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.
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]
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 ".
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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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 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 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 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.
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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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]
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
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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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).
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.
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 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.
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 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.
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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).
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
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.
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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
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.
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.
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
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
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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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.
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.