Answer:
Explanation:
From the given information:
(a)
the average magnetic flux through each turn of the inner solenoid can be calculated by the formula:
[tex]\phi _ 1 = B_1 A[/tex]
[tex]\phi _ 1 = ( \mu_o \dfrac{N_i}{l} i_1)(\pi ( \dfrac{d}{2})^2)[/tex]
[tex]\phi _ 1 = ( 4 \pi *10^{-7} \ T. m/A ) ( \dfrac{310}{20*10^{-2} \ m }) (0.130 \ A) ( \pi ( \dfrac{2.20*10^{-2} \ m }{2})^ 2[/tex]
[tex]\phi_1 = 9.625 * 10^{-8} \ Wb[/tex]
(b)
The mutual inductance of the two solenoids is calculated by the formula:
[tex]M = 23 *\dfrac{9.625*10^{-8} \ Wb}{0.130 \ A}[/tex]
M = [tex]1.703 *10^{-5}[/tex] H
(c)
the emf induced in the outer solenoid by the changing current in the inner solenoid can be calculate by using the formula:
[tex]\varepsilon = -N_o \dfrac{d \phi_1}{dt}[/tex]
[tex]\varepsilon = -M \dfrac{d i_1}{dt}[/tex]
[tex]\varepsilon = -(1.703*10^{-5} \ H) * (1800 \ A/s)[/tex]
[tex]\varepsilon = -0.030654 \ V[/tex]
[tex]\varepsilon = -30.65 \ V[/tex]
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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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
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 high-jumper clears the bar and has a downward velocity of - 5.00 m/s just before landing on an air mattress and bouncing up at 1.0 m/s. The mass of the high-jumper is 60.0 kg. What is the magnitude and direction of the impulse that the air mattress exerts on her
-- As she lands on the air mattress, her momentum is (m v)
Momentum = (60 kg) (5 m/s down) = 300 kg-m/s down
-- As she leaves it after the bounce,
Momentum = (60 kg) (1 m/s up) = 60 kg-m/s up
-- The impulse (change in momentum) is
Change = (60 kg-m/s up) - (300 kg-m/s down)
Magnitude of the change = 360 km-m/s
The direction of the change is up /\ .
The direction of a body or object's movement is defined by its velocity.In its basic form, speed is a scalar quantity.In essence, velocity is a vector quantity.It is the speed at which distance changes.It is the displacement change rate.
Solve the problem ?
Velocity is the pace and direction of an object's movement, whereas speed is the time rate at which an object is travelling along a path.In other words, velocity is a vector, whereas speed is a scalar value. We discuss the conceptive impulse in this puzzle.A high jumper weighing 60.0 kg sprints on the matrix at minus 6 meters per second in the downhill direction before falling to the mattress.her admirer.Speed drops to 0 meters/second.We must determine the impulse's size and presumed direction, which is upward and positive.The change in momentum is then equal to the impulse.The impulse therefore equals m times.the end velocity less the starting velocity.60.0kg times 0 minus minus 6 meters per second is the impulse, therefore.The impulse is 360 kilogram meters per second, or 360 newtons, to put it another way.The second is upward, and the direction.To learn more about magnitude refer
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In cricket how bowler and batsman use acceleration?
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.
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
New evidence increasingly emphasizes that __________.
g A mass of 2 kg is attached to a spring whose constant is 7 N/m. The mass is initially released from a point 4 m above the equilibrium position with a downward velocity of 10 m/s, and the subsequent motion takes place in a medium that imparts a damping force numerically equal to 10 times the instantaneous velocity. What is the differential equation for the mass-spring system.
Answer:
mass 20 times of an amazing and all its motion
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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You are trying to overhear a juicy conversation, but from your distance of 25.0 m, it sounds like only an average whisper of 25.0 dB. So you decide to move closer to give the conversation a sound level of 80.0 dB instead. How close should you come?
Answer:
r₂ = 1,586 m
Explanation:
For this problem we are going to solve it by parts, let's start by finding the sound intensity when we are 25 m
β = 10 log (I / I₀)
where Io is the sensitivity threshold 10⁻¹² W / m²
I₁ / I₀ = [tex]e^{\beta/10}[/tex]
I₁ = I₀ e^{\beta/10}
let's calculate
I₁ = 10⁻¹² e^{25/10}
I₁ = 1.20 10⁻¹¹ W / m²
the other intensity in exercise is
I₂ = 10⁻¹² e^{80/10}
I₂ = 2.98 10⁻⁹ W / m²
now we use the definition of sound intensity
I = P / A
where P is the emitted power that is a constant and A the area of the sphere where the sound is distributed
P = I A
the area a sphere is
A = 4π r²
we can write this equation for two points of the found intensities
I₁ A₁ = I₂ A₂
where index 1 corresponds to 25m and index 2 to the other distance
I₁ 4π r₁² = I₂ 4π r₂²
I₁ r₁² = I₂ r₂²
r₂ = √ (I₁ / I₂) r₁
let's calculate
r₂ = √ (1.20 10⁻¹¹ / 2.98 10⁻⁹) 25
r₂ = √ (0.40268 10⁻²) 25
r₂ = 1,586 m
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
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]
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
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 particle leaves the origin with a speed of 3 106 m/s at 38 degrees to the positive x axis. It moves in a uniform electric field directed along positive y axis. Find Ey such that the particle will cross the x axis at x
Answer:
If the particle is an electron [tex]E_y = 3.311 * 10^3 N/C[/tex]
If the particle is a proton, [tex]E_y = 6.08 * 10^6 N/C[/tex]
Explanation:
Initial speed at the origin, [tex]u = 3 * 10^6 m/s[/tex]
[tex]\theta = 38^0[/tex] to +ve x-axis
The particle crosses the x-axis at , x = 1.5 cm = 0.015 m
The particle can either be an electron or a proton:
Mass of an electron, [tex]m_e = 9.1 * 10^{-31} kg[/tex]
Mass of a proton, [tex]m_p = 1.67 * 10^{-27} kg[/tex]
The electric field intensity along the positive y axis [tex]E_y[/tex], can be given by the formula:
[tex]E_y = \frac{2 m u^2 sin \theta cos \theta}{qx} \\[/tex]
If the particle is an electron:
[tex]E_y = \frac{2 m_e u^2 sin \theta cos \theta}{qx} \\[/tex]
[tex]E_y = \frac{2 * 9.1 * 10^{-31} * (3*10^6)^2 *(sin38)( cos38)}{1.6*10^{-19} * 0.015} \\[/tex]
[tex]E_y = 3311.13 N/C\\E_y = 3.311 * 10^3 N/C[/tex]
If the particle is a proton:
[tex]E_y = \frac{2 m_p u^2 sin \theta cos \theta}{qx} \\[/tex]
[tex]E_y = \frac{2 * 1.67 * 10^{-27} * (3*10^6)^2 *(sin38)( cos38)}{1.6*10^{-19} * 0.015} \\[/tex]
[tex]E_y = 6.08 * 10^6 N/C[/tex]
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.
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
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
What is the on ohooke benden
er ord power
What is the main difference between work, power and energy
Answer:Work is the energy required to move an object from one point to another. while power is the energy transferred per unit time.
Which circuits are parallel circuits?
Answer:
The bottom two lines.
Explanation:
They need their own line of voltage quantity. A parallel circuit has the definition of 'two or more paths for current to flow through.' The voltage does stay the same in each line.
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.
Learn more about Impulse here:
https://brainly.com/question/15495020
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
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
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.
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
What will happen to an astronaut when the jets produce these four forces
An airplane is flying on a bearing of N 400 W at 500 mph. A strong jet-stream speed wind of 100 mph is blowing at S 500 W.
Required:
a. Find the vector representation of the plane and of the wind.
b. Find the resultant vector that represents the actual course of the plane.
c. Give the resulting speed and bearing of the plane.
Answer:
A. a (-321.393, 383.022) b (-76.40, -64.278)
B. (-397.991, 318.744)
C. a. resulting speed 509.9mph b. bearing of the plane = 51.6°
Explanation:
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 stationary 6-kg shell explodes into three pieces. One 4.0 kg piece moves horizontally along the negative x-axis. The other two fragments, each 1.0 kg, move in directions that make 60o angle above and below the positive x-axis and their speeds are 60 m/s each. What is the velocity of the 4.0-kg fragment
Answer:
-15 m/s
Explanation:
The computation of the velocity of the 4.0 kg fragment is shown below:
For this question, we use the correlation of the momentum along with horizontal x axis
Given that
Weight of stationary shell = 6 kg
Other two fragments each = 1.0 kg
Angle = 60
Speed = 60 m/s
Based on the above information, the velocity = v is
[tex]1\times 60 \times cos\ 60 + 1\times 60 \times cos\ 60 - 4\ v = 0[/tex]
[tex]\frac{60}{2} + \frac{60}{2} - 4\ v = 0[/tex]
[tex]v = \frac{60}{4}[/tex]
= -15 m/s