An alpha particle (q = +2e, m = 4.00 u) travels in a circular path of radius 4.29 cm in a uniform magnetic field with B = 1.53 T. Calculate (a) its speed, (b) its period of revolution, (c) its kinetic energy, and (d) the potential difference through which it would have to be accelerated to achieve this energy.
Answers
Answer:
A) speed = 3144287.425 m/s
B) period = 8.57x10^-8 s
C) KE = 3.302x10^-14 J
D) potential difference = 103187.5 V
Explanation:
Detailed explanation and calculation is shown in the image below.
Related Questions
Wile E. Coyote wants to launch Roadrunner into the air using a long lever asshown below. The lever starts at rest before the Coyote steps onto one end and pivots around its center of mass. Roadrunner is standing on the other end, which is held up by a small piece of ground underneath. The lever has mass 1.2 kg and length 16 m, the Roadrunner has mass 3.2 kg, and there is no friction at the pivot point of the lever.
a) Draw free body diagrams for the lever, Coyote, and Roadrunner. Make sure to label where all of the forces act on the lever’s free body diagram.
b) What is the minimal mass that Coyote must have to cause the lever to start rotating?
c) The coyote actually has a mass of 18 kg. If the roadrunner goes flying when the lever gets to a 45∘ angle from the horizontal, what is the roadrunner’s speed at that time, assuming the pivot is frictionless?
d) (Optional/challenge) What is the angular acceleration just after Coyote steps onto the lever?
Answers
Answer:
Explanation:
Find attached the solution
A small, solid cylinder with mass = 20 kg and radius = 0.10 m starts from rest and rotates without friction about a fixed axis through its center of mass. A string is wrapped around the circumference of the cylinder and pulled using a constant force F. The resulting angular acceleration of the cylinder is 5.0 rad/s2. What length of string has unwound after 4.0 s, in meters? (The moment of inertia of the cylinder is 1 half M R squared.)
Answers
Answer:
Explanation:
angular acceleration α = 5 rad /s ²
θ = 1/2 α t² , θ is angle of rotation , t is time .
= .5 x 5 x 4²
= 40 rad .
θ = l / r , θ is angle formed , l is length unwound , r is radius o wheel .
l = θ x r
= 40 x .1
= 4 m .
We found that the time after the mass passes through the equilibrium point at which the mass attains its extreme displacement from the equilibrium position is 3 4 seconds. Lastly, we must find the position of the mass at the instant the mass attains its extreme displacement from the equilibrium, which is x 3 4 . We note that the result is measured in feet. x(t) = −e−4t + 2te−4t
Answers
Answer:
A) X(0.5) = 0 ft
B) X(0.75) = 0.023 ft.
Explanation: Given that the result is measured in feet.
x(t) = −e−4t + 2te−4t
Factorizing the above equation lead to
x(t) = e^-4t( -1 + 2t )
The mass passes through the equilibrium position when x(t) = 0
0 = e^-4t( -1 + 2t )
-1 + 2t = 0
2t = 1
t = 0.5s
x(t) = e^-4t( -1 + 2t )
Substitute t = 0.5
x(0.5) = e^-4(0.5)(-1 + 2(0.5))
X(0.5) = 0
Also given that mass attains its extreme displacement from the equilibrium position is t = 3/ 4
= 0.75 seconds
x(t) = e^-4t( -1 + 2t )
x(t) = e^-4(0.75)( -1 + 2(0.75) )
X(0.75) = (e^-3)(0.5)
X(0.75) = 0.023 ft.
What is the wavelength in nm of a light whose first order bright band forms a diffraction angle of 30 degrees, and the diffraction grating has 700 lines per mm?
Answers
Answer:
The wavelength is 3500 nm.
Explanation:
d= [tex]\frac{1}{700 lines per mm} = 0.007mm = 7000 nm[/tex]
n= 1
θ= 30°
λ= unknown
Solution:
d sinθ = nλ
λ = [tex]\frac{7000 nm sin 30}{1}[/tex]
λ = 3500 nm
If you wish to observe features that are around the size of atoms, say 1 .5 x 100 m, with electromagnetic radiation, the radiation must have a wavelength of about the size of the atom itself. ะเ50% Part (a) If you had a nic that you would have to use?
Answers
The question is incomplete! Complete question along with answer and step by step explanation is provided below.
Question:
If you wish to observe features that around the size of atoms, say 1.5×10⁻¹⁰ m, with electromagnetic radiation, the radiation must have a wavelength about the size of the atom itself.
a) If you had a microscope which was capable of doing this, what would the frequency of electromagnetic radiation be, in hertz that you would have to use?
b) What type of electromagnetic radiation would this be?
Given Information:
Wavelength = λ = 1.5×10⁻¹⁰ m
Required Information:
a) Frequency = f = ?
b) Type of electromagnetic radiation = ?
Answer:
a) Frequency = f = 2×10¹⁸ Hz
b) Type of electromagnetic radiation = X-rays
Explanation:
a) The frequency of the electromagnetic radiation is given by
f = c/ λ
Where λ is the wavelength of the electromagnetic radiation and c is the speed of light and its value is 3×10⁸ m/s
f = 3×10⁸/1.5×10⁻¹⁰
f = 2×10¹⁸ Hz
Therefore, the frequency of the electromagnetic radiation would be 2×10¹⁸ Hz.
b)
The frequency range of X-rays is 3×10¹⁶ Hz to 3×10¹⁹ Hz
The frequency 2×10¹⁸ lies in that range, therefore, the type of electromagnetic radiation is X-rays
To observe atomic-sized features, we would need to use forms of radiation with similar wavelengths to the atom's size, such as X-rays. This is because observable detail with electromagnetic radiation is limited by wavelength; therefore, visible light cannot detect atoms due to their much smaller size.
To observe features that are around the size of atoms with electromagnetic radiation, the radiation must have a wavelength of about the size of the atom itself. In the area of physics, this concept is a key part of the study of wave optics.
If we imagine atoms to be about 1.5 x 10^-10 m (or 0.1 nm) in size, because wavelength and size must be on similar scales, we'd need to use forms of radiation with similar wavelengths.
This falls within the range of X-rays on the electromagnetic spectrum. Therefore, to observe atomic-sized features, an appropriate method would be X-ray microscopy or other forms of X-ray imaging.
It's important to note that visible light can never detect individual atoms because atoms are much smaller than visible light's wavelength, reinforcing the concept that observable detail is limited by wavelength.
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A Young's interference experiment is performed with blue-green laser light. The separation between the slits is 0.500 mm, and the screen is located 3.24 m from the slits. The first bright fringe is located 3.22 mm from the center of the interference pattern. What is the wavelength of the laser light?
Answers
Answer:
λ = 4.97 x 10⁻⁷ m = 497 nm
Explanation:
The formula for the distance of a bright fringe from center in Young's double slit experiment, is given by:
y = mλL/d
where,
y = distance of bright fring from center = 3.22 mm = 3.22 x 10⁻³ m
m = No. of bright fringe = 1
L = Distance between slits and screen = 3.24 m
d = slit separation = 0.5 mm = 0.5 x 10⁻³ m
λ = wavelength of light = ?
3.22 x 10⁻³ m = (1)(λ)(3.24 m)/(0.5 x 10⁻³ m)
(3.22 x 10⁻³ m)(0.5 x 10⁻³ m)/(3.24 m) = λ
λ = 4.97 x 10⁻⁷ m = 497 nm
A 6.41 $\mu C$ particle moves through a region of space where an electric field of magnitude 1270 N/C points in the positive $x$ direction, and a magnetic field of magnitude 1.28 T points in the positive $z$ direction. If the net force acting on the particle is 6.40E-3 N in the positive $x$ direction, calculate the magnitude of the particle's velocity. Assume the particle's velocity is in the $x$-$y$ plane.
Answers
Answer:
The particle's velocity is 212.15 m/s.
Explanation:
Given that,
Charge of particle, [tex]q=6.41\ \mu C=6.41\times 10^{-6}\ C[/tex]
The magnitude of electric field, E = 1270 N/C
The magnitude of magnetic field, B = 1.28 T
Net force, [tex]F=6.4\times 10^{-3}\ N[/tex]
We need to find the magnitude of the particle's velocity. the net force acting on the particle is given by Lorentz force as :
[tex]F=qE+qvB\\\\v=\dfrac{F-qE}{qB}\\\\v=\dfrac{6.4\times 10^{-3}-6.41\times 10^{-6}\times 1270}{6.41\times 10^{-6}\times 1.28}\\\\v=-212.15\ m/s[/tex]
So, the particle's velocity is 212.15 m/s.
What is the approximate size of the smallest object on the Earth that astronauts can resolve by eye when they are orbiting 250 km above the Earth? Assume λ = 506 nm and a pupil diameter is 4.90 mm. (In this problem, you may use the Rayleigh criterion for the limiting angle of resolution of an eye.)
Answers
Answer:
the approximate size of the smallest object on the Earth that astronauts can resolve by eye when they are orbiting 250 km above the Earth is y = 31.495 m
Explanation:
Using Rayleigh criterion for the limiting angle of resolution of an eye
[tex]\theta = \frac{1.22\lambda }{D } \\ \\ \theta = \frac{1.22*506 *10^{-9} }{4.90*10^{-3}m}[/tex]
[tex]\theta = 1.2598*10^{-4}[/tex] rad
[tex]\theta = 125.98*10^{-6} \ rad[/tex]
Thus; the separation between the two sources is expressed as:
[tex]\theta = \frac{y}{L} \\ \\ y = L \theta \\ \\ y = (250*10^3 )(125.98*10^{-6} \ rad) \\ \\ y = 31.495 \ m[/tex]
Thus; the approximate size of the smallest object on the Earth that astronauts can resolve by eye when they are orbiting 250 km above the Earth is y = 31.495 m
Final answer:
The approximate size of the smallest object that astronauts can resolve by eye when orbiting 250 km above the Earth is approximately 628 meters.
Explanation:
The approximate size of the smallest object on Earth that astronauts can resolve by eye when they are orbiting 250 km above the Earth can be determined using the Rayleigh criterion. According to the Rayleigh criterion, the resolution of the eye is determined by the angle of resolution, which is given by:
angle of resolution = 1.22 * (wavelength / pupil diameter)
Using the given values, we can calculate the angle of resolution:
angle of resolution = 1.22 * (506 nm / 4.90 mm) = 1.2611957657 x 10^-3 radians
To determine the approximate size of the smallest object, we need to find the linear size corresponding to this angular resolution at a distance of 250 km:
linear size = 2 * distance * tan(angle of resolution)
linear size = 2 * 250000 m * tan(1.2611957657 x 10^-3 radians) = 628.1079000855 m
Therefore, the approximate size of the smallest object that astronauts can resolve by eye when they are orbiting 250 km above the Earth is approximately 628 meters.
Two lasers are shining on a double slit, with slit separation d. Laser 1 has a wavelength of d/20, whereas laser 2 has a wavelength of d/15. The lasers produce separate interference patterns on a screen a distance 6.00 m away from the slits.Which laser has its first maximum closer to the central maximum?
Answers
Answer:
The first laser
Explanation:
Using the equation
ym = m*λ*L/d
so for the first max for each one m =1
so for laser 1
y = (d/20)*(6.0m)/d
the d cancel out so we have
y = 6.0m /20 = 0.3
for laser 2 we do the same thing
y = 6.0m/15 = 0.4
y ( laser1 ) < y ( laser2 )
(1) hence first maxima of laser1 is closest to central maxima
Δy = 0.4m -0 .3 m = 0.1m
I want to examine the surface of a planet that is completely covered by a thick layer of clouds all the time. What wavelength of electromagnetic radiation would I be smartest to use?
A. visible light
B. x-rays
C. radar waves
D. ultra-violet
E. none would work
Answers
Answer:C
Explanation:
Radar waves will be useful to examine the surface of the planet.
Radio waves are the shorter wavelength microwaves ranging approximately 1 cm.
Short wavelength causes large reflection from large objects. These waves are useful to detect smaller objects like water drops in the cloud.
They can penetrate clouds and that is why they are used in weather forecasting, air traffic control, etc.
The best wavelength of electromagnetic radiation to use when trying to examine a planet covered by a thick layer of clouds is radar waves, as they can penetrate the clouds and provide information about the surface beneath.
Explanation:If you want to examine the surface of a planet that is completely covered by a thick layer of clouds all the time, the smartest wavelength of electromagnetic radiation to use would be C. Radar waves. The reason is that radar waves, which are a type of radio wave, can penetrate through clouds and provide information about the surface beneath. Other types of electromagnetic radiation such as visible light, x-rays, and ultra-violet would be absorbed or scattered by the clouds, preventing you from seeing the surface.
For example, radar is used here on Earth for weather prediction because it can 'see' through clouds, providing details about storm structures. Similarly, space probes such as the Magellan mission to Venus used radar wavelengths to map the surface of Venus, a planet notoriously covered in thick clouds.
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To practice Problem-Solving Strategy 25.1 for charge interaction problems. A proton and an alpha particle are momentarily at rest at a distance rrr from each other. They then begin to move apart. Find the speed of the proton by the time the distance between the proton and the alpha particle doubles. Both particles are positively charged. The charge and the mass of the proton are, respectively, eee and mmm. The charge and the mass of the alpha particle are, respectively, 2e2e and 4m4m.
Answers
By using the principle of energy conservation and accounting for the charges and masses of the proton and alpha particle, we can establish that the speed of the proton when the distance between the particles doubles is given by the square root of four times Coulomb's constant times the charge of the proton squared, divided by the mass of the proton and the original distance.
Explanation:To solve the problem, we can apply the principle of energy conservation. In the initial state, the system has potential energy, which is due to the electrostatic interactions between the proton and the alpha particle. When the particles begin to move apart, some of this potential energy is converted into kinetic energy. Given that both particles are positively charged, the potential energy U of the interaction is given by Coulomb's law:
U = k * Q1 * Q2 / r
where Q1 and Q2 are the charges of the particles, r is the distance between them and k = 8.99 * 10^9 N * m^2/C^2 is Coulomb's constant. The alpha particle has 2e, and the proton has e, leading to the potential energy of the system being: U_initial = 2 * k * e^2 / r.
When the distance between the proton and the alpha particle doubles, the potential energy becomes: U_final = k * e^2 / (2r).
The change in potential energy (delta U) thus equals the kinetic energy of the proton (as the alpha particle is more massive, its kinetic energy can be neglected): delta U = U_initial - U_final = K_proton = (1/2) * m * v^2, where m and v are the mass and speed of the proton, respectively. Solving this equation for v, we get that the speed of the proton is: v = sqrt((4 * k * e^2)/(m * r)).
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After a party the host pours the remnants of several bottles of wine into a jug. He then inserts a cork with a 2.25-cm diameter into the jug, placing it in direct contact with the wine. He is amazed when he pounds the cork into place that the bottom of the jug, with a 16.5-cm diameter, breaks away. show answer No Attempt Find the force that the liquid exerts on the bottom of the bottle when the host is pounding the cork with a 150-N force. Ignore the force caused by the weight of the wine.
Answers
Answer:
F = 8066.67 N
Explanation:
The extra force exerted on the bottom of the jug can be expressed as the pressure generated by the cork multiplied by the area of the bottom. We can also obtain the Pressure P by dividing the force F1 applied to the cork by it's area A1.
Thus;
F = PA2 = (F1/A1) x (A2)
F = (F1/(πd1²/4)) x (πd2²/4)
π/4 will cancel out to give;
F = F1(d2/d1)²
F = 150(16.5/2.25)
F = 8066.67 N
Using Pascal's Principle, the force the host applies to the cork is transferred through the wine to the bottom of the jug. To calculate the force exerted on the jug's bottom, we find the cross-sectional areas of the cork and bottom of the jug and apply the principle that pressure remains constant throughout the fluid. The amplified force is then calculated from these values.
To calculate the force the liquid exerts on the bottom of the bottle when the host pounds the cork with a 150-N force, we first consider Pascal's Principle, which states that pressure applied to an enclosed fluid is transmitted undiminished to every part of the fluid and the walls of the containing vessel. The key to solving this problem is to recognize that the force exerted by pounding the cork is amplified by the liquid because the area at the bottom of the jug is larger than the area of the cork.
The force on the cork is given by:
F1 = 150 N, and the cross-sectional area of the cork, A1, can be calculated using the cork's diameter:
A1 = π*(d1/2)^2 = π*(2.25 cm/2)^2
The cross-sectional area of the bottom of the jug, A2, is:
A2 = π*(d2/2)^2 = π*(16.5 cm/2)^2
According to Pascal's Principle:
P1 = P2
where P1 is the pressure applied to the cork and P2 is the pressure on the bottom of the jug. Thus,
P1 = F1/A1
Therefore, the force on the bottom of the jug, F2, can be calculated:
F2 = P2*A2 = (F1/A1)*A2
By inserting the values and calculating F2, we determine the force that the liquid exerts on the bottom of the bottle.
A motorcycle daredevil plans to ride up a 2.0-m-high, 20° ramp, sail across a 10-m-wide pool filled with hungry crocodiles, and land at ground level on the other side. He has done this stunt many times and approaches it with confidence. Unfortunately, the motor-cycle engine dies just as he starts up the ramp. He is going 11 m/s at that instant, and the rolling friction of his rubber tires (coefficient 0.02) is not negligible. Does he survive, or does he become croco-dile food? Justify your answer by calculating the distance he travels through the air after leaving the end of the ramp
Answers
Answer:
He becomes a croco-dile food
Explanation:
From the question we are told that
The height is h = 2.0 m
The angle is [tex]\theta = 20^o[/tex]
The distance is [tex]w = 10m[/tex]
The speed is [tex]u = 11 m/s[/tex]
The coefficient of static friction is [tex]\mu = 0.02[/tex]
At equilibrium the forces acting on the motorcycle are mathematically represented as
[tex]ma = mgsin \theta + F_f[/tex]
where [tex]F_f[/tex] is the frictional force mathematically represented as
[tex]F_f =\mu F_x =\mu mgcos \theta[/tex]
where [tex]F_x[/tex] is the horizontal component of the force
substituting into the equation
[tex]ma = mgsin \theta + \mu mg cos \theta[/tex]
[tex]ma =mg (sin \theta + \mu cos \theta )[/tex]
making a the subject of the formula
[tex]a = g(sin \theta = \mu cos \theta )[/tex]
substituting values
[tex]a = 9.8 (sin(20) + (0.02 ) cos (20 ))[/tex]
[tex]= 3.54 m/s^2[/tex]
Applying " SOHCAHTOA" rule we mathematically evaluate that length of the ramp as
[tex]sin \theta = \frac{h}{l}[/tex]
making [tex]l[/tex] the subject
[tex]l = \frac{h}{sin \theta }[/tex]
substituting values
[tex]l = \frac{2}{sin (20)}[/tex]
[tex]l = 5.85m[/tex]
Apply Newton equation of motion we can mathematically evaluate the final velocity at the end of the ramp as
[tex]v^2 =u^2 + 2 (-a)l[/tex]
The negative a means it is moving against gravity
substituting values
[tex]v^2 = (11)^2 - 2(3.54) (5.85)[/tex]
[tex]v= \sqrt{79.582}[/tex]
[tex]= 8.92m/s[/tex]
The initial velocity at the beginning of the pool (end of ramp) is composed of two component which is
Initial velocity along the x-axis which is mathematically evaluated as
[tex]v_x = vcos 20^o[/tex]
substituting values
[tex]v_x = 8.92 * cos (20)[/tex]
[tex]= 8.38 m/s[/tex]
Initial velocity along the y-axis which is mathematically evaluated as
[tex]v_y = vsin\theta[/tex]
substituting values
[tex]v_y = 8.90 sin (20)[/tex]
[tex]= 3.05 m/s[/tex]
Now the motion through the pool in the vertical direction can mathematically modeled as
[tex]y = y_o + u_yt + \frac{1}{2} a_y t^2[/tex]
where [tex]y_o[/tex] is the initial height,
[tex]u_y[/tex] is the initial velocity in the y-axis
[tex]a_y[/tex] is the initial acceleration in the y axis with a constant value of ([tex]g = 9.8 m/s^2[/tex])
at the y= 0 which is when the height above ground is zero
Substituting values
[tex]0 = 2 + (3.05)t - 0.5 (9.8)t^2[/tex]
The negative sign is because the acceleration is moving against the motion
[tex]-(4.9)t^2 + (2.79)t + 2m = 0[/tex]
Solving using quadratic formula
[tex]\frac{-b \pm \sqrt{b^2 -4ac} }{2a}[/tex]
substituting values
[tex]\frac{-3.05 \pm \sqrt{(3.05)^2 - 4(-4.9) * 2} }{2 *( -4.9)}[/tex]
[tex]t = \frac{-3.05 + 6.9}{-9.8} \ or t = \frac{-3.05 - 6.9}{-9.8}[/tex]
[tex]t = -0.39s \ or \ t = 1.02s[/tex]
since in this case time cannot be negative
[tex]t = 1.02s[/tex]
At this time the position the motorcycle along the x-axis is mathematically evaluated as
[tex]x = u_x t[/tex]
x [tex]=8.38 *1.02[/tex]
[tex]x =8.54m[/tex]
So from this value we can see that the motorcycle would not cross the pool as the position is less that the length of the pool
If the horizontal distance x covered by the motorcyclist is greater than or equal to 10 meters, he survives; otherwise, he does not, From the given calculations, x = 8.21 m, so he becomes crocodile food.
According to question, the height of motorcycle daredevil is h = 2.0 m, the angle of the ramp is θ = 20°, the distance of the wide pool is w = 10 m, the speed of the motorcycle daredevil is u = 11 m/s, and the coefficient of static friction between the ramp and the tyres of the motorcycle daredevil is μ = 0.02.
At equilibrium, the forces acting on the motorcycle are equated as:
ma = mgsinθ + f
here, m = mass of the motorcycle, a is the acceleration of the motorcycle, g is the acceelration due to gravity (g = 9.8 m/s²), and f is the frictional force between the ramp and the tyres of the motorcycle daredevil mathematically represented using normal reaction N as
f =μ N = μ mgcosθ
substituting into the equation
ma = mg sinθ + μ mg cosθ
ma =mg (sinθ + μ cosθ)
⇒ a = g(sinθ + μ cosθ)
⇒ a = 9.8 (sin(20°) + (0.02) cos (20°))
⇒ a = 3.54 m/s²
Now, for the ramp, we can say that:
[tex]sin \theta = \frac{h}{l}[/tex]
[tex]\therefore l = \frac{h}{sin \theta}[/tex]
substituting the above values, we get:
[tex]l = \frac{2}{sin20^{\circ}}[/tex]
or, l = 5.85m
Using Newton equation of motion, the final velocity of the motorcycle at the end of the ramp, can be given as:
[tex]v^2 =u^2 + 2 (-a)l[/tex]
As the motorcycle is moving against gravity, the acceleration (a) is taken negative. From the above values, we get:
[tex]v^2 = (11)^2 - 2(3.54) (5.85)[/tex]
⇒ [tex]v= \sqrt{79.582} \hspace{0.8 mm} m/s[/tex]
or, v = 8.92m/s
The motion of the motorcycle is a projectile motion. The initial velocity at the beginning of the pool (end of ramp) is composed of two component which are the x, and the y components respectively.
Initial velocity along the x-axis which is [tex]v_x[/tex] = v cos 20°
substituting values, we get:
[tex]v_x[/tex] = 8.92 cos (20°)
or, v = 8.38 m/s
Initial velocity along the y-axis which [tex]v_y[/tex] = v sin 20°
substituting values, we get:
[tex]v_y[/tex] = 8.90 sin (20)
[tex]v_y[/tex] = 3.05 m/s
Now the motion through the pool in the vertical direction can mathematically modeled as
[tex]y = y_o + u_yt + \frac{1}{2}a_y t^2[/tex]
where y₀ is the initial height, [tex]u_y[/tex] is the initial velocity in the y-axis, and [tex]a_y[/tex] is the initial acceleration in the y-axis with a constant value of (g = 9.8 m/s^2)
At y = 0
0 = 2 + (3.05) t - 0.5 (9.8) t²
The negative sign is because the acceleration is moving against the motion:
- (4.9) t² + (2.79) t + 2m = 0
To solve for t, we can use the quadratic formula:
[tex]t = \frac{- b \hspace{0.5} \pm \sqrt{(b^2 - 4ac)}}{2a}[/tex]
In this case, a = 4.9, b = -2.79, and c = -2.
Plugging in these values, we get:
[tex]t = \frac{(2.79 \pm \sqrt{(-2.79)^2 - 4(4.9)(-2)}}{2(4.9)}[/tex]
Solving for both possible values of t, we get:
t ≈ [tex]\frac{(2.79 + 6.858)}{9.8}[/tex] ≈ 0.98 seconds
t ≈ [tex]\frac{(2.79 - 6.858)}{9.8}[/tex] ≈ -0.41 seconds
Since time cannot be negative,
∴ t ≈ 0.98 seconds
Hence displacement along the x-coordinate will be:
x = [tex]u_x[/tex] t
⇒ x =8.38 × 0.98 m
x = 8.21 m
So from this value we can see that the motorcycle would not cross the pool as the position is less that the length of the pool.
A disk of radius 1.6 m and mass 3.8 kg rotates about an axis through its center. A force of 18.4 N is applied tangentially to the edge of the hoop, causing it to rotate counterclockwise. Assuming the disk is initially at rest, after 3 s... a) what is the radial acceleration of a point halfway between the axis and the edge of the disk
Answers
Answer:
At the edge, angular acceleration = 3.025 rad/s2
At halfway = 13.11 rad/s2
Explanation:
Detailed explanation and calculation is shown in the image below
Answer:
263.8 m/s2
Explanation:
Assume this is a solid disk, we can find its moments of inertia:
[tex]I = mr^2/2 = 3.8*1.6^2/2 = 4.864 kgm^2[/tex]
The torque T generated by force F = 18.4N is:
[tex]T = Fr = 18.4*1.6 = 29.44 Nm[/tex]
So the angular acceleration of the disk according to Newton's 2nd law is:
[tex]\alpha = T/I = 29.44 / 18.4 = 6.05 rad/s^2[/tex]
If the disk starts from rest, then after 3s its angular speed is
[tex]\omega = \alpha \Delta t = 6.05*3 = 18.16 rad/s[/tex]
And so its radial acceleration at this time and half way from the center to the edge is:
[tex]a_r = \omega^2(r/2) = 18.16^2*(1.6/2) = 263.8 m/s^2[/tex]
Note that this value is the same anywhere
Potential Difference Across Axon Membrane The axoplasm of an axon has a resistance Rax. The axon's membrane has both a resistance (Rmem) and a capacitance (Cmem). A single segment of an axon can be modeled by a circuit with Rmem and Cmem in parallel with each other and in series with an open switch, a battery, and Rax. Imagine the voltage of the battery is ΔV. Part A We can model the firing of an action potential by the closing of the switch, which completes the circuit. Immediately after the switch closes, what is the potential difference across the membrane of this single segment?
Answers
Answer:
check the diagram in the attachment below.
After the switch closes, the voltage across the membrane will be zero. It is so due to the fact that the capacitor will be short circuited.
Explanation:
question
Potential Difference Across Axon Membrane The axoplasm of an axon has a resistance Rax. The axon's membrane has both a resistance (Rmem) and a capacitance (Cmem). A single segment of an axon can be modeled by a circuit with Rmem and Cmem in parallel with each other and in series with an open switch, a battery, and Rax. Imagine the voltage of the battery is ΔV. Part A We can model the firing of an action potential by the closing of the switch, which completes the circuit. Immediately after the switch closes, what is the potential difference across the membrane of this single segment?
ANSWER;
After the switch closes, the voltage across the membrane will be zero. It is so due to the fact that the capacitor will be short circuited.
Answer:
the potential difference across the membrane of this single segment is zero.
ΔVmem=0
Explanation:
When the switch is closed, the capacitor ([tex]C_{mem}\\[/tex]) will be short. It means the current will not flow through the resistor([tex]R_{mem[/tex]). Rather current will only flow through the capacitor as there is no resistance. Due to zero resistance, the voltage across the capacitor will also be zero. So, the potential difference across the membrane of this single segment is zero.
ΔVmem=0
Presbyopia is the tendency to gradually become far-sighted (hyperopic) as you age. If you have normal vision when you are young, you have a near point of 25 cm. A. If the distance between your eye's lens and retina is 1.71 cm, what is the focal length of your eye's lens when you look at an object at your near point? f = cm B. As you get older, suppose that the near point of your eye increases to 46 cm. What is the focal length of your eye's lens when you look at an object at your near point now? f = cm C. With your near point at 46 cm, what is the focal length of the corrective lens (placed directly in front of your eye's lens) which you would need to look at an object that is 25 cm in front of your eye? f = cm D. As you continue to age, the corrective lenses will no longer be sufficient to allow you to see an object that is 25 cm in front of your eye. When you are wearing your corrective lenses, suppose that you can now see objects only if they are no closer than 36 cm in front of your eye. What is the actual near point of your eye now? N = cm What is the focal length of your eye's lens that corresponds to this new near point?
Answers
Answer:
1.62cm
Explanation:
a) object distance do = 25cm
image distance di = 1.71cm
we have the relation
1/f = 1/di +1/do
plugging in the values we get
f = 1.6cm
b) di = 1.71cm
do = 46cm
that gives
f = 1.65cm
C) do = 25cm
di =46 cm
plugging in the values in the above relation we get
f = 16.2cm
D) do = 36cm
f = 16.2cm
plugging in the equation
1/di = 1/f -1/do
plugging in the values we get
di = 29.4 cm
the new near point is 29.4cm
taking do = 29.4cm
di = 1.71 cm
we get focal length of eye as
f = 1.62cm
Final answer:
Calculating the focal length for presbyopia involves using the lens formula with different object distances. Accommodation factors in changes in near point due to age. Corrective lenses' focal lengths can also be found using similar optical principles.
Explanation:
When considering presbyopia, which is the age-related tendency to become far-sighted or hyperopic, there are several aspects of optics that come into play. To calculate the focal length of an eye's lens (f), we use the lens formula 1/f = 1/di + 1/do, where di is the image distance equal to the lens-to-retina distance, and do is the object distance. In this scenario, di remains at 1.71 cm, and the near point alterations represent changes in do.
A. When the object is at the near point of 25 cm, the focal length is found using the lens formula with do=25 cm.
B. As the near point shifts to 46 cm due to aging, we recalculate the focal length with this new do.
C. To correct presbyopia so the person can see an object at 25 cm, we need to determine the focal length of a corrective lens that compensates for the eye's inability to focus on close objects. This can be done by assuming that the corrective lens will bring the object effectively to the new near point and using the lens formula accordingly.
D. If the corrected vision allows seeing objects only if they are no closer than 36 cm, we can find the actual near point of the eye, which would dictate the adjusted focal length of the eye's lens needed to accommodate this near point.
Consider a proton, a deuteron (nucleus of deuterium, i.e., Hydrogen-2), and an alpha particle (nucleus of Helium-4), all with the same speed. These particles enter a region of uniform magnetic field B, traveling perpendicular to B. What is the ratio of the deuteron's orbital radius to the proton's orbital radius?
Answers
Answer:
the ratio of the deuteron's orbital radius to the proton's orbital radius is 2 : 1
Explanation:
Detailed explanation and calculation is shown in the image below
Three blocks are pushed along a rough surface by a force with magnitude P, as shown above. Fc is the magnitude of the contact force between blocks 2 and 3 and Ff, Fn, and Fg are the magnitudes of friction, normal, and gravitational forces on block 3, respectively. Please draw a FBD.
Answers
Answer:
See the FBD attached
Explanation:
Since Ff, Fn, and Fg are the magnitudes of friction, normal, and gravitational forces on block 3, you can build the FBD (free bdy diagram) for block 3.
The FBD of the block 3 must include all the forces acting on the block 3.
The FBD is shown on the graph attached.
Observe this:
Fc is the contact force that block 2 exerts on block 3. Assuming block 2 is to the left of block 3, the force is toward the right.Ff is the friction force exerted by the ground on the block 3. Its direction is opposite to the movement. Assuming the block is moving to the right, the direction of Ff is to the left. Also, assuming the block is accelerating to the right, the magnitude of Ff is less than the magnitude of Fc, and the arrow that represents Ff is shorter than the arrow that represents Fc.Fn is the normal force on block 3 due to the contact with the ground. It points upward, as it opposes the gravitaional force. Assuming the block is not moving vertically, the magnitude of Fn is equal to the magnitude of Fg, and both arrows have the same length.Fg is the gravitational force: the pulling force of the Earth on the block. It is equal to the weight of the block.The Michelson-Morley experiment a) confirmed that time dilation occurs. b) proved that length contraction occurs. c) verified the conservation of momentum in inertial reference frames. d) supported the relationship between mass and energy. e) indicated that the speed of light is the same in all inertial reference frames.
Answers
Answer:
e) indicated that the speed of light is the same in all inertial reference frames.
Explanation:
In 18th century, many scientists believed that the light just like air and water needs a medium to travel. They called this medium aether. They believed that even the space is not empty and filled with aether.
Michelson and Morley tried to prove the presence and speed of this aether through an interference experiment in 1887. They made an interferometer in which light was emitted at various angles with respect to the supposed aether. Both along the flow and against the flow to see the difference in the speed of light. But they did not find no major difference and thus it became the first proof to disprove the theory of aether.
It thus proved that the speed of light remains same in all inertial frames.
Also, it became a base for the special theory of relativity by Einstein.
You want the current amplitude through a 0.450-mH inductor (part of the circuitry for a radio receiver) to be 1.90 mA when a sinusoidal voltage with an amplitude of 13.0 V is applied across the inductor.What frequency is required?
Answers
Answer:
Frequency required will be 2421.127 kHz
Explanation:
We have given inductance [tex]L=0.450H=0.45\times 10^{-3}H[/tex]
Current in the inductor [tex]i=1.90mA=1.90\times 10^{-3}A[/tex]
Voltage v = 13 volt
Inductive reactance of the circuit [tex]X_l=\frac{v}{i}[/tex]
[tex]X_l=\frac{13}{1.9\times 10^{-3}}=6842.10ohm[/tex]
We know that
[tex]X_l=\omega L=2\pi fL[/tex]
[tex]2\times 3.14\times f\times 0.45\times 10^{-3}=6842.10[/tex]
f = 2421.127 kHz
Two particles of masses m1m1m_1 and m2m2m_2 (m1
Answers
Answer:
the correct answer is: more than 5 meters of particular m1, but less than 10 m of the particle m₂
Explanation:
The center of mass, the point of a system where all external forces can be applied, is defined by
[tex]x_{cm}[/tex] = 1 /M ∑ [tex]x_{i} m_{i}[/tex]
where M is the total mass of the system, x_{i} m_{i} are the position and mass of each item in the system,
let's apply this equation to our case
the total mass is
M = m₁ + m₂
for the calculations we must fix a reference system, we will place it in the second mass (m₂)
x_{cm}= 1/M (m₁ d + m₂ 0)
where d is the distance between the two masses in this case d = 10 m
x_{cm} = m₁ / (m₁ + m₂) d
x_{cm} = m₁ / (m₁ + m₂) 10
as the mass m₁ <m₂
Let us analyze the answer if the masses sides were equal the center of mass would be x_{cm} = 5 m, but since m₁ <m₂ the center of mass must be closer to m₂.
Therefore the correct answer is: more than 5 meters of particular m1, but less than 10 m of the particle m₂
A turntable is off and is not spinning. A 0.8 g ant is on the disc and is 9 cm away from the center. The turntable is turned on and 0.8 s later it has an angular speed of 45 rpm. Assume the angular acceleration is constant and determine the following quantities for the ant 0.4 s after the turntable has been turned on. Express all quantities using appropriate mks units.
Answers
Complete Question
The complete is shown on the first uploaded image
Answer:
[tex]\alpha = 5.89 rad/s^2[/tex]
[tex]w__{0.4}}= 2.36 \ rad/s[/tex]
[tex]v= 0.212m/s[/tex]
[tex]a_t= 0.5301 m/s[/tex]
[tex]a_r = 0.499 m/s[/tex]
[tex]a = 0.7279 m/s[/tex]
[tex]F_{net}=5.823*10^{-4}N[/tex]
Explanation:
From the question we are told that
mass of the ant is [tex]m_a = 0.8g = \frac{0.8}{1000} = 0.00018kg[/tex]
The distance from the center is [tex]d = 9cm = \frac{9}{100} = 0.09m[/tex]
The angular speed is [tex]w = 45rpm = 45 * \frac{2 \pi }{60} = 1.5 \pi[/tex]
The time taken to attain angular acceleration of 45rpm [tex]t_1 = 0.8s[/tex]
The time taken is [tex]t_2 = 0.4 s[/tex]
The angular acceleration is mathematically represented as
[tex]\alpha = \frac{w}{t}[/tex]
[tex]= \frac{1.5}{0.8}[/tex]
[tex]\alpha = 5.89 rad/s^2[/tex]
The angular velocity at time t= 0.4s is mathematically represented as
[tex]w__{0.4s}} = \alpha * t_2[/tex] Recall angular acceleration is constant
[tex]= 5.89 * 0.4[/tex]
[tex]w__{0.4}}= 2.36 \ rad/s[/tex]
The linear velocity is mathematically represented as
[tex]v = w__{t_2}} * r[/tex]
[tex]= 2.36 * 0.09[/tex]
[tex]v= 0.212m/s[/tex]
The tangential acceleration is mathematically represented as
[tex]a_{t} = \alpha * r[/tex]
[tex]= 5.89 * 0.09[/tex]
[tex]a_t= 0.5301 m/s[/tex]
The radial acceleration is mathematically represented as
[tex]a_r = \frac{v^2}{r}[/tex]
[tex]= \frac{0.212^2}{0.09}[/tex]
[tex]a_r = 0.499 m/s[/tex]
The resultant velocity is mathematically represented as
[tex]a = \sqrt{a_t^2 + a_r^2}[/tex]
[tex]= \sqrt{0.53^2 + 0.499^2}[/tex]
[tex]a = 0.7279 m/s[/tex]
The net force is mathematically represented as
[tex]F_{net} = 0.0008 * 0.7279[/tex]
[tex]F_{net}=5.823*10^{-4}N[/tex]
A child is riding a tricycle along a straight horizontal road. The pedals are attached directly to the front wheel, which has a radius of 15 cm. The rear wheels are smaller and have a radius of 8 cm. There is no slip between the wheels and the road surface. If the child is pedaling at 16 rpm (revolutions per minute), then the angular speed of the rear wheels is closest to:
Answers
Answer:
Angular speed of rear wheel = 3.1425 rad/s
Explanation:
N1 = 16 rpm
w1 angular speed of big wheel = 2¶N/60 = (2 x 3.142 x 16)/60 = 1.676 rad/s
R1 = radius of big wheel = 15 cm = 15x10^-2 = 0.15 m
R2 = radius of small wheel = 8 cm = 0.08 m
w2 =?
Using w1R1 = w2R2
1.676 x 0.15 = w2 x 0.08
0.2514 = 0.08w2
w2 = 0.2514/0.08 = 3.1425 rad/s
The cross-sectional area of the output piston in a hydraulic device is 3.0 times the area of the input piston. For each 1.5 cm the input piston moves, how far does the output piston move
Answers
Answer:
For each 1.5 cm the input piston moves, then output piston will move 0.5 cm
Explanation:
Let cross-sectional area of the input piston = A
Let cross-sectional area of the output piston = 3A
pressure (P) = Force (F) / Area (A)
F = PA
For a constant gravitational force on the inlet and outlet piston;
P₁A₁ = P₂A₂
But pressure = ρgh
where;
ρ is density of water
g is acceleration due to gravity
h is the distance or height moved by the piston
(ρg)h₁A₁ = (ρg)h₂A₂
h₁A₁ = h₂A₂
Area of output piston = 3 times area of input piston
h₁A₁ = h₂(3A₁)
For each 1.5 cm the input piston moves, then output piston will move;
1.5A₁ = h₂(3A₁)
1.5 = 3h₂
h₂ = 1.5 / 3
h₂ = 0.5 cm
Thus, for each 1.5 cm the input piston moves, then output piston will move 0.5 cm
Albert is piloting his spaceship, heading east with a speed of 0.92 cc relative to Earth. Albert's ship sends a light beam in the forward (eastward) direction, which travels away from his ship at a speed cc. Meanwhile, Isaac is piloting his ship in the westward direction, also at 0.92 cc, toward Albert's ship
With what speed does Isaac see Albert's light beam pass his ship?
Answers
Answer:
He sees the light as 1c
Explanation:
According to relativity, the speed of light is the same in all inertial frame of reference.
If we were to add the velocities as applicable to a normal moving bodies, the relative speed of the light beam will exceed c which will break relativistic law since nothing can go past the speed of light.
When 450-nm light is incident normally on a certain double-slit system the number of interference maxima within the central diffraction maximum is 5. When 900-nm light is incident on the same slit system the number is: a. 9b. 2c. 3d. 10e. 5
Answers
Answer:
Explanation:
width of central diffraction maxima = 2 λD / d₁
λ is wave length of light , D is screen distance and d₁ is slit width
width of each interference fringe = λD / d₂ , d₂ is slit separation.
No of interference fringe in central diffraction fringe
= width of central diffraction maxima / width of each interference fringe
= 2 λD / d₁ x λDd₂ / λD
No = 2 d₂ / d₁
No = 5
5 = 2 d₂ / d₁
Since this number does not depend upon wavelength so it will remain the same
No of required fringe will be 5 .
right option
e ) 5.
When 900-nm light is incident on the same slit system the number is 5.
The given parameters:
First wavelength of the incident light, λ₁ = 450 nmThe number of interference maxima, n = 5Second wavelength, λ₂ = 900 nmThe number of interference fringe in central diffraction fringe is calculated as follows;
[tex]n = \frac{width \ of \ central \ diffraction\ maxima}{ width \ of \ each \ interference\ fringe}\\\\n = \frac{2\pi \lambda D/d_1}{2\pi \lambda D/d_2} \\\\n = \frac{2d_2}{d_1}[/tex]
The number of number of interference fringe is independent of the wavelength.
Thus, when 900-nm light is incident on the same slit system the number is 5.
Learn more about number of interference fringe here: https://brainly.com/question/4449144
A cylinder of mass m and a block of wood of mass m are both released from rest at a height h on an inclined plane at the same time. The cylinder rolls down the plane without slipping due to friction. The block slides down the incline. If they both experience the same frictional force, which has more total kinetic energy when they reach the bottom?
A. The cylinder
B. The block
C. they have the same total kinetic energy when they reach the bottom
D. there is not enough information to answer this question
Answers
Answer:
A
Explanation:
Because in block friction does not work hence case of cylinder has more total K.E
A bucket of water of mass 16.0kg is suspended by a rope wrapped around a windlass, that is a solid cylinder with diameter 0.350m with mass 11.4kg . The cylinder pivots on a frictionless axle through its center. The bucket is released from rest at the top of a well and falls a distance 10.1m to the water. You can ignore the weight of the rope. a) What is the tension in the rope while the bucket is falling? Take the free fall acceleration to be g = 9.80m/s^2 . b) With what speed does the bucket strike the water? Take the free fall acceleration to be g = 9.80m/s^2 . c) What is the time of fall? Take the free fall acceleration to be g = 9.80m/s^2 . d) While the bucket is falling, what is the force exerted on the cylinder by the axle? Take the free fall acceleration to be g = 9.80m/s^2 .
Answers
Answer:
Explanation:
We shall consider tension to be T.
For motion of bucket in downward direction
mg - T = ma , m is mass of bucket , a is linear acceleration of bucket
for rotational motion of wheel
Tortque by force Tr = I x α I is moment of inertia , α is angular acceleration , r is radius of the wheel.
Putting the values in the equation above
mg - I x α / r = ma
mg - 1/2 M r ²x a / r² = ma
mg - 1/2 M a = ma
mg = 1/2 M a + ma
a = mg / (m + .5 M)
= 16 X 9.8 / (16 +.5 x 11.4)
= 7.22 m /s²
mg - T = ma
T = m( g - a )
= 16 x ( 9.8 - 7.22 )
= 41.28 N.
b )
v² = u² + 2 a h , v is velocity when bucket falls by height h
= 0 + 2 x 7.22 x 10.1
= 145.84
v = 12.07
c )
v = u + at , t is time of fall
12.07 = 0 + 7.22 t
t = 1.67 s
d )
Force by axle on cylinder = T
= 41.28 N .
Which of the following careers would require a degree in physics
A. Art historian
B. War correspondent
C. Nuclear engineer
D. X-ray technician
Answers
Answer:
Nuclear engineer
Explanation:
Because it requires a lot of movements during that career.
A body with mass of 200 g is attached to the end of a spring that is stretched 20 cm by a force of 9 N. At time tequals0 the body is pulled 1 m to the left, stretching the spring, and set in motion with an initial velocity of 20 m/s to the left. (a) Find x(t) in the form Upper C cosine (omega 0 t minus alpha ). (b) Find the amplitude and the period of motion of the body.
Answers
The problem refers to an object undergoing simple harmonic motion, initially displaced and set in motion. The displacement and the period of motion can be calculated using information of spring constant, mass, initial velocity, and principles of oscillations.
Explanation:The question refers to a classic setup for simple harmonic motion (SHM), where a body of known mass is attached to a spring with a known force constant. The force constant is the ratio of the force applied to the spring and the resulting extension; we can calculate it as k = F/x = 9N/0.2m = 45 N/m. This accounts for the spring's stiffness.
As SHM is in effect here, the displacement of the body from the equilibrium position can be represented by the equation x(t) = A cos(ωt + α). In this case, the body is initially displaced 1m to the left (A = 1m) and given an initial velocity to the left (which determines α).
Angular frequency ω can be determined by the formula √(k/m), where k is the spring constant and m is the mass of the body (noting to convert grams to kilograms for SI units). Hence, ω = √(45 N/m / 0.2 kg) = 15 rad/s.
On the other hand, initial velocity is related to α by the equation v = -Aωsinα, thus α can be calculated as such. Here, it is worth mentioning that the amplitude is the maximum displacement (1m) and the period of motion, given by T = 2π/ω, refers to the time taken for one complete oscillation.
Learn more about Simple Harmonic Motion here:https://brainly.com/question/28208332
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