Answer:
A) Apparent Weight = 590 N
Explanation:
As we know that frequency is given as
[tex]f = \frac{1}{30}[/tex]
[tex]f = 0.033 Hz[/tex]
now the angular speed is given as
[tex]\omega = 2\pi f[/tex]
[tex]\omega = 2\pi(0.033) = 0.21 rad/s[/tex]
now at the top position we will have
[tex]mg - N = m\omega^2 R[/tex]
[tex]N = mg - m\omega^2 R[/tex]
[tex]N = 600 - \frac{600}{9.8}(0.21)^2(4.0)[/tex]
[tex]N = 590 N[/tex]
The woman's apparent weight at the top of the Ferris wheel is less than her actual weight due to the centripetal force experienced during the uniform circular motion. This apparent weight can be calculated by subtracting the centripetal force from the gravitational force, yielding an answer of 590 N.
Explanation:This problem revolves around the concepts of centripetal force and apparent weight in the context of uniform circular motion. The woman's apparent weight at the top of the Ferris wheel is less than her actual weight because of the centripetal force directed towards the center of the Ferris wheel.
To calculate the apparent weight, we should subtract the centripetal force from the gravitational force. The gravitational force is her actual weight, and since weight = mass * gravity, her mass equals 600N/9.8 m/s2 ~= 61.2 kg. The angular velocity of the Ferris wheel (ω) is 2π rad/30s since it makes one revolution every 30s. Using the formula centripetal force = m*ω2*r, we find that the centripetal force equals 61.2 kg * (2π rad/30s)2 * 4m = 10 N approximately.
Finally, the woman's apparent weight at the top is the gravitational force minus the centripetal force, or 600 N - 10 N, which equals 590 N. Therefore, the correct answer is A) 590 N.
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A 0.125-kg particle undergoes simple harmonic motion along the horizontal ????-axis between the points ????1=−0.299 m and ????2=0.387 m. The period of oscillation is 0.581 s. Find the frequency, ????, the equilibrium position, ????eq, the amplitude, ????, the maximum speed, ????max, the maximum magnitude of acceleration, ????max, the force constant, ????, and the total mechanical energy, ????tot.
A 0.293-kg particle undergoes simple harmonic motion along the horizontal x-axis between the points x1 = -0.285 m and x2 = 0.395 m. The period of oscillation is 0.641 s. Find the frequency, f, the equilibrium position, xeq, the amplitude, A, the maximum speed, vmax, the maximum magnitude of acceleration, amax, the force constant, k, and the total mechanical energy, Etot.
That being said,
f=1/T=1/0.641=1.56 s
xeq=(x1+x2)/2=(-0.285+0.395)/2=0.055 m
A=(x2-x1)/2=(0.395+0.285)/2=0.340 m
x(t)=Acos(wt+ϕ)
v(t)=-wAsen(wt+ϕ)
w=2πf=2*3.14*1.56=9.8 rad/s
vmax=-wA=9.8*0.340=-3.3 m/s
a(t)=-w^2Acos(wt+ϕ)
amax=-w^2A=96.04*0.340=-32.6 m/s^2
w=(k/m)^1/2 => k=mw^2=0.293*96.04=28.1 N/m
Etot=1/2kA^2
Etot=0.5*28.1*0.116=1.63 J
The characteristics of the simple harmonic movement allow to find the results for the questions about the movement of the mass with the spring are:
a) The frequency is: f = 1.72 Hz
b) The equilibrium position is: x₀ = 0.088 m
c) The amplitude is: A = 0.686 m
d) The maximum speed is: v = 7.42 m / s
e) the maximum acceleration is: a = 80.16 m / s²
f) The mechanical energy is: E = 3.44 J
Given parameters
Particle mass m = 0.125 kg Initial point x₀ = -0.299 m End point x_f = 0.387 m. Oscillation period T = 0.581 sTo find.
a) Frequency.
b) the equilibrium position.
c) The amplitude.
d) Maximum speed.
e) Maximum acceleration.
f) The total mechanical energy.
Simple harmonic motion is a periodic motion where the restoring force is proportional to the elongation. This movement is fully described by the expression
x = A cos (wt + Ф)
w² = [tex]\sqrt{ \frac{k}{m} }[/tex]
Where x is the displacement, A the amplitude of the movement, w the angular velocity, t the time, Ф a phase constant that meets the initial conditions, k the spring constant and m the mass attached to the spring.
a) Frequency and period are related.
[tex]f= \frac{1}{T}[/tex]
Let's calculate.
[tex]f = \frac{1}{0.581 }[/tex]
f = 1.72 Hz
b) The equilibrium position occurs at the midpoint of the movement
[tex]x_o = \frac{x_f + x_o}{2}[/tex]
[tex]x_o = \frac{0.387 + (-0.299)}{2}[/tex]
x₀ = 0.088 m
c) The amplitude is the maximum elongation of the spring
[tex]A= x_f - x_o[/tex]
A = 0.387 - (-0.299)
A = 0.686 m
d) The speed is defined by the variation of the position with respect to time.
[tex]v = \frac{dx}{dt}[/tex]
Let's make the derivatives
v = - A w cos (wt + Ф)
The angular velocity is related to the period.
[tex]w= \frac{\pi }{T} \\w= \frac{2 \pi }{0.581}[/tex]
w = 10.81 rad / s
The speed is maximum when the cosine is maximum ±1
v = A w
v = 0.686 10.81
v = 7.42 m / s
e) Acceleration is defined as the change in velocity over time.
[tex]a = \frac{dv}{dt}[/tex]
a = -A w² sin (wt + Ф)
The acceleration is maximum when the sine function is maximum ±1
a = A w²
a = 0.686 10.81²
a = 80.16 m / s²
f) The mechanical energy is calculate for the point of maximum elongation.
E = ½ k A²
k = w² m
Let's replace.
E = ½ w² m A²
Let's calculate.
E = ½ 10.81² 0.125 0.686²
E = 3.44 J
In conclusion, using the characteristics of simple harmonic motion we can find the results for the questions about the motion of the mass with the spring are:
a) The frequency is: f = 1.72 Hz
b) The equilibrium position is: x₀ = 0.088 m
c) The amplitude is: A = 0.686 m
d) The maximum speed is: v = 7.42 m / s
e) the maximum acceleration is: a = 80.16 m / s²
f) The mechanical energy is: E = 3.44 J
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The waste products of combustion leave the internal combustion engine through the: A. crankshaft, B. exhaust valve. C. cylinder. D. intake valve.
Answer:
B.Exhaust Value
Explanation:
The waste products of combustion leave the internal combustion engine through the Exhuast Value.
Answer:
Exhaust Value
Explanation:
A solenoid 10.0 cm in diameter and 70.0 cm long is made from copper wire of diameter 0.100 cm, with very thin insulation. The wire is wound onto a cardboard tube in a single layer, with adjacent turns touching each other. To produce a field of 5.10 mT at the center of the solenoid, what power must be delivered to the solenoid?
The power delivered to the solenoid is calculated by finding the current through the solenoid and its resistance, then applying the formula P=I²R. This involves applying known laws and formulas of electromagnetism.
Explanation:The power delivered to the solenoid can be calculated given the field strength inside the solenoid and the physical dimensions and properties of the solenoid and wire. The field strength inside a solenoid is given by the formula B = µon, where n is the number of loops per unit length. We first find n, and then we need to find the current I because the power P delivered to the solenoid is given by P=I²R, where R is the resistance of the wire.
The resistance R of the wire can be found using the formula R = ρL/A, where ρ is the resistivity of copper, L is the length of the wire, and A is the cross-sectional area of the wire. After calculating R and I, we can compute P. Therefore, it's paramount to apply known formulas and laws of electromagnetism to solve this. It's important to note that B is the magnetic field inside the solenoid, µo is the permeability of free space and n is number of turns per unit length.
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To produce a magnetic field of 5.10 mT at the center of a solenoid, a power of 61.4 W must be delivered. This is calculated by determining the number of turns, the required current, the resistance of the copper wire, and then using the power formula. The key parameters are the number of turns (700), current (3.61 A), resistance (4.71 Ω), and power (61.4 W).
To solve this problem, we'll need to follow these steps:
Calculate the number of turns of wire on the solenoid.
Determine the current needed to produce the required magnetic field.
Find the resistance of the wire.
Calculate the power delivered to the solenoid.
1. Calculate the number of turns of wire
The diameter of the wire is 0.100 cm, so the circumference of each turn is approximately equal to the solenoid's length (70.0 cm). The number of turns (N) is:
N = 70.0 cm / 0.100 cm = 700 turns
2. Determine the current needed
The formula for the magnetic field (B) inside a solenoid is given by:
B = μ₀ * (N/L) * I
where:
B = 5.10 mT = 5.10 x 10⁻³ T
μ₀ = 4π * 10⁻⁷ T⋅m/A (permeability of free space)
N = 700 turns
L = 70.0 cm = 0.70 m
Solving for the current (I):
5.10 x 10⁻³ = (4π x 10⁻⁷) * (700 / 0.70) * I
5.10 x 10⁻³ = (4π x 10⁻⁷) * 1000 * I
5.10 x 10⁻³ = (4π x 10⁻⁴) * I
I = 3.61 A
3. Find the resistance of the wire
The resistance (R) of the wire is given by:
R = ρ * (L/A)
where:
ρ = 1.68 x 10⁻⁸ Ω⋅m (resistivity of copper)
L = N * circumference of each turn
A = * (diameter/2)²
Calculating the total length (L) of the wire:
L = 700 * 0.100 * m = 219.91 m (approximately)
Calculating the cross-sectional area (A):
A = * (0.001 m / 2)²
A = 3.14 x (0.0005)²
A = 7.85 x 10⁻⁷ m²
Thus, the resistance (R) is:
R = 1.68 x 10⁻⁸ * (219.91 / 7.85 x 10⁻⁷)
R = 1.68 x 10⁻⁸ * (28.014 x 10⁷)
R = 4.71 Ω
4. Calculate the power delivered
Power (P) is given by:
P = I² * R
Substituting in the values:
P = (3.61)² * 4.71 = 13.032 * 4.71 = 61.4 W
Therefore, 61.4 W power must be delivered to the solenoid to produce the required magnetic field.
When an object is oscillating in simple harmonic motion in the vertical direction, its maximum speed occurs when the object A) is at its highest point B) is at its lowest point C) is at the equilibrium point D) has the maximum net force exerted on it E) has a position equal to its amplitude
Answer:
C) is at the equilibrium point
Explanation:
As we know that the equation of displacement of SHM executing object is given as
[tex]x = A sin(\omega t + \phi)[/tex]
now for velocity of the particle we will have
[tex]v = \frac{dx}{dt}[/tex]
now we will have
[tex]v = A\omega cos(\omega t + \phi)[/tex]
now if velocity is maximum then we will have
[tex]cos(\omega t + \phi) = \pm 1[/tex]
so at this situation we will have
[tex](\omega t + \phi) = N\pi[/tex]
now at this angle the value of
[tex]sin(\omega t + \phi) = 0[/tex]
so the position must be mean position at which the particle is at equilibrium position
In simple harmonic motion, an object oscillating vertically reaches its maximum speed at the equilibrium point. This is the position where the object would rest in the absence of force and is where the maximum velocity is attained during the oscillation.
Explanation:In the context of simple harmonic motion, an object oscillating in the vertical direction attains its maximum speed at the equilibrium point (C). This equilibrium point represents the position where the object would naturally rest in the absence of force. An object attached to a spring and placed on a frictionless surface serves as a straightforward example of a simple harmonic oscillator. When displaced from its equilibrium, this object executes simple harmonic motion with an amplitude A and a period T. The maximum velocity or speed is achieved as the object moves through the equilibrium point, irrespective of the direction. The maximum displacement from the equilibrium, referred to as amplitude, does not correlate with the object's speed.
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A glass window 0.33 cm thick measures 87 cm by 36 cm. How much heat flows through this window per minute if the inside and outside temperatures differ by 14°C? Express your answer using two significant figures.
Answer:
6.38 x 10^4 J
Explanation:
d = 0.33 cm = 0.33 x 10^-2 m, Area = 87 x 36 cm^2 = 0.87 x 0.36 m^2
ΔT = 14 degree C, t = 1 min = 60 second
K = 0.8 W / m K
Heat = K A ΔT t / d
H = 0.8 x 0.87 x 0.36 x 14 x 60 / (0.33 x 10^-2)
H = 6.38 x 10^4 J
The question is related to calculating the rate of heat conduction through a glass window in Physics. For a complete answer, the thermal conductivity of glass is needed, which is not provided in the question. With that value, the formula for heat conduction can be used to find the heat flow per minute.
Explanation:The subject of this question is Physics, and it applies to a High School grade level. To determine the rate of heat conduction through a window, we can use the formula for heat transfer through conduction:
Q = \(\frac{k \cdot A \cdot \Delta T \cdot t}{d}\)
where:
Q is the heat transferred,
k is the thermal conductivity,
A is the area,
\(\Delta T\) is the temperature difference,
t is the time, and
d is the thickness of the material.
However, the question does not provide the value of the thermal conductivity (k) for glass, which is necessary to calculate the rate of heat conduction. With the thermal conductivity value, we could then insert the given dimensions, temperature difference, and time into this formula to find the heat flow per minute. Without this value, the question cannot be fully answered as the data is incomplete. For a complete solution, thermal conductivity of the material is required.
A box rests on the back of a truck. The coefficient of friction between the box and the surface is 0.32. (3 marks) a. When the truck accelerates forward, what force accelerates the box? b. Find the maximum acceleration the truck can have before the box slides.
Answer:
Part a)
here friction force will accelerate the box in forward direction
Part b)
a = 3.14 m/s/s
Explanation:
Part a)
When truck accelerates forward direction then the box placed on the truck will also move with the truck in same direction
Here if they both moves together with same acceleration then the force on the box is due to friction force between the box and the surface of the truck
So here friction force will accelerate the box in forward direction
Part b)
The maximum value of friction force on the box is known as limiting friction
it is given by the formula
[tex]F = \mu mg[/tex]
so we have
[tex]F = ma = \mu mg[/tex]
now the acceleration is given as
[tex]a = \mu g[/tex]
[tex]a = (0.32)(9.8) = 3.14 m/s^2[/tex]
Final answer:
The force that accelerates the box when the truck moves forward is the force of static friction. The maximum acceleration the truck can have before the box slides can be found using the equation: max acceleration = coefficient of friction x acceleration due to gravity.
Explanation:
When the truck accelerates forward, the force that accelerates the box is the force of static friction between the box and the surface of the truck bed. This force opposes the motion of the box and allows it to stay in place on the truck.
The maximum acceleration the truck can have before the box slides can be found using the equation:
max acceleration = coefficient of friction x acceleration due to gravity
Using the given coefficient of friction between the box and the surface, you can substitute the value and solve for the maximum acceleration.
A fisherman notices that his boat is moving up and down periodically without any horizontal motion, owing to waves on the surface of the water. It takes a time of 2.80 s for the boat to travel from its highest point to its lowest, a total distance of 0.600 m. The fisherman sees that the wave crests are spaced a horizontal distance of 5.50 m apart.
Part A How fast are the waves traveling? Express the speed v in meters per second using three significant figures.
Answer:
Very fast!
Explanation:
The waves are traveling very fast. I know this because they're not traveling slow for sure.
If this is incorrect, they're probably traveling moderately fast.
Photons with wavelength 1 pm are incident on electrons. What is the frequency of the Compton-scattered photons at an angle of 60°?
Answer:
f = 1.354*10^{20} Hz
Explanation:
By conservation of linear momentum, wavelength shift due to collision of photon to electron is given by following formula
[tex]\lambda ^{'}-\lambda =\frac{h}{m_{o}c}(1-cos\theta )[/tex]
where h is plank constant = 6.626*10^{-34}
c = speed of light = 3*10^{8} m/s
scattered angle = 60 degree
m = rest mass of electron = 9*10^{-31}
[tex]\lambda ^{'}=10^{-12} +\frac{6.626*10^{-34}}{9*10^{-31}*3*10^{8}}(1-cos60^{o} )[/tex]
[tex]\lambda ^{'}= 2.215 pm[/tex]
we know that 1 pm = 10^{-12}m
[tex]f = \frac{c}{\lambda ^{'}}[/tex]
[tex]f = \frac{3*10^{8}}{2.215 *10^{-12}} = 1.354*10^{20} Hz[/tex]
f = 1.354*10^{20} Hz
A rock is thrown straight up. What is the net external force acting on the rock when it is at the top of its trajectory?
Answer:
Net external force acting on the rock when it is at the top of its trajectory is force due to gravity (mg).
Explanation:
The forces acting on a rock thrown up are force due to gravity and air resistance. Air resistance is directly proportional to velocity of rock, when velocity is zero air resistance is zero. When it is at the top of its trajectory its velocity is zero. So air resistance is also zero. Hence only gravitational force acts on the rock.
Net external force acting on the rock when it is at the top of its trajectory is force due to gravity (mg).
At the peak of its trajectory, the only significant force acting on a rock is the force of gravity. Thus, the net external force on the system is represented as -mg, where m is the mass of the rock and g is the acceleration due to gravity. At this point, the rock begins its descent due to the gravitational pull.
Explanation:The question is asking about the net external force acting on a rock when it is at the top of its trajectory. At the peak of its trajectory, the only significant force acting on the rock is the force of gravity (weight of the rock), assuming air resistance is negligible. Thus, the net external force on the system, if air resistance is neglected, is represented as -mg, where m is the mass of the rock and g is the acceleration due to gravity.
For example, if a rock is thrown straight upwards with an initial velocity, as it reaches the topmost point of its trajectory, the vertical velocity momentarily becomes zero before the rock starts descending again. At this point, the only major force acting on the rock is the gravitational pull exerted by the Earth, also known as the rock's weight.
It is important to note that this net external force (-mg) is what determines the acceleration of the object at the top of its trajectory. Simply put, even at the top of its path, the rock is still experiencing the force of gravity which pulls it back towards Earth, leading to its eventual fall back to the ground.
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A hunter shoots a bullet horizontally at a rock cliff wall that is 170 m away. She hears the sound of the bullet hitting the rock 0.95 seconds later. Knowing that the speed of sound was 340 m/s, what was the speed of the bullet?
Answer:
377.78 m/s
Explanation:
Let the speed of bullet is v.
Time taken by the sound is t' and time taken by the bullet is t.
Speed of sound = 340 m/s
Distance = 170 m
Time taken by the sound, t' = distance / speed of sound = 170 / 340 = 0.5 sec
time taken by the bullet, t = 0.95 - 0.5 = 0.45 sec
Speed of bullet = distance / time taken by bullet
Speed of bullet = 170 / 0.45 = 377.78 m/s
A circular coil that has 100 turns and a radius of 10.0 cm lies in a magnetic field that has a magnitude of 0.0650 T directed perpendicular to the coil. (a) What is the magnetic flux through the coil? (b) The magnetic field through the coil is increased steadily to 0.100 T over a time interval of 0.500 s. What is the magnitude of the emf induced in the coil during the time interval?
Answer:
(a) 0.204 Weber
(b) 0.22 Volt
Explanation:
N = 100, radius, r = 10 cm = 0.1 m, B = 0.0650 T, angle is 90 degree with the plane of coil, so theta = 0 degree with the normal of coil.
(a) Magnetic flux, Ф = N x B x A
Ф = 100 x 0.0650 x 3.14 x 0.1 0.1
Ф = 0.204 Weber
(b) B1 = 0.0650 T, B2 = 0.1 T, dt = 0.5 s
dB / dt = (B2 - B1) / dt = (0.1 - 0.0650) / 0.5 = 0.07 T / s
induced emf, e = N dФ/dt
e = N x A x dB/dt
e = 100 x 3.14 x 0.1 x 0.1 x 0.07 = 0.22 V
(a) The magnetic flux through a coil 0.204 Weber
(b) The emf induced in the coil during the time interval 0.22 Volt
What will be the magnetic flux and emf induced in the coil?It is given that
Number of the turns N = 100,
Radius of the coil r = 10 cm = 0.1 m,
The magnetic field B = 0.0650 T
Since the angle is 90 degrees with the plane of the coil, so [tex]\Theta[/tex]= 0 degrees with the normal coil.
(a) The Magnetic flux, will be given as
[tex]\rm\phi =N\times B\times A[/tex]
By putting the values
[tex]\phi =100\times0.0650\times3.14\times 0.1\times0.1[/tex]
[tex]\phi=0.204 \rm \ weber[/tex]
(b) The emf induced will to be given as
[tex]B_1=0.0650T\ \ B_2=0.1T\ \ dt=0.5s[/tex]
[tex]\dfrac{dB}{dt} =\dfrac{B_2-B_1}{dt} =\dfrac{0.1-0.650}{0.5} =0.07\dfrac{T}{s}[/tex]
induced emf,
[tex]e=N\dfrac{d\phi}{dt}[/tex]
[tex]e=N\times A \times \dfrac{dB}{dt}[/tex]
[tex]e=100\times 3.14\times0.1\times0.1\times0.07=0.22V[/tex]
Thus
(a) The magnetic flux through a coil 0.204 Weber
(b) The emf induced in the coil during the time interval 0.22 Volt
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A generator is built with 500 turns.The area of each turn is 0.02 m^2 , and the external magnetic field through the coils is 0.30 T. If a turbine rotates the coil at 960 rpm, what is the induced rms voltage?
Answer:
213.18 volt
Explanation:
N = 500, A = 0.02 m^2, B = 0.30 T, f = 960 rpm = 960 / 60 = 16 rps
The maximum value of induced emf is given by
e0 = N x B x A x ω
e0 = N x B x A x 2 x π x f
e0 = 500 x 0.3 x 0.02 x 2 x 3.14 x 16
e0 = 301.44 Volt
The rms value of induced emf is given by
[tex]e_{rms} = \frac{e_{0}}{\sqrt{2}}[/tex]
[tex]e_{rms} = \frac{301.44}{\sqrt{2}}[/tex]
e rms = 213.18 volt
Which of the following is created by fans? a) Air flow b) Water flow c) Wastewater flow d) Vacuum flow
Gas is confined in a tank at a pressure of 9.8 atm and a temperature of 29.0°C. If two-thirds of the gas is withdrawn and the temperature is raised to 83.0°C, what is the pressure of the gas remaining in the tank? 9.1e-21 Incorrect: Your answer is incorrect.
Answer:
3.9 atm
Explanation:
Ideal gas law states:
PV = nRT
where P is absolute pressure,
V is volume,
n is number of moles,
R is universal gas constant,
and T is absolute temperature.
The volume of the tank is constant, so we can say:
V₁ = V₂
Using ideal gas law to write in terms of P, n, R, and T:
n₁ R T₁ / P₁ = n₂ R T₂ / P₂
n₁ T₁ / P₁ = n₂ T₂ / P₂
Initially:
P₁ = 9.8 atm
T₁ = 29.0°C = 302.15 K
n₁ = n
Afterwards:
P₂ = P
T₂ = 83.0°C = 356.15 K
n₂ = n/3
Substituting:
n (302.15 K) / (9.8 atm) = (n/3) (356.15 K) / P
(302.15 K) / (9.8 atm) = (1/3) (356.15 K) / P
P = 3.85 atm
Rounding to 2 significant figures, P = 3.9 atm.
A blacksmith heats a 1.5 kg iron horseshoe to 500 C, containing 20 kg of water at 18 C then plunges it into a bucket What is the equilbrlum temperature? Express your answer using two significant figures.
Answer:
21.85 C
Explanation:
mass of iron = 1.5 kg, initial temperature of iron, T1 = 500 C
mass of water = 20 kg, initial temperature of water, T2 = 18 C
let T be the equilibrium temperature.
Specific heat of iron = 449 J/kg C
specific heat of water = 4186 J/kg C
Use the principle of caloriemetry
heat lost by the hot body = heat gained by the cold body
mass of iron x specific heat of iron x decrease in temperature = mass of water x specific heat of water x increase in temperature
1.5 x 449 x (500 - T) = 20 x 4186 x (T - 18)
336750 - 673.5 T = 83720 T - 1506960
1843710 = 84393.5 T
T = 21.85 C
A 1400-kg car is traveling with a speed of 17.7 m/s. What is the magnitude of the horizontal net force that is required to bring the car to a halt in a distance of 33.1 m?
Answer:
The magnitude of the horizontal net force is 13244 N.
Explanation:
Given that,
Mass of car = 1400 kg
Speed = 17.7 m/s
Distance = 33.1 m
We need to calculate the acceleration
Using equation of motion
[tex]v^2-u^2=2as[/tex]
Where, u = initial velocity
v = final velocity
s = distance
Put the value in the equation
[tex]0-(17.7)^2=2a\times 33.1[/tex]
[tex]a=\dfrac{-(17.7)^2}{33.1}[/tex]
[tex]a=-9.46\ m/s^2[/tex]
Negative sign shows the deceleration.
We need to calculate the net force
Using newton's formula
[tex]F = ma[/tex]
[tex]F =1400\times(-9.46)[/tex]
[tex]F=-13244\ N[/tex]
Negative sign shows the force is opposite the direction of the motion.
The magnitude of the force is
[tex]|F| =13244\ N[/tex]
Hence, The magnitude of the horizontal net force is 13244 N.
Determine the force between two long parallel wires, that are separated by a dis- tance of 0.1m the wire on the left has a current of 10A and the one on the right BT has a current of 15A, both going up. 31十131
Answer:
3 x 10^-4 N/m
Attractive
Explanation:
r = 0.1 m, i1 = 10 A, i2 = 15 A
The force per unit length between two parallel wires is given by
[tex]F = \frac{\mu _{0}}{4\pi }\times \frac{2i_{1}i_{2}}{r}[/tex]
[tex]F = \frac{10^{-7}\times 2\times 10\times 15}{0.1}[/tex]
F = 3 x 10^-4 N/m
Thus, the force per unit length between two wires is 3 x 10^-4 N/m, the force is attractive in nature because the direction of flow of current in both the wires is same.
When a 0.15 kg baseball is hit, its velocity changes from +17 m/s to -17 m/s.
(a) What is the magnitude of the impulse delivered by the bat to the ball?
N·s
(b) If the baseball is in contact with the bat for 1.5 ms, what is the magnitude of the average force exerted by the bat on the ball?
Final answer:
The magnitude of the impulse delivered by the bat to the ball is 5.1 kg·m/s. The magnitude of the average force exerted by the bat on the ball is 3.4 × 10^3 N.
Explanation:
To find the magnitude of the impulse delivered by the bat to the ball, we need to use the formula for impulse:
Impulse = change in momentum
Momentum is given by the product of mass and velocity (momentum = mass × velocity). The change in momentum is the final momentum minus the initial momentum (change in momentum = final momentum - initial momentum).
Given that the mass of the baseball is 0.15 kg, the initial velocity is +17 m/s, and the final velocity is -17 m/s, we can calculate the magnitude of the impulse:
Impulse = (0.15 kg × -17 m/s) - (0.15 kg × 17 m/s) = -5.1 kg·m/s
So, the magnitude of the impulse delivered by the bat to the ball is 5.1 kg·m/s.
To find the magnitude of the average force exerted by the bat on the ball, we can use the formula for average force:
Average force = impulse / time
Given that the time the ball is in contact with the bat is 1.5 ms (1.5 × 10^-3 s) and the magnitude of the impulse is 5.1 kg·m/s, we can calculate the magnitude of the average force:
Average force = 5.1 kg·m/s / (1.5 × 10^-3 s) = 3.4 × 10^3 N
Therefore, the magnitude of the average force exerted by the bat on the ball is 3.4 × 10^3 N.
A 9.6-g bullet is fired into a stationary block of wood having mass m = 4.95 kg. The bullet imbeds into the block. The speed of the bullet-plus-wood combination immediately after the collision is 0.591 m/s. What was the original speed of the bullet? (Express your answer with four significant figures.)
Answer:
Original speed of the bullet = 305.21 m/s
Explanation:
Here momentum is conserved.
Mass of bullet = 9.6 g = 0.0096 kg
Mass of wood = 4.95 kg
Let velocity of bullet be v.
Initial momentum = 0.0096 v
Final mass = 0.0096 + 4.95 = 4.9596 kg
Final velocity = 0.591 m/s
Final momentum = 4.9596 x 0.591 = 2.93 kgm/s
Equating both momentum
0.0096 v = 2.93
v = 305.21 m/s
Original speed of the bullet = 305.21 m/s
A driver parked his car on a steep hill and forgot to set the emergency brake. As a result, the car rolled down the hill and crashed into a parked truck. If the car was moving at 10 mph (4.5 m/s) when it hit the truck 7.0 seconds after it began to move, what was the car’s average acceleration while it rolled down the hill? Define the positive x direction to be down the hill.
Answer:0.642
Explanation:
Given
initially car is at rest i.e u=0
When car hits the truck it's velocity is 4.5 m/s
it hits the truck after 7 sec
Now average acceleration of car is =[tex]\frac{change\ in\ velocity}{time\ taken}[/tex]
average acceleration of car =[tex]\frac{\left ( final velocity\right )-\left ( initial velocity\right )}{time}[/tex]
average acceleration of car=[tex]\frac{\left ( 4.5\right )-\left ( 0\right )}{7}[/tex]
average acceleration of car=[tex]0.642 m/s^{2}[/tex]
Taking Down hill as positive x axis
average acceleration of car=[tex]0.642\hat{i}[/tex]
A star rotates in a circular orbit about the center of its galaxy. The radius of the orbit is 6.9 x 1020 m, and the angular speed of the star is 9.2 x 10-15 rad/s. How long (in years) does it take for the star to make one revolution around the center?
Answer:
2.16 x 10^7 years
Explanation:
R = 6.9 x 10^20 m
ω = 9.2 x 10^-15 rad/s
T = 2π / ω
T = ( 2 x 3.14 ) / ( 9.2 x 10^-15)
T = 6.826 x 10^14 second
T = 2.16 x 10^7 years
By using the star's given angular velocity and the formula for a revolution's time period, it can be calculated that it takes the star approximately 2.17 x 10^8 years to complete one revolution around the center of the galaxy.
Explanation:The timeframe in which a star completes one revolution around the center of a galaxy is referred to as a galactic year. In this scenario, the star's time period for one revolution, T, can be calculated using the formula T = 2π/ω, where ω represents the star's angular speed. So, T = 2π / (9.2 x 10^-15 rad/s). The resulting value is in seconds since the angular velocity was in rad/s. This can then be converted to years by using the relevant conversion factor: 1 year = 3.15 x 10^7 seconds.
By performing these calculations, we find that it takes the star about 2.17 x 10^8 years to complete one revolution around the center of the galaxy. It is important to note that this is a much larger timeframe than a human lifetime, meaning that humans have only seen a small fraction of a galactic year.
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Three balls are tossed with same initial speed from a fourth floor dormitory window. Ball A is launched 45 degrees above the horizontal, Ball B is launched 45 degrees below the horizontal, and Ball C is launched horizontally. Which ball hits the ground with the greatest speed
Answer:
All the balls hit the ground with same velocity.
Explanation:
Let the speed of throwing be u and height of building be h.
Ball A is launched 45 degrees above the horizontal
Initial vertical speed = usin45
Vertical acceleration = -g
Height = -h
Substituting in v² = u² +2as
v² = (usin45)² -2 x g x (-h)
v² = 0.5u² +2 x g x h
Final vertical speed² = 0.5u² +2 x g x h
Initial horizontal speed = final horizontal speed = ucos45
Final horizontal speed² = 0.5u²
Magnitude of final velocity
[tex]v=\sqrt{0.5u^2+2gh+0.5u^2}=\sqrt{u^2+2gh}[/tex]
Ball C is launched horizontally
Initial vertical speed = 0
Vertical acceleration = -g
Height = -h
Substituting in v² = 0² +2as
v² = 0² -2 x g x (-h)
v² = 2 x g x h
Final vertical speed² = 2 x g x h
Initial horizontal speed = final horizontal speed = u
Final horizontal speed² =u²
Magnitude of final velocity
[tex]v=\sqrt{2gh+u^2}=\sqrt{u^2+2gh}[/tex]
Ball B is launched 45 degrees below the horizontal
Initial vertical speed = usin45
Vertical acceleration = g
Height = h
Substituting in v² = u² +2as
v² = (usin45)² +2 x g x h
v² = 0.5u² +2 x g x h
Final vertical speed² = 0.5u² +2 x g x h
Initial horizontal speed = final horizontal speed = ucos45
Final horizontal speed² = 0.5u²
Magnitude of final velocity
[tex]v=\sqrt{0.5u^2+2gh+0.5u^2}=\sqrt{u^2+2gh}[/tex]
All the magnitudes are same so all the balls hit the ground with same velocity.
The ball C which was launched horizontally will hit the ground fastest.
Time of motion of the balls
With respect to angle of projection above the horizontal, the time of motion each ball is calculated as follows;
[tex]T = \frac{u sin\theta }{g}[/tex]
where;
θ is the angle of projection above the horizontalu is the initial velocityT is the time to hit the groundWhen θ is 45 degrees above the horizontal;
[tex]T = \frac{u \times sin(45)}{9.8}\\\\ T = 0.072u[/tex]
When θ is 45 degrees below the horizontal = 135 degrees above the horizontal
[tex]T = \frac{ u \times sin(135)}{9.8} \\\\T = 0.072 \ s[/tex]
When launched horizontally, θ = 90 degrees above the horizontal
[tex]T = \frac{u \times sin(90)}{9.8} \\\\T = 0.1 u[/tex]
Thus, the ball C which was launched horizontally will hit the ground fastest.
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A coil of self- induction 0.5 H, its ohmic resistance is 10 ⦠carrying a dc current of intensity 2 A, The voltage at its terminals isâ¦â¦ A. Zero
B. 20
C. 5
D. 0.2
Answer:
Option (B)
Explanation:
L = 0.5 H, I = 2 A, R = 10 Ohm
V = I R
In case of dc the inductor cannot works.
V = 2 x 10 = 20 V
Max (mass = 15 kg) is hanging from one end of a 13-m long bungee cord that has its other end fixed to a bridge above. The bungee cord has a circular cross section with a diameter of 2.5 cm, and a Young’s modulus of 17 MPa. What is the stress in the bungee cord due to Max’s weight?
The stress in the bungee cord due to the Max's weight is [tex]\fbox{\begin \\ 299511\text{ N}/\text{m}^{2}\end{minispace}}[/tex] or [tex]\fbox{\begin\\ 2.99\times10^{5}\text{ N}/\text{m}^{2}\end{minispace}}[/tex] or [tex]\fbox{\begin\\3.0\times10^{5}\text{ N}/\text{m}^{2}\end{minispace}}[/tex].
Further Explanation:
The stress developed in the bungee cord is the amount of tensile force developed inside the bungee cord due to the Max's weight.
Given:
The diameter d of the bungee cord is [tex]2.5\text{ cm}[/tex].
The mass m of Max is [tex]15\text{ kg}[/tex].
Concept:
The stress developed in the bungee cord is given by:
[tex]\fbox{\begin \sigma =\dfrac{F}{A}\end{minispace}}[/tex] ... (1)
Here, [tex]\sigma[/tex] is the stress developed in the bungee cord, [tex]F[/tex]is the force due to Max's weight and [tex]A[/tex] is the area of cross-section of the bungee cord.
The radius of the bungee cord will be the half of its diameter.
[tex]r=\dfrac{d}{2}[/tex]
Substitute [tex]2.5\text{ cm}[/tex] for [tex]d[/tex].
[tex]r= \dfrac{2.5\text{ cm}}{2}\\r=1.25 \text{ cm}[/tex]
Convert the radius of the bungee cord in meter.
[tex]r=\dfrac{1.25}{100} \text{ m}\\r=0.0125\text{ m}[/tex]
The area of cross-section of the bungee cord is given by:
[tex]A=\pi r^{2}[/tex]
Substitute [tex]0.0125\text{ m}[/tex] for [tex]r[/tex].
[tex]A=\pi (0.0125\text{ m})^{2}\\A=4.908\times10^{-4} m^{2}[/tex]
The force on the bungee cord due to Max's mass is the gravitational force acting on Max's body. The gravitational force acting on Max's body is given by:
[tex]F=mg[/tex]
Here, m is the mass of Max's body and g is acceleration due to gravity.
Consider the value of acceleration due to gravity on Earth be [tex]9.80\text{ m}/\text{s}^{2}[/tex].
Substitute [tex]15\text{ kg}[/tex] for m and [tex]9.80\text{ m}/\text{s}^{2}[/tex] for g in above expression.
[tex]F=(15\text{ kg})\times(9.80\text{ m}/\text{s}^{2})\\F=147\text{ N}[/tex]
Substitute [tex]147\text{ N}[/tex] for F and [tex]4.980\times10^{-4}\text{ m}^{2}[/tex] for A in equation (1).
[tex]\sigma=\dfrac{147\text{ N}}{4.908\times10^{-4}\text{ m}^{2}}\\\sigma=299511\text{ N}/\text{m}^{2}[/tex]
Thus, the stress developed in the bungee cord is [tex]\fbox{\begin \\ 299511\text{ N}/\text{m}^{2}\end{minispace}}[/tex] or [tex]\fbox{\begin\\ 2.99\times10^{5}\text{ N}/\text{m}^{2}\end{minispace}}[/tex] or [tex]\fbox{\begin\\3.0\times10^{5}\text{ N}/\text{m}^{2}\end{minispace}}[/tex].
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Answer Details:
Grade: Senior school
Subject: Physics
Chapter: Stress and Strain
Keywords:
Max, Max's weight, force, stress, area, 299511 N/m2, 299511 N/m^2, 2.99x10^5 N/m2, gravitational, radius, bungee, cord, weight, developed, mass, 3x10^5 N/m^2.
The stress in the bungee cord due to Max’s weight is 294,000 N/m².
The given parameters;
mass of Max, m = 15 kgdiameter of the cord, d = 2.5 cmradius of the cord, r = = 1.25 cm = 0.0125 mYoung's modulus of the rod, E = 17 MPaThe area of the bungee cord is calculated as follows;
[tex]A = \pi r^2\\\\A = \pi \times (0.0125)^2\\\\ A= 0.0005 \ m^2[/tex]
The weight of Max is calculated as follows;
[tex]F = mg\\\\F = 15 \times 9.8\\\\F = 147 \ N[/tex]
The stress in the bungee cord due to Max’s weight is calculated as;
[tex]\sigma = \frac{F}{A} \\\\\sigma = \frac{147}{0.0005} \\\\\sigma = 294,000 \ N/m^2[/tex]
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A 52.5-turn circular coil of radius 5.35 cm can be oriented in any direction in a uniform magnetic field having a magnitude of 0.455 T. If the coil carries a current of 25.3 mA, find the magnitude of the maximum possible torque exerted on the coil.
Answer:
5.43 x 10^-3 Nm
Explanation:
N = 52.5, radius, r = 5.35 cm = 0.0535 m, B = 0.455 T, I = 25.3 mA = 0.0253 A
Torque = N I A B Sin theta
Here, theta = 90 degree
Torque = 52.5 x 0.0253 x 3.14 x 0.0535 x 0.0535 x 0.455
Torque = 5.43 x 10^-3 Nm
Starting from rest, the boy runs outward in the radial direction from the center of the platform with a constant acceleration of 0.5 m/s2 . The platform rotates at a constant rate of 0.2 rad/s. Determine his velocity (magnitude) when t = 3 sec
Answer:
His velocity when t= 3 sec is V= 1.56 m/s
Explanation:
a= 0.5 m/s²
ω= 0.2 rad/s
t= 3 sec
Vr= a*t
Vr= 1.5 m/s
r= a*t²/2
r= 2.25m
Vt= w*r
Vt= 0.45 m/s
V= √(Vr²+Vt²)
V= 1.56 m/s
The boy's tangential velocity, after accelerating radially for 3 seconds at 0.5 m/s^2 while the platform rotates at 0.2 rad/s, is 1.5 m/s.
Explanation:To determine the boy's velocity (magnitude) at t = 3 sec while he runs outward radially from the center of a rotating platform, we must consider both his radial (tangential) velocity due to his acceleration and the rotational movement of the platform.
The boy starts from rest with a constant radial acceleration of 0.5 m/s2. Using the kinematic equation v = u + at (where u is the initial velocity, a is the acceleration, and t is the time), we can calculate his radial velocity after 3 seconds as:
vradial = 0 + (0.5 m/s2)(3 s) = 1.5 m/s
The platform's constant rotation rate is given as 0.2 rad/s. This rotational movement does not change the boy's radial (tangential) velocity directly, as the question only asks for his velocity in the radial direction. Therefore, the total magnitude of the boy's velocity after 3 seconds is 1.5 m/s tangentially.
A 66 g66 g ball is thrown from a point 1.05 m1.05 m above the ground with a speed of 15.1 ms/15.1 ms . When it has reached a height of 1.59 m1.59 m , its speed is 10.9 ms/10.9 ms . What was the change in the mechanical energy of the ball-Earth system because of air drag
Answer:
[tex]\Delta E = 8.20 - 4.95 = 3.25 J[/tex]
Explanation:
Initial total mechanical energy is given as
[tex]ME = U + KE[/tex]
here we will have
[tex]U = mgh[/tex]
[tex]U = (0.066)(9.81)(1.05) = 0.68 J[/tex]
also we have
[tex]KE = \frac{1}{2}mv^2[/tex]
[tex]KE = \frac{1}{2}(0.066)(15.1)^2[/tex]
[tex]KE = 7.52 J[/tex]
[tex]ME_i = 0.68 + 7.52 = 8.2 J[/tex]
Now similarly final mechanical energy is given as
[tex]U = mgh[/tex]
[tex]U = (0.066)(9.81)(1.59) = 1.03 J[/tex]
also we have
[tex]KE = \frac{1}{2}mv^2[/tex]
[tex]KE = \frac{1}{2}(0.066)(10.9)^2[/tex]
[tex]KE = 3.92 J[/tex]
[tex]ME_f = 1.03 + 3.92 = 4.95 J[/tex]
Now change in mechanical energy is given as
[tex]\Delta E = ME_i - ME_f[/tex]
[tex]\Delta E = 8.20 - 4.95 = 3.25 J[/tex]
A 2.07-kg fish is attached to the lower end of an unstretched vertical spring and released. The fish drops 0.131 m before momentarily coming to rest. (a) What is the spring constant of the spring? (b) What is the period of the oscillations of the fish? ?
Answer:
part a)
k = 310 N/m
part b)
T = 0.51 s
Explanation:
Part A)
As per work energy theorem we have
Work done by gravity + work done by spring = change in kinetic energy
[tex]mgx - \frac{1}{2}kx^2 = 0[/tex]
[tex](2.07)(9.8)(0.131) - \frac{1}{2}k(0.131)^2 = 0[/tex]
now we will have
[tex] k = 310 N/m[/tex]
Part B)
Time period of oscillation is given as
[tex]T = 2\pi\sqrt{\frac{m}{k}}[/tex]
[tex]T = 2\pi\sqrt{\frac{2.07}{310}}[/tex]
[tex]T = 0.51 s[/tex]
Can the dielectric permittivity of an insulator be negative?
An elevator car, with a mass of 450 kg is suspended by a single cable. At time = 0s, the elevator car is raised upward. The tension on the cable is constant at 5000N during the first 3 seconds of operation. Determine the magnitude of the velocity of the elevator at time = 3 seconds.
Answer:
33.33 m/s
Explanation:
m = 450 kg. T = 5000 N, t = 3 seconds,
let the net acceleration is a.
T = m a
a = 5000 / 450 = 11.11 m/s^2
u = 0 , v = ?
Let v be the velocity after 3 seconds.
Use first equation of motion
v = u + a t
v = 0 + 11.11 x 3 = 33.33 m/s