Answer:
2.1 m³
Explanation:
Boyle's Law states:
The absolute pressure exerted by a given mass of an ideal gas is inversely proportional to the volume it occupies if the temperature and amount of gas remain unchanged within a closed system (Wikipedia) }
We can express this relationship as
[tex]P \propto \dfrac{1}{V}[/tex]
where V is the volume under pressure P
This can be re-written as:
PV = a constant
We are given that [tex]$V = 1.5m^3$[/tex] when [tex]$P = 14N/m^2$[/tex]
So PV = 1.5 x 14 = 21 Nm
At [tex]10 N/m^2[/tex] the PV product must also be 21
So 10V' = 21 where V' = volume at [tex]\displaystyle 10 N/m^2[/tex]
Therefore V' = 21/10 = 2.1 m³
Drawing of circuit in series, with 6 lamps , 3 electrical equipment, 1 power source, and is opened
The circuit is drawn given below
When the key is 'on' then the current is flowing and all the lamps will glow
Resistance is used to resist the flow of current.
Ammeter is used to measure the current flowing in the circuit
Set the cannon to have an initial speed of 15 m/s. For which situation do you think the cannon ball will go highest: if it is set at a 60-degree angle, or if it is set at a 70-degree angle?
Question 4 options:
60 degree
70 degree
The cannon ball will travel the highest distance when the angle of projection is 70 degrees.
What is the maximum height of a projectile?The maximum height reached by a projectile is calculated using the following formula.
H = u²sin²θ/2g
where;
u is the initial velocity of the projectile θ is the angle of pojectiong is acceleration due to gravitywhen the angle of projection is 60 degrees;
H = (15² (sin 60)²) / (2 x 9.8)
H = 8.6 m
when the angle of projection is 70degrees;
H = (15² (sin 70)²) / (2 x 9.8)
H = 10.14 m
Thus, the cannon ball will travel the highest distance when the angle of projection is 70 degrees.
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What does the top pressure gauge in (Figure 1) read? Express your answer using two significant figures.
Digital manometers can be used to measure liquids, gases, and air, depending on the device. If the instruments have an internal memory, it works quite well.
What does the top pressure gauge indicate?Pressure Gauge in Psig Pressure as a percentage of the surrounding atmospheric pressure. The gauge pressure for this kind of pressure includes the pressure from the weight of the atmosphere because it is zero-referenced against atmospheric pressure.
These gauges typically have a measurement range of 0… 1 mbar to 0… 600 mbar and an accuracy class of 0.1 to 2.5. Although pressure gauges typically require very minimal upkeep, problems can happen.
Therefore, This tool can be used to measure pressure in a variety of locations. Measurements of negative pressure, overpressure, and differential pressure are all possible.
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Suppose the ball is thrown from the same height as in the PRACTICE IT problem at an angle of 29.0° below the horizontal. If it strikes the ground 47.8 m away, find the following. (Hint: For part (a), use the equation for the x-displacement to eliminate v0t from the equation for the y-displacement.)
The time that the ball took to take off at an angle of 29.0° is 1.13 s
When two lines meet at a point, they form an angle. The word "angle" refers to the length of the "opening" between these two rays. The symbol is used to denote it.
Radians, a unit of circularity or rotation, and degrees are the two most common units used to measure angles. In daily life, angles are present. For the design of highways, structures, and sports venues, engineers and architects use angles.
The motion along the x-direction is always uniform, so the position along the x-direction is given by
x(t) = (v₀cos∅)t
where,
v₀ is the initial speed,
cos∅ is the angle of launching,
t is the time.
We can re-write the formula as
v₀t = x(t)/cos∅ ...1
Along the y-direction the motion is accelerated (free-fall), so the equation for the vertical position is
y(t) = h = v₀sin∅t - 1/2gt² ...2
where,
h = 45.0 m is the height of the building
g = 9.8m/s² is the acceleration of gravity.
Substituting (1) into (2),
y(t) = h = x(t)tan∅ - 1/2gt²
And when this happens, the displacement along the x-axis is x = 47.8 m. Solving the equation for t and substituting the numbers, we find the time of flight:
[tex]t = \sqrt{\frac{2(h-xtan\theta)}{g} }[/tex]
[tex]t = \sqrt{\frac{2(45-47.8tan39^{\circ})}{9.8} }[/tex]
t = 1.13 s
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Full question
Suppose the ball is thrown from a 45m height building as in the PRACTICE IT problem at an angle of 29.0° below the horizontal. If it strikes the ground 47.8 m away, find the following. (Hint: For part (a), use the equation for the x-displacement to eliminate v0t from the equation for the y-displacement.)
a) the time of flight s
A car of mass m = 1090 kg is traveling down a θ = 11 degree incline. When the car's speed is v0 = 16 m/s, a mechanical failure causes all four of its brakes to lock. The coefficient of kinetic friction between the tires and road is μk = 0.45.
Calculate the distance the car travels down the hill L in meters until it comes to a stop at the end.
The distance the car of mass 1090 kg travels with a velocity of 16 m / s down the hill having a 11 degree decline until it comes to a stop at the end is 52 m
θ = 11°
m = 1090 kg
g = 9.8 m / s²
μ[tex]_{k}[/tex] = 0.45
W = m g
W = 1090 * 9.8
W = 10682 N
Resolving W into its horizontal and vertical components,
[tex]W_{y}[/tex] = W cos θ
[tex]W_{y}[/tex] = 10682 * cos 11°
[tex]W_{y}[/tex] = 10682 * 0.98
[tex]W_{y}[/tex] = 10468.36 N
[tex]W_{x}[/tex] = W sin θ
[tex]W_{x}[/tex] = 10682 * sin 11°
[tex]W_{x}[/tex] = 10682 * 0.19
[tex]W_{x}[/tex] = 2029.58 N
N = [tex]W_{y}[/tex]
N = 10468.36 N
F[tex]_{k}[/tex] = μ[tex]_{k}[/tex] N
F[tex]_{k}[/tex] = 0.45 * 10468.36
F[tex]_{k}[/tex] = 4710.76 N
∑ [tex]F_{x}[/tex] = m a
[tex]W_{x}[/tex] - F[tex]_{k}[/tex] = m a
2029.58 - 4710.76 = 1090 * a
a = - 2681.18 / 1090
a = - 2.46 m / s²
Acceleration is negative because the car is decelerating in positive direction.
v² = u² + 2 a s
0 = 16² + ( 2 * - 2.46 * s )
4.92 s = 256
s = 52 m
Therefore, the distance the car travels down the hill is 52 m
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A stone is projected upwards to an angle of 30 degrees to the horizontal from the top of a tower of height 100m and it hits the ground at a point Q. If the initial velocity of projection is 100m/s. Calculate the maximum height of the stone above the ground.
The maximum height of the stone above the ground is 227.55 m.
What is the maximum height reached by the stone?
The maximum height reached by the stone is calculated by applying the following kinematic equation.
H = u²sin²θ/2g
where;
u is the initial velocity of the stoneθ is the angle of projection of the stoneg is acceleration due to gravitySubstitute the given parameters and solve for the maximum height reached by the stone.
H = (100² x (sin 30)²)/(2 x 9.8)
H = 127.55 m
Height of the stone above the ground = 127.55 m + 100 m = 227.55 m
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What effect would that have on the force resulting in accordance to Newton's Third law ?
According to the Newton's Third law, the force sent back with equal and opposite in the reaction.
What is meant by Newton's Third law?According to Newton's third law, two objects exert a force of equal magnitude and opposite direction. This law represents a certain symmetry in nature: forces always occur in pairs, and one body cannot exert a force on another without experiencing a force on it. Newton's 3rd law of motion states that action and reaction are always equal but opposite in direction. Newton's third law that for every action in nature there is an equal and opposite reaction. If object A exerts a force on object B, object B also uses an alike and across-form force on object A.
So we can conclude that the force is in equal in magnitude but in opposite direction according to Newton's Third law.
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Can you please explain to me what answer would be correct and how?
Newton's third law states that; with every action, there is a direct and opposite reaction. As you walk across the flooor, you are applying a force due to your weight on the floor. The floor would apply an an equal force onto your feet which is in opposite direction and this force pushes you forward. This force is called th normal force. Thus, the correct option is
d)
The Big Dipper is an example of an observed pattern of stars called a ___________.
Answer:
The answer is the asterism.
Explanation:
An asterism is recognizable pattern of stars . some other commonly recognized asterisms are the little dipper ,orion's belt , and the teapot
The terminal side contains the point (-6, -8). Find sin θ.Question 1 options:.8.6-.8-.6
Answer:
The first option: 0.8
Explanation:
Let us draw the angle.
Now,
[tex]\sin \theta=\frac{opposite}{\text{hypotenuse}}[/tex]Now, using the Pythagoras theorem we find that
[tex]hypotenuse=\sqrt[]{6^2+8^2}[/tex][tex]\begin{gathered} hypotenuse=\sqrt[]{36+64^{}} \\ \Rightarrow hypotenuse=10 \end{gathered}[/tex]Therefore,
[tex]\sin \theta=\frac{opposite}{hypotenuse}=\frac{opposite}{10}[/tex]since for our angle, opposite = 8, we have
[tex]\sin \theta=\frac{8}{10}[/tex][tex]\boxed{\sin \theta=0.8.}[/tex]Therefore, the first choice (0.8) is the correct answer.
Best player in nba right now
Best player in NBA right now Giannis Antetokounmpo.
The National Basketball Association, NBA is a professional basketball league in North America. The league consists of 30 teams and is one of the predominant expert sports activities leagues within the united states of America and Canada. It's far the finest guys professional basketball league within the international.
To sell great aware device of technical education where excellence, relevance to market desires and participation with the aid of all stakeholders are prime and principal determinants.
The NBA is a 70-year-vintage employer born from innovation. It turned into June 1946 in the big apple city while Boston garden owner Walter Brown found out that primary ice hockey arenas, which were vacant maximum nights, will be used to host basketball games.
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Anton applies a force on a shopping cart and makes it move forward. What can be said about the forces acting on the shopping cart at the moment Anton applies the force? The forces acting on the shopping cart are...
Given that, Anton is applying a force on a cart making it move around.
The forces acting on the cart are gravitational force, normal force, frictional force, and the force applied by Anton.
The gravitational force will be acting downwards. From Newton's third law there will be a reaction force to the gravitational force and that is called the normal force. And the formal force will be acting upwards. These two forces are equal to each other in magnitude. And thus they will cancel each other out making the net force on the cart in the vertical direction zero.
The force applied by Anton will be acting forward and the frictional force will be opposing the movement of the cart. The net force in the horizontal direction will be directed forward.
From Newton's second law, the net force is equal to the product of the mass of the cart and the acceleration produced by the net force.
1. An ordinary light bulb is marked “60A, 120V”, its resistance is ______.
In order to calculate the resistance, we can divide the voltage and the current, according to the formula below:
[tex]R=\frac{V}{I}[/tex]Where R is the resistance (in ohm), V is the voltage (in Volt) and I is the current (in Ampere).
So, using V = 120 V and I = 60 A, we have:
[tex]R=\frac{120}{60}=2\text{ ohms}[/tex]Therefore the resistance is 2 ohms.
The carbon monoxide molecule (CO) consists of a carbon atom and an oxygen atoms separated by a distance of 1.13x10^-10 m, The mass mc of the carbon atom is 0.750 times the mass mo of the oxygen atom, or mc= -0.750mo , Determine the location of the center of the mass of this molecule relative to the carbon atom.
The location of the centre of the mass of the CO molecule relative to the carbon atom is 6.46 ×10⁻¹¹ m.
What is the centre of mass of the molecule?The Centre of Mass of that object is a position that is relative to an object or a system of objects. The average position of all parts of the system that are weighted to their masses is known as the Centre of Mass.
Given, the distance between the carbon and oxygen atoms, r = 1.13 ×10⁻¹⁰m.
The ratio of the mass of the carbon and oxygen:
Mc/Mo = 12/16 = 0.750
Consider that Xc is the position of carbon and Xo is the position of the oxygen atom in CO molecule.
The centre of the mass of the system is equal to the:
[tex]X_{co}=\frac{M_CX_C+M_OX_O}{M_C+M_O}[/tex]
[tex]X_{co}=\frac{(M_C/M_O)X_C+X_O}{M_C/M_O+1}[/tex]
As we are calculating the centre of the mass relative to the carbon atom.
Therefore Xc = 0 and Xo= 1.13 ×10⁻¹⁰m.
[tex]X_{CO} =\frac{1.13\times 10^{-10}}{0.750 +1}[/tex]
Xco = 6.46 ×10⁻¹¹ m
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I can not understand the closed circle and open circle. Please help!
First, take into account that the open circle means that the point is not included into the domain and the range. The closed point means that such a point is included into both domain and range.
In this case, althoug there is a open circle, there is another part of the curve on which the y value is included into the path.
Then, the range of the graph is:
ran f = [2,7]
Which is equivalent to all number from 2 to 7.
After being pushed away from the departure gate, an aircraft is standing still with the brakes applied and the engine running on very low. When the pilot releases the brakes he finds that the plane still does not start to move. Here are three explanations for why the plane does not move. Which one do you agree with?
Since he pilot releases the brakes and he finds that the plane still does not start to move, the explanation I agree with is option C:
When the pilot releases the brakes two forces act on it. The engines exert a force in one direction and the ground exerts a force on the wheels in the opposite direction. The plane does not move because the force exerted by the ground on the wheels is the same strength as the force of the engines.How do planes fly and stay in the air?The lift of the airplane is provided by the flow of air around the wings. Additionally, the wings' form contributes to lift. The airplane is being pulled toward Earth by weight. The weight of an airplane is distributed evenly from front to back during construction.
Note that the design of an airplane wing is intended to speed up airflow across the top of the wing. Air pressure decreases when it moves more quickly. Therefore, the pressure at the top of the wing is lower than that at the bottom. The pressure differential exerts a force on the wing, lifting it into the air.
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See full question below
After being pushed away from the departure gate, an aircraft is standing still with the brakes applied and the engine running on very low. When the pilot releases the brakes he finds that the plane still does not start to move. Here are three explanations for why the plane does not move. Which one do you agree with?
A. When a single force that is too weak acts on an object it will not move. When the pilot releases the brakes the only force acting on the plane is that of the
engines but because this force is too weak, the plane will not move.
B. When the pilot releases the brakes two forces act on it. The engines exert a force in one direction and the ground exerts a force on the wheels in the opposite direction. The plane does not move because the force exerted by the ground on the wheels is stronger than the force of the engines.
C. When the pilot releases the brakes two forces act on it. The engines exert a force in one direction and the ground exerts a force on the wheels in the opposite direction. The plane does not move because the force exerted by the ground on the wheels is the same strength as the force of the engines.
If a spring is stretched 4m from its starting length when 20n of force is applied, then how much work (in joules) is done by the spring when it is stretched 10 m?
ANSWER:
250 J
STEP-BY-STEP EXPLANATION:
F = 20N is required to stretch the spring by 4 meters
We know that the force is equal to:
[tex]F=k\cdot x[/tex]We solve for k (spring constant):
[tex]k=\frac{F}{x}=\frac{20}{4}=5\text{ N/m}[/tex]The work done in stretching the spring is given by the following equation (in this case the stretch is 10 meters:
[tex]\begin{gathered} W=\frac{1}{2}k\cdot x^2 \\ \text{ Replacing} \\ W=\frac{1}{2}\cdot5\cdot10^2 \\ W=250\text{ J} \end{gathered}[/tex]The work required is 250 joules.
How many hours would it take a plane to fly 1,500 miles if it was moving at a constant speed of 300 miles/hour. Include units!
Brainliest included
For a equilibrium shown. There are two strings are strong enough to withstand a maximum tension of 80 n. what is the largest value of w they can support
The largest value of w (in N)that they can support as shown in equilibrium.
What is equilibrium?
a State of balance between opposing forces or actions that is either static as in nobody acted on by forces whose resultant is zero or dynamic reaction when the rates of reaction in both differentions are equal.
Sol- the total force in x direction is given as,
€Fx=0
T1 sin 37°=T2 sin 53°
100×3/5 =T/2 ×4/5
T2=75 N
The given total force is-
€Fy=0
T1 cos 37°=T2 cos 53°+W
W=80×4/5 -75×3/5
W=19N
Thus
The largest value of W is 19N.
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A ball thrown vertically upwards at a speed 30mls.
a)Calculate ,how high a ball will travel when its speed is 20m/s?
b)Calculate the maximum height of the ball will reach ?
c)Find time the ball in air ?
Ball moving vertically upwards,
initial velocity = 30 m/s
final velocity = 0 m/s [ the ball stops after reaching a certain height ]
Applying v = u + at
0 = 30 + 10t [ We divide the course of the object into an uphill travel and a descending journey by assuming there is no friction and that g is equal to 10 m/s². ]
Therefore, t = 3 seconds
By symmetry, the descent will take the same amount of time.
Therefore, the object needs a total of 6 seconds before it returns.
using, H=(v²−u²)/2a
ultimate velocity = 0 m/s (it stops when maximum height is reached)
beginning speed = 20 m/s
a = −g = −9.8m/s² (acceleration owing to gravity, negative sign because it is acting downward)
Replace the values
Obtaining, H=20.408m
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Suppose the coefficient of kinetic friction between mA and the plane in the figure(Figure 1) is μk = 0.15, and that mA=mB=2.7kg. part aAs mB moves down, determine the magnitude of the acceleration of mA and mB, given θ = 30 ∘ .part b What smallest value of μk will keep the system from accelerating?
We are asked to determine the acceleration of the system. To do that we will consider the two masses to be a single system and we will use the following free-body diagram.
Where:
[tex]\begin{gathered} W_b=\text{ weight of B} \\ W_a=\text{ weight of A} \\ W_{ah}=\text{ component of the weight of A in the horizontal direction} \\ W_{av}=\text{ component of the weight of B in the vertical direction} \\ N=\text{ normal force} \\ F_f=\text{ force of friction} \end{gathered}[/tex]To determine the components of the weight of "A" we use the following triangle:
Therefore, the vertical component is given by:
[tex]W_{av}=W_a\cos30[/tex]Since the weight is the product of the mass and the acceleration we have;
[tex]W_{av}=m_ag\cos30[/tex]The horizontal component is given by:
[tex]W_{ah}=W_a\sin30[/tex]Substituting the formula for weight:
[tex]W_{ah}=m_ag\sin30[/tex]The friction force is given by:
[tex]F_f=\mu N[/tex]The normal force "N" is equivalent to the vertical component of the weight since there is no acceleration in that direction. Therefore, we have:
[tex]F_f=\mu W_{av}[/tex]Substituting the values:
[tex]F_f=\mu m_ag\cos30[/tex]Now, the net force is equivalent to the forces acting on the system. Therefore, the net force is:
[tex]F_{net}=m_bg-F_f-W_{ah}[/tex]Substituting we get:
[tex]F_{net}=m_bg-\mu m_ag\cos30-m_ag\sin30[/tex]According to Newton's second law we have that the net force is equal to the product of the mass and the acceleration:
[tex]F_{net}=ma[/tex]The total mass is the sum of the masses "A" and "B".
[tex]F_{net}=(m_a+m_b)a[/tex]Substituting the net force:
[tex]m_bg-\mu m_ag\cos30-m_ag\sin30=(m_a+m_b)a[/tex]Now, we divide both sides by the total mass:
[tex]\frac{m_bg-\mu m_ag\cos30-m_ag\sin30}{m_a+m_b}=a[/tex]Since we have that the masses are equal:
[tex]m_a=m_b=m[/tex]Substituting we get:
[tex]\frac{mg-\mu mg\cos(30)-mg\sin(30)}{m+m}=a[/tex]Adding the masses in the denominator:
[tex]\frac{mg-\mu mg\cos(30)-mg\sin(30)}{2m}=a[/tex]We can cancel out the "m":
[tex]\frac{g-\mu g\cos(30)-g\sin(30)}{2}=a[/tex]Now, we plug in the values:
[tex]\frac{9.8\frac{m}{s^2}-(0.15)(9.8\frac{m}{s^2})\cos30-9.8\frac{m}{s^2}\sin30}{2}=a[/tex]Solving the operations:
[tex]1.81\frac{m}{s^2}=a[/tex]Therefore, the acceleration is 1.81 meters per second square.
Part B. To determine the value of the coefficient of friction that will keep the system from accelerating we must go back to the formula for the acceleration:
[tex]\frac{g-\mu g\cos(30)- g\sin(30)}{2}= a[/tex]Now, we will set the acceleration to zero:
[tex]\frac{g-\mu g\cos(30)- g\sin(30)}{2}=0[/tex]Multiplying both sides by 2:
[tex]g-\mu g\cos(30)-g\sin(30)=0[/tex]Now, we can divide both sides by "g":
[tex]1-\mu\cos30-\sin30=0[/tex]Now, we solve for the coefficient of friction. We subtract 1 to both sides:
[tex]-\mu\cos(30)-\sin(30)=-1[/tex]Now, we add sin(30) to both sides:
[tex]-\mu\cos(30)=-1+\sin(30)[/tex]Now, we divide both sides by -cos(30):
[tex]\mu=\frac{-1+\sin(30)}{-\cos(30)}[/tex]Now, we solve the operations:
[tex]\mu=0.58[/tex]Therefore, the friction coefficient is 0.58
A cannon ball is shot at 10 m/s at an angle of 30 degrees. It takes 2.2 seconds for the ball to reach the ground. (A) How high is the cliff that the cannonball is shot from? (B) How far from the base of the cliff does the ball land?
A ) The height of cliff from which the cannonball is shot from = 12.72 m
B ) The distance from the base of the cliff = 19.14 m
- H = [tex]u_{y}[/tex] T - 1 / 2 g T²
T = Total time taken
g = Acceleration due to gravity
H = Height above the ground
[tex]u_{y}[/tex] = Y-component of initial velocity
T = 2.2 s
g = 9.8 m / s²
θ = 30°
u = 10 m / s
[tex]u_{y}[/tex] = u sin θ
[tex]u_{y}[/tex] = 10 * sin 30°
[tex]u_{y}[/tex] = 5 m / s
- H = ( 5 * 2.2 ) - ( 0.5 * 9.8 * 2.2 * 2.2 )
- H = 11 - 23.72
H = 12.72 m
R = [tex]u_{x}[/tex] T
[tex]u_{x}[/tex] = u cos θ
[tex]u_{x}[/tex] = 10 * cos 30°
[tex]u_{x}[/tex] = 8.7 m / s
R = 8.7 * 2.2
R = 19.14 m
Therefore,
A ) The height of cliff from which the cannonball is shot from = 12.72 m
B ) The distance from the base of the cliff = 19.14 m
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Find a model for simple harmonic motion satisfying the specified the conditions.
displacement (t=0) 0 centimeters
amplitude 8 centimeters
period 2 seconds
Answer: d = 8 sin (πt
Explanation:
when displacement is 0m in t = 0s, then it's a sine equation
∴ d = A sin (ωt + Ф)
when t = 0, then there's no initial phase
Period ⇒ 2π/ω = 2 ⇒ ω = π
∴ d = 8 sin (πt)
The contestant now pulls the block of ice with a rope over his shoulder at 25 degrees. Calculate the minimum force F he must exert to get the block moving. (b) What is its acceleration once it starts to move, if that force is maintained?
The minimum force that the contestant must exert to get the block moving is 53.33 N and the acceleration of the block is 0.874m/s² once it starts to move and the force is maintained.
When a force is applied at an angle, the force will have two components, one along the floor and one along the vertical.
If the force is F applied at an angle θ, then the vertical component of force is Fsinθ and the floor component is Fcosθ.
In out case θ = 25°,
A.) There will we a frictional force acting on the body which will be opposite to the component of the force which is along the floor, if the floor component just overcomes the static frictional force than the block will start to move.
So, we can write,
Friction force = Fcosθ
μN = Fcosθ
where,
μ is the coefficient of static friction,
N is the normal reaction by the floor on the block which would be equal to the mg,
F is the minimum force Applied.
Putting all the values,
(0.1)(10)(48) = (F)(cos25°)
48 = F(0.9)
F = 53.33 Newtons.
B.) When the block will start moving,
There will be a kinetic frictional force on it as well as a Fcosθ on it,
The friction will be opposite direction of the motion, Fcosθ will be in the direction of the motion,
So, we can write, The net force on the body
F = Fcosθ - Fr
F = Fcosθ - μN
F = Fcosθ - umgsinθ
When the box is moving the normal reaction N will be equal to the vertical component of the force,
Here, u is the coefficient of kinetic friction,
Putting all the values,
F = 48 - 0.03x10x48x0.42
F = 48 - 6.048
F = 41.95 N
we can also write,
F = ma
where a is the acceleration of the block and m is the mass of the block,
41.95 = 48a
a = 0.874m/s².
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A contestant in a winter games event pushes a 48 kg block of ice across a frozen lake with a rope over his shoulder at a 25 degree angle.
The coefficient of static friction is 0.1 and the coefficient of kinetic friction is 0.03.
A.) Calculate the minimum force F (in N) he must exert to get the block moving.
B.) What is its acceleration (in m/s^2) once it starts to move, if that force is maintained?
A current of 17.0A is maintained in a single circular loop of 2.00m circumference. A magnetic field of 0.800T is directed parallel to the plane of the loop. (a) Calculate the magnetic moment of the loop. [5] (b) What is the magnitude of the torque exerted on the loop by the magnetic field?
A. The moment of the loop is 5.4 x [tex]10^{-3} Am^{2}[/tex]
B. The torque exerted on the loop by the magnetic field is 4.32 [tex]10^{-3}[/tex] Nm.
Given that the current I in the circular loop is 17.0 mA and the circumference of the loop is 2.00 m.
The radius of the circular loop is calculated as given below.
2πr = 2
r = 2/2 x 3.14
r = 0.18m
The moment in the circular loop is calculated as given below.
M = in
M = 17 x [tex]10^{-3}[/tex] x 3.14 x 0.318²
M = 5.4 x [tex]10^{-3} Am^{2}[/tex].
The torque exerted on the loop by the magnetic field is given below:
T = MB
T = 5.4 x [tex]10^{-3}[/tex] x 0.800
T = 4.32 x [tex]10^{-3} Nm[/tex].
This is the torque of the current-carrying loop in a uniform magnetic field. This formula can be shown to be valid for loops of any shape. The loop carries a current I, has N turns, and the plane A and the normal to the loop each make an angle θ with the field B. The net force on the loop is zero. A magnetic field exerts a force on a straight wire through which an electric current flows. Apply torque to the loop of the wire carrying the current. A torque rotates an object around a fixed axis.
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Light hits a mirror at a 45° angle. It will be reflected at an angle _____.
equal to 45°
greater than 45°
less than 45°
Light hits a mirror at a 45° angle. It will be reflected at an angle equal to 45°
According to law of reflection, when a light ray falls on a surface that is smooth, the angle of reflection will be equal to the angle of incidence. The incident ray, the reflected ray and the normal to the surface all will lie in the same plane.
Reflection is the process in which a light ray gets bounced back after falling on a surface. The angle between in the incident ray and the normal to the surface is known as angle of incidence. The angle between in the reflected ray and the normal to the surface is known as angle of reflection.
Therefore, light hits a mirror at a 45° angle. It will be reflected at an angle equal to 45°
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What is the index of refraction of a refractive medium if the angle of incidence in air (n=1.0003) is 87 degrees and the angle of refraction is 10 degrees?
Given:
The refractive index of the air, n₁=1.0003
The angle of incidence, θ₁=87°
The angle of refraction, θ₂=10°
To find:
The refractive index of the medium.
Explanation:
From the snell's law,
[tex]n_1\sin\theta_1=n_2\sin\theta_2[/tex]Where n₂ is the refractive index of the medium.
On substituting the known values,
[tex]\begin{gathered} 1.0003\sin87\degree=n_2\sin10\degree \\ \implies n_2=\frac{1.0003\times\sin87\degree}{\sin10\degree} \\ =5.75 \end{gathered}[/tex]Final answer:
Thus the refractive index of the medium is 5.75
Which of the following methods can demonstrate cause and effect?
correlational
experimental
naturalistic observation
survey method
in some places, insect "zappers", with their blue lights are a familiar sight on a summer's night. these devices use a high voltage to electrocute insects. one such device uses an ac voltage of 4320V, which is obtained from a standard 120,0V outlet by means of a transformer. if the primary coil has 21 turns, how many turns are in the secondary coil?
For a device that electrocutes insects with an AC voltage of 4320V which is obtained from a standard 1200V transformer, If the primary coil has 21 turns, the number of turns in the secondary coil is 76 turns.
Number of turns and output voltage of the primary coil and secondary coil in a transformer are related as the ratio of number of turns in the primary and secondary coil is equal to the ratio of output voltage in the primary and secondary coil respectively.
It can be stated as N₁/N₂ = V₁/V₂, where N₁ and N₂ are the number of turns in the primary and secondary coil respectively, and V₁ and V₂ are the output voltage of primary and secondary coil respectively.
According to the question,
21/N₂ = 1200/4320
N₂ = 75.6
Hence the number of turns should be an integer value, the number of turns in secondary coil are 76 turns.
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How did Buddhism and the Maury and contribute to technology in ancient India?
During the Maurya empire, the Indian lifestyle and manner of existence had been deeply stimulated by Buddhism. Buddhism appealed to human beings of decreased castes because it emphasized individuals' course to enlightenment and salvation, which might be attained in this existence.
The Maurya Empire's political cohesion and inner peace endorsed the expansion of trade in India. at some stage in Ashoka's reign, authorities oversaw the building of essential roadways, and the Mauryan international network of exchange multiplied. India's exports to places like Bactria and Persia blanketed silk, textiles, and spices.
The Maurya Empire become centralized through the conquest of the Indo-Gangetic simple, and its capital town become located at Pataliputra (current Patna). outdoor this imperial middle, the empire's geographical volume became dependent on the loyalty of military commanders who managed the armed towns sprinkling it.
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