Answer:
The distance between the screen and slit is 1.08 m
Explanation:
Given that,
Wavelength = 681 nm
Width a= 0.109 mm
Number of fringe n = 14
Distance from the center of the central maximum d= 9.51 cm
We need to calculate the distance between the screen and slit
Using formula of distance
[tex]D=\dfrac{d\times a}{n\lambda}[/tex]
Where, a = width
d = distance from the center of the central maximum
[tex]\lambda[/tex] = wavelength
Put the value into the formula
[tex]D=\dfrac{9.51\times10^{-2}\times0.109\times10^{-3}}{14\times681\times10^{-9}}[/tex]
[tex]D = 1.08\ m[/tex]
Hence, The distance between the screen and slit is 1.08 m
A Ferris wheel has radius 4.0 m and makes one revolution every 30 s with uniform rotation. A woman who normally weighs 600 N is sitting on one of the benches attached at the rim of the wheel. What is her apparent weight (the normal force exerted on her by the bench) as she passes over the top of the Ferris wheel? A. 590 N
B. 600 N
C. 520 N
D. 0 N
Answer:
A) Apparent Weight = 590 N
Explanation:
As we know that frequency is given as
[tex]f = \frac{1}{30}[/tex]
[tex]f = 0.033 Hz[/tex]
now the angular speed is given as
[tex]\omega = 2\pi f[/tex]
[tex]\omega = 2\pi(0.033) = 0.21 rad/s[/tex]
now at the top position we will have
[tex]mg - N = m\omega^2 R[/tex]
[tex]N = mg - m\omega^2 R[/tex]
[tex]N = 600 - \frac{600}{9.8}(0.21)^2(4.0)[/tex]
[tex]N = 590 N[/tex]
The woman's apparent weight at the top of the Ferris wheel is less than her actual weight due to the centripetal force experienced during the uniform circular motion. This apparent weight can be calculated by subtracting the centripetal force from the gravitational force, yielding an answer of 590 N.
Explanation:This problem revolves around the concepts of centripetal force and apparent weight in the context of uniform circular motion. The woman's apparent weight at the top of the Ferris wheel is less than her actual weight because of the centripetal force directed towards the center of the Ferris wheel.
To calculate the apparent weight, we should subtract the centripetal force from the gravitational force. The gravitational force is her actual weight, and since weight = mass * gravity, her mass equals 600N/9.8 m/s2 ~= 61.2 kg. The angular velocity of the Ferris wheel (ω) is 2π rad/30s since it makes one revolution every 30s. Using the formula centripetal force = m*ω2*r, we find that the centripetal force equals 61.2 kg * (2π rad/30s)2 * 4m = 10 N approximately.
Finally, the woman's apparent weight at the top is the gravitational force minus the centripetal force, or 600 N - 10 N, which equals 590 N. Therefore, the correct answer is A) 590 N.
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A father racing his son has 1/4 the kinetic energy of the son, who has 1/3 the mass of the father. The father speeds up by 1.2 m/s and then has the same kinetic energy as the son. What are the original speeds of (a) the father and (b) the son?
Answer:
Explanation:
KE_s: Kinetic Energy Son
KE_f: Kinetic Energy Father.
Relationship
KE_f: = (1/4) KE_s
m_s: = (1/3) m_f
v_f: = velocity of father
v_s: = velocity of the son
Relationship
1/2 mf (v_f + 1.2)^2 = 1/2 m_s (v_s)^2 Multiply both sides by 2.
mf (v_f + 1.2)^2 = m_s * (v_s)^2 Substitute for the mass of the m_s
mf (v_f + 1.2)^2 = (m_f/3) * (v_s)^2 Divide both sides by father's mass
(v_f + 1.2)^2 = 1/3 * (v_s)^2 multiply both sides by 3
3*(v_f + 1.2)^2 = (v_s) ^2 Take the square root both sides
√3 * (v_f + 1.2) = v_s
Note
You should work your way through all the cancellations to find the last equation shown aboutWe have another step to go. We have to use the first relationship to get the final answer.KE_f = (1/4) KE_s Multiply by 4
4* KE_f = KE_s Substitute (again)
4*(1/2) m_f (v_f + 1.2)^2 = 1/2* (1/3)m_f *v_s^2 Divide by m_f
2* (v_f + 1.2)^2 = 1/6 * (v_s)^2 multiply by 6
12*(vf + 1.2)^2 = (v_s)^2 Take the square root
2*√(3* (v_f + 1.2)^2) = √(v_s^2)
2*√3 * (vf + 1.2) = v_s
Use the second relationship to substitute for v_s so you can solve for v_f
2*√3 * ( v_f + 1.2) = √3 * (v_f + 1.2) Divide by sqrt(3)
2(v_f + 1.2) = vf + 1.2
Edit
2vf + 2.4 = vf + 1.2
2vf - vf + 2.4 = 1.2
vf = 1.2 - 2.4
vf = - 1.2
This answer is not possible, but 2 of us are getting the same answer. The other person is someone whose math I would never question. She rarely makes an error. And I do mean rarely. Could you check to see that you have copied this correctly?
A rope pulls a 82.5 kg skier at a constant speed up a 18.7° slope with μk = 0.150. How much force does the rope exert?
Answer:
374 N
Explanation:
N = normal force acting on the skier
m = mass of the skier = 82.5
From the force diagram, force equation perpendicular to the slope is given as
N = mg Cos18.7
μ = Coefficient of friction = 0.150
frictional force is given as
f = μN
f = μmg Cos18.7
F = force applied by the rope
Force equation parallel to the slope is given as
F - f - mg Sin18.7 = 0
F - μmg Cos18.7 - mg Sin18.7 = 0
F = μmg Cos18.7 + mg Sin18.7
F = (0.150 x 82.5 x 9.8) Cos18.7 + (82.5 x 9.8) Sin18.7
F = 374 N
A 1.50-kg block is pushed against a vertical wall by means of a spring (k = 860 N/m). The coefficient of static friction between the block and the wall is 0.54. What is the minimum compression in the spring to prevent the block from slipping down?
Answer:
0.032 m
Explanation:
Consider the forces acting on the block
m = mass of the block = 1.50 kg
[tex]f_{s}[/tex] = Static frictional force
[tex]F_{n}[/tex] = Normal force on the block from the wall
[tex]F_{s}[/tex] = Spring force due to compression of spring
[tex]F_{g}[/tex] = Force of gravity on the block = mg = 1.50 x 9.8 = 14.7 N
k = spring constant = 860 N/m
μ = Coefficient of static friction between the block and wall = 0.54
x = compression of the spring
Spring force is given as
[tex]F_{s}[/tex] = kx
From the force diagram of the block, Using equilibrium of force along the horizontal direction, we get the force equation as
[tex]F_{n}[/tex] = [tex]F_{s}[/tex]
[tex]F_{n}[/tex] = kx eq-1
Static frictional force is given as
[tex]f_{s}[/tex] = μ [tex]F_{n}[/tex]
Using eq-1
[tex]f_{s}[/tex] = μ k x eq-2
From the force diagram of the block, Using equilibrium of force along the vertical direction, we get the force equation as
[tex]f_{s}[/tex] = [tex]F_{g}[/tex]
Using eq-2
μ k x = 14.7
(0.54) (860) x = 14.7
x = 0.032 m
A proton moves perpendicularly to a uniform magnetic field B with a speed of 1.5 × 107 m/s and experiences an acceleration of 0.66 × 1013 m/s 2 in the positive x direction when its velocity is in the positive z direction d the magnitude of the field. The elemental charge is 1.60 × 10−19 C . Answer in units of T.
Answer:
Magnetic field, B = 0.0045 T
Explanation:
It is given that,
Speed of the proton, [tex]v=1.5\times 10^7\ m/s[/tex]
Acceleration of the proton, [tex]a=0.66\times 10^{13}\ m/s^2[/tex]
Charge on proton, [tex]q=1.6\times 10^{-19}\ C[/tex]
The magnetic force is balanced by the force due to the acceleration of the proton as :
[tex]qvB=ma[/tex]
[tex]B=\dfrac{ma}{qv}[/tex]
[tex]B=\dfrac{1.67\times 10^{-27}\ kg\times 0.66\times 10^{13}\ m/s^2}{1.6\times 10^{-19}\ C\times 1.5\times 10^7\ m/s}[/tex]
B = 0.0045 T
So, the magnitude of magnetic field on the proton is 0.0045 T. Hence, this is the required solution.
A 400-turn circular coil (radius = 1.0 cm) is oriented with its plane perpendicular to a uniform magnetic field which has a magnitude that varies sinusoidally with a frequency of 90 Hz. If the maximum value of the induced emf in the coil is observed to be 4.2 V, what is the maximum value of the magnitude of the varying magne?
Answer:
The Magnetic field is 59.13 mT.
Explanation:
Given that,
Number of turns = 400
Radius = 1.0 cm
Frequency = 90 Hz
Emf = 4.2 V
We need to calculate the angular velocity
Using formula of angular velocity
[tex]\omega=2\pi f[/tex]
[tex]\omega=2\times3.14\times90[/tex]
[tex]\omega=565.2\ rad/s[/tex]
We need to calculate the magnetic flux
Relation between magnetic flux and induced emf
[tex]\epsilon=NA\omega B[/tex]
[tex]B=\dfrac{\epsilon}{NA\omega}[/tex]
Put the value into the formula
[tex]B=\dfrac{4.2}{400\times\pi\times(1.0\times10^{-2})^2\times 565.2}[/tex]
[tex]B=0.05913\ T[/tex]
[tex]B=59.13\ mT[/tex]
Hence, The Magnetic field is 59.13 mT.
A screwdriver is being used in a 13.5 T magnetic field. what maximum emf can be induced in V along its 10.5 cm length when it moves through the field at 0.85 m/s?
Answer:
EMF = 1.20 V
Explanation:
It is given that,
Magnetic field used by the screwdriver, B = 13.5 T
Length of screwdriver, l = 10.5 cm = 0.105 m
Speed with which it is moving. v = 0.85 m/s
We need to find the maximum EMF induced in the screwdriver. It is given by :
[tex]\epsilon=BLv[/tex]
[tex]\epsilon=13.5\ T \times 0.105\ m \times 0.85\ m/s[/tex]
[tex]\epsilon=1.20\ V[/tex]
So, the maximum emf of the screwdriver is 1.20 V. Hence, this is the required solution.
A 34.34 g sample of a substance is initially at 26.7 °C. After absorbing 2205 J of heat, the temperature of the substance is 152.1 °C. What is the specific heat (c) of the substance?
Answer:
The specific heat of the substance is c= 512.04 J/kg K
Explanation:
ΔQ= 2205 J
m= 0.03434 kg
ΔT= 125.4 ºC
ΔQ= m * c * ΔT
c= ΔQ / (m * ΔT)
c= 512.04 J/Kg K
Two 1.50-V batteries—with their positive terminals in the same direction—are inserted in series into the barrel of a flashlight. One battery has an internal resistance of 0.240 Ω, the other an internal resistance of 0.180 Ω. When the switch is closed, a current of 600 mA occurs in the lamp. (a) What is the bulb's resistance? Ω (b) What fraction of the chemical energy transformed appears as internal energy in the batteries? %
a. The resistance of the bulb is equal to 4.58 Ohms.
b. The fraction of the chemical energy transformed which appears as internal energy in the batteries is 8.4%.
Given the following data:
Voltage of battery = 1.5 VoltsInternal resistance A = 0.240 OhmsInternal resistance B = 0.180 OhmsCurrent = 600 mA = 0.6 Ampsa. To find the resistance of the bulb:
First of all, we would determine the total electromotive force (E) of the electric circuit:
[tex]E = 2 \times 1.50[/tex]
E = 3.0 Volts
Total internal resistance = [tex]0.240 + 0.180 = 0.42 \;Ohms[/tex]
Mathematically, electromotive force (E) is given by the formula:
[tex]E = V + Ir[/tex] ....equation 1.
Where:
E is the electromotive force (E).V is the the voltage or potential difference.I is the current.r is the internal resistance.According to Ohm's law, voltage is given by:
[tex]V = IR[/tex] ....equation 2
Substituting eqn 2 into eqn 1, we have:
[tex]E = IR + Ir\\\\E = I(R + r)\\\\R + r = \frac{E}{I} \\\\R = \frac{E}{I} - r\\\\R = \frac{3}{0.6} - 0.42\\\\R = 5 - 0.42[/tex]
Resistance, R = 4.58 Ohms
b. To determine what fraction of the chemical energy transformed appears as internal energy in the batteries:
First of all, we would determine the electromotive force (E) in the batteries.
[tex]E_B = IR\\\\E_B = 0.6 \times 0.42[/tex]
[tex]E_B[/tex] = 0.252 Volt
[tex]Percent = \frac{E_B}{E} \times 100\\\\Percent = \frac{0.252}{3} \times 100\\\\Percent = \frac{25.2}{3}[/tex]
Percent = 8.4 %
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The resistance of the bulb and the fraction of the chemical energy transformed into internal energy in the batteries can be determined using Ohm's law and the power dissipation formula. The total resistance in the series circuit is the sum of the individual resistances, and the total voltage is the sum of the individual voltages.
Explanation:The total voltage supplied by the batteries is the sum of their voltages, so Vtot = 1.5V + 1.5V = 3.0V. The total internal resistance of the batteries is the sum of their internal resistances, Rt = 0.240Ω + 0.180Ω = 0.420Ω. For the bulb's resistance, we can rearrange Ohm's law to R = V/I. Knowing the total voltage (3.0V) and the current (600 mA or 0.600 A), we can find the total resistance in the circuit.
Subtracting the total internal resistance gives us the resistance of the bulb. As for the second part of the question, the power dissipated in the internal resistances can be found using P = I²r for each battery, and then add these together. The total power supplied by the batteries is P = IV, using the total current and total voltage. The fraction of the chemical energy that appears as internal energy in the batteries is then Pinternal / Ptotal.
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Newton’s Second Law establishes the relationship between mass, net applied force, and acceleration given by F=ma. Consider a 4 kg box of holiday candy on a horizontal surface such as a table. There is a 10N applied force to the right and a 2N force to the left. The box accelerates with a magnitude of _________?
Answer:
a= 2 m/s^2
Explanation:
take to the right as positive
let Ftot be the total forces acting on the box , m be the mass of the box and a be the acceleration of the box.
Ftot = 10 - 2 = 8 N
and,
Ftot = ma
a = Ftot/m
= 8/4
= 2 m/s^2
therefore, the acceleration of the box is of magnitude of 2 m/s^2.
Suppose you are inside a smooth-flying [no bumps] jet plane moving at constant speed in a straight line across the Atlantic Ocean. Can you detect this motion [i.e., can you tell that you are moving] without directly or indirectly making use of anything outside of your own reference frame? If so, how could you detect this; and if not, why not?
Answer:
No we cannot perform any experiment that tells us weather we are moving or not at a constant speed.
Explanation:
Motion of any body is relative to other reference bodies. We can perceive motion only if our position with respect to a fixed object changes. This is the fundamental concept of reference frame in classical physics or Newtonian Physics that motion is always from a reference.
We choose a fixed body as reference and ,measure the distance we cover from this fixed point and also our speed with respect to this fixed point. The choice of frame of reference is completely dependent on the observer.
Since in the given case a reference cannot be established outside the plane thus we cannot detect our motion.
When two or more resistors are connected in parallel to a battery A) the voltage across each resistor is the same. B) the total current flowing from the battery equals the sum of the currents flowing through each resistor. C) the equivalent resistance of the combination is less than the resistance of any one of the resistors D) all of the other choices are true
Answer:
Here, for parallel resisitors
Option D) all of the other choices are are true
is correct.
Explanation:
In parallel connection:
1) Voltage across each element connected in parallel remain same.
2) Kirchhoff's Current Law (KCL), sum of the current entering and leaving the junction will be zero
3) The equivalent resistance of the elements connected in parallel is always less than the individual resistance of any resistor in the circuit.
The equivalent resistance in a parallel circuit is given by:
[tex]\frac{1}{R_{eq}} = \frac{1}{R_1} +\frac{1}{R_2} +......+ \frac{1}{R_n}[/tex]
Final answer:
When resistors are connected in parallel, the voltage across each resistor is the same, the total current flowing from the battery equals the sum of the currents flowing through each resistor, and the equivalent resistance of the combination is less than the resistance of any one of the resistors.
Explanation:
When two or more resistors are connected in parallel to a battery:
The voltage across each resistor is the same. This is because in a parallel circuit, all the resistors have the same potential difference across them.
The total current flowing from the battery equals the sum of the currents flowing through each resistor. In a parallel circuit, the total current is divided among the different resistors.
The equivalent resistance of the combination is less than the resistance of any one of the resistors. This is because adding resistors in parallel decreases the overall resistance of the circuit.
A playground tire swing has a period of 2.0 s on Earth. What is the length of its chain?
Answer:
Length of the chain, l = 0.99 m
Explanation:
Given that,
A playground tire swing has a period of 2.0 s on Earth i.e.
T = 2 s
We need to find the length of this chain. The relationship between the length and the time period is given by :
[tex]T=2\pi \sqrt{\dfrac{l}{g}}[/tex]
Where
l = length of chain
g = acceleration due to gravity
[tex]l=\dfrac{T^2g}{4\pi^2}[/tex]
[tex]l=\dfrac{(2)^2\times 9.8}{4\pi^2}[/tex]
l = 0.99 meters
So, the length of the chain is 0.99 meters. Hence, this is the required solution.
A uniform thin rod is hung vertically from one end and set into small amplitude oscillation. If the rod has a length of 2.6 m, this rod will have the same period as a simple pendulum of length ____ cm. Round your answer to the nearest whole number.
Answer:
Length of pendulum, l = 1.74 meters
Explanation:
The time period of simple pendulum is, [tex]T=2\pi\sqrt{\dfrac{l}{g}}[/tex]
Where
l is the length of simple pendulum
The time period of uniform thin rod is hung vertically from one end is, [tex]T=2\pi\sqrt{\dfrac{2l'}{3g}}[/tex]
l' is the length of uniform rod, l' = 2.6 m
It is given that the rod and pendulum have same time period. So,
[tex]2\pi\sqrt{\dfrac{l}{g}}=2\pi\sqrt{\dfrac{2l'}{3g}}[/tex]
After solving above expression, the value of length of the pendulum is, l = 1.74 meters. Hence, this is the required solution.
A mass of 0.14 kg is attached to a spring and set into oscillation on a horizontal frictionless surface. The simple harmonic motion of the mass is described by x(t) = (0.28 m)cos[(8 rad/s)t]. Determine the following. (a) amplitude of oscillation for the oscillating mass .
Answer:
The amplitude of oscillation for the oscillating mass is 0.28 m.
Explanation:
Given that,
Mass = 0.14 kg
Equation of simple harmonic motion
[tex]x(t)=(0.28\ m)\cos[(8\ rad/s)t][/tex]....(I)
We need to calculate the amplitude
Using general equation of simple harmonic equation
[tex]y=A\omega \cos\omega t[/tex]
Compare the equation (I) from general equation
The amplitude is 0.28 m.
Hence, The amplitude of oscillation for the oscillating mass is 0.28 m.
Two vectors A⃗ and B⃗ are at right angles to each other. The magnitude of A⃗ is 4.00. What should be the length of B⃗ so that the magnitude of their vector sum is 9.00?
Answer:
B= [tex]\sqrt{65}[/tex] ≅8.06
Explanation:
Using the Pythagorean theorem:
[tex]C^{2}[/tex]= [tex]A^{2}[/tex] + [tex]B^{2}[/tex]
where C represents the length of the hypotenuse and A and B the lengths of the triangle's other two sides, we can find out the lenght of B assuming the value of the hypotenuse being 9 and A being 4.
[tex]9^{2}[/tex]=[tex]4^{2}[/tex] + [tex]B^{2}[/tex]
81= 16+ [tex]B^{2}[/tex]
81-16= [tex]B^{2}[/tex]
B= [tex]\sqrt{65}[/tex] ≅8.06
The length of B is equal to 8.06 units
Data given;
A = 4.0B = ?C = 9.0Resolution of VectorsTo solve this question, we have to use the formula of finding resultant vectors
Since it's a right-angle triangle, let's use Pythagoras' theorem
[tex]C^2=A^2 + B^2\\9^2 = 4^2 + B^2\\b^2 = 9^2 - 4^2\\b^2 = 65\\b = \sqrt{65}\\b = 8.06[/tex]
From the calculation above, the length of B is equal to 8.06.
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Sand falls from an overhead bin and accumulates in a conical pile with a radius that is always twotwo times its height. Suppose the height of the pile increases at a rate of 33 cm divided by scm/s when the pile is 1010 cm high. At what rate is the sand leaving the bin at that instant?
Answer:
-423 m³/s
Explanation:
Volume of a cone is:
V = ⅓ π r² h
Given r = 2h:
V = ⅓ π (2h)² h
V = ⁴/₃ π h³
Taking derivative with respect to time:
dV/dt = 4π h² dh/dt
Given h = 1010 cm and dh/dt = 33 cm/s:
dV/dt = 4π (1010 cm)² (33 cm/s)
dV/dt ≈ 4.23×10⁸ cm³/s
dV/dt ≈ 423 m³/s
The pile is growing at 423 m³/s, so the bin is draining at -423 m³/s.
The rate at which the sand is leaving the bin at that instant is [tex]423\times 10^{6} cm^{3}/s[/tex].
GivenIt is given that the radius of a conical bin is two times its height and at the instant when the height of the bin is [tex]1010cm[/tex], the height of the pile increases at a rate of [tex]33 cm/s[/tex].
Volume of the binThe formula for the volume of a cone is given as,
[tex]V=\frac{1}{3}\pi r^{2}h[/tex]
Substitute [tex]r=2h[/tex] as per the question
[tex]V=\frac{4}{3}\pi h^{3}[/tex]
This is the volume of the conical bin.
Rate of change in the volume of the binTo find the rate of change in the volume of the bin, differentiate the expression for volume w.r.t. time using the chain rule as follows,
[tex]\frac{dV}{dt}=\frac{dV}{dh}\times \frac{dh}{dt}[/tex]
[tex]\frac{dV}{dt}=\frac{4}{3}\pi (3h^{2}) \times \frac{dh}{dt}\\\\\frac{dV}{dt}=4\pi h^{2} \times \frac{dh}{dt}\\[/tex]
Now, according to the question, at [tex]h=1010cm[/tex], [tex]\frac{dh}{dt}=33[/tex].
Substituting these values, the rate at which the sand is leaving the bin is,
[tex]\frac{dV}{dt}=4\pi (1010)^{2} \times 33\\\frac{dV}{dt}=423,025,503.90235\\\frac{dV}{dt}=423\times10^{6}cm^{3}/s[/tex]
So, the rate at which the sand leaves the conical bin at the given instant is [tex]423\times10^{6}cm^{3}/s[/tex]
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Light of wavelength 400 nm is incident on a single slit of width 15 microns. If a screen is placed 2.5 m from the slit. How far is the first minimum from the central maximum?
Answer:
0.0667 m
Explanation:
λ = wavelength of light = 400 nm = 400 x 10⁻⁹ m
D = screen distance = 2.5 m
d = slit width = 15 x 10⁻⁶ m
n = order = 1
θ = angle = ?
Using the equation
d Sinθ = n λ
(15 x 10⁻⁶) Sinθ = (1) (400 x 10⁻⁹)
Sinθ = 26.67 x 10⁻³
y = position of first minimum
Using the equation for small angles
tanθ = Sinθ = y/D
26.67 x 10⁻³ = y/2.5
y = 0.0667 m
The first minimum in a single-slit diffraction pattern of light with a wavelength of 400 nm incident on a single slit of width 15 microns, 2.5 m from the screen, is approximately 66.67 mm from the central maximum.
The question asks for the distance of the first minimum from the central maximum in a single-slit diffraction pattern.
The distance to the first minimum from the central maximum in a single-slit diffraction pattern is calculated using the formula: [tex]y = \frac{\lambda X D}{d}[/tex]
where y - is the distance from the central maximum to the first minimum on the screen (meters), λ (lambda) - is the wavelength of light (meters), D - is the distance between the slit and the screen (meters), and d - the width of the slit (meters)
Given λ (lambda) = 400 nm = [tex]400 X 10^{-9} m[/tex], D = 2.5 m, d = 15 microns = [tex]15 X 10^{-6} m[/tex]
Calculation:
[tex]y = \frac{(400 X 10^{-9} m) X (2.5 m)}{(15 X 10^{-6} m)}\\y = 0.0667 \ meters (or 66.7 \ millimeters)[/tex]
The first minimum is 66.7 millimeters from the central maximum.
A baseball approaches home plate at a speed of 44.0 m/s, moving horizontally just before being hit by a bat. The batter hits a pop-up such that after hitting the bat, the baseball is moving at 53.0 m/s straight up. The ball has a mass of 145 g and is in contact with the bat for 2.20 ms. What is the average vector force the ball exerts on the bat during their interaction?
Explanation:
It is given that,
Speed of the baseball, u = 44 m/s
Speed of the baseball, v = 53 m/s
Mass of the ball, m = 145 g = 0.145 kg
Time of contact between the ball and the bat, t = 2.2 ms = 0.0022 s
[tex]F=ma[/tex]
[tex]F=\dfrac{mv}{t}[/tex]
[tex]F_1=\dfrac{0.145\ kg\times 44\ m/s}{0.0022\ s}[/tex]
F₁ = 2900 N...........(1)
[tex]F=ma[/tex]
[tex]F=\dfrac{mv}{t}[/tex]
[tex]F_2=\dfrac{0.145\ kg\times 53\ m/s}{0.0022\ s}[/tex]
F₂ = 3493.18 N.........(2)
In average vector form force is given by :
[tex]F=F_1+F_2[/tex]
[tex]F=(2900i+(-3493.18)\ N[/tex]
[tex]F=(2900i-3493.18j)\ N[/tex]
Hence, this is the required solution.
Alpha particles of charge q = +2e and mass m = 6.6 × 10-27 kg are emitted from a radioactive source at a speed of 1.6 × 107 m/s. What magnetic field strength would be required to bend them into a circular path of radius r = 0.18 m?
Answer:
Magnetic field, B = 1.84 T
Explanation:
It is given that,
Charge on alpha particle, q = +2e = [tex]3.2\times 10^{-19}\ C[/tex]
Mass of alpha particle, [tex]m=6.6\times 10^{-27}\ kg[/tex]
Speed of alpha particles, [tex]v=1.6\times 10^7\ m/s[/tex]
We need to find the magnetic field strength required to bend them into a circular path of radius, r = 0.18 m
So, [tex]F_m=F_c[/tex]
[tex]F_m\ and\ F_c[/tex] are magnetic force and centripetal force respectively
[tex]qvB=\dfrac{mv^2}{r}[/tex]
[tex]B=\dfrac{mv}{qr}[/tex]
[tex]B=\dfrac{6.6\times 10^{-27}\ kg\times 1.6\times 10^7\ m/s}{3.2\times 10^{-19}\ C\times 0.18\ m}[/tex]
B = 1.84 T
So, the value of magnetic field is 1.84 T. Hence, this is the required solution.
The magnetic field strength required to bend them into a circular path is 1.83 T.
Force of the emitted chargesThe magnetic force on the emitted charge is given as;
F = qvB
The centripetal force of the emitted charge is given as;
F = mv²/r
Magnetic field strengthThe magnetic field strength required to bend them into a circular path is calculated as follows;
qvB = mv²/r
[tex]B = \frac{mv}{rq}[/tex]
[tex]B = \frac{6.6 \times 10^{-27} \times 1.6 \times 10^7}{(2\times 1.6 \times 10^{-19} ) \times 0.18} \\\\B = 1.83 \ T[/tex]
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A meteoroid is traveling east through the atmosphere at 18. 3 km/s while descending at a rate of 11.5 km/s. What is its speed, in km/s?
Answer:
The speed of meteoroid is 21.61 km/s in south-east.
Explanation:
Given that,
A meteoroid is traveling through the atmosphere at 18.3 km/s. while descending at a rate of 11.5 km/s it means 11.5 km/s in south.
We need to draw a diagram
Using Pythagorean theorem
[tex]AC^2=AB^2+BC^2[/tex]
[tex]AC^2=(18.3)^3+(11.5)^2[/tex]
[tex]AC=\sqrt{(18.3)^2+(11.5)^2}[/tex]
[tex]AC=21.61\ km/s[/tex]
Hence, The speed of meteoroid is 21.61 km/s in south-east.
The speed of the meteoroid is calculated using the Pythagorean theorem and is approximately 21.62 km/s.
To calculate the speed (magnitude of the velocity), the equation is: speed = √(horizontal velocity)² + (vertical velocity)².
Thus, the speed = √(18.3 km/s)² + (11.5 km/s)² = √(335.29 + 132.25) km²/s² = √467.54 km²/s² = 21.62 km/s.
The meteoroid's speed through the atmosphere is approximately 21.62 km/s.
Two identical loudspeakers are some distance apart. A person stands 5.80 m from one speaker and 3.90 m from the other. What is the fourth lowest frequency at which destructive interference will occur at this point? The speed of sound in air is 343 m/s.
Answer:
f = 632 Hz
Explanation:
As we know that for destructive interference the path difference from two loud speakers must be equal to the odd multiple of half of the wavelength
here we know that
[tex]\Delta x = (2n + 1)\frac{\lambda}{2}[/tex]
given that path difference from two loud speakers is given as
[tex]\Delta x = 5.80 m - 3.90 m[/tex]
[tex]\Delta x = 1.90 m[/tex]
now we know that it will have fourth lowest frequency at which destructive interference will occurs
so here we have
[tex]\Delta x = 1.90 = \frac{7\lambda}{2}[/tex]
[tex]\lambda = \frac{2 \times 1.90}{7}[/tex]
[tex]\lambda = 0.54 m[/tex]
now for frequency we know that
[tex]f = \frac{v}{\lambda}[/tex]
[tex]f = \frac{343}{0.54} = 632 Hz[/tex]
A 4.0 × 102-nm thick film of kerosene (n = 1.2) is floating on water. White light is normally incident on the film. What is the visible wavelength in air that has a maximum intensity after the light is reflected? Note: the visible wavelength range is 380 nm to 750 nm.
Answer:
the visible wavelength is 480 nm
Explanation:
Given data
thick film = 4.0 × 10² nm
n = 1.2
wavelength range = 380 nm to 750 nm
to find out
the visible wavelength in air
solution
we know that index of water is 1 and kerosene is 1.2
we can say that when light travel reflected path difference is = 2 n t
and for maximum intensity it will be k × wavelength
so it will be 2 n t = k × wavelength
2 × 1.2 × 4.0 × 10² = k × wavelength
wavelength = 2 × 1.2 × 4.0 × 10² / k
here k is 2 for visible
so wavelength = 2 × 1.2 × 4.0 × 10² / 2
wavelength = 480 nm
the visible wavelength is 480 nm
Suppose you are an astronaut on a spacewalk, far from any source of gravity. You find yourself floating alongside your spacecraft but 10 m away, with no propulsion system to get back to it. In your tool belt you have a hammer, a wrench, and a roll of duct tape. How can you get back to your spacecraft?
a. Move like you are flying to the spaceship
b. Move like you are swimming to the spaceship
c. Throw the items away from the spaceship
d. Throw the items to the spaceship.
Answer:
c. Throw the items away from the spaceship.
Explanation:
By the Principle of action and reaction yu can get back to your spacecraft throwing the items away from the spaceship.
Answer:
c. Throw the items away from the spaceship
Explanation:
The propulsion that throwing items away from the spaceship will propel you to the opposite direction in which you are throwing the objects, this means that if you throw the items away from the spaceship you will be getting force of rpopulsion towards the spaceship.
A proton is placed in a uniform electric field and then released. Then an electron is placed at this same point and released. Which of the following is correct?
The 2 particles experience the same magnitude of force, experience the same magnitude of acceleration, and move in different directions.
The 2 particles experience the same magnitude of force, experience different magnitudes of acceleration, and move in different directions.
The 2 particles experience different magnitudes of force, experience different magnitudes of acceleration, and move in different directions.
The 2 particles experience the same magnitude of force, experience different magnitudes of acceleration, and move in the same direction.
None of the above.
Answer:
option (b)
Explanation:
Let the electric field is given by E.
mass of proton = mp
mass of electron = me
acceleration of proton = ap
acceleration of electron = ae
Charge on both the particle is same but opposite in nature.
The force on proton = q E
The force on electron = - q E
acceleration of proton, ap = q E / mp
acceleration of electron, ae = - q E / me
We observe that the force is same in magnitude but opposite in direction, acceleration is also different and opposite in direction.
Final answer:
The electron and proton experience the same magnitude of force but different accelerations due to their mass difference, and move in opposite directions because of their opposite charges.
Explanation:
When a proton and an electron are placed in a uniform electric field and released, they both experience the same magnitude of force, because they have equal and opposite charges of the same magnitude. However, their accelerations differ due to their masses. The electron has a much smaller mass compared to the proton, and according to Newton's second law (F = ma), a given force will produce a larger acceleration on an object with a smaller mass.
Therefore, while the magnitudes of the forces are the same, the electron will experience a greater magnitude of acceleration than the proton. Finally, they move in opposite directions because the electric field exerts a force in the direction of the field on positive charges and in the opposite direction on negative charges. Therefore, the electron moves in the opposite direction to the proton when released in the same electric field.
If one gram of matter could be completely converted into energy, the yield would be
A. 0.51 MeV.
B. 931 MeV.
C. 3 × 1013 J.
D. 9 × 1013 J.
E. 9 × 1016 J.
Answer:
Energy, [tex]E=9\times 10^{13}\ J[/tex]
Explanation:
It is given that,
Mass of matter, m = 1 g = 0.001 Kg
If this matter is completely converted into energy, we need to find the yield. The mass of an object can be converted to energy as :
[tex]E=mc^2[/tex]
c = speed of light
[tex]E=0.001\ kg\times (3\times 10^8\ m/s)^2[/tex]
[tex]E=9\times 10^{13}\ J[/tex]
So, the energy yield will be [tex]9\times 10^{13}\ J[/tex]. Hence, this is the required solution.
A 78 kg skydiver can be modeled as a rectangular "box" with dimensions 24 cm × 35 cm × 170 cm . If he falls feet first, his drag coefficient is 0.80.What is his terminal speed if he falls feet first? Use ? = 1.2 kg/^m3 for the density of air at room temperature.
Answer:
The terminal speed of his is 137.68 m/s.
Explanation:
Given that,
Mass of skydiver = 78 kg
Area of box[tex]A =24\times35=840\ cm[/tex]
Drag coefficient = 0.80
Density of air [tex]\rho= 1.2\times kg/m^3[/tex]
We need to calculate the terminal velocity
Using formula of drag force
[tex]F_{d} = \dfrac{1}{2}\rho v^2Ac[/tex]
Where,
[tex]\rho[/tex] = density of air
A = area
C= coefficient of drag
Put the value into the formula
[tex]78\times9.8=\dfrac{1}{2}\times1.2\times v^2\times24\times10^{-2}\times35\times10^{-2}\times0.80[/tex]
[tex]v^2=\dfrac{2\times78\times9.8}{1.2\times24\times10^{-2}\times35\times10^{-2}\times0.80}[/tex]
[tex]v=\sqrt{\dfrac{2\times78\times9.8}{1.2\times24\times10^{-2}\times35\times10^{-2}\times0.80}}[/tex]
[tex]v=137.68\ m/s[/tex]
Hence, The terminal speed of his is 137.68 m/s.
The terminal speed of the skydiver with dimensions as a rectangular box as he falls feet first is 137.68 m/s.
What is the terminal speed?Terminal speed of a body is the maximum speed, which is achieved by the object when it fall through a fluid.
In the case of terminal velocity, the force of gravity becomes equal to the sum of the drag force and buoyancy force due to fluid on body.
Terminal velocity can be find out as,
[tex]v=\sqrt{\dfrac{2mg}{\rho AC_d}}[/tex]
Here, (m) is the mass, (g) is gravitational force, ([tex]\rho[/tex]) is the density of fluid, (A) is the project area and ([tex]C_d[/tex]) is the drag coefficients.
It is given that, the mass of the skydiver is 78 kg The dimensions of the skydiver is s 24 cm × 35 cm × 170 cm.
The coefficient of drag is 0.80 and the density of air is 1.2 kg/m³.
Put the values in the above formula as,
[tex]v=\sqrt{\dfrac{2\times78\times9.8}{1.2\times0.24\times0.35\times 0.8}}\\v=137.68\rm m/s[/tex]
Thus the terminal speed of the skydiver as he falls feet first is 137.68 m/s.
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The 630 nm light from a helium neon laser irradiates a grating. The light then falls on a screen where the first bright spot is separated from the central maxim by 0.51m. Light of another wavelength produces its first bright spot at 0.39 m from its central maximum. Determine the second wavelength.
Answer:
The second wavelength is 482 nm.
Explanation:
Given that,
Wavelength = 630 nm
Distance from central maxim = 0.51 m
Distance from central maxim of another wavelength = 0.39 m
We need to calculate the second wavelength
Using formula of width of fringe
[tex]\beta=\dfrac{\lambda d}{D}[/tex]
Here, d and D will be same for both wavelengths
[tex]\lambda[/tex] = wavelength
[tex]\beta [/tex] = width of fringe
The width of fringe for first wavelength
[tex]\beta_{1}=\dfrac{\lambda_{1} d}{D}[/tex]....(I)
The width of fringe for second wavelength
[tex]\beta_{2}=\dfrac{\lambda_{2} d}{D}[/tex]....(II)
Divided equation (I) by equation (II)
[tex]\dfrac{\beta_{1}}{\beta_{2}}=\dfrac{\lambda_{1}}{\lambda_{2}}[/tex]
[tex]\lambda_{2}=\dfrac{630\times10^{-9}\times0.39}{0.51}[/tex]
[tex]\lambda_{2}=4.82\times10^{-7}[/tex]
[tex]\lambda=482\ nm[/tex]
Hence, The second wavelength is 482 nm.
To find the second wavelength, we can use the formula for the wavelength of light from a diffraction grating. In this case, we know the first wavelength is 630 nm, the first bright spot is separated from the central maximum by 0.51 m, and we need to find the second wavelength.
Explanation:To find the second wavelength, we can use the formula for the wavelength of light from a diffraction grating: wavelength = (m * d * sin(theta)) / n. In this case, we know the first wavelength is 630 nm, the first bright spot is separated from the central maximum by 0.51 m, and we need to find the second wavelength. The m and n values are the same for both cases, so we can set up the equation:
630 nm = (m * 0.51 m * sin(theta)) / n
wavelength = (m * 0.51 m * sin(theta)) / n
Next, we can solve for the second wavelength by rearranging the equation:
wavelength = (630 nm * n) / (m * 0.51 m * sin(theta))
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A hunter is standing on flat ground between two vertical cliffs that are directly opposite one another. He is closer to one cliff than the other. He fires a gun and, after a while, hears three echoes. The second echo arrives 1.76 s after the first, and the third echo arrives 1.38 s after the second. Assuming that the speed of sound is 343 m/s and that there are no reflections of sound from the ground, find the distance (in m) between the cliffs.
Answer:
Distance=538.51m
Explanation:
The echo is heard after covering double distance .
Therefore 2d=(t1+t2)×speed.
2d={(1.76+1.38)×343}
2d=743.02
d=1077.02÷2
=538.51m
A man, a distance d=3~\text{m}d=3 m from a target, throws a ball at an angle \theta= 70^\circθ=70 ∘ above the horizontal. If the initial speed of the ball is v=5~\text{m/s}v=5 m/s, what height hh does the ball strike the building?
Answer:
The ball doesn't strike the building because it strikes the ground at d=1.62 meters.
Explanation:
V= 5 m/s < 70º
Vx= 1.71 m/s
Vy= 4.69 m/s
h= Vy * t - g * t²/2
clearing t for the flying time of the ball:
t= 0.95 s
d= Vx * t
d= 1.62 m