Answer:
could u give a more longer understanding
What is the common temperature of the final equilibrium state of the system? A 10.0 kg block of ice at 0.0°C is placed into 10 kg water bath at 90.0C in an isolated container. The system eventually comes to equilibrium at a common temperature. Ignore the effect of the container.
Answer:
The equilibrium temperature is 45°C
Explanation:
Given;
mass of block of ice, m₁ = 10 kg
mass of water, m₂ = 10 kg
initial temperature of the ice, t₁ = 0.0°C
initial temperature of water, t₂ = 90.0°C
let the equilibrium temperature = T
Apply the principles of mixture;
Heat lost by the water = heat gained by the ice
m₂c(90 - T) = m₁c(T - 0)
m₂(90 - T) = m₁(T - 0)
10(90 - T) = 10(T)
90 - T = T
90 = 2T
T = 90 / 2
T = 45°C
Therefore, the equilibrium temperature is 45°C
The corona is to the sun as ___ to earth
A. The atmosphere is
B. The oceans are
C. The seasons are
D. The moon is
A magnet was cut in half to form the pieces shown. Which image best represents the original magnet?
IT'S OPTION A!!
Answer : The correct option is (A).
Explanation :
As we know that a magnet is an object which produces an invisible magnetic field and attract metals like iron, nickel, cobalt.
The magnets have two poles that is a north pole and a south pole.
If you take a magnet and cut a magnet in half then two pieces would separate and form new sets of poles. That means, when we cut a magnet of south-north pole then it gives two pieces of magnet like south-north and south-north pole.
From the given options we conclude that the option A is the representation of original image of magnet.
Hence, the correct option is (A).
Answer:
A
Explanation:
IMAGE
A boat with a weight of 547 newtons is floating in a harbor. What is the buoyant force on the boat?
A. 55.7 newtons
B. 5,371.5 newtons
C. 273.5 newtons
D. 547 newtons
Answer:
a
Explanation:
its correct
which statement below is FALSE? *
1 point
Gravity acts on any object with mass.
The closer objects are, the stronger the gravitational force between them.
The farther you are from the center of Earth, the less gravitational force you feel.
Gravitational force depends on what the object is made of.
Answer:
the farther you are from earth
Explanation:
It is this because when you get into space there is less gravity
(b) You start to move in the direction you found in part (a) at a speed of 3 cm/sec. How fast is the concentration changing
Complete Question
The concentration of salt in a fluid at (x,y,z) is given by F(x,y,z)=x^2+y^4+2x^2z^2 mg/cm3 You are at the point (1,1,1).
a. In which direction should you move if you want the concentration to increase the fastest?
I keep getting <5,2,8> for this answer and it says it is incorrect
You start to move in the direction you found in part (a) at a speed of 4 cm/sec. How fast is the concentration changing?
Answer:
a
[tex]\vec \Delta F (1 ,1 , 1) = < 6 , 4 , 4 >[/tex]
b
[tex] M = 2 \sqrt{ 17 }[/tex]
Explanation:
From the question we are told that
The equation is [tex]F(x,y,z)=x^2+y^4+2x^2z^2[/tex]
differentiating with respect to x
[tex]F_x (x, y, z) = 2x + 4xz^2[/tex]
differentiating with respect to y
[tex]F_y (x, y, z) = 4y^3[/tex]
differentiating with respect to z
[tex]F_z (x, y, z) = 4x^2z[/tex]
Gnerally the rate of change of the salt concentration is mathematically represented as
[tex]\vec \Delta F (x ,y , z) = <F_x , F_y , F_z >[/tex]
=> [tex]\vec \Delta F (x ,y , z) = <2x + 4xz^2 ,4y^3 , 4x^2z >[/tex]
At the point (1,1,1)
[tex]\vec \Delta F (1 ,1 , 1) = <2(1) + 4(1)(1)^2 ,4(1)^3 , 4(1)^2(1) >[/tex]
=> [tex]\vec \Delta F (1 ,1 , 1) = < 6 , 4 , 4 >[/tex]
generally rate at which the concentration is changing is mathematically represented as
[tex]M= \sqrt{ 6^2 +4^2 + 4^2}[/tex]
=> [tex]M= \sqrt{ 36 + 16 + 16}[/tex]
=> [tex]M= \sqrt{68}[/tex]
=> [tex]M = \sqrt{ 4 * 17 }[/tex]
=> [tex]M= 2 \sqrt{ 17 }[/tex]
A train travels 6.1m/s (E)The train’s destination is 134 km(E), how long until it arrives?
Answer:
The time taken is [tex]t = 6.102 \ hours [/tex]
Explanation:
From the question we are told that
The velocity of the train is [tex]v_t = 6.1 \ m/s[/tex]
The distance covered by the train is [tex]D = 134 \ km = 134000 \ m[/tex]
Generally the time taken is mathematically represented as
[tex]t = \frac{ 134000}{ 6.1 }[/tex]
=> [tex]t = \frac{ 134000}{ 6.1 }[/tex]
=> [tex]t = 21967.21 \ sec [/tex]
Converting to hours
[tex]t = 21967.21 = \frac{21967.21 }{3600} = 6.102 \ hours [/tex]
If your mass is 72 kg, your textbook's mass is 3.7 kg, and you and your
textbook are separated by a distance of 0.33 m, what is the gravitational
force between you and your textbook? Newton's law of gravitation is
Gm;m2 The gravitational constant G is 6.67 x 10-11 N·m2/kg2.
F gravity
O A. 4.94 x 10-7N
B. 2.45 x 103 N
O C. 5.38 x 10-8N
O D. 1.63 x 10-7N
Answer:
The answer is 1.63 x 10 -7 N
Explanation:
An appropriate solution is 1.63 x 10 -7 N
Gravitational pressure method: F=G{frac{m_1m_2}{r^2}}
Newton's law of gravitation is Gm;m2
The gravitational regular G is 6.67 x 10-11 N·m2/kg2.
What is gravitational pressure?
Definitions of gravitational pressure. The force of attraction among all masses inside the universe; especially the appeal of the earth's mass for bodies close to its floor.
What law is Newton's regulation of gravitation?Newton's regulation of typical gravitation states that two bodies in an area pull on every other with a force proportional to their hundreds and the space among them. For large objects orbiting each other—the moon and Earth, for instance—means they without a doubt exert important force on each other.
Learn more about gravitation here: https://brainly.com/question/72250
#SPJ2
PLEASE HELP GIVING 15 P
Answer:
can you make the pic more clear
i think it's D
In some cases, neither of the two equations in the system will contain a variable with a coefficient of 1, so we must take a further step to isolate it. Let's say we now have
3C+4D = 5
2C+5D = 2
None of these terms has a coefficient of 1. Instead, we'll pick the variable with the smallest coefficient and isolate it. Move the term with the lowest coefficient so that it's alone on one side of its equation, then divide by the coefficient. Which of the following expressions would result from that process?
a. C=53−43D
b. C=1−52D
c. D=25−25C
d. D=54−34C
Answer:
a) C = (5-4D)/3
Explanation:
Given the simultaneous equation
3C+4D = 5 .... 1
2C+5D = 2 .... 2
In order to isolate one of the variables, we will make one of the variables in any of the equation.
Using equation 1:
3C+4D = 5
Make C the subject of the formula:
Subtract 4D from both sides of the equation.
3C+4D-4D= 5-4D
3C = 5-4D
Divide both sides by 3:
3C/3 = (5-4D)/3
C = (5-4D)/3
Hence the expression that would result from the process is C = (5-4D)/3
Please Help!!
A stone is thrown horizontally with an initial speed of 10m/s from the edge of the cliff. A stop watch measures the stone’s trajectory time from the top of the hill to the bottom to be 6.7s. What is the height of the cliff?
Answer:
h = 219.96 m
Explanation:
Speed of the stone with which it was thrown horizontally, v = 10 m/s
We need to find the height of the cliff if the stone’s trajectory time from the top of the hill to the bottom to be 6.7s.
It means we need to find the distance covered by the stone. As the horizontal speed of the stone is given , it means there is no vertical motion in the stone,u'=0
Using second equation of motion,
[tex]d=u't+\dfrac{1}{2}at^2[/tex]
Put u'=0 and a=g
So,
[tex]d=\dfrac{1}{2}gt^2\\\\\text{Putting all the values to find d}\\\\d=\dfrac{1}{2}\times 9.8\times 6.7^2\\\\d=219.96\ m[/tex]
So, the height of the cliff is 219.96 m.
Blood pressure is usually measured by wrapping a closed air-filled jacket equipped with a pressure gage around the upper arm of a person at the level of the heart. Using a mercury manometer and a stethoscope, the systolic pressure (the maximum pressure when the heart is pumping) and the diastolic pressure (the minimum pressure when the heart is resting) are measured in mmHg. The systolic and diastolic pressures of a healthy person are about 120 mmHg and 80 mmHg, respectively, and are indicated as 120/80. Express both of these gage pressures in kPa, psi, and meter water column.
Take the densities of water and mercury as 1000 kg/m3 and 13,600 kg/m3, respectively.
The high and low pressures, in kPa, are kPa and kPa, respectively.
The high and low pressures, in psi, are psi and psi, respectively.
The high and low pressures, in meter water column, are m and m, respectively.
Answer:
a) High and low pressures are 15.999 kilopascals and 10.666 kilopascals, respectively.
b) High and low pressures are 2.320 pounds per square inch and 1.547 pounds persquare inch, respectively.
c) High and low pressures are 1.632 meters water column and 1.088 meters water column, respectively.
Explanation:
a) High and low pressures in kilopascals:
101.325 kPa equals 760 mm Hg, then, we can obtain the values by a single conversion:
[tex]p_{high} = 120\,mm\,Hg\times \frac{101.325\,kPa}{760\,mm\,Hg}[/tex]
[tex]p_{high} = 15.999\,kPa[/tex]
[tex]p_{low} = 80\,mm\,Hg\times \frac{101.325\,kPa}{760\,mm\,Hg}[/tex]
[tex]p_{low} = 10.666\,kPa[/tex]
High and low pressures are 15.999 kilopascals and 10.666 kilopascals, respectively.
b) High and low pressures in pounds per square inch:
14.696 psi equals 760 mm Hg, then, we can obtain the values by a single conversion:
[tex]p_{high} = 120\,mm\,Hg\times \frac{14.696\,psi}{760\,mm\,Hg}[/tex]
[tex]p_{high} = 2.320\,psi[/tex]
[tex]p_{low} = 80\,mm\,Hg\times\frac{14.696\,psi}{760\,mm\,Hg}[/tex]
[tex]p_{low} = 1.547\,psi[/tex]
High and low pressures are 2.320 pounds per square inch and 1.547 pounds persquare inch, respectively.
c) High and low pressures in meter water column in meters water column:
We can calculate the equivalent water column of a mercury column by the following relation:
[tex]\frac{h_{w}}{h_{Hg}} = \frac{\rho_{Hg}}{\rho_{w}}[/tex]
[tex]h_{w} = \frac{\rho_{Hg}}{\rho_{w}}\times h_{Hg}[/tex] (Eq. 1)
Where:
[tex]\rho_{w}[/tex], [tex]\rho_{Hg}[/tex] - Densities of water and mercury, measured in kilograms per cubic meter.
[tex]h_{w}[/tex], [tex]h_{Hg}[/tex] - Heights of water and mercury columns, measured in meters.
If we know that [tex]\rho_{w} = 1000\,\frac{kg}{m^{3}}[/tex], [tex]\rho_{Hg} = 13600\,\frac{kg}{m^{3}}[/tex], [tex]h_{Hg, high} = 0.120\,m[/tex] and [tex]h_{Hg, low} = 0.080\,m[/tex], then we get that:
[tex]h_{w, high} = \frac{13600\,\frac{kg}{m^{3}} }{1000\,\frac{kg}{m^{3}} } \times 0.120\,m[/tex]
[tex]h_{w, high} = 1.632\,m[/tex]
[tex]h_{w, low} = \frac{13600\,\frac{kg}{m^{3}} }{1000\,\frac{kg}{m^{3}} } \times 0.080\,m[/tex]
[tex]h_{w, low} = 1.088\,m[/tex]
High and low pressures are 1.632 meters water column and 1.088 meters water column, respectively.
why exactly do people use simple
machines like a pulley
An object increases its velocity from 22 m/s to 36 m/s in 5.0 s. What is the average velocity of the object?
Answer:
[tex]a=2.8\ m/s^2[/tex]
Explanation:
Given that,
Initial velocity of an object, u = 22 m/s
Final velocity of an object, v = 36 m/s
Time, t = 5 s
It can be assumed to find the average acceleration of the object instead of average velocity.
The change in velocity per unit time is equal to average acceleration of an object. It can be given by :
[tex]a=\dfrac{v-u}{t}\\\\a=\dfrac{36\ m/s-22\ m/s}{5}\\\\\a=\dfrac{14}{5}\ m/s^2\\\\a=2.8\ m/s^2[/tex]
So, the acceleration of the object is [tex]2.8\ m/s^2[/tex].
A box is sliding along a frictionless surface and gets to a ramp. Disregarding friction, how fast should the box be going on the ground in order to slide up the ramp to a height of 2.5 meters, where it stops? (Use g = 9.8 m/s2.)
Answer:
7.0 m/s
Explanation: I just did it
PLEASE HELP
At a circus, a human cannonball is shot from a cannon at 15m/s at an angle of 40° above horizontal. She
leaves the cannon 2m off the ground and lands in a net 1m off the ground. What height does she reach?
How much ground distance does she cover?
Answer:
Explanation:
[tex]a_{x}=0[/tex] [tex]v_{xo}= v_{o}cos(delta)t[/tex] [tex]a_{y}=-g[/tex] [tex]v_{yo}=sin[/tex]θ
X-direction | Y-direction
[tex]x=x_o+v_{xo}t[/tex] ⇒ [tex]x=v_{xo}cos(delta)t[/tex] |
You want to know how tall a cliff is, so you drop a rock off the edge. You estimate that
the rock has a mass of about 15 kg. You hear a splash 1.00 seconds later. How tall is the
cliff?
Analysing the Question:
Neglecting air resistance
We are given:
mass (m) = 15 kg
time taken by the rock to reach the water (t) = 1 second
height of the cliff (h) = h m
initial velocity of the rock (u) = 0 m/s [ the rock was 'dropped' and not 'thrown' downwards]
acceleration of the rock due to gravity (a) = 9.8 m/s²
Finding the height to the cliff:
from the seconds equation of motion
s = ut + 1/2*at²
replacing the variables
h = (0)(1) + 1/2 * (9.8) *(1)²
h = 4.9 m
Therefore, the cliff is 4.9 m high
Nate throws a basketball straight downward, letting it bounce once before Watching it. We can ignore air
resistance.
What is true about the acceleration and velocity of the ball on its way down (while in the air)?
Assume that upward is the positive direction,
Choose 1 answer:
Acceleration increases and velocity is negative and decreasing.
Acceleration remains constant and velocity is negative and decreasing,
Acceleration decreases and velocity is positive and increasing.
Acceleration remains constant and velocity is negative and increasing.
Answer:
D. Acceleration remains constant and velocity is negative and decreasing.
Explanation:
Khan Academy
Your friend has been hired to design the interior of a special executive express elevator for a new office building. This elevator has all the latest safety features and will stop with an acceleration of g/4 in the case of an emergency. The management would like a decorative lamp hanging from the unusually high ceiling of the elevator. He designs a lamp which has three sections which hang one directly below the other. Each section is attached to the previous one by a single thin wire, which also carries the electric current. The lamp is also attached to the ceiling by a single wire. Each section of the lamp weighs 10.0 N. Because the idea is to make each section appear that it is floating on air without support, he wants to use the thinnest wire possible. Unfortunately the thinner the wire, the weaker it is.
Required:
Calculate the force on each wire in case of an emergency stop.
Answer:
The force on each wire is
[tex]T_1 = 12.5 \ N [/tex]
[tex] T_2 = 25 \ N [/tex]
[tex] T_3 = 50 \ N [/tex]
Explanation:
From the question we are told that
The acceleration at which the elevator will stop is [tex]a = \frac{g}{4}[/tex]
The weight of each section of the wire is [tex]W = \ 10 \ N[/tex]
Generally [tex]W_1 = W_2 = W_3[/tex] here [tex]W_1 , W_2 , W_3[/tex] are weight at each section
Generally considering the first section, the force acting along the y-axis is mathematically represented as
[tex]\sum F_y_1 = T_1 - W_1 = m * a[/tex]
Here [tex]T_ 1[/tex] represents the tension on the wire at the first section while [tex]W_1[/tex] represents the weight of the lamp at the first section
So
[tex]T_1 - 10 = m * \frac{g}{4}[/tex]
=> [tex]T_1 - 10 = \frac{W_1}{4}[/tex]
=> [tex]T_1 - 10 = \frac{10}{4}[/tex]
=> [tex]T_1 = 12.5 \ N [/tex]
Generally considering the second section, the force acting along the y-axis is mathematically represented as
[tex]\sum F_y_2 = T_2 -T_1- W_2 = m * a[/tex]
=> [tex] T_2 - T_1- 10 = m * \frac{g}{4}[/tex]
=> [tex] T_2 - 12.5- 10 = \frac{W_2}{4}[/tex]
=> [tex] T_2- 12.5- 10 = \frac{10}{4}[/tex]
=> [tex] T_2 = 25 \ N [/tex]
Generally considering the third section, the force acting along the y-axis is mathematically represented as
[tex]\sum F_y_3 = T_3- T_2 -T_1- W_3 = m * a[/tex]
[tex] T_3 - T_2 - T_1- 10 = m * \frac{g}{4}[/tex]
[tex] T_3 - 25 - 12.5- 10 = \frac{W_2}{4}[/tex]
[tex] T_3 - 25 - 12.5- 10 = \frac{10}{4}[/tex]
[tex] T_3 = 50 \ N [/tex]
help pls
I just need an explanation to why it’s that answer
Answer:
The area under the speed - time graph denotes the distance travelled by the object
In the given graph, we just have to think about the first 4 seconds, we also notice that the velocity at 4 seconds is 20 m/s
The distance travelled by the object in 4 seconds is the area of the triangle in the graph with a base of 4 units and height of 20 units (image included)
Distance Travelled = Area of triangle = 1/2 * base * height
Distance Travelled = 1/2 * 4 * 20
Distance travelled = 4 * 10
Distance travelled = 40 m
Give reasons:
1. We use tongs to a catch a burning piece of a coal.
2. We use a pulley to draw water from a well.
3. A second class lever multiplies the effort more than other levers.
Explanation:
1) 1 : to Prevent a burning we use tongs
2:it can burn our skin hoa it's help
2) 1: because pulley helps in water draw from a well
3) 1: second class lever has mechanical advantage more than one as load is in between fulcrum an effort making the effort arm longer than the load arm
A bird has a mass of 0.8 kg. Calculate the weight of the bird.
Answer:
the mass and wieght means same
Plzzz hurry!! Pick 3 answer choices
Andrea ate a bowl of cereal for breakfast. Then, she played a game of soccer with her friends. During the game, Andrea became hot and started to sweet. Which three statements provide evidence of conservation of energy in this scenario?
A.The chemical energy in the cereal transformed to mechanical that Andrea used to play soccer.
B.The thermal energy of Andreas body caused the kinetic energy of surrounding air particles to increase.
C.The chemical energy in the cereal transformed to thermal energy that increased Andreas body temperature.
D.The mechanical energy of Andreas legs was created by the collision of her feet with the ground.
Answer:
A B and C are correct I think
The three statements that gives the proof with related to the conversation of energy should be option A, option B, and option C.
What is a conversation of energy?
It is the principle of physics where the energy of the bodies should be interacted and remained the same. In this, the chemical energy in the cereal that shifted to the mechanical should be considered. The thermal energy resulted the kinetic energy so that there is the increment of the air particles. And, the chemical energy in the cereal should be shifted to the thermal energy should raise the temperature of the body.
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A skater glides off a frozen pond onto a patch of ground at a speed of 1.8 m/s. Here she is slowed at a constant rate of 3.00 m/s. How fast is the skater moving when she has slid 0.37 m across the ground?
Answer:
1.01 m/s
Explanation:
Given
initial speed, u = 1.8m/s
acceleration, a = -3.00m/s² (it's negative because the skater slowed down)
distance, s = 0.37m
Required
Determine the final velocity (v)
This will be solved using the following equation of motion
[tex]v\² = u\² + 2as[/tex]
Substitute values for u, a and s
[tex]v\² = 1.8\² + 2 * -3 * 0.37[/tex]
[tex]v\² = 3.24 - 2.22[/tex]
[tex]v\² = 1.02[/tex]
Take square roots
[tex]v = \sqrt{1.02[/tex]
[tex]v = 1.00995049384[/tex]
[tex]v = 1.01 m/s[/tex] (Approximated)
The final speed of the skater when she has slid 0.37 m is 1.0 m/s.
The given parameters:
Speed of the skater, v = 1.8 m/sacceleration of the skater = 3.0 m/s²Displacement of the skater = -0.37 mThe final speed of the skater when she has slid 0.37 m is calculated as follows;
[tex]v_f^2 = v_i^2 + 2ad\\\\v_f ^2 = 1.8^2 + 2(3)(-0.37)\\\\v_f^2 = 1.02\\\\v_f = \sqrt{1.02} \\\\v_f = 1.0 \ m/s[/tex]
Thus, the final speed of the skater when she has slid 0.37 m is 1.0 m/s.
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What is the best description of the relationship between emerging scientific ideas and open-mindedness?
HURRY !
Answer: C ON EDGE 2021
Explanation:
A student wants to determine the density of titanium. The student measures the mass of a solid chunk of titanium, but when he drops it into the water to determine the volume, some water splashes out of the cylinder. Will this cause the calculated density to be higher or lower than it should be? Explain!
Answer:
Lower than it should be.
Explanation:
Hello.
In this case, since the density is computed by:
[tex]\rho =\frac{m}{V}[/tex]
And we obtain the volume of the solid by substracting the mass of the water and the solid minus the mass of water:
[tex]V_{solid}=V_{solid\ with \ water}-V_{water}[/tex]
If some water is splashed out of the cylinder, the volume of water will be lower than originally measured, it means that the volume of the solid will be higher than real. In such a way, since the density is in an inversely proportional relationship with volume, as the volume of the solid is wrongly increased, therefore the density of the solid will be lower than in should be.
Regards.
Different satellites orbit the earth with a vast range of altitudes, from just a couple hundred km, all the way to tens of thousands of km above the surface. The international space station (ISS) is in a low earth orbit, just 400km above the surface (you can see it with the naked eye at sunset and sunrise as a bright, moving dot). At this altitude, the acceleration due to gravity has a value of 8.69.
A. What is the speed of the ISS?
B. What is the orbital period (T) of the ISS in minutes?
Answer:
Find the answer in the explanation.
Explanation:
Given that the international space station (ISS) is in a low earth orbit, just 400km above the surface. And at this altitude, the acceleration due to gravity has a value of 8.69.
A.) Since the radius of the earth is equal to 6400 km. At it to 400km
The radius of the orbit = 6400 + 400
The radius = 6800 km.
Where orbital speed = sqrt ( g × r )
Orbital speed = sqrt ( 6800 × 8.69 )
Orbital speed = 243.09 km/h
B.) Orbital period acceleration to Kepler third law of planetary motion state that:
The square of the period is directly proportional to the cube of the radius.
T^2 = (4π^2 /GM) r^3
T^2=(4π^2/ 6.67×10^11 × 6.0 × 10^24)r^3
T^2 = (4π^2/ 4.002^36) × 6800^3
T^2 = 9.865×10^-36 × 6800^3
T^2 = 3.10 × 10^-15
T = 5.57 × 10^-8 hours.
Convert the hours to minutes by multiplying it by 60
T = 3.4 × 10^-6 minutes.
Since the radius of the earth is equal to 6400 km. At it to 400km
The radius of the orbit = 6400 + 400
The radius = 6800 km.
Where:
Orbital speed = sqrt ( g × r )Orbital speed = sqrt ( 6800 × 8.69 )Orbital speed = 243.09 km/hThe speed of the ISS is 243.09Km/h.
Part B:
The orbital period (T) of the ISS in minutes is :
T^2 = (4π^2 /GM) r^3T^2=(4π^2/ 6.67×10^11 × 6.0 × 10^24)r^3T^2 = (4π^2/ 4.002^36) × 6800^3T^2 = 9.865×10^-36 × 6800^3T^2 = 3.10 × 10^-15T = 5.57 × 10^-8 hours.Convert the hours to minutes by multiplying it by 60
T = 3.4 × 10^-6 minutes.
The orbital period (T) of the ISS in minutes is 3.4 × 10^-6 minutes.
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A 200-loop coil of cross sectional area 8.5 cm2 lies in the plane of the page. An external magnetic field of 0.060 T is directed out of the plane of the page. The external field decreases to 0.020 T in 12 milliseconds. What is the magnitude of the total change in the external magnetic flux enclosed by the coil
Answer:
Δϕ = -2.89 x 10^-4 Wb
Explanation:
Given that A 200-loop coil of cross sectional area 8.5 cm2 lies in the plane of the page. An external magnetic field of 0.060 T is directed out of the plane of the page. The external field decreases to 0.020 T in 12 milliseconds. What is the magnitude of the total change in the external magnetic flux enclosed by the coil
Let ΔB = Change in magnetic field
A = cross sectional area = 8.5 cm^2
t = time = 12ms
Flux change = Δϕ = (ΔB*A) / t
Δϕ = [(0.02 - 0.06)x(8.5 x 10^-2)^2 ] / 0.012
Δϕ = (-0.04 x 7.225^-3) / 0.012
Δϕ = -2.89 x 10^-4 Wb
Therefore, the magnitude of the total change in the external magnetic flux enclosed by the coil is -2.89 x 10^-4 Wb
Which of the following best describes the accuracy of a measurement?
A.
the precision of a measuring tool
B.
the amount of uncertainty involved in a measurement
C.
the degree to which the measurement approaches the quantity's true value
D.
the degree to which repeated measurements agree
Answer:
l think the answer is C. part
The degree to which the measurement approaches the quantity's true value
is what best describes the accuracy of a measurement.
When a quantity is measured, the value is usually unknown by the individual
taking the measurement . This should however be done with a very good
tool/equipment to eliminate inaccuracy.
The accuracy of a quantity is usually measured by how close the difference
is between the measurement taken and the actual value of the quantity.
The different in value gotten should always be very small.
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One strategy in a snowball at a high angle over level ground. While your opponent is watching this first snowball, you throw a second one at low angle times to arrive before or at the same time as the first one . Assume both snowballs are throw with a speed of 33.2 m/s. The first o e is thrown at an angle of 57 degrees with respect to the horizontal. At what angle should the second snowball be throw to arrive at the same point as the first? Answer in units of degrees
Answer:
33°
Explanation:
We are given;
Speed at which both snowballs are thrown; v = 33.2 m/s
Angle at which snowballs are thrown with respect to the horizontal; θ = 57°
Now,we want to find out the angle at which the second snowball should be thrown in order to arrive at the same point as the first.
To calculate this angle, we will use complementary angle concept.
Now, because the target is in the same place, there will be two launch angles that will make the snow ball to be placed on the target.
The is calculated from;θ1 = 45° - (57° - 45°) = 33°