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.
The wavelength corresponding to light with a frequency of 4 x 10^14 Hz is 1.33m 0.075 mm 7500 nm 750 nm
Answer:
Wavelength of light is 750 nm.
Explanation:
It is given that,
Frequency of light, [tex]\nu=4\times 10^{14}\ Hz[/tex]
The relationship between the wavelength and the frequency of light is given by :
[tex]c=\nu\times \lambda[/tex]
Where
c = speed of light
[tex]\nu[/tex] = frequency of light
[tex]\lambda[/tex] = wavelength of light
[tex]\lambda=\dfrac{c}{\nu}[/tex]
[tex]\lambda=\dfrac{3\times 10^8\ m/s}{4\times 10^{14}\ Hz}[/tex]
[tex]\lambda=7.5\times 10^{-7}\ m[/tex]
[tex]\lambda=750\ nm[/tex]
Hence, the correct option is (d) "750 nm".
The wavelength corresponding to light with a frequency of 4 x 10^14 Hz is found using the formula λ = c/f and equals 750 nm, falling within the visible spectrum.
Explanation:The wavelength corresponding to a frequency of 4 x 10^14 Hz can be calculated using the equation c = λf, where c is the speed of light (3.0 × 10^8 m/s), λ is the wavelength in meters, and f is the frequency in hertz (Hz). To find the wavelength, we rearrange the equation to λ = c/f. Plugging in the values, we get λ = (3.0 × 10^8 m/s) / (4 x 10^14 Hz) which equals 750 nm. Therefore, the wavelength corresponding to light with a frequency of 4 x 10^14 Hz is 750 nm, which falls within the range of visible light wavelengths (400 nm to 750 nm).
The rate constant for this second‑order reaction is 0.760 M−1⋅s−1 at 300 ∘C. A⟶products How long, in seconds, would it take for the concentration of A to decrease from 0.750 M to 0.330 M?
Answer:
2.23 s
Explanation:
For a second-order reaction:
1 / [A] = 1 / [A]₀ + kt
Given [A] = 0.330 M, [A]₀ = 0.750 M, and k = 0.760 M⁻¹s⁻¹:
1 / 0.330 = 1 / 0.750 + 0.760t
t = 2.23
It would take 2.23 seconds.
Final answer:
To find the time it takes for a second-order reaction concentration to decrease from 0.750 M to 0.330 M at the given rate constant, we use the integrated second-order rate law. After substituting the given values and solving for time, we find that it takes approximately 1.20 seconds.
Explanation:
To calculate the time taken for the concentration of A to decrease from 0.750 M to 0.330 M in this second-order reaction, we can use the integrated second-order rate law which is given by:
[tex]\( \frac{1}{[A]_t} - \frac{1}{[A]_0} = k \cdot t \)[/tex]
Where [tex]\([A]_t\)[/tex] is the concentration of A at time t, [tex]\([A]_0\)[/tex] is the initial concentration of A, k is the rate constant, and t is the time.
Plugging in the values, we get:
[tex]\( \frac{1}{0.330 \ M} - \frac{1}{0.750 \ M} = 0.760 \ M^{-1}s^{-1} \cdot t \)[/tex]
After solving for t, we find that the time required is approximately 1.20 seconds. Note that the units of k given [tex](\( M^{-1}s^{-1} \))[/tex] are consistent with a second-order reaction, which confirms the correctness of the approach.
Suppose you apply a force of "20 N to a 0.25-meter-long wrench attached to a bolt in a direction perpendicular to the bolt. Determine the magnitude of the torque when the force is applied at an angle of 120degrees°" to the wrench.
Answer:
Torque is defined as the force that produces rotation around the axis of an object. Torque is also known as the moment of force. It's S.I unit is Nm.
Formula of torque is:
τ= rxF(sinθ)
Given Data:
Force = 20 N
r = 0.25 m
θ = 120°
Solution:
τ= rxF(sinθ)
= 0.25 x 20 (sin 120)
= 2.903 Nm
Thus, the total torque applied on the wrench is 2.903 Nm.
The magnitude of the torque when the force is applied at an angle of 120 degrees to the wrench is 4.33 N.m.
Explanation:To determine the magnitude of the torque when a force is applied at an angle to a wrench, we can use the formula:
Torque = Force x Distance x sin(angle)
In this case, the force is 20 N, the distance is 0.25 meters, and the angle is 120 degrees. Plugging in these values, we get:
Torque = 20 N x 0.25 m x sin(120°)
Torque = 20 N x 0.25 m x 0.866
Torque = 4.33 N.m
Therefore, the magnitude of the torque when the force is applied at an angle of 120 degrees to the wrench is 4.33 N.m.
A bullet moving at a speed of 152 m/s passes through a plank of wood at 128m/s. Another bullet moving at 97m/s passes through the same plank at what speed?
Answer:
The speed of the plank is 81.68 m/s
Explanation:
Given that,
Speed of bullet = 152 m/s
Speed of wood = 128 m/s
Speed of another bullet = 97 m/s
We need to calculate the speed of plank
Using conservation of momentum
[tex]m_{1}u_{1}=(m_{1}+m_{2})v[/tex]
Where,
u = initial velocity
v = final velocity
[tex]152m_{1}=(m_{1}+m_{2})128[/tex]....(I)
[tex]m_{1}97=(m_{1}+m_{2})v[/tex]....(II)
From equation(I) and equation(II)
[tex]\dfrac{152}{97}=\dfrac{128}{v}[/tex]
[tex]= \dfrac{128\times97}{152}[/tex]
[tex]v=81.68\ m/s[/tex]
Hence, The speed of the plank is 81.68 m/s
A 25.0−gsample of an alloy at 93.00°Cis placed into 50.0 gof water at 22.00°Cin an insulated coffee-cup calorimeter with a heat capacity of 9.20 J/K.If the final temperature of the system is 31.10°C,what is the specific heat capacity of the alloy?
Answer:
1.23 J/(g °C)
Explanation:
[tex]m_{w}[/tex] = mass of water = 50 g
[tex]c_{w}[/tex] = specific heat of water = 4.186 J/(g °C)
[tex]T_{wo}[/tex] = Initial temperature of water = 22.00 °C
[tex]m_{a}[/tex] = mass of alloy = 25 g
[tex]c_{a}[/tex] = specific heat of alloy = ?
[tex]T_{ao}[/tex] = Initial temperature of alloy = 93.00 °C
[tex]T_{f}[/tex] = Final equilibrium temperature = 31.10 °C
Using conservation of heat
Heat Lost by alloy = Heat gained by water
[tex]m_{a}c_{a}(T_{ao}-T_{f})[/tex] = [tex]m_{a}c_{a}(T_{f} - T_{ao})[/tex]
(25) [tex]c_{a}[/tex] (93 - 31.10) = (50) (4.186) (31.10 - 22)
[tex]c_{a}[/tex] = 1.23 J/(g °C)
How much work would a child do while puling a 12-kg wagon a distance of 3m with a 22 N force directed 30 degrees with respect to the horizontal? (A) 82J (B) 52J (C) 109J (D) 95J
Answer:
The work done will be 57.15 J
Explanation:
Given that,
Mass = 12 kg
Distance = 3 m
Force = 22 N
Angle = 30°
We need to calculate the work done
The work done is defined as,
[tex]W = Fd\cos\theta[/tex]
Where, F = force
d = displacement
Put the value into the formula
[tex]W=22\times3\times\cos30^{\circ}[/tex]
[tex]W=22\times3\times\dfrac{\sqrt{3}}{2}[/tex]
[tex]W = 57.15\ J[/tex]
Hence, The work done will be 57.15 J
The process in which heat flows by the mass movement of molecules from one place to another is snown as (2 pts) A) conduction B) radiation. C) convection. D) all of the above E) none of the above
Answer:
The process in which heat flows by the mass movement of molecules from one place to another is C) convection.
Hope this helps :)
The process in which heat flows by the mass movement of molecules from one place to another is convection.
Heat transfer by convectionHeat transfer by convection is a method of heat transfer that involves the mass movement of molecules from one place to another.
Other methods of heat transferHeat can also be transfered in the following methods;
Conduction, andRadiationThus, the process in which heat flows by the mass movement of molecules from one place to another is convection.
Learn more about convection here: https://brainly.com/question/893656
#SPJ2
Numerous engineering and scientific applications require finding solutions to a set of equations. Ex: 8x + 7y = 38 and 3x - 5y = -1 have a solution x = 3, y = 2. Given integer coefficients of two linear equations with variables x and y, use brute force to find an integer solution for x and y in the range -10 to 10.
To find an integer solution to given linear equations, iterate through integer values for x and y within the range of -10 to 10 and substitute them into the equations to check for solutions.
Explanation:Finding Integer Solutions for Linear EquationsTo find an integer solution for the set of given linear equations using brute force, you would substitute integer values for x and y within the specified range of -10 to 10. The equations given as examples, such as 7y = 6x + 8, 4y = 8, and y + 7 = 3x, are all linear equations that can be rearranged into the standard form y = mx + b, where m is the slope and b is the y-intercept.
In practice, you would start with -10 for x, solve for y in each equation, and check if y is an integer. Repeat incrementally until you reach 10 for x. If you find a pair of x and y values that satisfy both equations, you've found a solution. Remember to verify your solution by substiting the found values back into the original equations to ensure they hold true.
For example, if the equations were 2x + 3y = 6 and 4x - y = 10, you would try each pair of x and y between -10 and 10 until you find a pair that works for both equations. This brute force method is not the most efficient, but it will eventually yield an integer solution, if one exists within the given range.
Learn more about Integer Solutions for Linear Equations here:https://brainly.com/question/30709147
#SPJ12
The solution that satisfies both equations is x = 3 and y = 2.
In many engineering and scientific applications, finding solutions to systems of linear equations is crucial. Given the linear equations 8x + 7y = 38 and 3x - 5y = -1, we are tasked with finding integer solutions for x and y in the range of -10 to 10.
Steps to Solve Using Brute Force :
Identify the range for x and y, which is -10 to 10.Check each pair of (x, y) values within the given range to see if they satisfy both equations.We will systematically check all integer pairs (x, y) within the specified range:
For x = -10, iterate y from -10 to 10 and check if both 8x + 7y = 38 and 3x - 5y = -1 are satisfied.Repeat this process for each integer value of x from -10 to 10.After performing these steps, we find the solution: x = 3 and y = 2. Both equations are satisfied by this pair because:
8(3) + 7(2) = 24 + 14 = 38 and 3(3) - 5(2) = 9 - 10 = -1
Thus, x = 3 and y = 2 is the correct solution.
The energy of a photon is proportional to its a) amplitude. d) wave number, k-2m/A c) velocity. b) frequency
Answer:
Frequency
Explanation:
Photons are the packet of energy. They are massless and chargeless particles. They travel in the vacuum with the speed of light. The energy of photon is given by :
[tex]E=h\nu[/tex]
Where
h = Planck's constant
[tex]\nu[/tex] = frequency of photon
Or [tex]E=\dfrac{hc}{\lambda}[/tex]
c = speed of light
[tex]\lambda[/tex] = wavelength of photon
From the above equation, it is clear that the energy of photon is directly proportional to its frequency.
At a sand and gravel plant, sand is falling off a conveyor and onto a conical pile at a rate of 6 cubic feet per minute. The diameter of the base of the cone is approximately three times the altitude. At what rate is the height of the pile changing when the pile is 2 feet high? (Hint: The formula for the volume of a cone is V = 1 3 πr2h.)
Answer:
2/(3π) ft/min ≈ 0.212 ft/min
Explanation:
Volume of a cone is:
V = ⅓ π r² h
The diameter is three times the altitude, so:
2r = 3h
r = 3/2 h
Substituting:
V = ⅓ π (3/2 h)² h
V = ⅓ π (9/4 h²) h
V = ¾ π h³
Taking derivative with respect to time:
dV/dt = 9/4 π h² dh/dt
Given dV/dt = 6 and h = 2:
6 = 9/4 π (2)² dh/dt
6 = 9π dh/dt
dh/dt = 2/(3π)
dh/dt ≈ 0.212 ft/min
n electric motor rotating a workshop grinding wheel at a rate of 94 rev/min is switched off. Assume constant negative angular acceleration of magnitude 1.70 rad/s. (a) How long does it take for the grinding wheel to stop? (b) Through how many radians has the wheel turned during the interval found in (a)? rad
Answer:
a) It takes 5.79 seconds for the grinding wheel to stop.
b) In an interval of 5.79 seconds it rotates 28.48 rad.
Explanation:
a) We have equation of motion v = u + at
Here v = 0 rad/s, a = -1.70 rad/s², u = 94 rev/min = 9.84 rad/s
Substituting
0 = 9.84 - 1.70 x t
t = 5.79 seconds.
It takes 5.79 seconds for the grinding wheel to stop.
b) We have equation of motion v² = u² + 2as
v = 0 rad/s, a = -1.70 rad/s², u = 9.84 rad/s
Substituting
0² = 9.84² - 2 x 1.70 x s
s = 28.48 rad
So in an interval of 5.79 seconds it rotates 28.48 rad.
A tennis ball is shot vertically upward in an evacuated chamber inside a tower with an initial speed of 20.0 m/s at time t = 0 s. Approximately how long does it take the tennis ball to reach its maximum height?
Using the physics of projectile motion, it can be calculated that a tennis ball shot vertically upward with an initial speed of 20 m/s will take approximately 2.04 seconds to reach its maximum height.
Explanation:The subject of this question is the physics of motion, specifically projectile motion. The tennis ball is being shot vertically upward, so its motion can be considered as projectile motion. The time it takes for the ball to reach its maximum height can be calculated by using the vertical motion properties of projectiles. We know that the upwards velocity of the ball when released is 20 m/s and the acceleration due to gravity is -9.8 m/s² (indicating the ball is slowing as it rises, eventually down to 0 m/s at its highest point).
Now, to calculate time, we can use this formula for acceleration: 'v = u + gt'. Here, v is final vertical velocity, u is initial vertical velocity, g is acceleration due to gravity and t is the time. When the ball reaches its maximum height, its final vertical velocity will be 0. So we transform the formula to 't= (v-u)/g', and inputting the known values, '-(0 - 20 m/s)/-9.8 m/s², gives us a time of approximately 2.04 seconds
Thus, it will take approximately 2.04 seconds for the tennis ball to reach its maximum height.
Learn more about Projectile Motion here:https://brainly.com/question/29545516
#SPJ3
How large must the coeffficient of static friction be between the tires and the road if a car is to round a level curve of radius 120 m at a speed of 120 km/h Express your answer using two significant figures.
Answer:
Coefficient of static friction required = 0.94
Explanation:
Coefficient of static friction required [tex]=\frac{v^2}{gR}[/tex]
Velocity, v = 120 km/h = 33.33 m/s
g = 9.81 m/s²
R = 120 m
Coefficient of static friction required[tex]=\frac{33.33^2}{9.81\times 120}=0.94[/tex]
Coefficient of static friction required = 0.94
The man fires an 80 g arrow so that it is moving at 80 m/s when it hits and embeds in a 8.0 kg block resting on ice. How far will the block slide on the ice before stopping? A 7.1 N friction force opposes its motion.
Answer:
The block will be slide 0.36 m on the ice.
Explanation:
Given that,
Mass of arrow m₁= 80 g
Velocity of arrow u₁= 80 m/s
Mass of block m₂= 8.0 kg
Force F = 7.1 N
Using conservation of momentum
[tex]m_{1}u_{1}=m_{2}v_{2}[/tex]
[tex]80\times10^{-3}\times80=8.0\times v[/tex]
[tex]v =\dfrac{80\times10^{-3}\times80}{8.0}[/tex]
[tex]v = 0.8\ m/s[/tex]
The work done is equal to the change in kinetic energy
[tex]W=\Delta KE[/tex]
[tex]W=\dfrac{1}{2}mv^2[/tex]
[tex]W=\dfrac{1}{2}\times8.0\times0.8^2[/tex]
[tex]W=2.56\ J[/tex]
We know that,
The work is defined as,
[tex]W = F\cdot d[/tex]
[tex]d = \dfrac{W}{F}[/tex]
[tex]d=\dfrac{2.56}{7.1}[/tex]
[tex]d =0.36\ m[/tex]
Hence, The block will be slide 0.36 m on the ice.
The solution to the question is found by applying the principles of conservation of momentum to calculate the velocity of the block after collision and then using the work-energy theorem to find the distance the block slides on ice.
Explanation:The calculation to find the answer to your question requires the usage of conservation of momentum and the theory of work. Firstly, we apply conservation of momentum, using the formula: initial momentum = final momentum. Momentum, p=m*v, where m is mass and v is velocity. The initial momentum is the momentum of the arrow just before it hits the block, equal to (0.080 kg * 80 m/s). The block of ice initially is at rest, so it has no momentum.
Post-collision, the arrow and block move together, so the final momentum is (8.080 kg * V), with V being the velocity we wish to calculate. Set the initial and final momentum equal and solve for V.
Now we have the velocity of the block and arrow post-collision. The block slides until brought to rest by friction. Here we use the work-energy theory, where the work is equal to the change in kinetic energy. The work done by the friction force is (friction force * distance), and the change in kinetic energy is (1/2)*m*V^2 - 0. Solve for distance to find the answer.
Learn more about Physics - Conservation of Momentum and Work-Energy Theorem here:https://brainly.com/question/36673923
#SPJ3
A skydiver is falling vertically downward toward the earth with a constant velocity. What can we conclude from this? O His velocity is 9.8 m/s downward. The net force on him is equal to his weight. O His acceleration is not zero but it cannot be determined without more data O His acceleration is 9.8 m/s2 dwnward O The net force on him is zero.
Answer:
The net force on him is zero.
Explanation:
The velocity of skydiver is constant.
As we know that the acceleration is rate of change in velocity. So, here velocity os constant it means acceleration of skydiver is zero.
According to Newton's second law
Force acting on a body is equal to the product of mass and velocity of the body.
As acceleration is zero that means the net force acting on the body is zero.
Calculating work for different springs Calculate the work required to stretch the following springs 0.5 m from their equilibrium positions. Assume Hooke’s law is obeyed. a. A spring that requires a force of 50 N to be stretched 0.2 m from its equilibrium position. b. A spring that requires 50 J of work to be stretched 0.2 m from its equilibrium position.
Answer:
Part a)
U = 31.25 J
Part b)
U = 312.5 J
Explanation:
Part A)
A spring that requires a force of 50 N to be stretched 0.2 m from its equilibrium position.
So here we have
[tex]F = kx[/tex]
[tex]50 = k(0.2)[/tex]
k = 250 N/m
now the energy stored in the spring is given by
[tex]U = \frac{1}{2}kx^2[/tex]
[tex]U = \frac{1}{2}(250)(0.5)^2[/tex]
[tex]U = 31.25 J[/tex]
Part B)
A spring that requires 50 J of work to be stretched 0.2 m from its equilibrium position.
So here we know the formula of spring energy as
[tex]U = \frac{1}{2}kx^2[/tex]
[tex]50 = \frac{1}{2}k(0.2)^2[/tex]
[tex]k = 2500 N/m[/tex]
now by the formula of energy stored in spring
[tex]U = \frac{1}{2}kx^2[/tex]
[tex]U = \frac{1}{2}(2500)(0.5)^2[/tex]
[tex]U = 312.5 J[/tex]
Final answer:
The work required to stretch the first spring 0.5 m from its equilibrium position, given a spring constant of 250 N/m, is 31.25 J. For the second spring, with a spring constant of 2500 N/m, the work required is 312.5 J.
Explanation:
To calculate the work required to stretch the springs mentioned in the student's question, we use Hooke's law which is given by W = ½ k x², where W is the work done, k is the spring constant, and x is the displacement of the spring from its equilibrium position.
Part a
Given that a force of 50 N stretches the spring 0.2 m, we first determine the spring constant using F = k x:
k = F / x = 50 N / 0.2 m = 250 N/m.
Then, we calculate the work done to stretch the spring 0.5 m from its equilibrium position using W = ½ k x² = ½ × 250 N/m × (0.5 m)² = 31.25 J.
Part b
If 50 J of work is required to stretch the spring 0.2 m, the spring constant can be obtained by rearranging the work formula:
50 J = ½ k (0.2 m)², solving for k, yields k = 50 J / (0.5 × (0.2 m)²) = 2500 N/m.
Finally, calculating the work needed to stretch the spring 0.5 m from its equilibrium using W = ½ k x² = ½ × 2500 N/m × (0.5 m)² = 312.5 J.
A 125-kg astronaut (including space suit) acquires a speed of 2.50 m/s by pushing off with her legs from a 1900-kg space capsule. (a) What is the change in speed of the space capsule? (b) If the push lasts 0.600 s, what is the average force exerted by each on the other? As the reference frame, use the position of the capsule before the push. (c) What is the kinetic energy of each after the push?
(a) 0.165 m/s
The total initial momentum of the astronaut+capsule system is zero (assuming they are both at rest, if we use the reference frame of the capsule):
[tex]p_i = 0[/tex]
The final total momentum is instead:
[tex]p_f = m_a v_a + m_c v_c[/tex]
where
[tex]m_a = 125 kg[/tex] is the mass of the astronaut
[tex]v_a = 2.50 m/s[/tex] is the velocity of the astronaut
[tex]m_c = 1900 kg[/tex] is the mass of the capsule
[tex]v_c[/tex] is the velocity of the capsule
Since the total momentum must be conserved, we have
[tex]p_i = p_f = 0[/tex]
so
[tex]m_a v_a + m_c v_c=0[/tex]
Solving the equation for [tex]v_c[/tex], we find
[tex]v_c = - \frac{m_a v_a}{m_c}=-\frac{(125 kg)(2.50 m/s)}{1900 kg}=-0.165 m/s[/tex]
(negative direction means opposite to the astronaut)
So, the change in speed of the capsule is 0.165 m/s.
(b) 520.8 N
We can calculate the average force exerted by the capsule on the man by using the impulse theorem, which states that the product between the average force and the time of the collision is equal to the change in momentum of the astronaut:
[tex]F \Delta t = \Delta p[/tex]
The change in momentum of the astronaut is
[tex]\Delta p= m\Delta v = (125 kg)(2.50 m/s)=312.5 kg m/s[/tex]
And the duration of the push is
[tex]\Delta t = 0.600 s[/tex]
So re-arranging the equation we find the average force exerted by the capsule on the astronaut:
[tex]F=\frac{\Delta p}{\Delta t}=\frac{312.5 kg m/s}{0.600 s}=520.8 N[/tex]
And according to Newton's third law, the astronaut exerts an equal and opposite force on the capsule.
(c) 25.9 J, 390.6 J
The kinetic energy of an object is given by:
[tex]K=\frac{1}{2}mv^2[/tex]
where
m is the mass
v is the speed
For the astronaut, m = 125 kg and v = 2.50 m/s, so its kinetic energy is
[tex]K=\frac{1}{2}(125 kg)(2.50 m/s)^2=390.6 J[/tex]
For the capsule, m = 1900 kg and v = 0.165 m/s, so its kinetic energy is
[tex]K=\frac{1}{2}(1900 kg)(0.165 m/s)^2=25.9 J[/tex]
The astronaut's push off leads to a change in the speed of the space capsule of -0.164 m/s. The average force exerted is approximately 416.67N. The final kinetic energy of the astronaut and the space capsule are 156.25J and 25628.1J respectively.
Explanation:This problem covers the principle of conservation of momentum. The astronaut and the space capsule constitute a closed system, where the total momentum before and after the push must be equal.
(a) Using the principle of conservation of momentum (initial momentum = final momentum), we can calculate the change in speed of the space capsule when the astronaut pushes off. We start with the equation m1V1 + m2V2 = 0, where m1 is the astronaut's mass and V1 is her speed, and m2 is the space capsule's mass and V2 is its velocity. Solving for V2 gives us a change in speed of the space capsule of -0.164m/s (which is in the opposite direction to the astronaut's motion).
(b) The average force exerted can be calculated by changing momentum over time (Force = Change in momentum / Time). Here we obtain approximately 416.67N.
(c) The final kinetic energy of each object is K = 1/2*m*v^2. For the astronaut, this is approximately 156.25J and for the space capsule, this is 25628.1J.
Learn more about Conservation of Momentum here:https://brainly.com/question/33316833
#SPJ3
A potter's wheel moves uniformly from rest to an angular speed of 0.20 rev/s in 32.0 s. (a) Find its angular acceleration in radians per second per second. rad/s2 (b) Would doubling the angular acceleration during the given period have doubled final angular speed?
a. The wheel accelerates uniformly, so its constant acceleration is equal to the average acceleration:
[tex]\alpha=\dfrac{0.20\frac{\rm rev}{\rm s}-0}{32.0\,\rm s}=0.0063\dfrac{\rm rev}{\mathrm s^2}[/tex]
b. Yes. Since
[tex]\alpha=\dfrac{\Delta\omega}{\Delta t}=\dfrac\omega{\Delta t}[/tex]
then multiplying [tex]\alpha[/tex] by 2 means we double the change in angular speed, but the wheel starts from rest so only the final angular speed [tex]\omega[/tex] gets doubled.
A 245-kg merry-go-round in the shape of a uniform, solid, horizontal disk of radius 1.50 m is set in motion by wrapping a rope about the rim of the disk and pulling on the rope. What constant force would have to be exerted on the rope to bring the merry-go-round from rest to an angular speed of 0.600 rev/s in 2.00 s? (State the magnitu
Answer:
F = 345.45 N
Explanation:
Angular acceleration of the disc is given as rate of change in angular speed
it is given by formula
[tex]\alpha = \frac{d\omega}{dt}[/tex]
[tex]\alpha = \frac{2\pi(0.600)}{2}[/tex]
[tex]\alpha = 1.88 rad/s^2[/tex]
now we know that moment of inertia of the solid uniform disc is given as
[tex]I = \frac{1}{2}mR^2[/tex]
[tex]I = \frac{1}{2}245(1.50)^2[/tex]
[tex]I = 275.625 kg m^2[/tex]
now we have an equation for torque as
[tex]\Tau = I\alpha[/tex]
[tex]r F = 275.625(1.88)[/tex]
[tex]F = \frac{275.625(1.88)}{1.50}[/tex]
[tex]F = 345.45 N[/tex]
A double threaded right handed worm gear transmits 15 hp at 1150 rpm. The pitch of the worm is 0.75 inches and pitch diameter of 3 inches. The pressure angle is 14 1/2° and the coefficient of friction is 0.12. Find the worm gear efficiency.
Answer:
not that high, because worms are never that efficient
The efficiency of a worm gear transmitted at a power of 15HP and at 1150 rpm with a pitch of 0.75 inches and pitch diameter of 3 inches is 77%.
What is Worm gear?
A worm gear is a type of gear that utilizes a spiral-threaded shaft to drive a toothed wheel. One of the six simple machines is the vintage worm gear. A worm gear is essentially a screw butted up against what appears to be a typical spur gear with slightly curved and inclined teeth.
Given: Power of gear(P) = 15 hp,
Torque (N) = 1150 rpm;
The pitch of gear (p) = 0.75;
Pitch diameter (D₁) = 3 inches;
Pressure angle (α) = 14 1 / [tex]2^{0}[/tex];
Friction coefficient (μ) = 0.12
If m is the module of gear, z is the no of teeth, γ is the worm lead angle and l is the length of gear then
l = [tex]\pi[/tex]mz
l = 238.75 (m = 1 / p = 2)
And tanγ = 1 / [tex]\pi[/tex]D₁
γ = 44.92
Now If the efficiency is η then ;
η =(cosα - μtanγ) / (cosα + μtanγ)
η = 77%
Therefore, the efficiency is 77%
To know more about worm gear:
https://brainly.com/question/28593004
#SPJ2
How can one distinguish breccia from conglomerate? a. Breccia contains angular clasts and conglomerate contains rounded clasts. b. Breccia contains rounded clasts and conglomerate contains angular clasts. c. Different minerals cement breccia than cement conglomerate. d. Breccia contains sand and conglomerate contains silt.
a.Breccia contains angular clasts and conglomerate contains rounded clasts.
Breccia and conglomerate are rocks that can primarily be distinguished by the shape of their clasts; breccia has angular fragments while the fragments in conglomerate are rounded.
Explanation:One can distinguish between breccia and conglomerate based on the nature of their clasts. Breccia contains angular clasts, meaning its fragments have not been smoothed or rounded and retain sharp, rough edges. On the other hand, conglomerate contains rounded clasts.
The fragments within a conglomerate have been weathered and eroded over time, resulting in more rounded and smooth shapes. Both of these sedimentary rocks are formed from fragments of igneous rock or the shells of living organisms, but their difference lies in the shape and wear of their components.
Learn more about Rocks here:https://brainly.com/question/19930528
#SPJ12
A 50.0 mg sample of an unknown radioactive substance was placed in storage and its mass measured periodically. After 19.7 days, the amount of radioactive substance had decreased to 3.13 mg. What is the half-life of the unknown radioactive substance?
Explanation:
This problem can be solved using the Radioactive Half Life Formula:
[tex]A=A_{o}.2^{\frac{-t}{h}}[/tex] (1)
Where:
[tex]A=3.13mg[/tex] is the final amount of the material
[tex]A_{o}=50mg[/tex] is the initial amount of the material
[tex]t=19.7days[/tex] is the time elapsed
[tex]h[/tex] is the half life of the material (the quantity we are asked to find)
Knowing this, let's substitute the values and find [tex]h[/tex] from (1):
[tex]3.13mg=(50mg)2^{\frac{-19.7days}{h}}[/tex] (2)
[tex]\frac{3.13mg}{50mg}=2^{\frac{-19.7days}{h}}[/tex] (3)
Applying natural logarithm in both sides:
[tex]ln(\frac{3.13mg}{50mg})=ln(2^{\frac{-19.7days}{h}})[/tex] (4)
[tex]-2.77=-\frac{19.7days}{h}ln(2)[/tex] (5)
Clearing [tex]h[/tex]:
[tex]h=\frac{-19.7days}{-2.77}(0.693)[/tex] (6)
Finally:
[tex]h=4.928days[/tex]
The half-life is the period required for the quantity of a radioactive substance to become half its original quantity. By using given information and the half-life formula, we can calculate the half-life of the unknown radioactive substance.
Explanation:The half-life of a radioactive substance is the time period required for the quantity of the substance to reduce to half its initial amount due to radioactive decay. This problem involves using the half-life formula: T = (t/log_2) * (log(N0/N)), where T is the half-life, t is the elapsed time, N0 is the initial quantity of the substance, and N is the quantity after time t.
From the question, we know N0 = 50.0mg, N = 3.13mg, and t = 19.7 days. Substituting these values into the formula allows us to calculate the half-life, which will give the answer.
Learn more about Half-Life here:https://brainly.com/question/37906890
#SPJ3
What is magnetic flux? Explain Faraday's law of Induction.
Final answer:
Magnetic flux is a representation of the magnetic field passing through an area—it's calculated using the field strength, area vector, and the cosine of their angle. Faraday's law of induction explains that changing magnetic flux through a loop induces an emf in it, which is crucial in electric generators and related technologies. Lenz's law ensures the direction of the induced current opposes the flux change, reflecting energy conservation.
Explanation:
Magnetic Flux and Faraday's Law of Induction
Magnetic flux is essentially a measure of the quantity of magnetic field lines passing through a given area. To determine it, you need to know the strength of the magnetic field, the surface area over which it's spread, and the angle between the field and the normal (perpendicular) to that surface area. Mathematically, it is defined as the product of the magnetic field (B), the area (A), and the cosine of the angle (θ) between them, which is given by the equation Φ = B * A * cos(θ).
Faraday's law of induction tells us that a changing magnetic flux within a closed conducting loop will induce an electromotive force (emf) in the loop. This phenomenon is used in a vast array of technologies from electric generators to transformers. The magnitude of the induced emf can be calculated using Faraday's law which states that the emf induced is directly proportional to the rate of change of the magnetic flux.
Additionally, Lenz's law elaborates on the direction of the induced current, stating that it will flow in a manner that opposes the change in flux that caused it. This is a manifestation of the conservation of energy.
Application Example
An example of Faraday's law would be a coil of wire exposed to a changing magnetic field. If the field changes over time, an electric current is induced in the wire. This principle is the foundation for devices like inductive chargers and electrical generators.
Traveling at an initial speed of 1.5 x 106 m/s, a proton enters a region of constant magnetic field of magnitude 1.5 T. If the proton's initial velocity vector makes an angle of 30 with the magnetic field, compute the proton's speed 4 s after entering the magnetic field. 0 A. 5.0 x 105 m/s O B. 7.5 x 105 m/s ° C. 1.5 x 106 m/s 0 D. 3.0 x 106 m/s
Final answer:
The proton's speed after 4 seconds will remain the same as its initial speed of 1.5 x 10^6 m/s.
Explanation:
To find the proton's speed 4 seconds after entering the magnetic field, we need to apply the right-hand rule. The magnetic force on a charged particle moving in a magnetic field is given by the equation F = qvBsin(θ), where F is the force, q is the charge, v is the velocity, B is the magnetic field, and θ is the angle between the velocity and the magnetic field. Since the proton's initial velocity vector makes an angle of 30° with the magnetic field, the force acting on the proton will be perpendicular to its velocity. Therefore, it will not change the proton's speed, but rather cause it to move in a circular path. As a result, the proton's speed after 4 seconds will remain the same as its initial speed of 1.5 x 10^6 m/s.
A person desires to reach a point that is 2.17 km from her present location and in a direction that is 29.6° north of east. However, she must travel along streets that are oriented either north-south or east-west. What is the minimum distance she could travel to reach her destination?
She needs to move [tex]x[/tex] km in the east-west direction and [tex]y[/tex] km in the north-south direction so that
[tex]\sqrt{x^2+y^2}=2.17[/tex]
and
[tex]\tan29.6^\circ=\dfrac yx[/tex]
Solve the system to get
[tex]x=1.89\,\mathrm{km}[/tex]
[tex]y=1.07\,\mathrm{km}[/tex]
An athlete swings a ball, connected to the end of a chain, in a horizontal circle. The athlete is able to rotate the ball at the rate of 8.13 rev/s when the length of the chain is 0.600 m. When he increases the length to 0.900 m, he is able to rotate the ball only 6.04 rev/s. (a) Which rate of rotation gives the greater speed for the ball? 6.04 rev/s 8.13 rev/s (b) What is the centripetal acceleration of the ball at 8.13 rev/s? m/s2 (c) What is the centripetal acceleration at 6.04 rev/s? m/s2
(a) 6.04 rev/s
The speed of the ball is given by:
[tex]v=\omega r[/tex]
where
[tex]\omega[/tex] is the angular speed
r is the distance of the ball from the centre of the circle
In situation 1), we have
[tex]\omega=8.13 rev/s \cdot 2\pi = 51.0 rad/s[/tex]
r = 0.600 m
So the speed of the ball is
[tex]v=(51.0 rad/s)(0.600 m)=30.6 m/s[/tex]
In situation 2), we have
[tex]\omega=6.04 rev/s \cdot 2\pi = 37.9 rad/s[/tex]
r = 0.900 m
So the speed of the ball is
[tex]v=(37.9 rad/s)(0.900 m)=34.1 m/s[/tex]
So, the ball has greater speed when rotating at 6.04 rev/s.
(b) [tex]1561 m/s^2[/tex]
The centripetal acceleration of the ball is given by
[tex]a=\frac{v^2}{r}[/tex]
where
v is the speed
r is the distance of the ball from the centre of the trajectory
For situation 1),
v = 30.6 m/s
r = 0.600 m
So the centripetal acceleration is
[tex]a=\frac{(30.6 m/s)^2}{0.600 m}=1561 m/s^2[/tex]
(c) [tex]1292 m/s^2[/tex]
For situation 2 we have
v = 34.1 m/s
r = 0.900 m
So the centripetal acceleration is
[tex]a=\frac{v^2}{r}=\frac{(34.1 m/s)^2}{0.900 m}=1292 m/s^2[/tex]
Final answer:
The rate of rotation that gives the greater speed for the ball is 8.13 rev/s. The centripetal acceleration of the ball at 8.13 rev/s is 39.43 m/s^2, while the centripetal acceleration at 6.04 rev/s is 32.90 m/s^2.
Explanation:
The rate of rotation that gives the greater speed for the ball is 8.13 rev/s. The speed of the ball is directly proportional to the rate of rotation. So, a higher rate of rotation will result in a greater speed for the ball.
The centripetal acceleration of the ball at 8.13 rev/s can be calculated using the formula:
centripetal acceleration = (angular velocity)^2 * radius
Plugging in the values:
centripetal acceleration = (8.13 rev/s)^2 * 0.6 m = 39.43 m/s^2
The centripetal acceleration of the ball at 6.04 rev/s can be calculated in the same way:
centripetal acceleration = (6.04 rev/s)^2 * 0.9 m = 32.90 m/s^2
You drive a race car around a circular track of radius 300 m at a constant spe 108 km/h. Your acceleration will be: A) 39.9 m/s B) 9.8 m/s C) 3.0 m/s D) 0.36 m/s
Answer:
C
Explanation:
Centripetal acceleration is:
a = v² / r
First, convert km/h to m/s:
108 km/h × (1000 m / km) × (1 h / 3600 s) = 30 m/s
Therefore, the acceleration is:
a = (30 m/s)² / 300 m
a = 3 m/s²
Answer:
C
Explanation:
Positive acceleration describes an increase in speed; negative acceleration describes a decrease in speed.
While doing her crossfit workout, Yasmeen holds an 7.0 kg weight at arm's length, a distance of 0.57 m from her shoulder joint. What is the torque about her shoulder joint due to the weight if her arm is horizontal? A 30 N m B. 4.0 N m C. 43N-m D. 39 N m
Answer:
D. 39 N m
Explanation:
m = mass of the weight used in crossfit workout = 7.0 kg
Force due to the weight used is given as
F = mg
F = (7.0) (9.8)
F = 68.6 N
d = distance of point of action of weight from shoulder joint = 0.57 m
τ = Torque about the shoulder joint due to the weight
Torque about the shoulder joint due to the weight is given as
τ = F d
Inserting the values
τ = (68.6) (0.57)
τ = 39 Nm
A cooling tower for a nuclear reactor is to be constructed in the shape of a hyperboloid of one sheet. The diameter at the base is 260 m and the minimum diameter, 500 m above the base, is 220 m. Find an equation for the tower. (Assume the position of the hyperboloid is such that the center is at the origin with its axis along the z-axis, and the minimum diameter at the center.)
Answer:
r² / 110² − 3z² / 1375² = 1
Explanation:
The equation of a hyperboloid (which is a hyperbola rotated about the z axis or conjugate axis) that is centered at the origin is:
x² / a² + y² / b² − z² / c² = 1
If the cross sections are circular rather than elliptical, then a = b.
(x² + y²) / a² − z² / c² = 1
Or, if you prefer cylindrical coordinates:
r² / a² − z² / c² = 1
We know that at z = 0, r = 110. And at z = -500, r = 130.
110² / a² − 0 = 1
130² / a² − (-500)² / c² = 1
Solving:
a² = 110²
c² = 1375² / 3
Plugging in:
r² / 110² − 3z² / 1375² = 1
A figure skater is rotating at a rate of 200 revolutions per minute. What is the angular speed, in units of rad/s? 20.9 12000 200 3.33
Answer:
200 revolutions per minute = 20.9 rad/s
Explanation:
It is given that, a figure skater is rotating at a rate of 200 revolutions per minute. It is the angular velocity of the skater. We have to convert it into radian/second.
Revolution per minute can be converted to radian per minute as :
Since, 1 radian/second = 60/2π revolutions per minute
So, 1 revolution/minute = 2π/60 radian/second
200 revolution/minute = 200 × 2π/60 radian/second
200 revolution/minute = 20.9 radian/second
Hence, this is the required solution.