Answer:
d
Explanation:
Gravitational force is a type of force that pulls objects to the Earth's surface.
What is double-slit experiment?
The double-slit experiment shows that both matter and light can exhibit properties of conventionally defined waves and particles.
The double-slit experiment is a part of a class of "double path" experiments in which a wave is split into two separate waves that later combine to form a single wave (the wave is typically composed of many photons and is better known as a wave front, which should not be confused with the wave properties of the individual photon).
Isaac Newton's corpuscular theory of light, which had previously prevailed as the accepted explanation of light transmission in the 17th and 18th centuries, was defeated by double-slit experiment , which was conducted in the early 1800s.
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If the range of a projectile is and 256√3 m in the maximum height reached is 64 m. calculate the angle of projection
The angle of projection given that the range is 256√3 m and the maximum height reached is 64 m is 30°
Data obtained from the questionRange (R) = 256√3 mMaximum height (H) = 64 mAcceleration due to gravity (g) = 9.8 m/s²Angle of projection (θ) = ?How to determine the angle of projectionR = u²Sine(2θ) / g
256√3 = u²Sine(2θ) / 9.8
Cross multiply
256√3 × 9.8 = u²Sine(2θ)
Divide both sides by Sine(2θ)
u² = 256√3 × 9.8 / Sine(2θ)
H = u²Sine²θ / 2g
64 = [256√3 × 9.8 / Sine(2θ)] × [Sine²θ / 2 × 9.8]
64 = [256√3 / Sine(2θ)] × [Sine²θ / 2]
Recall
Sine²θ = SineθSineθ
Sine2θ = 2SineθCosθ
Thus,
64 = [256√3 / 2SineθCosθ] × [SineθSineθ / 2]
64 = 256√3 × Sineθ / 4Cosθ
Recall
Sineθ / Cosθ = Tanθ
Thus,
64 = 256√3 / 4 × Tanθ
Divide both side by 256√3 / 4
Tanθ = 64 ÷ 256√3 / 4
Tanθ = 0.5774
Take the inverse of Tan
θ = Tan⁻¹ 0.5774
θ = 30°
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As shown in (Figure 1), a layer of water covers a slab of material X in a beaker. A ray of light traveling upwards follows the path indicated.
a) Using the information on the figure, find the index of refraction of material X .
b) Find the angle the light makes with the normal in the air.
Answer: the index of refraction of material X will be 1.09 and the angle the light makes with the normal in the air is 81.25°.
Explanation: To find the answer, we need to know the Snell's law.
What is Snell's law of refraction? Using this, how to solve the problem?The Snell's law for refraction can be written as,[tex]\frac{sin (i)}{sin(r)} =\frac{n_r}{n_i}[/tex]
where, i is the incident angle, r is the refracted angle, n is the refractive index.
As we know that the refractive index of water is 1.33For the first case, incident angle from the picture is 65°, and the refracted angle is 48°. Thus, the refractive index of the medium X will be,[tex]\frac{sin 65}{sin48} =\frac{n_w}{n_X} \\n_X=\frac{1.33*sin48}{sin65} =1.09\\[/tex]
In the second case, incident angle is 48° and we have to find the refracted angle r for the air.As we know that the refractive index of air is 1.Thus, the refracted angle will be,[tex]\frac{sin 48}{sin r}=\frac{1}{1.33} \\\\ sin(r)=\frac{1.33*sin 48}{1}=0.988\\\\r=sin^{-1}(0.988)=81.25 degrees.[/tex]
Thus, we can conclude that, the index of refraction of material X will be 1.09 and the angle the light makes with the normal in the air is 81.25°.
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The light makes an angle of 81.25° with the normal in the air and the index of refraction of material X will be 1.09.
In order to determine the solution, we must understand Snell's law.
What is the refraction law of Snell? How can the issue be resolved with this?One way to express Snell's law for refraction is as follows:[tex]\frac{sin(i)}{sin(r)}=\frac{n_r}{n_i}[/tex]
where the refractive index is n and the incidence angle is i. The refracted angle is r.
As is well known, water has a refractive index of 1.33.The incidence angle from the image in the first case is 65°, while the refracted angle is 48°. Consequently, the medium X's refractive index will be,[tex]n_X=\frac{n_w*sin 48}{sin65} =1.09[/tex]
The incident angle in the second scenario is 48°, and we must determine the refracted angle r for the air.Since we now know that air has a refractive index of 1, so that the refracted angle is,[tex]sin(r)=n_w*sin48=0.988\\r=sin^{-1}(0.988)=81.25 degrees.[/tex]
As a result, we can infer that the material X will have an index of refraction of 1.09 and that the angle the light makes with the normal in the air is 81.25°.
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The specific heat of copper is 0.385 J/g°C.
How much heat is needed to raise the temperature of 6.00 g of copper by 15.0°C?
35.0 J
90.0 J
234 J
34.7 J
The amount of heat needed to raise the temperature of 6.00 g of copper by 15.0°C is 34.65J (option A).
How to calculate amount of heat?The amount of heat absorbed or released by a substance can be calculated using the following formula:
Q = mc∆T
Where;
Q = quantity of heat absorbed or releasedm = mass of the substancec = specific heat capacity∆T = change in temperatureQ = 6 × 0.385 × 15
Q = 90 × 0.385
Q = 34.65J
Therefore, the amount of heat needed to raise the temperature of 6.00 g of copper by 15.0°C is 34.65J.
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Two coils are wound around the same cylindrical form. When the current in the first coil is decreasing at a rate of -0.240 A/s, the induced emf in the second coil has a magnitude of 1.60×10−3 V.
a) What is the mutual inductance of the pair of coils?
b) If the second coil has 30 turns, what is the flux through each turn when the current in the first coil equals 1.25 A ?
c) If the current in the second coil increases at a rate of 0.365 A/s , what is the magnitude of the induced emf in the first coil?
The mutual inductance of the pair of coils is 6.67 x 10⁻³ H.
The the flux through each turn is 2.78 x 10⁻⁴ Tm².
The magnitude of the induced emf in the first coil is 2.435 x 10⁻³ V.
Mutual inductance of the coilM = -NΦ/I
M = emf/I
M = -(1.6 x 10⁻³)/(-0.24)
M = 6.67 x 10⁻³ H
Flux in the second coilM = NΦ/I
MI = NΦ
Φ = MI/N
Φ = (6.67 x 10⁻³ x 1.25)/(30)
Φ = 2.78 x 10⁻⁴ Tm²
Induced emf in the first coilemf = MI
emf = 6.67 x 10⁻³ x 0.365
emf = 2.435 x 10⁻³ V
Thus, the mutual inductance of the pair of coils is 6.67 x 10⁻³ H.
The the flux through each turn is 2.78 x 10⁻⁴ Tm².
The magnitude of the induced emf in the first coil is 2.435 x 10⁻³ V.
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A step down transformer with an input voltage of 220V decreases the voltage to half of the input .There is a current flowing of 15A in the primary coil. Find the output current
The output current for the given step down transformer is 30A.
What is the step down transformer?
A step down transformer is a passive device that converts high voltage power to low voltage power, while the output current is higher than the input current. They are used in power adaptors and rectifiers to decrease the voltage to the desired level. It works according to Faraday's law of Electromagnetic induction.
The current in the windings of a step down transformer is inversely proportional to the voltage in windings as:
Input voltage / Output voltage = Output current / Input current
220 / 110 = Outout current / 15
Output current = 30A
Hence, the output current (30A) obtained is higher than the given input current (15A).
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A computer monitor accelerates electrons between two plates and sends them at high speed to form an image on the screen. If the elec- trons gain 4.1 * 10-15 J of kinetic energy as they go from one accelerat- ing plate to the other, what is the voltage between the plates?
The voltage between the plates is 3.9 × [tex]10^-^3[/tex] V.
The work-energy theorem states that the delta in the equation equals the change in kinetic energy plus the change in potential energy. Here, a charge's potential energy is expressed as qV, where V is the position's electric potential. The greater the change in voltage per unit distance, the greater the electric field.
The kinetic energy of the electrons = 4.1 × [tex]10^-^1^5[/tex] J
Charge of the electron = 1.602 × [tex]10^-^1^9[/tex] coulomb
Using,
ΔU = q × ΔV
4.1 × [tex]10^-^1^5[/tex] = 1.602 × [tex]10^-^1^9[/tex] × ΔV
ΔV = 3.9 × [tex]10^-^3[/tex] V
Therefore, the voltage between the plates is 3.9 × [tex]10^-^3[/tex] V.
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Three parallel wires each carry current I in the directions shown in (Figure 1). The separation between adjacent wires is d.
a) Calculate the magnitude of the net magnetic force per unit length on the top wire.
b) Calculate the magnitude of the net magnetic force per unit length on the middle wire.
c) Calculate the magnitude of the net magnetic force per unit length on the bottom wire.
(a) The magnitude of the net magnetic force per unit length on the top wire is μI²/πd.
(b) The magnitude of the net magnetic force per unit length on the middle wire is zero.
(c) The magnitude of the net magnetic force per unit length on the bottom wire is 3μI²/4πd.
Force per unit length
The magnitude of the net magnetic force per unit length on the top wire is calculated as follows;
F₁/L = (μI₁/2π) x (I₂/d + I₃/d)
F₁/L = (μI/2π) x (I/d + I/d)
F₁/L = (μI/2π) x (2I/d)
F₁/L = μI²/πd
The magnitude of the net magnetic force per unit length on the middle wire is calculated as follows;
F₂/L = (μI₂/2π) x (I₃/d - I₁/d)
F₂/L = (μI/2π) x (I/d - I/d) = 0
The magnitude of the net magnetic force per unit length on the middle bottom is calculated as follows;
F₃/L = (μI₂/2π) x (I₁/d + I₂/d)
F₃/L = (μI/2π) x (I/2d + I/d)
F₃/L = (μI/2π) x (3I/2d)
F₃/L = 3μI²/4πd
Thus, the magnitude of the net magnetic force per unit length on the top wire is μI²/πd.
The magnitude of the net magnetic force per unit length on the middle wire is zero.
The magnitude of the net magnetic force per unit length on the bottom wire is 3μI²/4πd.
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A copper transmission cable 50.0 km long and 10.0 cm in diameter carries a current of 105 A. What is the potential drop across the cable? Let ρcopper = 1.72 × 10—8 Ω • m.
A) 5.75 V
B) 5.48 V
C) 11.5 V
D) 16.9 V
5.48 V is the potential drop across the cable for a copper transmission cable of 50.0 km long and 10.0 cm in diameter carries a current of 105 A
Ohm's Law states that the potential drop is determined by the equation: V = IR, where I is the current and R is the wire resistance.
R=PL/A
Under the assumption that all physical parameters and temperatures remain constant, Ohm's law asserts that the voltage across a conductor is directly proportional to the current flowing through it.
Only when the given temperature and the other physical variables remain constant does Ohm's law apply. Increasing the current causes the temperature to rise in some components. The filament of a light bulb serves as an illustration of this, where the temperature increases as the current increases. Ohm's law cannot be applied in this situation. The filament of the lightbulb defies Ohm's Law.
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In 1656, the Burgmeister (mayor) of the town of Magdeburg, Germany, Otto Von Guericke, carried out a dramatic demonstration of the effect resulting from evacuating air from a container. It is the basis for this problem. Two steel hemispheres of radius 0.430 m (1.41 feet) with a rubber seal in between are placed together and air pumped out so that the pressure inside is 15.00 millibar. The atmospheric pressure outside is 940 millibar.
1. Calculate the force required to pull the two hemispheres apart. [Note: 1 millibar=100 N/m2. One atmosphere is 1013 millibar = 1.013×105 N/m2 ]
2. Two equal teams of horses, are attached to the hemispheres to pull it apart. If each horse can pull with a force of 1450N (i.e., about 326 lbs), what is the minimum number of horses required?
The force required to pull the two hemispheres apart is 4.2×10⁴ N and 29 number of horses are needed to pull these hemispheres apart.
What's the expression of force in terms of pressure?Mathematically, force = pressure/areaTotal area of the two hemispheres = 4π×(0.43)²= 2.3 m²Total pressure on the hemispheres= 15 milibar (directed inward) + 940 milibar (atmospheric pressure) = 955 milibar=955×100 N/m²= 9.55×10⁴ N/m²
Force on the hemispheres= 9.55×10⁴/2.3 = 4.2×10⁴ NWhat's the minimum number of horses required to get 4.2×10⁴ N of force, if each horse can pull with a force of 1450N?No. of horses required to separate the hemispheres = 4.2×10⁴/1450 = 29
Thus, we can conclude that the 29 horses are needed to pull the two hemispheres with a force of 4.2×10⁴ N.
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A bullet of mass 50g moving with an initial speed of 500m/s penetrates a wall and comes to rest at in 0.2seconds. calculate the deceleration of the bullet over the 0.2second.
The deceleration of the bullet over 0.2 second, given the data from the question is –2500 m/s²
What is acceleration?This is defined as the rate of change of velocity which time. It is expressed as
a = (v – u) / t
Where
a is the acceleration
v is the final velocity
u is the initial velocity
t is the time
NOTE: Deceleration is the opposite of acceleration
With the above equation for acceleration, we can obtain the deceleration of the bullet. Details below:
How to determine the deceleration of the bulletInitial velocity (u) = 500 m/sFinal velocity (v) = 0 m/sTime (t) = 0.2 sDecelration (a) =?a = (v – u) / t
a = (0 – 500) / 0.2
a = –500 / 0.2
a = –2500 m/s²
Thus, the deceleration of the bullet is –2500 m/s²
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37 POINTS PLEASE HELP MEE. The distance from A to B consists of an uphill section and a downhill section. A cyclist rides from A to B and then returns to A for a period of 4 hours and 30 minutes. When traveling as well as when returning the uphill speed is 15 km / h and the speed at the downhill is 20 km / h. Do you calculate the length of the distance from A to B?
The distance from A to B is 39.375 Km.
What is the total distance travelled by the cyclist?
The total distance travelled by the cyclist is given by the formula below:
Total distance = average speed * total time takenTotal distance = (15 + 20)/2 * 4.5
Total distance = 78.75 Km
Thus, distance from A to B = 78.75/2 = 39.375 Km
In conclusion, the total distance travelled is determined from the average speed and total time taken.
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What are the balanced forces for someone in a parachute?
A. Gravity and air resistance
B. Centripetal and air resistance
C. Gravity and centripetal
D. Gravity and Earth
A flat circular coil carrying a current of 8.80 A has a magnetic dipole moment of 0.194 A⋅m2 to the left. Its area vector A⃗ is 4.0 cm2 to the left.
a) How many turns does the coil have?
b) An observer is on the coil's axis to the left of the coil and is looking toward the coil. Does the observer see a clockwise or counterclockwise current?
c) If a huge 45.0 T external magnetic field directed out of the paper is applied to the coil, what magnitude of torque results?
d) What direction of torque results?
Hi there!
a)
We can use the equation for the magnetic dipole moment to solve for the number of turns:
[tex]\mu_m = NIA\vec{n}[/tex]
[tex]\mu_m[/tex] = Magnetic dipole moment (0.194 Am²)
N = Number of loops (?)
A = Area of loop (4.0 cm²)
[tex]\vec{n}[/tex] denotes the area vector, or the normal line perpendicular to the area.
We first need to convert cm² to m² using dimensional analysis.
[tex]4.0 cm^2 * \frac{0.01m}{1 cm} * \frac{0.01 m}{1cm} = 0.0004 m^2[/tex]
Rearranging the equation to solve for 'N':
[tex]N = \frac{\mu_m}{IA}\\\\N = \frac{0.194}{(8.8)(0.0004)} = \boxed{55.11 \text{ turns}}[/tex]
**Since we cannot have part of a turn, the coil has about 55 turns.
b)
For this, we can use the Right-Hand-Rule for current. Looking at the coil from the left with your curled fingers going around the coil with the fingertips pointing through and to the left in the direction of the magnetic moment, your thumb points in the COUNTERCLOCKWISE direction.
c)
Now, let's use the equation for the torque produced by a magnetic field:
[tex]\tau = \mu_m \times B[/tex]
This is a cross-product, but since our magnetic field is perpendicular to the magnetic moment, we can disregard it.
Plugging in the values for the magnetic moment and the magnetic field:
[tex]\tau = 0.194 * 45 = \boxed{8.73 Nm}[/tex]
d)
Using the other RHR (current, field, force), the coil will spin about its vertical axis in the field. In more detail, if you look at the coil from the left-hand side with its opening towards you, from this perspective, the left of the coil will come towards you, and the right side of the coil will move away.
A 269 kg weather balloon is designed to lift a 2910 kg package. What volume should the balloon have after being inflated with helium in order that the total load can be lifted? (Assume standard temperature and pressure, at which the density of air is ρair = 1.29 kg/m3 and the density of helium is ρHe = 0.179 kg/m3.)
The volume of the helium balloon in order to lift the weight is 17,760m³.
To find the answer, we need to know about the buoyant force.
What's the buoyant force?When a lighter object is kept in a higher density medium, it experiences a force along upward by that medium. This is buoyant force.Mathematically, buoyant force= density × volume of the object×gWhat's the volume of helium required to lift the 269kg weather balloon and 2910kg package?To lift the weight, the buoyant force must equal to the weight.If V is the volume of helium, buoyant force= V×0.179×gSo, V×0.179×g = (269+2910)g=> V= 3179/0.179 = 17,760m³
Thus, we can conclude that the volume of the helium balloon in order to lift the weight is 17,760m³.
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A 269 kg weather balloon is designed to lift a 2910 kg package. What volume should the balloon have after being inflated with helium in order that the total load can be lifted? (Assume standard temperature and pressure, at which the density of air is ρ[tex]_{air}[/tex] = 1.29 kg//[tex]m^{3}[/tex] and the density of helium is ρ[tex]_{He}[/tex] = 0.179 kg/[tex]m^{3}[/tex].)
As, per the buoyancy force,the volume that the balloon should have is 2863[tex]m^{3}[/tex]
What is buoyancy force?
Air buoyancy is the upward force exerted on an object by the air that is displaced by object. Air buoyancy is responsible for the buoyancy created by the displaced air.
[tex]F_{b}[/tex] = -Vρg , where V= volume of the object
ρ = density of the object
g = acceleration due to gravity
[tex]F_{b}[/tex] = buoyant force
The buoyancy force must be equal to the total load lifted
ρ[tex]_{He}[/tex] × V × g + 269 + 2910 = ρ[tex]_{air}[/tex] × V × g
0.179 × V + 3179 = 1.29 V
0.179V + 3179 = 1.29V
0.179V- 1.29V = - 3179
1.11V = 3179
On solving , we get
V = 2863 [tex]m^{3[/tex]
Therefore, the volume that the balloon should have is 2863[tex]m^{3}[/tex].
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An astronaut on a spacewalk 200km above Earth drops a hammer (mass 2kg), which goes into orbit about the Earth (radius 6,400km) How long does it take the hammer to orbit the Earth?
Answer:
See below
Explanation:
T = period of orbit = sqrt ( 4 pi^2 r^3 / (G Me) )
G = gravitational constant 6.6743 x 10^-11 m^3 / (kg-s^2)
Me = mass of earth = 6 x 10^24 kg
r = radius = 6600 km = 6,600,000 m
plug in the values to find T = 5323.75 seconds
(check my math)
M1 is a spherical mass (25.0 kg) at the origin. M2 is also a spherical mass (10.6 kg) and is located on the x-axis at x = 94.8 m. At what value of x would a third mass with a 19.0 kg mass experience no net gravitational force due to M1 and M2?
The point where m3 experiences a zero net gravitational force due to M1 and m2 is 57.42 m.
Position of the third mass
m1<------(x)------> m3 <-----------(94.8 m - x)-------->m2
a point, x, where m3 experiences a zero net gravitational force due to M1 and m2;
Force on m3 due to m1 = Force on m3 due to m2
Gm1m3/d² = Gm2m3/r²
m1/d² = m2/r²
where;
d is the distance between m1 and m3 = xr is the distance between m3 and m2 = 94.8 - xm1/(x²) = m2/(94.8 - x)²
m1(94.8 - x)² = m2x²
(94.8 - x)² = (m2/m1)x²
(94.8 - x)² = (10.6/25)x²
(94.8 - x)² = 0.424x²
(94.8 - x)² = (0.651)²x²
94.8 - x = 0.651x
94.8 = 1.651x
x = 94.8/1.651
x = 57.42 m
Thus, the point where m3 experiences a zero net gravitational force due to M1 and m2 is 57.42 m.
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A yellow train of mass 100 kg is moving at 8 m/s toward an orange train of mass 200 kg traveling in the opposite direction on the same track at a speed of 1 m/s. What is the initial momentum of the yellow and orange trains combined?
A. 200 kgm/s
B. 1000 kgm/s
C. 800 kgm/s
D. 600 kgm/s
The initial momentum of the yellow and the orange train is 1000kgm/s.
Momentum is the product of the mass and velocity of any object.
Momentum is denoted by P.
Momentum P = mv , where m = mass and v = velocity.
Given:Mass of the orange train = 200kg
Velocity of the orange train = 1m/s
So, the momentum of the orange train will be,
∴ P = mv
P = 200 x 1
P = 200 kgm/s
∴ The initial momentum of the orange train is 200kgm/s.
Mass of the yellow train = 100kg
Velocity of the yellow train = 8m/s
So, the momentum of the yellow train will be,
∴ P = mv
P = 100 x 8
P = 800 kgm/s
∴ The initial momentum of the yellow train is 800kgm/s.
Therefore, the initial momentum of the yellow and the orange train is 1000kgm/s.
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Two long, straight, parallel wires, 10.0 cm apart carry equal 4.00-A currents in the same direction, as shown in (Figure 1).
a) Find the magnitude of the magnetic field at point P1 , midway between the wires.
b) What is its direction?
c) Find the magnitude of the magnetic field at point P2 , 25.0 cm to the right of P1 .
d) What is its direction?
e) Find the magnitude of the magnetic field at point P3 , 20.0 cm directly above P1 .
f) What is its direction?
(a) The magnitude of the magnetic field at point P1 , midway between the wires is 1.005 x 10⁻⁴ T and the direction will be out of the page.
(b) The magnitude of the magnetic field at point P2 , 25.0 cm to the right of P1 is 2.67 x 10⁻⁶ T and the direction is into the page.
(c) The magnitude of the magnetic field at point P3 , 20.0 cm directly above P1 is 3.88 x 10⁻⁶ T and the direction is downwards.
Magnetic field midway between the wiresB = μ/2π[I₁/0.5r + I₂/0.5r]
B = (μ/2π) x (I/0.5r + I/0.5r)
B = (μ/2π) x (2I/0.5r)
B = μI/0.5r
B = 2μI/r
where;
I is current in the wiresr is the distance between the wiresB = (2 x 4π x 10⁻⁷ x 4)/(0.1)
B = 1.005 x 10⁻⁴ T
The direction of the magnetic field is out of the page.
Magnetic field at 25 cm right of P1B = μI/2πd
d = 5 cm + 25 cm = 30 cm
B = (4π x 10⁻⁷ x 4)/(2π x 0.3)
B = 2.67 x 10⁻⁶ T
The direction of the magnetic field is into the page towards P1.
Magnetic field at 20 above P1B = μI/2πd
d = √(20² + 5²)
d = 20.62 cm
B = (4π x 10⁻⁷ x 4)/(2π x 0.2062)
B = 3.88 x 10⁻⁶ T
The direction of the magnetic field is downwards towards P1.
Thus, the magnitude of the magnetic field at point P1 , midway between the wires is 1.005 x 10⁻⁴ T and the direction will be out of the page.
The magnitude of the magnetic field at point P2 , 25.0 cm to the right of P1 is 2.67 x 10⁻⁶ T and the direction is into the page.
The magnitude of the magnetic field at point P3 , 20.0 cm directly above P1 is 3.88 x 10⁻⁶ T and the direction is downwards.
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A wire carrying a 25.0 A current bends through a right angle. Consider two 2.00 mm segments of wire, each 3.00 cm from the bend (Figure 1).
a) Find the magnitude of the magnetic field these two segments produce at point P , which is midway between them.
b) Find the direction of the magnetic field at point P
The magnitude of the magnetic field and the direction of the magnetic field at point P is mathematically given as
B=1.9*10^{-5}T
To determine the magnetic field direction, use the right-hand -rule on the page magnetic field is going.
What is the magnitude of the magnetic field these two segments produce at point P, which is midway between them.?Generally, the equation for Biot savant law is mathematically given as
[tex]B=(\frac{u}{4\pi}*{\frac{Ilsin\theta}{r^2})[/tex]
Their net field is
Bn=2B
[tex]Bn=2* B=(\frac{u}{4\pi}*{\frac{Ilsin\theta}{r^2})[/tex]
Hence
[tex]B=(\frac{4*\p *10^{-7}}{4\pi}*{\frac{(30)(2*10^{-3}sin45)}{\sqrt{(3*10^{-2})^2+((3*10^{-2})^2)}/2})[/tex]
B=1.9*10^{-5}T
In conclusion, To determine the magnetic field direction, use the right-hand -rule on the page magnetic field is going.
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Answer:
The magnitude of the magnetic field at the point P is [tex]1.57*10^{-5}T[/tex] and the field is pointing into the page.
Explanation:
The general form of a similar question to this is:
[tex]\vec{B} = \frac{\mu _{0} }{4\pi } * \oint \frac{Id\vec{l} \times \hat{r}}{r^{2} }[/tex]
where [tex]\vec{B}[/tex] is the vector of the Magnetic Field, [tex]\mu _{0}[/tex] is the Free Space Permeability Constant (equal to [tex]4\pi * 10^{-7} \frac{N}{A^2}[/tex]), [tex]I[/tex] is the current, and [tex]r[/tex] is the distance from the segment to the point P. (I will get to the [tex]d\vec{l} \times \hat{r}[/tex] term in a bit)
This equation is fairly complicated. Luckily, it can be simplified by looking at the magnitude and direction separately.
The first thing to simplify is the cross product. Due to the fact that a cross product can be simplified from [tex]\vec{x} \times \vec{y}[/tex] to [tex]xy\sin(\theta)[/tex], where [tex]\theta[/tex] is the angle between the 2 vectors, and [tex]\hat{r}[/tex] is the unit vector of [tex]r[/tex] (i.e. [tex]\hat{r} = \vec{r}/r[/tex]) we can simplify [tex]d\vec{l} \times \hat{r}[/tex] to just [tex]dl \sin(\theta)[/tex].
Next, we will look at the integral. In this scenario, everything will function as a constant, so we can essentially ignore it.
Finally, [tex]\frac{\mu_{0}}{4\pi}[/tex] simplifies down to [tex]10^{-7}[/tex].
This gives us our new equation for the Magnetic Field produced by a single segment at a point:
[tex]B = \frac{Il\sin\theta}{r^{2}}*10^{-7}[/tex]
Now we need to find values for [tex]r[/tex] and [tex]\theta[/tex]. Luckily, we are dealing with a 45-45-90 triangle with sides of [tex]1.5 cm[/tex]. This means the distance [tex]r[/tex] is [tex](1.5\sqrt2)cm[/tex]! Similarly, because it is a 45-45-90 triangle, our [tex]\theta[/tex] is [tex]45\textdegree[/tex]!
Now we can start plugging things in:
[tex]B = \frac{(25A)(2*10^{-3}m)\sin(45\textdegree)}{(1.5\sqrt2*10^{-2}m)^2}*10^{-7}\frac{N}{A^2}[/tex]
[tex]B = 7.86^{-6} \frac{N}{A}[/tex] or [tex]T[/tex]
This is the magnitude due to only one single segment. In order to find the total field, we need to know the direction of the field due to each segment.
Finding the direction is really easy. Just use the right hand rule. Point your thumb in the direction of the current and curl the rest of your fingers around an imaginary pole. The direction your fingers point is the direction of the field. In this case, the field lines due to the segments point into the page in the 4th quadrant (the origin is the bend). This means that at point P, both segments induce the same field in the same direction. Therefore, we can take our value from before and double it, giving us our final answer:
[tex]B = 1.57*10^{-5} T[/tex]; into the page.
A circular loop of wire with radius 0.0410 m and resistance 0.169 Ω is in a region of spatially uniform magnetic field, as shown in the following figure (Figure 1). The magnetic field is directed out of the plane of the figure. The magnetic field has an initial value of 7.78 T and is decreasing at a rate of -0.605 T/s.
a) Is the induced current in the loop clockwise or counterclockwise?
b) What is the rate at which electrical energy is being dissipated by the resistance of the loop?
(a) The induced current in the loop will be counterclockwise.
(b) The rate at which electrical energy is being dissipated by the resistance of the loop is 0.012 W.
Direction of the currentThe induced current in the loop will be counterclockwise to the direction of magnetic field.
Emf induced in the loopemf = -NdФ/dt
emf = -NBA/dt
where;
A is area of the loop
A = πr² = π(0.041)² = 5.28 x 10⁻³ m²
emf = -(-0.605 - 7.78) x 5.28 x 10⁻³
emf = 8.385 x 5.28 x 10⁻³
emf = 0.0442 V
Rate of energy dissipationP = emf²/R
P = (0.0442)²/0.169
P = 0.012 W
Thus, the rate at which electrical energy is being dissipated by the resistance of the loop is 0.012 W.
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What does it mean when we say two objects are in equilibrium
An object is considered to be in a condition of equilibrium when it is balanced with regard to all external forces.
Equilibrium:
An object is considered to be in equilibrium if both its angular acceleration and the acceleration of its center of mass are equal to zero. In layman's terms: The item must either be at rest or moving at a constant speed if it is not accelerating because F = ma (force = mass x acceleration). Even in motion, a body can be in equilibrium. This kind of equilibrium is referred to as a dynamic equilibrium.
A weight suspended by a spring or a brick laying on a flat surface is an example. The equilibrium is unstable if the force with the smallest deviation tends to increase the displacement. As an example, imagine a ball bearing on the edge of a razor blade.
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A uniform meter stick of mass 0.20 kg is pivoted at the 40 cm mark. Where should one hang a mass of 0.50 kg to balance the stick?
Answer:
36 cm
Explanation:
Mass of stick; m1 = 0.20kg at midpoint.
Total length; L=1.0 m
Pivot at 0.40m
Atached mass m2 = 0.50kg
Applying rotational equilibrium we have;
Ʈnet = 0
(m1g) • r1 = (m2g) • r2
(0.2) (0.1m) = (0.5)(x)
x = 0.04m =4cm
measured away from 40cm mark gives a position on the stick of; 40cm - 4cm = 36 cm
A fisherman and his young son are in a boat on a small pond. Both are wearing life jackets. The son is holding a large helium filled balloon by a string. Consider each action below independently and indicate whether the level of the water in the pond, Rises, Falls, is Unchanged or Cannot tell.
The son pops the helium balloon.
The fisherman knocks the tackle box overboard and it sinks to the bottom.
The son finds a cup and bails some water out of the bottom of the boat
The fisherman lowers himself in the water and floats on his back.
The fisherman lowers the anchor and it hangs one foot above the bottom of the pond.
The son gets in the water, looses his grip on the string, letting the balloon escape upwards.
Based on the weight of the objects in the water, the water level in the given scenario is as follows;
unchangedrisesfallsunchangedrisesrisesWhat changes will be observed in each of the given scenario?The rise or fall of a fluid when an object is placed in it is determined by the density, mass, and volume of the object.
According to Archimedes' principle, the upthrust or upward force acting on a body fully or partially immersed in a fluid, is equal to the weight of the fluid displaced.
Considering the given situations:
The son pops the helium balloon - water level is unchanged because the weight of the balloon is negligible.Fisherman knocks the tackle box overboard and it sinks to the bottom - water level will rise since the tackle box has significant weightSon finds a cup and bails some water out of the bottom of the boat - water level will fall since some volume of water is being removed and added to the boat.Fisherman lowers himself in the water and floats on his back - water level is unchanged because no additional weight is added to the waterFisherman lowers the anchor and it hangs one foot above the bottom of the pond - water level will rise since the anchor will displace some volume of water equal to its weight in the pondSon gets in the water, looses his grip on the string, letting the balloon escape upwards- water level will rise since the weight of the balloon and string are significant.In conclusion, the rise in water level depends on the weight of the objects.
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A mechanic pushes a 2.30 ✕ 103-kg car from rest to a speed of v, doing 4,800 J of work in the process. During this time, the car moves 30.0 m. Neglecting friction between car and road, find v and the horizontal force exerted on the car.
(a) the speed v
______m/s
(b) the horizontal force exerted on the car (Enter the magnitude.)
_______N
The horizontal force applied is 160 N while the velocity is 2.03 m/s.
What is the speed of the car?The work done by the car is obtained as the product of the force and the distance;
W = F x
F = ?
x = 30.0 m
W = 4,800 J
F = 4,800 J/30.0 m
F = 160 N
But F = ma
a = F/m
a = 160 N/2.30 ✕ 10^3-kg
a= 0.069 m/s
Now;
v^2 = u^2 + 2as
u = 0/ms because the car started from rest
v = √2as
v = √2 * 0.069 * 30
v = 2.03 m/s
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What do an electron and a neutron have in common?
Answer:
To give light yo people and the neutron are to give you health care
Q2. A force of 1000N is experienced by a charge in a field of strength of 250NC. Find the value of the charge establishing the field.
Answer:
4 C
Explanation:
The strength of an electric field can be defined as: [tex]E = \frac{f}{q}[/tex] where f=force and q=charge, and e=strength
Since we're given the strength and the force we can simply rearrange the equation so that we solve for Q:
Original Equation:
[tex]E=\frac{f}{q}[/tex]
Multiply both sides by q
[tex]E*q = f[/tex]
Divide both sides by E
[tex]q=\frac{f}{e}[/tex]
So now plug the known values into the equation
[tex]q=\frac{1000 N}{250 N/C}[/tex]
Simplify:
[tex]q = 4 c[/tex]
A projectile leaves the ground at an angle of 60° the horizontal.Its kinetic energy is E.Neglecting air resistance, find in terms of E its kinetic energy at the highest point of the motion
The kinetic energy of the projectile at the highest point of its motion will be E/4.
What is Projectile Motion?When a projectile will be thrown obliquely near the surface of the earth, it travels a curved path under uniform acceleration directed toward the center of the Earth. The path of a particle is called a projectile while the motion of a projectile is projectile motion.
Given, the angle of projection with horizontal, [tex]\theta = 60 ^o[/tex]
Consider that 'E' is the initial value of the kinetic energy of the projectile.
The equation for the initial kinetic energy is : [tex]E =\frac{1}{2}mu^2[/tex]
where m is the mass of the given projectile.
The component of the velocity of the projectile in the horizontal direction:
uₓ = u cosθ
uₓ = u cos 60°
uₓ = u/2
From the equation of motion: v = u +at
v = (u/2) + (0) t
v = u/2
The final kinetic energy of the projectile:
[tex]E_f = \frac{1}{2}mv^2[/tex]
[tex]E_f = \frac{1}{2}m(\frac{u}{2} )^2[/tex]
[tex]E_f = \frac{1}{4} (\frac{1}{2}mu^2 )[/tex]
[tex]E_f = \frac{E}{4}[/tex]
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Calculate the amount of air in a room 6m long, 5m wide and 3mm high.
Answer:
0.09kg of air
Explanation:
The dimensions of the room are given
change the height to meters by dividing it by thousand.
For the volume multiplying the length,width and height (all should be in the same unit most suitable being meters).
Volume refers to the amount of space inside a box or a object.
The amount of air is equal to the volume.
Answer:
90 m^3
Explanation:
Volume of the room:
6 m * 5 m * 3 m = 90 m^3 <=====( I changed 3mm to 3 m)
if 3mm is not a typo mistake
volume becomes ( 3 mm = .003 m)
6 m * 5 m * .003 m = .09 m^3 ( though unlikely )