Answer:
The magnetic field at a distance of 19.8 cm from the wire is 1.591 mT
Explanation:
Given;
first magnetic field at first distance, B₁ = 2.50 mT
first distance, r₁ = 12.6 cm = 0.126 m
Second magnetic field at Second distance, B₂ = ?
Second distance, r₂ = ?
Magnetic field for a straight wire is given as;
[tex]B = \frac{\mu I}{2 \pi r}[/tex]
Where:
μ is permeability
B is magnetic field
I is current flowing in the wire
r distance to the wire
[tex]Let \ \frac{\mu I}{2\pi} \ be \ constant; = K\\\\B = \frac{K}{r} \\\\K = Br\\\\B_1r_1 = B_2r_2\\\\B_2 =\frac{B_1r_1}{r_2} \\\\B_2 = \frac{2.5*10^{-3} *0.126}{0.198} \\\\B_2 = 1.591 *10^{-3}\ T\\\\B_2 = 1.591 \ mT[/tex]
Therefore, the magnetic field at a distance of 19.8 cm from the wire is 1.591 mT
Sophie throws a tennis ball down from a height of 1.5 m at an angle of 450 with respect to vertical. She drops another tennis ball from the same height. Use the Energy Interaction Model to predict which ball will hit the ground with greater speed.
Given that,
Height =1.5 m
Angle = 45°
We need to find the greater speed of the ball
Using conservation of energy
[tex]P.E_{i}+K.E_{f}=P.E_{f}+K.E_{f}[/tex]
[tex]mgh+\dfrac{1}{2}mv_{i}^2=mgh+\dfrac{1}{2}mv_{f}^2[/tex]
Here, initial velocity and final potential energy is zero.
[tex]mgh=\dfrac{1}{2}mv_{f}^2[/tex]
Put the value into the formula
[tex]9.8\times1.5=\dfrac{1}{2}v_{f}^2[/tex]
[tex]v_{f}^2=2\times9.8\times1.5[/tex]
[tex]v_{f}=\sqrt{2\times9.8\times1.5}[/tex]
[tex]v_{f}=5.42\ m/s[/tex]
Hence, the greater speed of the ball is 5.42 m/s.
The exit nozzle in a jet engine receives air at 1200 K, 150 kPa with negligible kinetic energy. The exit pressure is 80 kPa, and the process is reversible and adiabatic. Use constant specific heat at 300 K to find the
Complete question:
The exit nozzle in a jet engine receives air at 1200 K, 150 kPa with negligible kinetic energy. The exit pressure is 80 kPa, and the process is reversible and adiabatic. Use constant specific heat at 300 K to find the exit velocity.
Answer:
The exit velocity is 629.41 m/s
Explanation:
Given;
initial temperature, T₁ = 1200K
initial pressure, P₁ = 150 kPa
final pressure, P₂ = 80 kPa
specific heat at 300 K, Cp = 1004 J/kgK
k = 1.4
Calculate final temperature;
[tex]T_2 = T_1(\frac{P_2}{P_1})^{\frac{k-1 }{k}[/tex]
k = 1.4
[tex]T_2 = T_1(\frac{P_2}{P_1})^{\frac{k-1 }{k}}\\\\T_2 = 1200(\frac{80}{150})^{\frac{1.4-1 }{1.4}}\\\\T_2 = 1002.714K[/tex]
Work done is given as;
[tex]W = \frac{1}{2} *m*(v_i^2 - v_e^2)[/tex]
inlet velocity is negligible;
[tex]v_e = \sqrt{\frac{2W}{m} } = \sqrt{2*C_p(T_1-T_2)} \\\\v_e = \sqrt{2*1004(1200-1002.714)}\\\\v_e = \sqrt{396150.288} \\\\v_e = 629.41 \ m/s[/tex]
Therefore, the exit velocity is 629.41 m/s
A wire with mass 90.0g is stretched so that its ends are tied down at points 88.0cm apart. The wire vibrates in its fundamental mode with frequency 80.0Hz and with an amplitude of 0.600cm at the antinodes.a) What is the speed of propagation of transverse waves in the wire?b) Compute the tension in the wire.
Answer:
a) V = 140.8 m/s
b) T = 2027.52 N = 2.03 KN
Explanation:
a)
The formula for the speed of the wave is given as follows:
f₁ = V/2L
V = 2f₁L
where,
V = Speed of Wave = ?
f₁ = Fundamental Frequency = 80 Hz
L = Length of Wire = 88 cm = 0.88 m
Therefore,
V = (2)(80 Hz)(0.88 m)
V = 140.8 m/s
b)
Another formula for the speed of wave is:
V = √T/μ
V² = T/μ
T = V²μ
where,
T = Tension in String = ?
μ = Linear Mass Density of Wire = Mass of Wire/L = 0.09 kg/0.88 m
μ = 0.1 kg/m
Therefore,
T = (140.8 m/s)²(0.1 kg/m)
T = 2027.52 N = 2.03 KN
An electron moves at a speed of 1.0 x 104 m/s in a circular path of radius 2 cm inside a solenoid. The magnetic field of the solenoid is perpendicular to the plane of the electron’s path. Calculate (a) the strength of the magnetic field inside the solenoid and (b) the current in the solenoid if it has 25 turns per centimeter.
Answer:
(a) B = 2.85 × [tex]10^{-6}[/tex] Tesla
(b) I = I = 0.285 A
Explanation:
a. The strength of magnetic field, B, in a solenoid is determined by;
r = [tex]\frac{mv}{qB}[/tex]
⇒ B = [tex]\frac{mv}{qr}[/tex]
Where: r is the radius, m is the mass of the electron, v is its velocity, q is the charge on the electron and B is the magnetic field
B = [tex]\frac{9.11*10^{-31*1.0*10^{4} } }{1.6*10^{-19}*0.02 }[/tex]
= [tex]\frac{9.11*10^{-27} }{3.2*10^{-21} }[/tex]
B = 2.85 × [tex]10^{-6}[/tex] Tesla
b. Given that; N/L = 25 turns per centimetre, then the current, I, can be determined by;
B = μ I N/L
⇒ I = B ÷ μN/L
where B is the magnetic field, μ is the permeability of free space = 4.0 ×[tex]10^{-7}[/tex]Tm/A, N/L is the number of turns per length.
I = B ÷ μN/L
= [tex]\frac{2.85*10^{-6} }{4*10^{-7} *25}[/tex]
I = 0.285 A
An electron moving in a direction perpendicular to a uniform magnetic field at a speed of 1.6 107 m/s undergoes an acceleration of 7.0 1016 m/s2 to the right (the positive x-direction) when its velocity is upward (the positive y-direction). Determine the magnitude and direction of the field.
Answer:
B = 0.024T positive z-direction
Explanation:
In this case you consider that the direction of the motion of the electron, and the direction of the magnetic field are perpendicular.
The magnitude of the magnetic force exerted on the electron is given by the following formula:
[tex]F=qvB[/tex] (1)
q: charge of the electron = 1.6*10^-19 C
v: speed of the electron = 1.6*10^7 m/s
B: magnitude of the magnetic field = ?
By the Newton second law you also have that the magnetic force is equal to:
[tex]F=qvB=ma[/tex] (2)
m: mass of the electron = 9.1*10^-31 kg
a: acceleration of the electron = 7.0*10^16 m/s^2
You solve for B from the equation (2):
[tex]B=\frac{ma}{qv}\\\\B=\frac{(9.1*10^{-31}kg)(7.0*10^{16}m/s^2)}{(1.6*10^{-19}C)(1.6*10^7m/s)}\\\\B=0.024T[/tex]
The direction of the magnetic field is found by using the right hand rule.
The electron moves upward (+^j). To obtain a magnetic forces points to the positive x-direction (+^i), the direction of the magnetic field has to be to the positive z-direction (^k). In fact, you have:
-^j X ^i = ^k
Where the minus sign of the ^j is because of the negative charge of the electron.
Then, the magnitude of the magnetic field is 0.024T and its direction is in the positive z-direction
A depiction of a famous scientific experiment is given. Consider how the beam changes when the magnet is off compared to when the magnet is on. A bell-shaped evacuated glass tube with a narrow end and a wide end is connected to a battery at the narrow end. In the center of the tube there is a negatively charged plate above the tube, a positively charged plate below the tube, and a magnet with the field turned off. A beam originating at the narrow end of the tube travels toward the wide end of the tube. With the magnetic field turned off, the beam path bends toward the positively charged plate and ends at the lower half of the wide end of the tube. A bell-shaped evacuated glass tube with a narrow end and a wide end is connected to a battery at the narrow end. In the center of the tube there is a negatively charged plate above the tube, a positively charged plate below the tube, and a magnet with the field turned n. A beam originating at the narrow end of the tube travels toward the wide end of the tube. With the magnetic field turned on, the beam path travels in a straight path to the center of the wide end of the tube. What type of beam was used in this experiment?
Answer:
The beam used is a negatively charged electron beam with a velocity of
v = E / B
Explanation:
After reading this long statement we can extract the data to work on the problem.
* They indicate that when the beam passes through the plates it deviates towards the positive plate, so the beam must be negative electrons.
* Now indicates that the electric field and the magnetic field are contracted and that the beam passes without deviating, so the electric and magnetic forces must be balanced
[tex]F_{e} = F_{m}[/tex]
q E = qv B
v = E / B
this configuration is called speed selector
They ask us what type of beam was used.
The beam used is a negatively charged electron beam with a velocity of v = E / B
An airplane flies between two points on the ground that are 500 km apart. The destination is directly north of the origination of the flight. The plane flies with an air speed of 120 m/s. If a constant wind blows at 10.0 m/s due west during the flight, what direction must the plane fly relative to north to arrive at the destination? Consider: east to the right, west to the left, north upwards and south downwards
Answer:
θ = 4.78º
with respect to the vertical or 4.78 to the east - north
Explanation:
This is a velocity compound exercise since it is a vector quantity.
The plane takes a direction, the air blows to the west and the result must be to the north, let's use the Pythagorean theorem to find the speed
v_fly² = v_nort² + v_air²
v_nort² = v_fly² + - v_air²
Let's use trigonometry to find the direction of the plane
sin θ = v_air / v_fly
θ = sin⁻¹ (v_air / v_fly)
let's calculate
θ = sin⁻¹ (10/120)
θ = 4.78º
with respect to the vertical or 4.78 to the north-east
The index of refraction for a certain type of glass is 1.645 for blue light and 1.609 for red light. A beam of white light (one that contains all colors) enters a plate of glass from the air, nair≈1, at an incidence angle of 38.55∘. What is the absolute value of ????, the angle in the glass between blue and red parts of the refracted beams?
Answer:
blue θ₂ = 22.26º
red θ₂ = 22.79º
Explanation:
When a light beam passes from one material medium to another, it undergoes a deviation from the path, described by the law of refraction
n₁ sin θ₁ = n₂ sin θ₂
where n₁ and n₂ are the incident and transmitted media refractive indices and θ are the angles in the media
let's apply this equation to each wavelength
λ = blue
in this case n₁ = 1, n₂ = 1,645
sin θ₂ = n₁/ n₂ sin₂ θ₁
let's calculate
sin θ₂ = 1 / 1,645 sint 38.55
sin θ₂ = 0.37884
θ₂ = sin⁻¹ 0.37884
θ₂ = 22.26º
λ = red
n₂ = 1,609
sin θ₂ = 1 / 1,609 sin 38.55
sin θ₂ = 0.3873
θ₂ = sim⁻¹ 0.3873
θ₂ = 22.79º
the refracted rays are between these two angles
A force in the negative x-direction is applied for 27 ms to a 0.4 kg mass initially moving at 14 m/s in the x-direction. The force varies in magnitude and delivers an impulse with a magnitude of 32.4 N-s. What is the mass's velocity in the x-direction
Answer:
-67 m/s
Explanation:
We are given that
Mass of ball,m=0.4 kg
Initial speed,u=14 m/s
Impulse,I=-32.4 N-s
Time,t=27 ms=[tex]27\times 10^{-3} s[/tex]
We have to find the mass's velocity in the x- direction.
We know that
[tex]Impulse=mv-mu[/tex]
Substitute the values
[tex]-32.4=0.4v-0.4(14)[/tex]
[tex]-32.4+0.4(14)=0.4 v[/tex]
[tex]-26.8=0.4v[/tex]
[tex]v=\frac{-26.8}{0.4}=-67m/s[/tex]
Assume you have a rocket in Earth orbit and want to go to Mars. The required change in velocity is ΔV≈9.6km/s . There are two options for the propulsion system --- chemical and electric --- each with a different specific impulse. Recall that the relationship between specific impulse and exhaust velocity is: Vex=g0Isp Using the Ideal Rocket Equation and setting g0=9.81m/s2 , calculate the propellant fraction required to achieve the necessary ΔV for each of propulsion system. Part 1: Cryogenic Chemical Propulsion First, consider a cryogenic chemical propulsion system with Isp≈450s . Enter the required propellant fraction as a proportion with at least 2 decimal places (i.e., enter 0.25 to represent 25%): incorrect Part 2: Electric Propulsion Next, consider an electric propulsion system with Isp≈2000s . Enter the required propellant fraction as a proportion with at least 2 decimal places (i.e., enter 0.25 to represent 25%):
Answer: Part 1: Propellant Fraction (MR) = 8.76
Part 2: Propellant Fraction (MR) = 1.63
Explanation: The Ideal Rocket Equation is given by:
Δv = [tex]v_{ex}.ln(\frac{m_{f}}{m_{e}} )[/tex]
Where:
[tex]v_{ex}[/tex] is relationship between exhaust velocity and specific impulse
[tex]\frac{m_{f}}{m_{e}}[/tex] is the porpellant fraction, also written as MR.
The relationship [tex]v_{ex}[/tex] is: [tex]v_{ex} = g_{0}.Isp[/tex]
To determine the fraction:
Δv = [tex]v_{ex}.ln(\frac{m_{f}}{m_{e}} )[/tex]
[tex]ln(MR) = \frac{v}{v_{ex}}[/tex]
Knowing that change in velocity is Δv = 9.6km/s and [tex]g_{0}[/tex] = 9.81m/s²
Note: Velocity and gravity have different measures, so to cancel them out, transform km in m by multiplying velocity by 10³.
Part 1: Isp = 450s
[tex]ln(MR) = \frac{v}{v_{ex}}[/tex]
ln(MR) = [tex]\frac{9.6.10^{3}}{9.81.450}[/tex]
ln (MR) = 2.17
MR = [tex]e^{2.17}[/tex]
MR = 8.76
Part 2: Isp = 2000s
[tex]ln(MR) = \frac{v}{v_{ex}}[/tex]
ln (MR) = [tex]\frac{9.6.10^{3}}{9.81.2.10^{3}}[/tex]
ln (MR) = 0.49
MR = [tex]e^{0.49}[/tex]
MR = 1.63
The force a spring exerts on a body is a conservative force because:
a. a spring always exerts a force parallel to the displacement of the body.
b. the work a spring does on a body is equal for compressions and extensions of equal magnitude.
c. the net work a spring does on a body is zero when the body returns to its initial position.
d. the work a spring does on a body is equal and opposite for compressions and extensions of equal magnitude.
e. a spring always exerts a force opposite to the displacement of the body.
Answer:
c. the net work a spring does on a body is zero when the body returns to its initial position
Explanation:
A force is conservative when the net work done over any path that returns to the initial position is zero. Choice C matches that definition.
An ideal spring of the kind used in physics problems has the characteristic that it applies the same force at the same distance always. So any work required to extend or compress the spring is reversed when the reverse motion takes place.
A hockey puck slides off the edge of a horizontal platform with an initial velocity of 28.0 m/shorizontally in a city where the acceleration due to gravity is 9.81 m/s 2. The puck experiences no significant air resistance as it falls. The height of the platform above the ground is 2.00 m. What is the angle below the horizontal of the velocity of the puck just before it hits the ground
Answer:
θ = 12.60°
Explanation:
In order to calculate the angle below the horizontal for the velocity of the hockey puck, you need to calculate both x and y component of the velocity of the puck, and also you need to use the following formula:
[tex]\theta=tan^{-1}(\frac{v_y}{v_x})[/tex] (1)
θ: angle below he horizontal
vy: y component of the velocity just after the puck hits the ground
vx: x component of the velocity
The x component of the velocity is constant in the complete trajectory and is calculated by using the following formula:
[tex]v_x=v_o[/tex]
vo: initial velocity = 28.0 m/s
The y component is calculated with the following equation:
[tex]v_y^2=v_{oy}^2+2gy[/tex] (2)
voy: vertical component of the initial velocity = 0m/s
g: gravitational acceleration = 9.8 m/s^2
y: height
You solve the equation (2) for vy and replace the values of the parameters:
[tex]v_y=\sqrt{2gy}=\sqrt{2(9.8m/s^2)(2.00m)}=6.26\frac{m}{s}[/tex]
Finally, you use the equation (1) to find the angle:
[tex]\theta=tan^{-1}(\frac{6.26m/s}{28.0m/s})=12.60\°[/tex]
The angle below the horizontal is 12.60°
The angle below the horizontal of the velocity of the puck just before it hits the ground is 12.60°.
Given the following data:
Initial velocity = 28.0 m/s Acceleration due to gravity = 9.81 [tex]m/s^2[/tex]Displacement (height) = 2.00 meters.To find the angle below the horizontal of the velocity of the puck just before it hits the ground:
First of all, we would determine the horizontal and vertical components of the hockey puck.
For horizontal component:
[tex]V_y^2 = U_y^2 + 2aS\\\\V_y^2 = 0^2 + 2(9.81)(2)\\\\V_y^2 = 39.24\\\\V_y = \sqrt{39.24} \\\\V_y = 6.26 \; m/s[/tex]
For vertical component:
[tex]V_x = U_x\\\\V_x = 28.0 \;m/s[/tex]
Now, we can find the angle by using the formula:
[tex]\Theta = tan^{-1} (\frac{V_y}{V_x} )[/tex]
Substituting the values, we have:
[tex]\Theta = tan^{-1} (\frac{6.26}{28.0} )\\\\\Theta = tan^{-1} (0.2236)\\\\\Theta = 12.60[/tex]
Angle = 12.60 degrees.
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A 300 g bird flying along at 6.2 m/s sees a 10 g insect heading straight toward it with a speed of 35 m/s (as measured by an observer on the ground, not by the bird). The bird opens its mouth wide and enjoys a nice lunch.
Required:
What is the bird's speed immediately after swallowing?
Answer:
The velocity of the bird is [tex]v_f = 4.87 \ m/s[/tex]
Explanation:
From the question we are told that
The mass of the bird is [tex]m_1 = 300 \ g = 0.3 \ kg[/tex]
The initial speed of the bird is [tex]u_1 = 6.2 \ m/s[/tex]
The mass of the insect is [tex]m_2 = 10 \ g = 0.01 \ kg[/tex]
The speed of the insect is [tex]u_ 2 =-35 \ m/s[/tex]
The negative sign is because it is moving in opposite direction to the bird
According to the principle of linear momentum conservation
[tex]m_1 u_1 + m_2 u_2 = (m_1 + m_2 )v_f[/tex]
substituting values
[tex](0.3 * 6.2 ) + (0.01 * (-35)) = (0.3 + 0.01 )v_f[/tex]
[tex]1.51 = 0.31 v_f[/tex]
[tex]v_f = 4.87 \ m/s[/tex]
The Final velocity of Bird = 4.87 m/s
Mass of the bird = 300 g = 0.3 kg
Velocity of bird = 6.2 m/s
Momentum of Bird = Mass of bird [tex]\times[/tex] Velocity of Bird = 0.3 [tex]\times[/tex] 6.2 = 1.86 kgm/s
Mass of the insect = 10 g = 0.01 kg
Velocity of insect = - 35 m/s
Momentum of the Insect = Mass of Insect [tex]\times[/tex] Velocity of Insect = - 0.35 kgm/s
According to the law of conservation of momentum We can write that
In the absence of external forces on the system , the momentum of system remains conserved in that particular direction.
The bird opens the mouth and enjoys the free lunch hence
Let the final velocity of bird is [tex]v_f[/tex]
Initial momentum of the system = Final momentum of the system
1.86 -0.35 = [tex]v_f[/tex] ( 0.01 + 0.3 )
1.51 = [tex]v_f[/tex] 0.31
[tex]v_f[/tex] = 4.87 m/s
The Final velocity of Bird = 4.87 m/s
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What will happen to an astronaut when the jets produce these four forces
A car travels 2500 m in 8 minutes. Calculate the speed at which the car travelled
Answer:
5.95m/s to 2 decimal places
Explanation:
In physics speed is measured in metres per second so convert 8mins to seconds
8x60=420 seconds
The formula needed:
Speed (m/s)= Distance (m)/Time (s)
2500/420=5.95m/s
A disk between vertebrae in the spine is subjected to a shearing force of 640 N. Find its shear deformation taking it to have the shear modulus of 1.00 109 N/m2. The disk is equivalent to a solid cylinder 0.700 cm high and 4.30 cm in diameter.
Answer:
3.08*10^-6 m
Explanation:
Given that
Total shearing force, F = 640 N
Shear modulus, S = 1*10^9 N/m²
Height of the cylinder, L = 0.7 cm
Diameter of the cylinder, d = 4.3 cm
The solution is attached below.
We have our shear deformation to be 3.08*10^-6 m
A hockey puck on a frozen pond is given an initial speed of 20.0 m/s. If the puck always remains on the ice and slides 115 m before coming to rest, determine the coefficient of kinetic friction between the puck and ice.
Answer:
μ_k = 0.1773
Explanation:
We are given;
Initial velocity;u = 20 m/s
Final velocity;v = 0 m/s (since it comes to rest)
Distance before coming to rest;s = 115 m
Let's find the acceleration using Newton's second law of motion;
v² = u² + 2as
Making a the subject, we have;
a = (v² - u²)/2s
Plugging relevant values;
a = (0² - 20²)/(2 × 115)
a = -400/230
a = -1.739 m/s²
From the question, the only force acting on the puck in the x direction is the force of friction. Since friction always opposes motion, we see that:
F_k = −ma - - - (1)
We also know that F_k is defined by;
F_k = μ_k•N
Where;
μ_k is coefficient of kinetic friction
N is normal force which is (mg)
Since gravity acts in the negative direction, the normal force will be positive.
Thus;
F_k = μ_k•mg - - - (2)
where g is acceleration due to gravity.
Thus,equating equation 1 and 2,we have;
−ma = μ_k•mg
m will cancel out to give;
-a = μ_k•g
μ_k = -a/g
g has a constant value of 9.81 m/s², so;
μ_k = - (-1.739/9.81)
μ_k = 0.1773
The coefficient of kinetic friction between the hockey puck and ice is equal to 0.178
Given the following data:
Initial speed = 20 m/sFinal velocity = 0 m/s (since it came to rest)Distance = 115 mScientific data:
Acceleration due to gravity = 9.8 [tex]m/s^2[/tex]To determine the coefficient of kinetic friction between the hockey puck and ice:
First of all, we would calculate the acceleration of the hockey puck by using the third equation of motion.
[tex]V^2 = U^2 + 2aS\\\\0^2 =20^2 + 2a(115)\\\\-400=230a\\\\a=\frac{-400}{230}[/tex]
Acceleration, a = -1.74 [tex]m/s^2[/tex]
Note: The negative signs indicates that the hockey puck is slowing down or decelerating.
From Newton's Second Law of Motion, we have:
[tex]\sum F_x = F_k + F_n =0\\\\F_k =- F_n\\\\\mu mg =-ma\\\\\mu = \frac{-a}{g}\\\\\mu = \frac{-(-1.74)}{9.8}\\\\\mu = \frac{1.74}{9.8}[/tex]
Coefficient of kinetic friction = 0.178
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charged particles from the solar winds ultimately cause ___. a. the earth to maintain it's magnetic field b. the earth to change shape c. the auroras d. strong winds on earth
Answer:
The auroras C.
Explanation:
A 2.4-kg ball falling vertically hits the floor with a speed of 2.5 m/s and rebounds with a speed of 1.5 m/s. What is the magnitude of the impulse exerted on the ball by the floor
Answer:
9.6 Ns
Explanation:
Note: From newton's second law of motion,
Impulse = change in momentum
I = m(v-u).................. Equation 1
Where I = impulse, m = mass of the ball, v = final velocity, u = initial velocity.
Given: m = 2.4 kg, v = 2.5 m/s, u = -1.5 m/s (rebounds)
Substitute into equation 1
I = 2.4[2.5-(-1.5)]
I = 2.4(2.5+1.5)
I = 2.4(4)
I = 9.6 Ns
The magnitude of impulse will be "9.6 Ns".
According to the question,
Mass,
m = 2.4 kgFinal velocity,
v = 2.5 m/sInitial velocity,
u = -1.5 m/sBy using Newton's 2nd law of motion, we get
→ Impulse, [tex]I = m(v-u)[/tex]
By substituting the values, we get
[tex]= 2.4[2.5-(1.5)][/tex]
[tex]= 2.4(2.5+1.5)[/tex]
[tex]= 2.4\times 4[/tex]
[tex]= 9.6 \ Ns[/tex]
Thus the above answer is right.
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A total charge of 62 nC is uniformly distributed throughout a non-conducting sphere with a radius of 5.00 cm. The electric potential at r = 15.0 cm , relative to the potential far away, is:________
Answer:
2790 J/C
Explanation:
charge on sphere Q = 62 nC = [tex]62*10^{-9} C[/tex]
radius of the sphere r = 5.0 cm = 0.05 m
distance away from reference point d = 15.0 cm = 0.15 m
total distance of charge relative reference point R = r + d = 0.05 + 0.15 = 0.2 m
electric potential V is given as
[tex]V = \frac{kQ}{R}[/tex]
where k = Coulumb's constant = [tex]9*10^{9}[/tex] kg⋅m³⋅s⁻⁴⋅A⁻²
[tex]V = \frac{9*10^{9} * 62*10^{-9} }{0.2}[/tex] = [tex]\frac{9*62}{0.2}[/tex]
V = 2790 J/C
Select the correct answer. When waves travel through water, why do the moving water particles continue to return to their starting position? A. Waves transfer matter, but not energy. B. Waves transfer both matter and energy. C. Waves transfer energy, but not matter. D. Waves don’t transfer matter or energy.
Waves transfer energy, but not matter. (C)
Answer:
C. Waves transfer energy not matter
Explanation:
wave is a disturbance
Suppose that when you move the north pole of a bar magnetic toward a coil it induces a positive current in the coil. The strength of the field produced by an electromagnetic can be controlled electronically. Suppose you place a coil near the north pole of an electromagnet and increase the field while keeping everything stationary. Which one of the following will happen? a) A positive current will be induced in the coilb) A negative current will be induced in the coil c) No current will be induced in the coil since there is no relative motion.
Answer:
a) A positive current will be induced in the coil
Explanation:
Electromagnetic induction is the induction of an electric field on a conductor due to a changing magnetic field flux. The change in the flux can be by moving the magnet relative to the conductor, or by changing the intensity of the magnetic field of the magnet. In the case of this electromagnets, the gradual increase in the the electromagnet's field strength will cause a flux change, which will in turn induce an electric current on the coil.
According to Lenz law, the induced current acts in such a way as to negate the motion or action that is producing it. A positive current will be induced on the coil so as to repel any form of attraction between the north pole of the electromagnet and the coil. This law obeys the law of conservation of energy, since work has to be done to move the move them closer to themselves.
An object with a mass m slides down a rough 37° inclined plane where the coefficient of kinetic friction is 0.20. If the plane is 10 m long and the mass starts from rest, what will be its speed at the bottom of the plane?
Answer: 9.312 m/s
Explanation:
The friction force (opposite to the motion) is Fa = μ*m*g*cos(α) with μ = kinetic friction. The force that makes the motion is
F = m*g*sin(α).
The Newton's law gives:
F - Fa = m*a
m*g*sin(α) - μ*m*g*cos(α) = m*a
g*sin(α) - μ*g*cos(α) = a so a = 4.335 m/s²
It's a uniformly accelerated motion:
Space
S = 0.5*a*t²
10 = 0.5*a*t²
=> t = 2.148 s
Velocity
V = a*t = 9.312 m/s.
We have that the speed at the bottom of the plane is
[tex]v-9.3m/s[/tex]
From the question we are told that:
Angle of slide [tex]\theta =3.7 \textdegree[/tex]
Coefficient of kinetic friction [tex]\mu= 0.20[/tex]
Length [tex]L=10m[/tex]
Generally, the equation for acceleration along the slide is mathematically given by
[tex]a=gsin \theta-\mu cos\theta[/tex]
[tex]a=(9.8sin37-0.20*9.8*cos37[/tex]
[tex]a=4.33m/s^2[/tex]
Therefore
Velocity v is is mathematically given by
[tex]v=\sqrt{2as}[/tex]
[tex]v=\sqrt{2*4.33*10}[/tex]
[tex]v-9.3m/s[/tex]
In conclusion
The speed at the bottom of the plane is
[tex]v-9.3m/s[/tex]
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New evidence increasingly emphasizes that __________.
From a height of 40.0 m, a 1.00 kg bird dives (from rest) into a small fish tank containing 50.5 kg of water. Part A What is the maximum rise in temperature of the water if the bird gives it all of its mechanical energy
Answer:
0.00185 °C
Explanation:
From the question,
The potential energy of the bird = heat gained by the water in the fish tank.
mgh = cm'(Δt)................... Equation 1
Where m = mass of the bird, g = acceleration due to gravity, h = height, c = specific heat capacity of water, m' = mass of water, Δt = rise in temperature of water.
make Δt the subject of the equation
Δt = mgh/cm'............... Equation 2
Given: m = 1 kg, h = 40 m, m' = 50.5 kg
constant: g = 9.8 m/s², c = 4200 J/kg.K
Substitute into equation 2
Δt = 1(40)(9.8)/(50.5×4200)
Δt = 392/212100
Δt = 0.00185 °C
How much electrical energy is used by a 75 W laptop that is operating for 12
minutes?
"1 watt" means 1 joule of energy per second.
75 W means 75 joules/sec .
Energy = (75 Joule/sec) x (12 min) x (60 sec/min)
Energy = (75 x 12 x 60) (Joule-min-sec / sec-min)
Energy = 54,000 Joules
How many diffraction maxima are contained in a region of the Fraunhofer single-slit pattern, subtending an angle of 2.12°, for a slit width of 0.110 mm, using light of wavelength 582 nm?
Answer:
6
Explanation:
We are given that
[tex]\theta=2.12^{\circ}[/tex]
Slid width,a=0.110 mm=[tex]0.11\times 10^{-3} m[/tex]
[tex]1mm=10^{-3} m[/tex]
Wavelength,[tex]\lambda=582 nm=582\times 10^{-9}[/tex] m
[tex]1nm=10^{-9} m[/tex]
We have to find the number of diffraction maxima are contained in a region of the Fraunhofer single-slit pattern.
[tex]asin\theta=\frac{2N+1}{2}\lambda[/tex]
Using the formula
[tex]0.11\times 10^{-3}sin(2.12)=\frac{2N+1}{2}(582\times 10^{-9})[/tex]
[tex]2N+1=\frac{0.11\times 10^{-3}sin(2.12)\times 2}{582\times 10^{-9}}[/tex]
[tex]2N+1=13.98[/tex]
[tex]2N=13.98-1=12.98[/tex]
[tex]N=\frac{12.98}{2}\approx 6[/tex]
Hence, 6 diffraction maxima are contained in a region of the Fraunhofer single-slit pattern
A place-kicker must kick a football from a point 36.0 m (about 40 yards) from the goal. Half the crowd hopes the ball will clear the crossbar, which is 3.05 m high. When kicked, the ball leaves the ground with a speed of 21.6 m/s at an angle of 50.0° to the horizontal.
Required:
By how much does the ball clear or fall short (vertically) of clearing the crossbar?
Answer:
The difference is height is [tex]\Delta h =6.92 \ m[/tex]
Explanation:
From the question we are told that
The distance of ball from the goal is [tex]d = 36.0 \ m[/tex]
The height of the crossbar is [tex]h = 3.05 \ m[/tex]
The speed of the ball is [tex]v = 21.6 \ m/s[/tex]
The angle at which the ball was kicked is [tex]\theta = 50 ^o[/tex]
The height attained by the ball is mathematically represented as
[tex]H = v_v * t - \frac{1}{2} gt^2[/tex]
Where [tex]v_v[/tex] is the vertical component of velocity which is mathematically represented as
[tex]v_v = v * sin (\theta )[/tex]
substituting values
[tex]v_v = 21.6 * (sin (50 ))[/tex]
[tex]v_v = 16.55 \ m/s[/tex]
Now the time taken is evaluated as
[tex]t = \frac{d}{v * cos(\theta )}[/tex]
substituting value
[tex]t = \frac{36}{21.6 * cos(50 )}[/tex]
[tex]t = 2.593 \ s[/tex]
So
[tex]H = 16.55 * 2.593 - \frac{1}{2} * 9.8 * (2.593)^3[/tex]
[tex]H = 9.97 \ m[/tex]
The difference in height is mathematically evaluated as
[tex]\Delta h = H - h[/tex]
substituting value
[tex]\Delta h = 9.97 - 3.05[/tex]
[tex]\Delta h =6.92 \ m[/tex]
Identify the following as combination, decomposition, replacement, or ion exchange reactions: Al(s) + 3 Cl2(g) → 2 AlCl3(s) Ca(OH)2(aq) + H2SO4(aq) → CaSO4(aq) + 2 H2O(l
Answer:
2 Al(s) + 3Cl₂(g) → 2AlCl₃(s)
This is a combination reaction.
Ca(OH)₂(aq) + H₂SO₄(aq) → CaSO₄(aq) + 2H₂O(l)
This is a replacement reaction.
Explanation:
A combination reaction is a reaction in which two reagents are combined into one product. The reaction has the following general form:
A + B → AB
where A and B represent any two chemical substances.
2 Al(s) + 3Cl₂(g) → 2AlCl₃(s)
This is a combination reaction because a single compound forms from two or more reacting species.
Double Substitution, Double Displacement or Metastasis Reactions are those in which two elements found in different compounds exchange their positions forming two new compounds. These chemical reactions do not present changes in the number of oxidation or relative load of the elements. So they are not considered redox reactions.
The solvent of the double displacement reactions usually is water and the reagents and products are usually ionic compounds (cations or anions are exchanged), although they can also be acids or bases.
In general, this type of reaction can be expressed as:
AB + CD ⇒ AD + CD
In the reaction:
Ca(OH)₂(aq) + H₂SO₄(aq) → CaSO₄(aq) + 2H₂O(l)
This is a replacement reaction because it is a double replacement reaction in which the ions are exchanged to form new compounds.
Three blocks are placed in contact on a horizontal frictionless surface. A constant force of magnitude F is applied to the box of mass M. There is friction between the surfaces of blocks 2M and 3M so the three blocks accelerate together to the right.
Which block has the smallest net force acting on it?
A) M
B) 2M
C) 3M
D) The net force is the same for all three blocks Submit
Answer:
A) M
Explanation:
The three blocks are set in series on a horizontal frictionless surface, whose mutual contact accelerates all system to the same value due to internal forces as response to external force exerted on the box of mass M (Newton's Third Law). Let be F the external force, and F' and F'' the internal forces between boxes of masses M and 2M, as well as between boxes of masses 2M and 3M. The equations of equilibrium of each box are described below:
Box with mass M
[tex]\Sigma F = F - F' = M\cdot a[/tex]
Box with mass 2M
[tex]\Sigma F = F' - F'' = 2\cdot M \cdot a[/tex]
Box with mass 3M
[tex]\Sigma F = F'' = 3\cdot M \cdot a[/tex]
On the third equation, acceleration can be modelled in terms of F'':
[tex]a = \frac{F''}{3\cdot M}[/tex]
An expression for F' can be deducted from the second equation by replacing F'' and clearing the respective variable.
[tex]F' = 2\cdot M \cdot a + F''[/tex]
[tex]F' = 2\cdot M \cdot \left(\frac{F''}{3\cdot M} \right) + F''[/tex]
[tex]F' = \frac{5}{3}\cdot F''[/tex]
Finally, F'' can be calculated in terms of the external force by replacing F' on the first equation:
[tex]F - \frac{5}{3}\cdot F'' = M \cdot \left(\frac{F''}{3\cdot M} \right)[/tex]
[tex]F = \frac{5}{3} \cdot F'' + \frac{1}{3}\cdot F''[/tex]
[tex]F = 2\cdot F''[/tex]
[tex]F'' = \frac{1}{2}\cdot F[/tex]
Afterwards, F' as function of the external force can be obtained by direct substitution:
[tex]F' = \frac{5}{6}\cdot F[/tex]
The net forces of each block are now calculated:
Box with mass M
[tex]M\cdot a = F - \frac{5}{6}\cdot F[/tex]
[tex]M\cdot a = \frac{1}{6}\cdot F[/tex]
Box with mass 2M
[tex]2\cdot M\cdot a = \frac{5}{6}\cdot F - \frac{1}{2}\cdot F[/tex]
[tex]2\cdot M \cdot a = \frac{1}{3}\cdot F[/tex]
Box with mass 3M
[tex]3\cdot M \cdot a = \frac{1}{2}\cdot F[/tex]
As a conclusion, the box with mass M experiments the smallest net force acting on it, which corresponds with answer A.