Answer:
22000 N
Explanation:
Convert velocity to SI units:
98 km/h × (1000 m / km) × (1 h / 3600 s) = 27.2 m/s
Draw a free body diagram. There are three forces acting on the car. Normal force perpendicular to the bank, gravity downwards, and friction parallel to the bank.
I'm going to assume the friction force is pointed down the bank. If I get a negative answer, that'll just mean it's actually pointed up the bank.
Sum of the forces in the radial direction (+x):
∑F = ma
N sin θ + F cos θ = m v² / r
Sum of the forces in the y direction:
∑F = ma
N cos θ - F sin θ - W = 0
To solve the system of equations for F, first solve for N and substitute.
N = (W + F sin θ) / cos θ
Substituting:
((W + F sin θ) / cos θ) sin θ + F cos θ = m v² / r
(W + F sin θ) tan θ + F cos θ = m v² / r
W tan θ + F sin θ tan θ + F cos θ = m v² / r
W tan θ + F (sin θ tan θ + cos θ) = m v² / r
W tan θ + F sec θ = m v² / r
F sec θ = m v² / r - W tan θ
F = m v² cos θ / r - W sin θ
F = m (v² cos θ / r - g sin θ)
Given that m = 1900 kg, θ = 11°, v = 27.2 m/s, and r = 55 m:
F = 1900 ((27.2)² cos 11 / 55 - 9.8 sin 11)
F = 21577 N
Rounding to two sig-figs, you need at least 22000 N of friction force.
A 1.5-kg object has a velocity of 5j m/s at t = 0. It is accelerated at a constant rate for five seconds after which it has a velocity of (6i + 12j ) m/s. What is the magnitude of the resultant force acting on the object during this time interval?
Answer:
2.76 N
Explanation:
m = mass of the object = 1.5 kg
v₀ = initial velocity at t = 0, = 0 i + 5 j
v = final velocity of the object at t = 5, = 6 i + 12 j
t = time interval = 5 sec
a = acceleration of the object = ?
Acceleration of the object is given as
[tex]a = \frac{v - v_{o}}{t}[/tex]
inserting the values
a = ((6 i + 12 j) - (0 i + 5 j))/5
a = (6 i + 7 j)/5
a = 1.2 i + 1.4 j
magnitude of the acceleration is given as
|a| = √((1.2)² + (1.4)²)
|a| = 1.84 m/s²
magnitude of the resultant force is given as
|F| = m |a|
|F| = (1.5) (1.84)
|F| = 2.76 N
To find the magnitude of the resultant force, we use Newton's second law of motion. Evaluating the acceleration at 2.0s gives a magnitude of 24.8 m/s^2. Using the formula F = ma, the magnitude of the resultant force is 37.2 N.
Explanation:To find the magnitude of the resultant force acting on the object during the time interval, we need to use Newton's second law of motion, which states that the force is equal to the mass of the object multiplied by its acceleration.
First, we need to find the acceleration of the object. We can use the formula:
a(t) = 5.0i + 2.0tj - 6.0t^2 km/s^2
By evaluating a(2.0 s), we get a magnitude of 24.8 m/s^2.
Now, we can use Newton's second law:
F = ma
Substituting the values, we get:
F = 1.5 kg * 24.8 m/s^2
F = 37.2 N
Therefore, the magnitude of the resultant force acting on the object during this time interval is 37.2 N.
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Which of the following is NOT an example of an assembly process? a) Handling b) Fitting c) Automated robot d) Orientation
Answer: I think it is D
Explanation: It is the only one that does not make sense
Final answer:
The correct answer is 'c) Automated robot' because it is a tool used within the assembly process, not a process itself like handling, fitting, or orientation.
Explanation:
The question is asking which of the provided options is not an example of an assembly process. An assembly process involves the steps required to put together parts to make a complete product. The options given are:
Handling - the act of manipulating components in preparation for assembly.
Fitting - the process of putting parts together, which is certainly part of assembly.
Automated robot - often used in assembly to perform repetitive tasks more efficiently.
Orientation - this typically refers to aligning parts in the correct position for assembly.
Based on these definitions, automated robot is not an example of an assembly process but is rather a tool that might be used in the process. Therefore, the correct answer is 'c) Automated robot'
An 800-N billboard worker stands on a 4.0-m scaffold supported by vertical ropes at each end. scaffold weighs 500-N and the worker stands 1.0 m from one end, what is the tension in the rope farther from the worker? (a) 1300 N (b) 1800 N (c) 450 N (d) None of these.
Explanation and answer:
This problem is best answered by drawing a figure as a first step.
ABC is the scaffold.
A downward force of 500N is applied downwards at 1m from end A.
The weight of 800N is exerted by the scaffold uniformly distributed between A & C.
At A and C, ropes are attached to support the load.
Let Fc=tension in rope passing through C.
Take moments about A:
Fc = (500N * 1m +800N*(3+1)/2m / 4m
= (500 Nm + 1600Nm) / 4m
= 2100 Nm / 4m
= 525 N
A pendulum of length L=36.1 cm and mass m=168 g is released from rest when the cord makes an angle of 65.4 degrees with the vertical.
(a) How far does the mass fall (y-displacement) before reaching its lowest point?
(b) How much work is done by gravity as it falls to its lowest point?
(c) How much work is done by the string tension as it falls to its lowest point?
(a) -0.211 m
At the beginning the mass is displaced such that the length of the pendulum is L = 36.1 cm and the angle with the vertical is
[tex]\theta=65.4^{\circ}[/tex]
The projection of the length of the pendulum along the vertical direction is
[tex]L_y = L cos \theta = (36.1 cm)(cos 65.4^{\circ})=15.0 cm[/tex]
the full length of the pendulum when the mass is at the lowest position is
L = 36.1 cm
So the y-displacement of the mass is
[tex]\Delta y = 15.0 cm - 36.1 cm = -21.1 cm = -0.211 m[/tex]
(b) 0.347 J
The work done by gravity is equal to the decrease in gravitational potential energy of the mass, which is equal to
[tex]\Delta U = mg \Delta y[/tex]
where we have
m = 168 g = 0.168 kg is the mass of the pendulum
g = 9.8 m/s^2 is the acceleration due to gravity
[tex]\Delta y = 0.211 m[/tex] is the vertical displacement of the pendulum
So, the work done by gravity is
[tex]W=(0.168 kg)(9.8 m/s^2)(0.211 m)=0.347 J[/tex]
And the sign is positive, since the force of gravity (downward) is in the same direction as the vertical displacement of the mass.
(c) Zero
The work done by a force is:
[tex]W=Fd cos \theta[/tex]
where
F is the magnitude of the force
d is the displacement
[tex]\theta[/tex] is the angle between the direction of the force and the displacement
In this situation, the tension in the string always points in a radial direction (towards the pivot of the pendulum), while the displacement of the mass is tangential (it follows a circular trajectory): this means that the tension and the displacement are always perpendicular to each other, so in the formula
[tex]\theta=90^{\circ}, cos \theta = 0[/tex]
and so the work done is zero.
The pendulum falls 0.212 m, the work done by gravity is 0.349 J, and the work done by string tension is 0 J.
Explanation:The first part of the question asks for the vertical displacement (y-displacement) of the pendulum. The length of the pendulum is the hypotenuse of a right triangle, and the vertical displacement is the adjacent side, so we can use the cosine function to solve: y = L*cos(θ). Plugging in the given values: y = 0.361 m * cos(65.4 degrees) = 0.149 m. So the fall is the length of the pendulum minus this displacement: 0.361 m - 0.149 m = 0.212 m.
The second part of the question asks for the work done by gravity. The work done by gravity is equal to the weight of the pendulum times the vertical distance it falls (Work = m*g*y), or 0.168 kg * 9.8 m/s² * 0.212 m = 0.349 J.
The final part of the question asks for the work done by string tension. The tension force always acts perpendicular to the direction of displacement, meaning it does no work on the pendulum, as work is defined as force times the displacement in the direction of the force. Therefore, the work done by the tension in the string is 0 J.
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A current carrying circular loop of wire lies flat on a table top. When viewed from above, the current moves around the loop in a counterclockwise sense. What is the direction of the magnetic field caused by this current, inside the loop? The magnetic field A) circles the loop in a counterclockwise direction. B) points straight up. C) points straight down. D) circles the loop in a clockwise direction. E) points toward the east.
The magnetic field direction, when current moves in a circular loop in a counterclockwise sense, points straight up. The right-hand rule is used to determine the orientation of the magnetic field.
Explanation:When current moves in a circular path, it generates a magnetic field. Using the right-hand rule, where your thumb points to the direction of the current and your fingers curl in the direction of the magnetic field, it can be determined that when looking from above, if the current moves in a counterclockwise sense, the magnetic field direction will be pointing straight up from the loop. Hence, the correct answer is B) points straight up.
This rule, also referred to as RHR-2 (Right Hand Rule-2), is widely applied to illustrate the magnetic fields induced around current carrying conductors. The net force on a current-carrying loop of any plane shape in a uniform magnetic field is zero, but it induces a distinctive orientation of the magnetic field in its surroundings.
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The correct answer is B) points straight up. Using the right-hand rule for a counterclockwise current in a circular loop, the magnetic field inside the loop points straight up.
To determine the direction of the magnetic field inside a current-carrying circular loop, we use the right-hand rule. In this case, the current moves counterclockwise when viewed from above, so follow these steps:
Point your right thumb in the direction of the current.
Notice how your fingers curl in the direction of the magnetic field.
Since the current flows counterclockwise, the right-hand rule shows that the magnetic field lines point straight up inside the loop. Therefore, the correct answer is B) points straight up.
In a super-heater (A) pressure rises, temperature drops (B) pressure rises, temperature remains constant (C) pressure remains constant and temperature rises (D) both pressure and temperature remains constant
Answer:
i believe that it is d
Explanation:
In a super heater, the temperature of the steam rises while the pressure remains constant. This process helps to remove the last traces of moisture from the saturated steam.
Explanation:In a super heater, the conclusion is that option (C) pressure remains constant and temperature rises is the correct choice. A super heater is a device used in a steam power plant to increase the temperature of the steam, above its saturation temperature. The function of the super heater is to remove the last traces of moisture (1 to 2%) from the saturated steam and to increase its temperature above the saturation temperature. The pressure, however, remains constant during this process because the super heater operates at the same pressure as the boiler.
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A transformer is used to change: 1) power 2) frequency 3) capacitance 4) none of the above
4)none of the above.
A transformer is an electrical device that allows to increase or decrease from one voltage level to another in an AC circuit while conserving the power.
It is basically composed of two coils magnetically coupled in a core of ferromagnetic material increasing its permeability and effectiveness.
Answer: None of the above
Explanation:
How much time would a force of 646 N need to stop a 82 kg object moving at 41 m/s (1 decimal place and no spaces between answer and units).
Answer:
The time is 5.21 s.
Explanation:
Given that,
Force F = 646 N
Mass m = 82 kg
Velocity v = 41 m/s
We need to calculate the acceleration
Using formula of force
F= ma
[tex]a = \dfrac{F}{m}[/tex]
[tex]a = \dfrac{646}{82}[/tex]
[tex]a =7.87\ m/s^2[/tex]
We need to calculate the time
Now, using equation of motion
[tex]v = u+at[/tex]
[tex]41=0+7.87 t[/tex]
[tex]t = \dfrac{41}{7.87}[/tex]
[tex]t = 5.21\ s[/tex]
Hence, The time is 5.21 s.
An airplane flies eastward and always accelerates at a constant rate. At one position along its path it has a velocity of 34.3 m/s, it then flies a further distance of 40100 m and afterwards its velocity is 47.5 m/s. Find the airplane\'s acceleration and calculate how much time elapses while the airplane covers those 40100 m.
Explanation:
We'll need two equations.
v² = v₀² + 2a(x - x₀)
where v is the final velocity, v₀ is the initial velocity, a is the acceleration, x is the final position, and x₀ is the initial position.
x = x₀ + ½ (v + v₀)t
where t is time.
Given:
v = 47.5 m/s
v₀ = 34.3 m/s
x - x₀ = 40100 m
Find: a and t
(47.5)² = (34.3)² + 2a(40100)
a = 0.0135 m/s²
40100 = ½ (47.5 + 34.3)t
t = 980 s
An 80.0 g sample of a gas was heated from 25 ∘C25 ∘C to 225 ∘C.225 ∘C. During this process, 346 J of work was done by the system and its internal energy increased by 6085 J.6085 J. What is the specific heat of the gas?
The specific heat of the gas is 0.381 J/g°C.
Explanation:The specific heat of a substance is the amount of heat energy required to raise the temperature of 1 gram of the substance by 1 degree Celsius. To find the specific heat of the gas, we can use the equation:
Q = mass * specific heat * change in temperature
In this case, the mass of the gas is 80.0g, the change in temperature is 225°C - 25°C = 200°C, and the energy absorbed by the gas is 6085J. Plugging these values into the equation, we can solve for the specific heat:
6085J = 80.0g * specific heat * 200°C
Specific heat = 6085J / (80.0g * 200°C) = 0.381 J/g°C
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The specific heat of the given gas can be calculated using the formula c = Q / (m * ΔT). By substitifying Q = 6431J (sum of work done by the system and increase in internal energy), m = 80g (mass of gas), and ΔT = 200°C (change in temperature), we find that the specific heat of the gas is roughly 0.40 J/g°C
Explanation:The specific heat of a substance is the heat required to raise the temperature of 1 gram of the substance by 1 degree Celsius. For the given gas, we first need to calculate the heat absorbed which is the sum of the work done by the system and the increase in its internal energy, which equals 346J + 6085J = 6431J. The following information is given, mass m = 80.0g and the change in temperature ΔT = 200°C (225 °C - 25°C).
The formula for specific heat is:
Q = m*c*ΔT
where:
Q is the heat energy absorbed (or released),
m is the mass of the substance,
c is the specific heat capacity, and
ΔT is the change in temperature.
Solving this equation for the specific heat (c), we find:
c = Q / (m * ΔT).
So, substituting the given values into this formula, we have:
c = 6431J / (80.0g * 200°C = 0.40 J/g °C which will be the specific heat of the gas.
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A satellite weighs 104 N at ground control. What best approximates the acceleration it experiences in orbit at an altitude of twice the earth's radius ifFpa = GME2thm, where r is the distance separating the centers of mass of the satellite and the Earth? O A. 111 m/s2 O B. 2.5 m/s O C. 1.1 m/s2 ○ D. 0 m/s
Answer:
[tex]a_c = 1.1 m/s^2[/tex]
Explanation:
As we know that net force on the Satellite due to gravity will provide it centripetal force
so we can say here
[tex]F_g = F_c[/tex]
[tex]\frac{GMm}{r^2} = ma_c[/tex]
now we will have
acceleration given by the equation
[tex]a_c = \frac{GM}{r^2}[/tex]
now we have
r = R + 2R = 3R
[tex]a_c = \frac{GM}{9R^2}[/tex]
also we know that acceleration due to gravity on the surface of earth is given as
[tex]g = \frac{GM}{R^2} = 9.8 m/s^2[/tex]
so the acceleration of satellite is given as
[tex]a_c = \frac{9.8}{9} = 1.1 m/s^2[/tex]
Final answer:
The acceleration of a satellite in orbit at an altitude of twice Earth's radius is best approximated using the formula for gravitational acceleration, g = GME / r². Plugging in the values and simplifying, the acceleration is approximately 2.45 m/s², with option B (2.5 m/s²) being the closest approximation.
Explanation:
Calculating the Acceleration of a Satellite in Orbit
In order to best approximate the acceleration a satellite experiences in orbit at an altitude of twice the Earth's radius, we must apply Newton's law of universal gravitation and the formula for gravitational acceleration, g = GME / r². Given that the satellite weighs 104 N at ground control which reflects the gravitational force at Earth's surface, we use this formula to find its gravitational acceleration in orbit. The relevant equation provided, Fpa = GME2thm/r, appears to be a typo, but our main equation for gravitational acceleration does not require mass, as it cancels out during the calculation:
g = GME / r²
Since we're given that the altitude is twice the Earth's radius, the distance r from Earth's center to the satellite in orbit is 3RE (1RE for Earth's radius and 2RE for the altitude above Earth). Considering this, we can calculate:
g = (6.674 x 10-11 m³/kg s²) (5.972 x 1024 kg) / (3 x 6.371 x 106 m)²
After simplifying, we get an acceleration of approximately 2.45 m/s². Therefore, option B, 2.5 m/s², best approximates the acceleration the satellite experiences in orbit at this altitude.
What is the power of a motor that can accelerate a 1700 kg car from rest to 30.0 m S in 5.0 seconds?
Answer:
The power of the motor is 153000 watts.
Explanation:
It is given that,
Mass of the car, m = 1700 kg
Initially, it is at rest, u = 0
Final velocity of the car, v = 30 m/s
Time taken, t = 5 s
We need to find the power of a motor. Work done per unit time is called power of the motor. We know that the change in kinetic energy is equal to the work done i.e.
[tex]P=\dfrac{W}{t}=\dfrac{\Delta E}{t}[/tex]
[tex]P=\dfrac{\dfrac{1}{2}mv^2}{t}[/tex]
[tex]P=\dfrac{\dfrac{1}{2}\times 1700\ kg\times (30\ m/s)^2}{5\ s}[/tex]
P = 153000 watts
So, the power of the motor is 153000 watts. Hence, this is the required solution.
what is periodicity?
Hello There!
"Periodicity" are types of trends that are seen in element properties.
This is the same thing as periodic trends. These are patterns that are present in the periodic table. These trends show different properties of elements and how characteristics increase or decrease.
A 39-kg girl is bouncing on a trampoline. During a certain interval after leaving the surface of the trampoline her kinetic energy decreases to 160 J from 490 J. How high does she rise during this interval? Neglect air resistance.
Answer:
The distance is 0.86 m.
Explanation:
Given that,
Mass = 39 kg
Initial kinetic energy, [tex]K.E_{i} = 160\ J[/tex]
Final kinetic energy, [tex]K.E_{f} = 490\ J[/tex]
We need to calculate the work done
According to work energy theorem
[tex]W = \Delta K.E[/tex]
[tex]W=K.E_{f}-K.E_{i}[/tex]...(I)
Work done is the product of the force and displacement.
[tex]W = mgh[/tex]....(II)
From equation (I) and (II)
[tex]K.E_{f}-K.E_{i}=mgh[/tex]
[tex]490-160=39\times9.8\times h[/tex]
[tex]h = 0.86\ m[/tex]
Hence, The distance is 0.86 m.
How far away was a lightning strike if thunder is heard 7.20 seconds after the flash is seen? (Assume that sound traveled at 350.0m/s during the storm)
Answer: 2.52 Kilometers
Explanation:
We know that the formula to calculate speed is given by :-
[tex]\text{Speed}=\dfrac{\text{Distance}}{\text{Time}}\\\\\Rightarrow\text{Distance}=\text{Speed}\times\text{Time}[/tex]
Given: The speed of sound during the storm : [tex]V=3350.0m/s[/tex]
The time taken by sound to travel : [tex]t= 7.20\text{ seconds}[/tex]
Then , the distance traveled by sound :-
[tex]\text{Distance}=350\times7.2=2520\text{ meters}=2.52\text{ kilometers}[/tex]
To find the distance of a lightning strike heard 7.20 seconds after the flash, multiply the sound's speed (350.0 m/s) by the time delay (7.20 s), resulting in a distance of 2520 meters.
We use the speed of sound. Since sound traveled at 350.0 m/s during the storm, we multiply the time it took to hear the thunder by the speed of sound to calculate the distance:
Distance = Speed of Sound × Time
Substituting in the given values:
Distance = 350.0 m/s × 7.20 s = 2520 meters
Therefore, the lightning struck approximately 2520 meters away.
An object is thrown vertically upward at 27.1 m/s. The velocity of the object 3.4 seconds later is ____ m/s. Round your answer to the nearest tenth. Do not use scientific notation. Take up as positive and down as negative.
Answer:
-6.2 m/s (downward)
Explanation:
The velocity of an object thrown vertically upward is given by:
[tex]v= u + at[/tex]
where:
u is the initial velocity
a = g = -9.8 m/s^2 is the acceleration due to gravity
t is the time
In this problem,
u = 27.1 m/s
t = 3.4 s
So, the velocity after 3.4 s is
[tex]v=27.1 m/s + (-9.8 m/s^2)(3.4 s)=-6.2 m/s[/tex]
and the negative sign means the velocity points downward.
A particle moves in the xy plane with constant acceleration. At time zero, the particle is at x = 6 m, y = 8.5 m, and has velocity ~vo = (9 m/s) ˆı + (−2.5 m/s) ˆ . The acceleration is given by ~a = (4.5 m/s 2 ) ˆı + (3 m/s 2 ) ˆ . What is the x component of velocity after 3.5 s? Answer in units of m/s.
The x component of velocity after 3.5 seconds is 24.75 m/s, calculated using the equation v = v0 + at with the given initial velocity and constant acceleration.
To calculate the x component of velocity after 3.5 seconds for a particle under constant acceleration, we use the equation:
v = v0 + at
Here v is the final velocity, v0 is the initial velocity, a is the acceleration, and t is the time elapsed. Given v0 = 9 m/s in the x direction, and acceleration a = 4.5 m/[tex]s^2[/tex] in the x direction, and time t = 3.5 s, the calculation is:
v = 9 m/s + (4.5 x (3.5))
The x component of velocity after 3.5 seconds is:
v = 9 m/s + 15.75 m/s
v = 24.75 m/s
A 320.9 ng sample of an unknown radioactive substance was placed in storage and its mass measured periodically. After 47 days the amount of radioactive substance had decreased to 40.11 ng. How many half‑lives of the unknown radioactive substance have occurred?
Answer:
3 half life of the unknown radioactive substance have occurred.
Explanation:
Mass of sample = 320.9 ng
Mass after 1 half life = 0.5 x 320.9 = 160.45 ng
Mass after 2 half life = 0.5 x 160.45 = 80.225 ng
Mass after 3 half life = 0.5 x 80.225 =40.11 ng
So 3 half life of the unknown radioactive substance have occurred.
The unknown radioactive substance went through approximately 3 half-lives in the 47 days period.
Explanation:The subject of your question is related to radioactive decay, particularly the concept of a half-life, which is a term in physics used to describe the time it takes for half the atoms in a sample to decay. In your case, the initial mass of your substance was 320.9 ng and it decreased to 40.11 ng after 47 days. To figure out how many half-lives have occurred, we need to understand that with each half-life, the quantity of the substance halve its original mass.
A useful method to answer your question is to divide the final amount by the starting amount and then take the logarithm base 2 of the result. This will give us the number of times the amount halved, which is the number of half-lives. In your case, the calculation is like this: number of half lives = log2(320.9 ng / 40.11 ng) = approx. 3 half-lives.
This means that in 47 days, approximately 3 half‑lives of the unknown radioactive substance have occurred, based on the given data. It's important to note that because this is an approximate result, the true half-life of the substance might be slightly smaller or larger.
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An object is traveling such that it has a momentum of magnitude 23.3 kg.m/s and a kinetic energy of 262 J. Determine the following. (a) speed of the object in meters per second. (b) mass of the object in kilograms.
Explanation:
It is given that,
Momentum of an object, p = 23.3 kg-m/s
Kinetic energy, E = 262 J
(a) Momentum is given by, p = mv
23.3 = mv...........(1)
Kinetic energy is given by, [tex]E=\dfrac{1}{2}mv^2[/tex]
m = mass of the object
v = speed of the object
[tex]E=\dfrac{1}{2}\times (mv)\times v[/tex]
[tex]262=\dfrac{1}{2}\times 23.3\times v[/tex]
v = 22.48 m/s
(2) Momentum, p = mv
[tex]m=\dfrac{p}{v}[/tex]
[tex]m=\dfrac{23.3\ kg-m/s}{22.48\ m/s}[/tex]
m = 1.03 Kg
Hence, this is the required solution.
An object with momentum 23.3 kg.m/s and kinetic energy of 262J is traveling at a speed of approximately 30.21 m/s and its mass is approximately 0.771 kg based on the physics principles of kinetic energy and momentum.
Explanation:The question involves the physics concepts of momentum and kinetic energy. We are given the momentum (p) of 23.3 kg.m/s and the kinetic energy (KE) of 262 J of an object.
(a) The formula for kinetic energy is KE = 0.5 * m * v^2, where m is the mass of the object and v is its velocity. We can rearrange to find v = sqrt((2*KE) / m). The mass can be obtained from the momentum formula, p = m * v, hence m = p / v. Substituting the second equation into the first gives v = sqrt((2 * KE * v) / p), which simplifies to v = sqrt((2 * 262 J) / 23.3 kg.m/s) = 30.21 m/s.
(b) With the velocity calculated in (a), the mass of the object can now be found by rearranging the momentum formula to m = p / v = 23.3 kg.m/s / 30.21 m/s = 0.771 kg.
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If the net torque is zero, what does this imply about the clockwise and counterclockwise torques.
When the net torque is zero, there is no overall force causing the object to turn clockwise or counterclockwise.
An object is said to be in a condition of rotational equilibrium if there is no net torque operating on it. In other words, the object is keeping a constant angular velocity while not suffering any rotational acceleration. The object has a constant angular momentum when the net torque is zero.
If the net torque is zero, the item is being pulled in a direction that is equal to the sum of all the clockwise and anticlockwise torques pulling on it. To put it another way, any propensity the object has to spin in one direction is counterbalanced by an equal tendency to rotate in the opposing direction.
Hence, When the net torque is zero, there is no overall force causing the object to turn clockwise or counterclockwise.
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Final answer:
When the net torque is zero, it means that the clockwise and counterclockwise torques are balanced.
Explanation:
When the net torque is zero, it implies that the clockwise and counterclockwise torques balance each other out. In other words, the total torque in the clockwise direction is equal in magnitude but opposite in sign to the total torque in the counterclockwise direction.
Suppose a 50-turn coil lies in the plane of the page in a uniform magnetic field that is directed into the page. The coil originally has an area of 0.25 m2. It is stretched to have no area in 0.1s. What is the direction and magnitude of the induced emf if the uniform magnetic field has a strength of 1.5T?
Answer:
EMF = 187.5 volts
Explanation:
As per Faraday's law of electromagnetic induction we know that rate of change in the flux will induce EMF
so we can say
[tex]EMF = -\frac{d\phi}{dt}[/tex]
[tex]EMF = \frac{\phi_i - \phi_f}{\Delta t}[/tex]
now we will have
[tex]EMF = \frac{NBA - 0}{\Delta t}[/tex]
[tex]EMF = \frac{50(1.5)(0.25)}{0.10}[/tex]
[tex]EMF = 187.5 Volts[/tex]
So the induced EMF in the coil will be 187.5 Volts
Ammonia can be synthesized according to the equilibrium reaction shown below. If the concentrations of the reactants and products were measured and found to be 0.50 M (N2), 3.00 M (H2), and 1.98 M (NH3), what is the value of the reaction quotient? N2(g) + 3H2(g) --> 2NH3(g) Kc = 0.291 (this reaction is reversible and undergoes equilbrium)
The reaction quotient (Q) for the given reaction, calculated using initial concentrations of the reactants and the product, is approximately 0.092. This value suggests the reaction will move forward, producing more NH3 to reach equilibrium.
Explanation:The reaction quotient, commonly referred to as 'Q', is a value used to determine the direction in which a reaction will proceed. It is calculated similarly to the equilibrium constant but uses the initial concentrations instead. For this reaction, the equation for Q would be [NH3]^2 / ([N2] * [H2]^3) based on the balanced chemical equation.
The initial concentrations given in the question are 0.50 M for N2, 3.00 M for H2, and 1.98 M for NH3. To find Qc, substitute these concentrations into our Q equation to get (1.98)^2 / (0.50 * 3.00^3), which simplifies to approximately 0.092.
If Qc < Kc, the reaction will proceed in the forward direction to reach equilibrium, so in this case, since our Qc (0.092) is less than Kc (0.291), the reaction will produce more NH3 to reach equilibrium.
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Electric power is to be generated by installing a hydraulic turbine-generator at a site 120 m below the free surface of a large water reservoir that can supply water at a rate of 1900 kg/s steadily. Determine the power generation potential
Answer:
P = 2234400 Watt
Explanation:
Power generation is given as rate of work done
Here the turbine generator is located at 120 m below the free surface
So here rate of work done is given as rate of potential energy
[tex]Power = \frac{dw}{dt}[/tex]
[tex]Power = \frac{mgh}{t}[/tex]
so we have
[tex]Power = (\frac{dm}{dt})gh[/tex]
now we have
[tex]Power = 1900(9.8)(120) = 2234400 Watt[/tex]
so the power generation potential will be 2234400 Watt
What is the difference between center of gravity and center of mass?
A car is traveling at 53.0 mi/h on a horizontal highway. (a) If the coefficient of static friction between road and tires on a rainy day is 0.104, what is the minimum distance in which the car will stop?
Answer:
902 ft
Explanation:
First convert mi/h to ft/s.
53.0 mi/h × (5280 ft / mi) × (1 h / 3600 s) = 77.7 ft/s
Sum of the forces on the car in the y direction:
∑F = ma
N - W = 0
N = mg
Sum of the forces on the car in the x direction:
∑F = ma
-F = ma
-Nμ = ma
Substituting;
-mgμ = ma
-gμ = a
Acceleration is constant, so:
v² = v₀² + 2a(x - x₀)
(0 ft/s)² = (77.7 ft/s)² + 2(-32.2 ft/s² × 0.104)(x - 0)
x = 902 ft
The minimum stopping distance is 902 ft.
Question Part Points Submissions Used A car is traveling at 53.0 km/h on a flat highway. (a) If the coefficient of friction between road and tires on a rainy day is 0.115, what is the minimum distance in which the car will stop? Incorrect: Your answer is incorrect. m (b) What is the stopping distance when the surface is dry and the coefficient of friction is 0.575? m
(a) 95.9 m
The initial velocity of the car is
[tex]u=53.0 km/h = 14.7 m/s[/tex]
The car moves by uniformly accelerated motion, so we can use the SUVAT equation:
[tex]v^2 - u^2 = 2ad[/tex]
where
v = 0 is the final velocity
d is the stopping distance of the car
a is the acceleration of the car
The force of friction against the car is
[tex]F_f = - \mu mg[/tex]
where
[tex]\mu=0.115[/tex] is the coefficient of friction
m is the mass of the car
[tex]g = 9.8 m/s^2[/tex] is the acceleration due to gravity
According to Newton's second law, the acceleration is
[tex]a=\frac{F}{m}=\frac{-\mu mg}{m}=-\mu g[/tex]
Substituting into the previous equation:
[tex]v^2 - u^2 = -2\mu g d[/tex]
and solving for d:
[tex]d=\frac{v^2 -u^2}{-2\mu g}=\frac{0-(14.7 m/s)^2}{-2(0.115)(9.8 m/s^2)}=95.9 m[/tex]
(b) 19.1 m
This time, the coefficient of friction is
[tex]\mu = 0.575[/tex]
So the acceleration due to friction is:
[tex]a=-\mu g = -(0.575)(9.8 m/s^2)=-5.64 m/s^2[/tex]
And substituting into the SUVAT equation:
[tex]v^2 - u^2 = 2ad[/tex]
we can find the new stopping distance:
[tex]d=\frac{v^2 -u^2}{-2a}=\frac{0-(14.7 m/s)^2}{2(-5.64 m/s^2)}=19.1 m[/tex]
The magnetic field produced by a long straight current-carrying wire is A) inversely proportional to the current in the wire and proportional to the distance from the wire. B) inversely proportional to both the current in the wire and the distance from the wire. C) proportional to both the current in the wire and the distance from the wire D) proportional to the current in the wire and inversely proportional to the distance from the wire. E) independent of both the current in the wire and the distance from the wire.
Answer:
The magnetic field produced by a long straight current-carrying wire is directly proportional to the current in the wire and inversely proportional to the distance from the wire.
Explanation:
The magnetic field produced by a long straight current-carrying wire is given by :
[tex]B=\dfrac{\mu_0I}{2\pi d}[/tex]............(1)
Where
[tex]\mu_o[/tex] = permeability of free space, [tex]\mu_o=4\pi\times 10^{-7}\ T-m/A[/tex]
I = current flowing in the wire
d = distance from wire
From equation (1), it is clear that the magnetic field produced by a long straight current-carrying wire is directly proportional to the current flowing and inversely proportional to the distance from the wire. So, the correct option is (d).
The magnetic field produced by a long straight current-carrying wire is proportional to the current in the wire and inversely proportional to the distance from the wire, demonstrated by Ampere's Law and the right-hand rule.
Explanation:The magnetic field produced by a long straight current-carrying wire is D) proportional to the current in the wire and inversely proportional to the distance from the wire. This relationship is described by Ampere's Law and the formula B = μI/2πr, where B is the magnetic field strength, μ is the permeability of free space, I is the current in the wire, and r is the distance from the wire. This suggests that as the current increases, the magnetic field strength increases, and as the distance from the wire increases, the magnetic field strength decreases.
Furthermore, the direction of the magnetic field is given by the right-hand rule. If you point the thumb of your right hand in the direction of the current, your fingers will curl in the direction of the magnetic field loops, which form concentric circles around the wire.
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A spherical capacitor is formed from two concentric, spherical, conducting shells separated by vacuum. The inner sphere has radius 16 cm and the capacitance is 113 pF. What is the radius of the outer sphere? (Give your answer in decimal using cm as unit)
Answer:
19 cm
Explanation:
inner radius, a = 16 cm, C = 113 p F = 113 x 10^-12 F
Let b cm be the radius of outer sphere.
The formula of capacitance of spherical capacitor is given by
C = 4π∈0 a b / (b - a)
a b / ( b - a) = C / 4π∈0
a b / ( b - a ) = 113 x 10^-12 x 9 x 10^9
a b / ( b - a ) = 1.017
16 x 10^-2 x b x 10^-2 = 1.017( b - 16) x 10^-2
0.16 b = 1.017 b - 16.272
0.857 b = 16.272
b = 19 cm
A ladder 10 ft long rests against a vertical wall. If the bottom of the ladder slides away from the wall at a rate of 1.1 ft/s, how fast is the angle between the ladder and the ground changing when the bottom of the ladder is 6 ft from the wall? (That is, find the angle's rate of change when the bottom of the ladder is 6 ft from the wall.)
Explanation:
If the distance between the bottom of the ladder and the wall is x, then:
cos θ = x / 10
Taking derivative with respect to time:
-sin θ dθ/dt = 1/10 dx/dt
Substituting for θ:
-sin (acos(x / 10)) dθ/dt = 1/10 dx/dt
Given that x = 6 and dx/dt = 1.1:
-sin (acos(6/10)) dθ/dt = 1/10 (1.1)
-0.8 dθ/dt = 0.11
dθ/dt = -0.1375
The angle is decreasing at 0.1375 rad/s.
The brightest star in the night sky in the northern hemisphere is Sirius. Its distance from Earth is estimated to be 8.7 light years. A light year is the distance light travels in one year. Light travels at a speed of 3.00 × 108 m/s. Calculate the distance from Earth to Sirius in miles. (1 mi = 5280 ft) g
Answer:[tex]5.11(10)^{13}miles[/tex]
Explanation:
A light year is a unit of length and is defined as "the distance a photon would travel in vacuum during a Julian year at the speed of light at an infinite distance from any gravitational field or magnetic field. "
In other words: It is the distance that the light travels in a year.
This unit is equivalent to [tex]5.879(10)^{12}miles[/tex], which mathematically is expressed as:
[tex]1Ly=5.879(10)^{12}miles[/tex]
Doing the conversion:
[tex]8,7Ly.\frac{5.879(10)^{12}miles}{1Ly}=5.11(10)^{13}miles[/tex] This is the distance from Earth to Sirius in miles.
Final answer:
The distance from Earth to Sirius is approximately 50 trillion miles.
Explanation:
To calculate the distance from Earth to Sirius in miles, we need to convert the light year measurement to miles. We know that 1 light year is the distance light travels in one year, which we can calculate by multiplying the speed of light by the number of seconds in a year. Here's the step-by-step calculation:
Speed of light = 3.00 × 10^8 m/s
Number of seconds in a year = 365 days * 24 hours * 60 minutes * 60 seconds = 31,536,000 seconds
Distance in meters = speed of light * number of seconds in a year = 3.00 × 10^8 m/s * 31,536,000 seconds
Distance in miles = distance in meters / (5280 ft/mi * 12 in/ft)
Plugging in the values into the equation, we get:
Distance in miles = (3.00 × 10^8 m/s * 31,536,000 seconds) / (5280 ft/mi * 12 in/ft)
Simplifying the equation, the distance from Earth to Sirius is approximately 50 trillion miles.