113 J=(17.4 N)×(6.51 m)
(Question) A force of 57.1 Newtons is applied for 0.977 meters; what work was done on the object?
(Answer) 55.8 J
(Question) A car applies a force of 36.8 Newtons for a 668-meter loop; what was the work done?
(Answer) 0J
(Question) A 8.43-Newton force was applied for an unknown displacement; however, it is known that 376 joules of work was done. What is the value of the mystery displacement?
(Answer) 44.6 m
Explanation:
just did the quick check UwU
3. If you puth 350 ON on a 45kg box to move it in the same direction as the push, while using 4462.53 of work. How far did you move it?
Given,
The applied force, F=350.0 N
The mass of the box, m=45 kg
The work done by the applied force, W=4462.5 J
The work is given by the product of the applied force and the distance through which the force is applied.
Thus the work done by the applied force on the box is given by,
[tex]W=F\times d[/tex]Where d is the distance covered by the box.
On rearranging the above equation,
[tex]d=\frac{W}{F}[/tex]On substituting the known values,
[tex]\begin{gathered} d=\frac{4462.5}{350} \\ =12.75\text{ m} \end{gathered}[/tex]Thus the box move for a distance of 12.75 m
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
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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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
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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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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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
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
When an object is placed at a distance greater than twice the focal length of the lens, the real image produced is ___________.1) inverted and smaller than the object2) inverted and large than the object3) inverted and the same size as the object4) upright and the same size as the object
When an object is placed at a distance greater than twice the focal length of the lens, the real image produced by lens is inverted and smaller than the object, which means option (1) is correct.
An amusement park game requires you to knock down a wood post by throwing a ball at it. You can choose between a tennis ball and a sticky clay ball of equal mass.Assume you can trow either balls with the same speed and accuracy. Which one would you choose:a.It will be all the same.b.The tennis ball.c.The clay ball.
We know that the impulse is defined by the change in momentum, that is:
[tex]Impulse=\Delta\vec{p}[/tex]Now, we know that we can throw the balls with the same speed and accuracy which means that both balls have the same initial momentum. Now, since the tennis ball will bounce from the wood and the clay ball will stick on it this means that the final momentum of the tennis ball is greater; this means that the change in momentum of the tennis ball will be greater and hence the impulse it exerts on the wood is greater.
Therefore, you should choose the tennis ball.
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
A 20.0-N box rests on a 50.0-N box on a perfectly smooth horizontal floor. When a horizontal 15.0-N pull to the right is exerted on the lower box (see figure), both boxes move together what is the net force on the top block
The net force on the top block is 33.54 N.
What is net force?
Net force is the sum or resultant of two or more forces acting on a body.
Also, the net force is the force which is the sum of all the forces acting on an object simultaneously.
The sum of the upward forces is calculated as follows;
F(upward) = 50 N - 20 N = 30 N
The sum of the horizontal forces is calculated as follows;
F(horizontal) = 15 N
The net force or resultant force on the top block is calculated as follows;
F(net) = √Fy² + Fx²
where;
Fy is the sum of the vertical forcesFx is the sum of the horizontal forcesF(net) = √Fy² + Fx²
F(net) = √[(30)² + (15)²]
F(net) = 33.54 N
Thus, the net force or resultant force on the top block is determined as 33.54 N.
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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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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
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.
this is a 3 part question33) The human head can be considered as a 3.3-kg cranium protecting a 1.5-kg brain, with a small amount of cerebrospinal fluid that allows the brain to move a little bit inside the cranium. Suppose a cranium at rest is subjected to a force of 2800 N for 6.5 ms in the forward direction. (a) What is the finalspeed of the cranium? (b) The back of the cranium then collides with the back of the brain, which is still at rest, and the two move together. What is their final speed? (c) The cranium now hits an external object and suddenly comes to rest, but the brain continues to move forward. If the front of the brain interacts with the front of the cranium over a period of 15 m/s before coming to rest, what average force is exerted on the brain by the cranium?
Given data:
* The force applied is 2800 N.
* The time for which the force applied is 6.5 ms.
* The mass of the cranium is 3.3 kg.
Solution:
(a). The momentum of the cranium in the terms of the force applied is,
[tex]p=Ft[/tex]where F is the force applied, and t is the time,
Substituting the known values,
[tex]\begin{gathered} p=2800\times6.5\times10^{-3} \\ p=18200\times10^{-3} \\ p=18.2kgms^{-1} \end{gathered}[/tex]The velocity of the cranium from the momentum is,
[tex]\begin{gathered} mv=p \\ v=\frac{p}{m} \end{gathered}[/tex]where m is the mass of cranium including the brain inside it,
Substituting the known values,
[tex]\begin{gathered} v=\frac{18.2}{3.3+1.5} \\ v=3.79\text{ m/s} \end{gathered}[/tex]Thus, the final speed of the cranium is 3.79 m/s.
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
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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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]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
