Answer:
0.8 m
Explanation:
For "6 N" force :
F = magnitude of the force = 6 N
r = moment arm = 0.4 m
Torque due to "6 N" force is given as
τ = r F
τ = (0.4) (6)
τ = 2.4 Nm
For " 15 N" force :
F' = magnitude of the force = 15 N
r' = moment arm = ?
τ' = Torque = 5 τ = 5 x 2.4 = 12 Nm
Torque due to "15 N" force is given as
τ' = r' F'
12 = r' (15)
r' = 0.8 m
So the moment arm for "15 N" force is 0.8 m
A uniform metal bar is 5.00 m long and has mass 0.300 kg. The bar is pivoted on a narrow support that is 2.00 m from the left-hand end of the bar. What distance x from the left-hand end of the bar should an object with mass 0.900 kg be suspended so the bar is balanced in a horizontal position?
we have that from the Question, it can be said that distance x from the left-hand end of the bar is
x=1.834m
From the Question we are told
A uniform metal bar is 5.00 m long and has mass 0.300 kg. The bar is pivoted on a narrow support that is 2.00 m from the left-hand end of the bar. What distance x from the left-hand end of the bar should an object with mass 0.900 kg be suspended so the bar is balanced in a horizontal position?
Generally the equation for Balanced torque is mathematically given as
T=0.3/5*3
T=0.18kg
Therefore
The clockwise torque is
T_c=0.18*1.5g
And
The anti-clockwise torque is
T=0.3/5*2g
T=0.12
Hence
0.18*1.5g=0.3/5*2g+0.9kxg
Therefore
x=1.834m
Therefore
distance x from the left-hand end of the bar is
x=1.834m
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The position of the 0.9 kg mass to balance the metal bar is 1.83 m.
The given parameters;
length of the metal bar, L = 5 mmass of the metal bar, m = 0.3 kgpivot distance = 2 mThe center of gravity of the metal bar = 2.5 m
A sketch of weight on the metal bar;
|-----------2 m-----------|--- 0.5 m-----|
0---------------------------Δ--------------------------------------------------
P ↓|-------- x----------| ↓
0.9 kg 0.3 kg
Take moment about the pivot point;
0.9x = 0.3(0.5)
0.9x = 0.15
[tex]x = \frac{0.15}{0.9} \\\\x = 0.167 \ m[/tex]
2 m - P = 0.167 m
P = 2m - 0.167 m
P = 1.83 m
Thus, the position of the 0.9 kg mass to balance the metal bar is 1.83 m.
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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.
Choose the answer choice that BEST completes the following sentence. About four million tons of __________________ are converted to energy in the Sun’s core every second. Oxygen Matter Atmosphere Iron
Answer:
Option (2)
Explanation:
The sun is a large astronomical body where there occurs the process of nuclear fusion. This process is responsible for the occurrence of two important things. Firstly, it helps in the conversion o f hydrogen atoms into helium, that fuels the energy of the sun, and secondly, it helps in the continuous conversion of matter into energy, that reaches the earth's surface and on which the living organisms are directly dependent on.
Thus, there occurs conversion of about 4 million tons of matter into energy every second.
Therefore, the correct answer is option (2).
If at a particular instant and at a certain point in space the electric field is in the +x-direction and has a magnitude of 4.50 V/m , what is the magnitude of the magnetic field of the wave at this same point in space and instant in time?
Answer:
1.5 x 10^-8 Tesla
Explanation:
E = 4.5 V/m
The relation between electric field and the magnetic field in electromagnetic wave is given by
c = E / B
Where, c be the velocity of light in vacuum, e be the electric field and B be the magnetic field
B = E / c
B = 4.5 / ( 3 x 10^8)
B = 1.5 x 10^-8 Tesla
The magnetic field of an electromagnetic wave at a specific point can be calculated using the known electric field and the speed of light. In this case, the magnitude of the magnetic field is 1.5 x 10^-8 Tesla (T).
Explanation:In the context of an electromagnetic wave, the electric and magnetic fields are linked through Maxwell's Equations. Given that an electromagnetic wave propagates in free space, the ratio of the amplitudes of the electric field (E) to the magnetic field (B) is always equal to the speed of light (c), according to the equation c = E/B. Thus, when the magnitude of the electric field (E) is known, the magnetic field (B) can be calculated as B=E/c. In this case, the Electric field (E) is 4.50 V/m. Hence, using the value of speed of light c = 3x10^8 m/sec, we calculate the magnetic field (B) as follows: B = 4.50 V/m / 3x10^8 m/sec = 1.5 x 10^-8 Tesla (T).
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A ball rises vertically to a height h and returns to its original point of projection. What is the work done by the gravitational force?
The work done by the gravitational force on a ball that rises to a height and then returns to its original point is zero, as the work done during the ascent is equivalent but opposite to the work done during the descent.
Explanation:The work done by the gravitational force on an object that rises to a height and then returns to its original point is zero. This is because the work done by the gravitational force when the ball rises is equal in magnitude but opposite in direction to the work done when the ball falls back down.
When the ball ascends, the work done by the gravitational force is negative since the displacement (upwards) is opposite to the direction of gravitational force (downwards). The formula to calculate the work done in this phase is W = -mg * h, where 'm' is the mass of the ball, 'g' is the acceleration due to gravity and 'h' is the height.
During the ball's descent, the work done is positive as the displacement (downwards) is in the same direction as the direction of gravitational force (downwards). The work done in this phase can be calculated using the same formula W = mg * h (noting that 'h' is a negative height as the ball is descending). When you add these together, the total work done by gravity over the full journey of the object is zero.
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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]
A traveling electromagnetic wave in a vacuum has an electric field amplitude of 69.1 V/m . Calculate the intensity ???? of this wave. Then, determine the amount of energy ???? that flows through area of 0.0247 m^2 over an interval of 13.1 s, assuming that the area is perpendicular to the direction of wave propagation.
Answer:
The intensity of this wave and energy is 6.3385 N/m² and 2.0509 J.
Explanation:
Given that,
Electric field amplitude E₀= 69.1 V/m
Area A= 0.0247 m²
Time t= 13.1 s
We need to calculate the intensity
Using formula of intensity
[tex]S=\dfrac{1}{2}c\epsilon_{0}E_{0}^2[/tex]
Where, c = speed of light
Put the value into the formula
[tex]S=\dfrac{1}{2}\times3\times10^{8}\times8.85\times10^{-12}\times(69.1)^2[/tex]
[tex]S=6.3385\ N/m^2[/tex]
(b). We need to calculate the energy
Using formula of energy
[tex]E=SAt[/tex]
Where, A = area
t = time
S = intensity
Put the value into the formula
[tex]E =6.3385\times0.0247\times13.1[/tex]
[tex]E =2.0509\ J[/tex]
Hence, The intensity of this wave and energy is 6.3385 N/m² and 2.0509 J.
An electron moving at 5.06 103 m/s in a 1.23 T magnetic field experiences a magnetic force of 1.40 10-16 N. What angle does the velocity of the electron make with the magnetic field? There are two answers between 0° and 180°. (Enter your answers from smallest to largest.)
Answer:
[tex]8.1^{\circ}, 171.9^{\circ}[/tex]
Explanation:
The magnitude of the magnetic force exerted on the moving electron is:
[tex]F=qvB sin \theta[/tex]
where here we have
[tex]F=1.40\cdot 10^{-16} N[/tex] is the magnitude of the force
[tex]q=1.6\cdot 10^{-19} C[/tex] is the magnitude of the electron charge
B = 1.23 T is the magnetic field intensity
[tex]\theta[/tex] is the angle between the direction of the electron's velocity and the magnetic field
Solving the equation for [tex]\theta[/tex], we find:
[tex]sin \theta = \frac{F}{qvB}=\frac{1.40\cdot 10^{-16}N}{(1.6\cdot 10^{-19} C)(5.06\cdot 10^3 m/s)(1.23 T)}=0.141[/tex]
which gives the following two angles:
[tex]\theta = 8.1^{\circ}\\\theta = 180^{\circ}-8.1^{\circ} = 171.9^{\circ}[/tex]
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.
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.
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.
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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.
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).
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
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.
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A charge -353e is uniformly distributed along a circular arc of radius 5.30 cm, which subtends an angle of 48°. What is the linear charge density along the arc?
Answer:
- 1.3 x 10⁻¹⁵ C/m
Explanation:
Q = Total charge on the circular arc = - 353 e = - 353 (1.6 x 10⁻¹⁹) C = - 564.8 x 10⁻¹⁹ C
r = Radius of the arc = 5.30 cm = 0.053 m
θ = Angle subtended by the arc = 48° deg = 48 x 0.0175 rad = 0.84 rad (Since 1 deg = 0.0175 rad)
L = length of the arc
length of the arc is given as
L = r θ
L = (0.053) (0.84)
L = 0.045 m
λ = Linear charge density
Linear charge density is given as
[tex]\lambda =\frac{Q}{L}[/tex]
Inserting the values
[tex]\lambda =\frac{-564.8\times 10^{-19}}{0.045}[/tex]
λ = - 1.3 x 10⁻¹⁵ C/m
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.
An ideal spring has a spring constant (force constant) of 2500 N/m, is stretched 4.0 cm, How much elastic potential energy does it possess? (A) 3J (B) 0.00J (C) 1J (D) 2J (E) 4J
Answer:
Elastic potential energy, E = 2 J
Explanation:
It is given that,
Spring constant of the spring, k = 2500 N/m
The spring is stretched to a distance of 4 cm i.e. x = 0.04 m
We have to find the elastic potential energy possessed by the spring. A spring possessed elastic potential energy and it is given by:
[tex]E=\dfrac{1}{2}kx^2[/tex]
[tex]E=\dfrac{1}{2}\times 2500\ N/m\times (0.04\ m)^2[/tex]
E = 2 Joules.
Hence, the correct option is (d) " 2 Joules ".
What is an electric motor? Explain its operation. 2-3 sentence
Answer:
An electric motor is a device that changes electrical energy into mechanical energy. This change occurs due to the interaction between the magnetic field of magnets and the magnetic field due to the electric current in the loop. The interaction between the two produces a torque that makes the loop rotate on a shaft.
An electric motor is electric machinery that converts electrical energy supplied to it to mechanical energy.
It works on the principle of applying a magnetic field in electromagnetism.
A current-carrying loop is exposed to a magnetic field which results in torque.
The torque formed rotates the coil which is then followed by rotation of propellers as the current passes through the loop.
Explanation
An electric motor consist of a rotor- a moving part, commutator, brushes, axle, field-magnet, power supply and a stator- a part going around the rotor.
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%
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A 10 kg plank 3 meters in length extends off the edge of a pirate ship so that only 0.5 m remains on the deck. This is held in place by a 200 kg crate of rum sitting on top of the end of the plank on the deck. A 70 kg pirate is being forced to walk the plank. How far does he get from the deck before the plank tips and he falls?
Answer:
sorry I've never took in this class before I will not be able to help you
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 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.
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 ball with a mass of 275 g is dropped from rest, hits the floor and rebounds upward. If the ball hits the floor with a speed of 2.40 m/s and rebounds with a speed of 1.70 m/s, determine the following. (a) magnitude of the change in the ball's momentum (Let up be in the positive direction.)
Answer:
Change in momentum is 1.1275 kg-m/s
Explanation:
It is given that,
Mass of the ball, m = 274 g = 0.274 kg
It hits the floor and rebounds upwards.
The ball hits the floor with a speed of 2.40 m/s i.e. u = -2.40 m/s (-ve because the ball hits the ground)
It rebounds with a speed of 1.7 m/s i.e. v = 1.7 m/s (+ve because the ball rebounds in upward direction)
We have to find the change in the ball's momentum. It is given by :
[tex]\Delta p=p_f-p_i[/tex]
[tex]\Delta p=m(v-u)[/tex]
[tex]\Delta p=0.275\ kg(1.7\ m/s-(-2.4\ m/s))[/tex]
[tex]\Delta p=1.1275\ kg-m/s[/tex]
So, the change in the momentum is 1.1275 kg-m/s
The magnitude of the change in the ball's momentum when rebounding off the floor is 1.1275 kg·m/s, accounting for the change in direction during impact.
Explanation:To determine the magnitude of the change in the ball's momentum, you should first consider the initial and final momenta of the ball. Momentum is calculated as the product of mass and velocity. When the ball hits the floor, it has a downward momentum of (mass × velocity before hitting the floor). After rebounding, it has an upward momentum of (mass × velocity after rebounding). Since the problem states that up is in the positive direction, you will have to take into account the change in direction when calculating the change in momentum.
To calculate the magnitude of the change in momentum (Δp), you use the formula Δp = p_final - p_initial. Plugging in the values:
p_initial = mass × velocity before hitting = 0.275 kg × (-2.40 m/s) = -0.66 kg·m/s
p_final = mass × velocity after rebounding = 0.275 kg × 1.70 m/s = 0.4675 kg·m/s
Δp = p_final - p_initial = 0.4675 kg·m/s - (-0.66 kg·m/s) = 1.1275 kg·m/s
The negative sign for p_initial indicates that it was directed downwards. The magnitude of the change in momentum is simply the absolute value of Δp, which is 1.1275 kg·m/s.
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In an electroplating process, copper (ionic charge +2e, atomic weight 63.6 g/mol) is deposited using a current of 10.0 A. What mass of copper is deposited in 10.0 minutes? Avogadro's number is 6.022 × 1023 molecules/mol and e = 1.60 × 10-19 C.
Answer:
1.974 g
Explanation:
Electrochemical equivalent of copper, z = 0.000329 g/C
I = 10 A
t = 10 minutes = 10 x 60 = 600 seconds
By the use of Farady's law of electrolysis
m = z I t
m = 0.000329 x 10 x 600
m = 1.974 g
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.
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How long will be required for an object to go from a speed of 22m/s to a speed of 27m/s if the acceleration is 5.93m/s^2 ?
Answer:
Required time, t = 0.84 seconds
Explanation:
It is given that,
Initial speed of an object, u = 22 m/s
Final velocity of an object, v = 27 m/s
Acceleration, a = 5.93 m/s²
We have to find the time required for an object to go a speed of 22 m/s to a speed of 27 m/s. It can be solved by using first equation of motion as:
[tex]v=u+at[/tex]
Where
t = time
[tex]t=\dfrac{v-u}{a}[/tex]
[tex]t=\dfrac{27\ m/s-22\ m/s}{5.93\ m/s^2}[/tex]
t = 0.84 seconds
Hence, the time required for an object is 0.84 seconds.
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