Explanation:
The number of charges flowing per second inside the conductor is called the current through the conductor. Mathematically, it is given by :
[tex]I=\dfrac{Q}{t}[/tex]
Where
Q is the charge
t is time
Since, Q = ne
n is number of electrons
e is the charge on one electron
So, to find the current through the conductor we must know the number of electrons moving per second through it. Hence, this is the required solution.
A microscope has an objective lens with a focal length of 1.82 mm. The lenses are 47.8 cm apart What should the focal length of the eyepice be if I want the microscope to have an angular magnifica tion of -2000?
Answer:
focal length of eye lens is 3.28 cm
Explanation:
As we know that the magnification of the microscope is given by the formula
[tex]M = \frac{L}{f_o}\frac{D}{f_e}[/tex]
now we will have
[tex]L = 47.8 cm[/tex]
D = 25 cm
M = 2000
[tex]f_o = 1.82 mm = 0.182 cm[/tex]
now we need to solve above equation to find focal length of eye lens
so here we will have
[tex]f_e = \frac{L}{f_o}\frac{D}{M}[/tex]
[tex]f_e = \frac{47.8}{0.182}\frac{25}[2000}[/tex]
[tex]f_e = 3.28 cm[/tex]
A person is standing on a completely frictionless surface. Can they walk on this surface? Explain! What could they do to move across the surface? Explain!
Answer:
Explanation:
We cannot walk on a surface which has no friction.
To move across the surface, take a stone and throw it in the opposite direction of motion so that you get a reaction in the direction of motion and then you move across the surface.
Determine the position in which a solid cylindrical block of wood of diameter 0.3 m and length 0.4 m will float in water. Take specific gravity of wood as 0.5
Answer:
0.2 m
Explanation:
Diameter = 0.3 m
radius, r = 0.15 m
Length, H = 0.4 m
density of wood, d = 0.5 g/cm^3 = 500 kg/m^3
density of water, d = 1000 kg/m^3
Let h be the depth of cylinder immersed in water.
By the principle of floatation.
Buoyant force = Weight of cylinder
Volume immeresed x density of water x g = Volume of cylinder x density of wood x g
A x h x 1000 x g = A x H x 500 x g
1000 h = 500 x 0.4
h = 0.2 m
As you move away from a positive charge distribution, the electric field:
Always decreases
Always increases
Must change
None of the above
please explain this thoroughly!
Answer:
The electric field always decreases.
Explanation:
The electric field due to a point charge is given by :
[tex]E=\dfrac{kq}{r^2}[/tex]
Where
k = electric constant
q = charge
r = distance from the charge
It is clear from the above equation that as the distance from the charge particle increases the electric field decreases. As you move away from a positive charge distribution, the electric field always decreases. Hence, the correct option is (c) "Always decreases".
Using a 681 nm wavelength laser, you form the diffraction pattern of a 0.109 mm wide slit on a screen. You measure on the screen that the 14th dark fringe is 9.51 cm away from the center of the central maximum. How far is the screen located from the slit?
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 gas of helium atoms (each of mass 6.65 × 10-27 kg) are at room temperature of 20.0°C. What is the de Broglie wavelength of the helium atoms that are moving at the root-mean-square speed? (h = 6.626 × 10-34 J · s, the Boltzmann constant is 1.38 × 10-23 J/K)
Answer:
The de Broglie wavelength of the helium atoms is [tex]7.373\times10^{-11}\ m[/tex].
Explanation:
Given that,
Mass [tex]M=6.65\times10^{-27}\ kg[/tex]
Temperature = 20.0°C
We need to calculate the root-mean square speed
Using formula of root mean square speed
[tex]v_{rms}=\sqrt{\dfrac{3kTN_{A}}{M}}[/tex]
Where, N = Avogadro number
M = Molar mass
T = Temperature
k = Boltzmann constant
Put the value into the formula
[tex]v_{rms}=\sqrt{\dfrac{3\times1.38\times10^{-23}\times293\times6.022\times10^{23}}{4\times10^{-3}}}[/tex]
[tex]v_{rms}=1351.37\ m/s[/tex]
We need to calculate the de Broglie wavelength
Using formula of de Broglie wavelength
[tex]P=\dfrac{h}{\lambda}[/tex]
[tex]mv=\dfrac{h}{\lambda}[/tex]
[tex]\lambda=\dfrac{6.626\times10^{-34}}{6.65\times10^{-27}\times1351.37}[/tex]
[tex]\lambda=7.373\times10^{-11}\ m[/tex]
Hence, The de Broglie wavelength of the helium atoms is [tex]7.373\times10^{-11}\ m[/tex].
Final answer:
The de Broglie wavelength of helium atoms moving at the root-mean-square speed is approximately 4.779 × 10^-10 meters.
Explanation:
To find the de Broglie wavelength of helium atoms moving at the root-mean-square speed, we can use the equation:
λ = h / (m * v)
where λ is the de Broglie wavelength, h is Planck's constant (6.626 × 10-34 J · s), m is the mass of the helium atom (6.65 × 10-27 kg), and v is the root-mean-square speed.
The root-mean-square speed of helium atoms at room temperature can be found using the equation:
v = √(3 * k * T / m)
where k is the Boltzmann constant (1.38 × 10-23 J/K) and T is the temperature in Kelvin (20.0 + 273 = 293 K).
Plugging the values into the equations and solving for λ:
λ = (6.626 × 10-34 J · s) / (6.65 × 10-27 kg * √(3 * 1.38 × 10-23 J/K * 293 K / 6.65 × 10-27 kg))
λ = 4.779 × 10-10 m
Therefore, the de Broglie wavelength of the helium atoms moving at the root-mean-square speed is approximately 4.779 × 10-10 meters.
A ray of light traveling in air is incident on the flat surface of a piece of glass at an angle of 65.9° with respect to the normal to the surface of the glass. If the ray refracted into the glass makes an angle of 34.8° with respect to the normal, what is the refractive index of the glass?
Answer:
[tex]n_{glass}[/tex] = 1.6
Explanation:
[tex]\theta _{i}[/tex] = Angle of incidence = 65.9°
[tex]\theta _{r}[/tex] = Angle of refraction = 34.8°
[tex]n_{air}[/tex] = Index of refraction of air = 1
[tex]n_{glass}[/tex] = Index of refraction of glass = ?
Using Snell's law
[tex]n_{air}[/tex] Sin[tex]\theta _{i}[/tex] = [tex]n_{glass}[/tex] [tex]\theta _{r}[/tex]
(1) Sin65.9 = [tex]n_{glass}[/tex] Sin34.8
[tex]n_{glass}[/tex] = 1.6
Since the index of refraction of air is 1, the refractive index of the glass is 1.6 approximately
Refractive IndexRefractive Index is the measure of refraction or bending when light passes from one medium to another.
Given that a ray of light traveling in air is incident on the flat surface of a piece of glass at an angle of 65.9° with respect to the normal to the surface of the glass. If the ray refracted into the glass makes an angle of 34.8° with respect to the normal, the refractive index of the glass can be calculated with the formula below
n = sin i / sin r
Where
n = refractive indexi = angle of incidencer = angle of refractionSubstitute all the parameters
n = sin 65.9 / sin 34.8
n = 0.913 / 0.5707
n = 1.599
n = 1.6 approximately
Therefore, the refractive index of the glass is 1.6 approximately.
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Blades of an airplaneâs engine rotate with an initial speed is 40 rev/min. Assuming constant angular acceleration of magnitude 8 rad/s^2 . (a) how long does it take for the blades to reach the angular speed Ïf = 120 rev/ min?
(b)Through how many radians does a blade turn during the time found in (a).
Note : Write down the detail process
Answer:
Part a)
t = 1.05 s
Part b)
[tex]\theta = 8.78 rad[/tex]
Explanation:
Initial angular speed is given as
[tex]\omega_i = 40 rev/min = 0.66 rev/s[/tex]
[tex]\omega_i = 2\pi (0.66) = 4.19 rad/s[/tex]
angular acceleration is given as
[tex]\alpha = 8 rad/s^2[/tex]
now we have
part a)
final angular speed = 120 rev/min
[tex]\omega_f = 2\pi(\frac{120}{60} rev/s)[/tex]
[tex]\omega_f = 12.57 rad/s[/tex]
now by kinematics we have
[tex]\omega_f = \omega_i + \alpha t[/tex]
[tex]12.57 = 4.19 + 8 t[/tex]
[tex] t = 1.05 s[/tex]
Part b)
Angle turned by the blades is given by
[tex]\theta = \omega_i t + \frac{1}{2}\alpha t^2[/tex]
[tex]\theta = 4.19(1.05) + \frac{1}{2}(8)(1.05)^2[/tex]
[tex]\theta = 8.78 rad[/tex]
A ball is thrown from the top of one building toward a tall building 50 m away. The initial velocity of the ball is 20 m/s at 40° above the horizontal. How far above or below its original level will the ball strike the opposite wall?
Answer:
Ball hit the tall building 50 m away below 10.20 m its original level
Explanation:
Horizontal speed = 20 cos40 = 15.32 m/s
Horizontal displacement = 50 m
Horizontal acceleration = 0 m/s²
Substituting in s = ut + 0.5at²
50 = 15.32 t + 0.5 x 0 x t²
t = 3.26 s
Now we need to find how much vertical distance ball travels in 3.26 s.
Initial vertical speed = 20 sin40 = 12.86 m/s
Time = 3.26 s
Vertical acceleration = -9.81 m/s²
Substituting in s = ut + 0.5at²
s = 12.86 x 3.26 + 0.5 x -9.81 x 3.26²
s = -10.20 m
So ball hit the tall building 50 m away below 10.20 m its original level
The distance the ball will strike the opposite wall is 32.79 m.
Time of motion of the ball
The time of motion of the ball from the given height is calculated as follows;
h = vsinθ(t) + ¹/₂gt²
50 = 20 x sin(40)t + 0.5(9.8)t²
50 = 12.86t + 4.9t²
4.9t² + 12.86t - 50 = 0
solve the quadratic equation using formula method,
t = 2.14 s
Horizontal distance of the ballThe horizontal distance of the ball from the initial position is calculated as follows;
X = vcosθ(t)
X = 20 x cos(40) x 2.14
X = 32.79 m
Thus, the distance the ball will strike the opposite wall is 32.79 m.
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Cart A, with a mass of 0.4 kg, travels on a horizontal air track at 6 m/s and hits cart B, which has a mass of 0.8 kg and is initially at rest. After the collision, cart A has a velocity of 2 m/s in the opposite direction. What is the final kinetic energy of cart B
Answer:
6.4 J
Explanation:
For cart A
mA = 0.4 kg, uA = 6 m/s, vA = - 2 m/s
For cart B
mB = 0.8 kg, uB = 0, vB = ?
use the law of conservation of momentum
momentum of system before coliision = momentum of system after collision
mA uA + mB uB = mA vA + mB vB
0.4 x 6 + 0 = 0.4 x (- 2) + 0.8 x vB
vB = 4 m/s
Kinetic energy of Cart B after collision = 1/2 mB vB^2
= 1/2 x 0.8 x 4 x 4 = 6.4 J
Two parallel wires are separated by 6.65 cm, each carrying 3.25 A of current in the same direction. (a) What is the magnitude of the force per unit length between the wires?
Answer:
The magnitude of the force per unit of lenght between the wires are of F/L= 3.17 * 10⁻⁵ N/m.
Explanation:
d=0.0665m
I1=I2= 3.25A
μo= 4π * 10⁻⁷ N/A²
F/L= (μo * I1 * I2) / (2π * d)
F/L= 3.17 * 10⁻⁵ N/m
A projectile is launched with initial speed v and angle ? over level ground. The projectile's maximum height H and horizontal range D are related by the equation D = 1.8H, what was the launch angle of the projectile?
Answer:
65.75 deg
Explanation:
v = initial speed of launch of projectile
θ = initial angle of launch
H = maximum height of the projectile
maximum height of the projectile is given as
[tex]H=\frac{v^{2}Sin^{2}\theta }{2g}[/tex] eq-1
D = horizontal range of the projectile
horizontal range of the projectile is given as
[tex]D=\frac{v^{2}Sin{2}\theta }{g}[/tex] eq-2
It is also given that
D = 1.8 H
using eq-1 and eq-2
[tex]\frac{v^{2}Sin{2}\theta }{g} = (1.8) \frac{v^{2}Sin^{2}\theta }{2g}[/tex]
[tex]Sin{2}\theta = (1.8) \frac{Sin^{2}\theta }{2}[/tex]
[tex]2 Sin\theta Cos\theta= (0.9) Sin^{2}\theta[/tex]
[tex]2 Cos\theta = (0.9) Sin\theta[/tex]
tanθ = 2.22
θ = 65.75 deg
An electron in a magnetic field has a cyclotron frequency of 3.0x 10^12 Hz What is the magnetic fleld strength? (The mass of an electron is 9.1 x 10^-31 kg and the charge of an electron is 16x 10^-19C) A) 0.0093 T B) 110 T C) 12T D) 0.084 T
Answer:
The magnetic field strength, B = 110 T
Explanation:
It is given that,
Cyclotron frequency, [tex]\nu=3\times 10^{12}\ Hz[/tex]
We need to find the magnetic field strength. The formula for cyclotron frequency is given by :
[tex]\nu=\dfrac{qB}{2\pi m}[/tex]
B is the magnetic field strength
q and m are the charge and mass of electron.
[tex]B=\dfrac{2\pi m\nu}{q}[/tex]
[tex]B=\dfrac{2\pi\times 9.1\times 10^{-31}\ kg\times 3\times 10^{12}\ Hz}{1.6\times 10^{-19}\ C}[/tex]
B = 107.20 T
or
B = 110 T (approx)
So, the magnetic field strength of the electron is 110 T. Hence, this is the required solution.
Using the charge, mass, and cyclotron frequency of the electron, the magnetic field strength is calculated to be approximately 0.084 T.
Explanation:The frequency of the electron in the magnetic field, called the cyclotron frequency, can be used to find the strength of the magnetic field. The equation for cyclotron frequency is ƒ = qB/2πm, wherein ƒ is the frequency, q is the charge, B is the magnetic field, and m is the mass of the electron. In this case, to solve for B (magnetic field), we can rearrange the equation to B = 2πmƒ/q.
Substituting the given values, ƒ = 3.0 x 10^12 Hz, q = 16 x 10^-19 C, and m = 9.1 x 10^-31 kg, into the formula, and performing the proper calculation, the magnetic field strength is found to be approximately 0.084 T, which corresponds to the option (D).
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Two resistors connected in series have an equivalent resistance of 724.5 Ω, when they are connected in parallel, their equivalent resistance is 135.2 Ω. Find the resistance of each resistor.
Answer:
544.65 ohm and 179.84 ohm
Explanation:
Hello
Let
Resistor 1 (R1) and Resistor 2 (R2)
in series [tex]R_{eq} =R1+R2[/tex]
in parallel
[tex]\frac{1}{R_{eqp} } =\frac{1}{R1} +\frac{1}{R2}\\\\R1=R_{eqs} -R2\\\\R1 =724.5-R2 (equation 1)\\ R_{eqp}}=\frac{R1*R2}{R1+R2}\\\\\ 135.2 = \frac{R1*R2}{R1+R2}(equation 2)\\\\\\\\replacing 1 in \ 2\\\\probema data[/tex]
[tex]135.2=\frac{(724.5-R2)(R2)}{724.5-R2+R2} }\\135.2=\frac{(-R_{2} ^{2}+724.5R2) }{724.5}\\ 135.2*724.5=-R_{2} ^{2}+724.5R_{2}\\\\R_{2} ^{2} -724.5R_{2} +97952.4 =0\\R_{2} =\frac{724.5 \±\sqrt{(-724.5^{2})-4(1)(97952.4) } }{2(1)}\\R_{2} =\frac{724.5\±\sqrt{(133092.25) } }{2(1)}\\\\R_{2} =\frac{724.5+364.81810}{2} \\R_{2} =544.65 ohm\\\\R_{2} =\frac{724.5-364.81810}{2} =179.84 ohm[/tex]\\\\
let R2=544.65 and replace in equation 1
R1=724.5-544.65
R1=179.85
so, the resistors are 544.65 ohm and 179.84 ohm
Have a great day
An asteroid is moving along a straight line. A force acts along the displacement of the asteroid and slows it down. The asteroid has a mass of 3.5× 104 kg, and the force causes its speed to change from 6600 to 5700m/s. (a) What is the work done by the force? (b) If the asteroid slows down over a distance of 1.5× 106 m determine the magnitude of the force.
Try this option; answers are marked with red colour: a) 19.215 GJ; b) 12.81 kN.
All the details are in the attached picture.
A sled is pulled along a level path through snow by a rope. A 20-lb force acting at an angle of 40° above the horizontal moves the sled 90 ft. Find the work done by the force. (Round your answer to the nearest whole number.)
Answer:
Work one by the force, W = 1378.87 lb-ft
Explanation:
It is given that,
Force acting on the sled, F = 20 lb
Angle between the force and the horizontal, θ = 40°
Distance moved, d = 90 ft
We need to find the work done by the force. We know that the work done can be calculated as :
[tex]W=Fd\ cos\theta[/tex]
[tex]W=20\ lb\times 90\ ft\ cos(40)[/tex]
W = 1378.87 lb-ft
So, the work done by the force is 1378.87 lb-ft. Hence, this is the required solution.
If the gravitational force between two masses located a distance 100 m apart is F, what would be the force between these objects if the distance between them increases to 800 m?
Answer:
[tex] \frac{F}{64}[/tex]
Explanation:
m = product of masses of two objects
r = distance between the two objects = 100 m
F = initial force between the two object
r' = new distance between the two objects = 800 m
F' = new force between the two objects = ?
k = constant of proportionality
initial force between the two object is given as
[tex]F=\frac{km}{r^{2}}[/tex] eq-1
new force between the two objects is given as
[tex]F'=\frac{km}{r'^{2}}[/tex] eq-2
Dividing eq-2 by eq-1
[tex]\frac{F'}{F}=\frac{r^{2}}{r'^{2}}[/tex]
Inserting the values
[tex]\frac{F'}{F}=\frac{100^{2}}{800^{2}}[/tex]
[tex]F' = \frac{F}{64}[/tex]
A boy flies a kite with the string at a 30∘ angle to the horizontal. The tension in the string is 4.5 N. Part A Part complete How much work does the string do on the boy if the boy stands still?
Answer:
Work done is zero
Explanation:
given data
Angle of kite with horizontal = 30 degree
tension in the string = 4.5 N
WE KNOW THAT
Work = force * distance
horizontal force = [tex]Tcos\theta = 4.5*cos30 = 3.89 N[/tex]
DISTANCE = 0 as boy stands still. therefore
work done = 3.89 *0 = 0
A soccer player swings his leg forward to kick the ball. If his leg has a moment of inertia of 0.9 kg m2, and is initially at rest, how much torque must he generate with his hip flexor muscle in order to reach an angular velocity of -8 rad/s in 0.5 seconds?
Answer:
14.4 Nm
Explanation:
Moment of Inertia, I = 0.9 kg m^2, w0 = 0, w = 8 rad/s, t = 0.5 second
Use first equation of motion for rotational motion
w = w0 + α t
where, α be the angular acceleration
8 = 0 + α x 0.5
α = 16 rad/s^2
Now Torque = Moment of inertia x angular acceleration
τ = I x α
τ = 0.9 x 16 = 14.4 Nm
A truck tire rotates at an initial angular speed of 21.5 rad/s. The driver steadily accelerates, and after 3.50 s the tire's angular speed is 28.0 rad/s. What is the tire's angular acceleration (in rad/s2) during this time?
Given:
initial angular speed, [tex]\omega _{i}[/tex] = 21.5 rad/s
final angular speed, [tex]\omega _{f}[/tex] = 28.0 rad/s
time, t = 3.50 s
Solution:
Angular acceleration can be defined as the time rate of change of angular velocity and is given by:
[tex]\alpha = \frac{\omega_{f} - \omega _{i}}{t}[/tex]
Now, putting the given values in the above formula:
[tex]\alpha = \frac{28.0 - 21.5}{3.50}[/tex]
[tex]\alpha = 1.86 m/s^{2}[/tex]
Therefore, angular acceleration is:
[tex]\alpha = 1.86 m/s^{2}[/tex]
A circular swimming pool has a diameter of 8 meters. The sides are 3 meters high and the depth of the water is 1.5 meters. How much work is required to pump all of the water over the side? Your answer must include the correct
The amount of work required to pump all of the water over the side of the pool is 471,238.9 Newton Meters or Joules.
Explanation:The work done to pump water out of a pool involves the concept of physics specifically related to potential energy, gravity, and volume. The work done to move a certain volume of water is given by the formula: Work = Weight x Height.
First, we need to find the volume of the water in the pool. The pool's shape resembles a cylinder, and the volume is given by the formula for a cylinder: Volume = pi × (diameter/2)²× height. Given a diameter of 8 meters and a height of 1.5 meters, the volume to be moved is pi * (8/2)² × 1.5 = 48pi cubic meters.
The weight of this water can be calculated by multiplying its volume by its density. The density of water is 1000 kg/m³. Therefore, the weight of the water is Volume x Density x g (acceleration due to gravity), which is 48pi × 1000 × 9.8 = 471,238.9 kg×m²/s² or Newton Meter (Nm) which is the unit of work.
So, the amount of work required to pump all the water over the side is 471,238.9 Nm or Joules (J).
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The work required to pump all the water over the side of a circular swimming pool of diameter 8 meters and water depth 1.5 meters is calculated using the formula Work = Force x Distance. The Force required is given by the weight of the water, which depends on its volume and density. The result is about 353,429.16 Joules.
Explanation:To calculate the work required to pump water out of a swimming pool, we can use the formula for work: Work = Force x Distance. The force required is equal to the weight of the water which depends on the volume of the water and its density.
First, let's calculate the volume of the water in the pool. Given that it's a circular pool with a diameter of 8 meters, the radius is 4 meters. The depth of the water is 1.5 meters. So, volume (V) = πr²h = π×(4m)²×(1.5m) = 24π cubic meters.
The density of water is 1000 kg/m³. Therefore, the weight of water = Volume x Density x Gravity = 24π m³ ×1000 kg/m³ × 9.8 m/s² = 235619.44 kg×m/s², or Newtons. This is the force we need to overcome to lift the water.
The distance that we want to lift this mass is the depth of the pool, assuming we are pumping the water over the side of the pool which is 1.5 meters high. So, Work = Force x Distance = 235619.44 N×1.5m = 353429.16 Joules.
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A parachutist who weighs 200lbs is falling at 120 miles/hour when his parachute opens. His speed is reduced to 15 miles/hour in a vertical distance of 120ft. What force did the parachute exert on the jumper?
Answer:
F = 3482.9 N
Explanation:
Change in velocity of the Parachutist is given as
[tex]v_f = 15 mph = 6.675 m/s[/tex]
[tex]v_i = 120 mph = 53.4 m/s[/tex]
now it is given as
[tex]\Delta v = v_f - v_i [/tex]
[tex]\Delta v = 120 - 15 = 105 mph[/tex]
[tex]\Delta v = 46.7 m/s[/tex]
now the acceleration of the parachutist is given as
[tex]a = \frac{v_f^2 - v_i^2}{2d}[/tex]
distance moved by the parachutist is given as
[tex]d = 120 ft = 36.576 m[/tex]
now we have
[tex]a = \frac{6.675^2 - 53.4^2}{2(36.576)}[/tex]
[tex]a = - 38.4m/s^2[/tex]
Now the mass of parachutist is given as
[tex]m = 200 lb = 90.7 kg[/tex]
now we have
[tex]F = ma[/tex]
[tex]F = (90.7 kg)(38.4) = 3482.9 N[/tex]
Consider two objects on the Moon's surface that can just be resolved by one of the 10.0 m telescopes at the Keck Observatory. What is the separation of the two objects? Assume the resolving power is only limited by diffraction effects. The distance from the Earth to the Moon is 3.8 x 105km, assume the wavelength of the light is 550 nm.
Answer: 25 m
Explanation:
The resolving power of a telescope is given by:
[tex] \theta = 1.22\frac{\lambda}{diameter}[/tex]
Diameter of the telescope = 10.0 m
Wavelength of the telescope = 550 nm = 550 ×10⁻⁹m
Thus, the resolving power of one of telescopes at the Keck Observatory is:
[tex] \theta = 1.22 \frac{550\times 10^{-9}}{10.0} =671 \times 10^{-10} rad = 3.85 \times 10^{-6} deg[/tex]
Also, [tex] tan\theta = \frac {\text {separation between two objects}}{\text{distance to the objects}} [/tex]
separation between objects = tan (3.85×10⁻⁶)° × 3.8×10⁵km = 2.5 × 10⁻²km =25 m
A proton moves with a speed of 3.60 106 m/s horizontally, at a right angle to a magnetic field. What magnetic field strength is required to just balance the weight of the proton and keep it moving horizontally? (The mass and charge of the proton are 1.67 ✕ 10−27 kg and 1.60 ✕ 10−19 C, respectively.) B = T
Answer:
The magnetic field strength is required [tex] 2.84\times10^{-14}\ T[/tex]
Explanation:
Given that,
Speed of proton[tex]v = 3.60\times10^{6}\ m/s[/tex]
Mass of proton[tex]m_{p}=1.67\times10^{-27}\ kg[/tex]
Charge[tex]q =1.60\times10^{-19}\ C[/tex]
When a proton moves horizontally, at a right angle to a magnetic field .
Then, the gravitational force balances the magnetic field
[tex]mg=Bqv\sin\theta[/tex]
[tex]B = \dfrac{mg}{qv}[/tex]
Here, [tex]\theta = 90^{\circ}[/tex]
Where, B = magnetic field
q = charge
v = speed
Put the value into the formula
[tex]B = \dfrac{1.67\times10^{-27}\times9.8}{1.60\times10^{-19}\times3.60\times10^{6}}[/tex]
[tex]B = 2.84\times10^{-14}\ T[/tex]
Hence, The magnetic field strength is required [tex] 2.84\times10^{-14}\ T[/tex]
Final answer:
To balance the gravitational force on a proton moving at a right angle through a magnetic field, the required field strength is found using the magnetic force formula, resulting in a necessary field strength of 2.86 x 10⁻²³ T.
Explanation:
The student is asking about the magnetic force required to counteract the gravitational force on a proton moving horizontally through a magnetic field. To solve this, we need to use the formula F = qvB, where F is the magnetic force, q is the charge of the proton, v is the velocity of the proton, and B is the magnetic field strength we wish to find.
First, we calculate the weight of the proton using W = mg, where m is the mass of the proton and g is the acceleration due to gravity (9.81 m/s²). This weight is the force we aim to balance with the magnetic force.
Now, let's calculate the weight of the proton: W = (1.67 × 10⁻²⁷ kg) × (9.81 m/s²) = 1.64 × 10⁻²⁶ N.
To keep the proton moving horizontally, the magnetic force needs to equal the proton's weight. So we set F to W and solve the equation for B:
B = W/(qv) = (1.64 × 10⁻²⁶ N) / ((1.60 × 10⁻²⁹ C) × (3.60 × 10⁶ m/s))
B = 2.86 × 10⁻² T
Therefore, a magnetic field strength of 2.86 × 10⁻² Tesla is required to just balance the weight of the proton and keep it moving horizontally.
A sound wave from a siren has an intensity of 100.0 point, what the sound intensity level of the sound?
Answer:
Sound intensity level of the sound = 140 dB
Explanation:
We have expression for sound intensity level
[tex]L=10log_{10}\left ( \frac{I}{I_0}\right )[/tex]
We have sound intensity, I = 100 W/m²
I₀ = 10⁻¹² W/m²
We need to find sound intensity level
Substituting
[tex]L=10log_{10}\left ( \frac{I}{I_0}\right )=10log_{10}\left ( \frac{100}{10^{-12}}\right )=10x14=140dB[/tex]
Sound intensity level of the sound = 140 dB
Electric field intensity, E can be defined as the electric (Coulomb) force on a unit test (1C) charge. Find the electric field intensity at a distance, d from the center of a +Q charged proton. use k as Coulombe constant.
Explanation:
Electric field intensity E is defined as the electric (Coulomb) force on a unit test (1 C) charge. Mathematically, it is given by :
[tex]E=\dfrac{F}{q}[/tex]
The electric force is given by :
[tex]F=k\dfrac{qQ}{d^2}[/tex]
Where
Q and q are electric charges
d is the distance between charges
The electric field intensity at a distance d from the center is given by :
[tex]E=\dfrac{k\dfrac{qQ}{d^2}}{q}[/tex]
So,
[tex]E=\dfrac{kQ}{d^2}[/tex]
Hence, this is the required solution.
An elevator is moving downward when someone presses the emergency stop button. The elevator comes to rest a short time later. Give the signs for the velocity and the acceleration of the elevator after the button has been pressed but before the elevator has stopped. Enter the correct sign for the elevator's velocity and the correct sign for the elevator's acceleration, separated by a comma. For example, if you think that the velocity is positive and the acceleration is negative, then you would enter ,- . If you think that both are zero, then you would enter 0,0 .
Answer:
Velocity: +ve, Acceleration: -ve
Explanation:
Here I've considered downward direction as positive direction.
Answer:
The answer is -,+ that is minus, plus
Explanation:
In the question, the elevator was described as moving downward, therefore its direction is negative. (-)
From the question we could tell the elevator is decelerating, so the acceleration vector should be pointing upward, in contrast with the motion of the elevator.(+)
VELOCITYVelocity is a vector quantity that indicates how fast an object is moving and in what direction, it has to do with and object’s displacement, time, and direction. The SI unit of velocity is meter per second (m/s). It should not be confused with speed which is a scalar quantity and measures on distance moved without stating what direction it moves.
For instance, it would not be enough to say that the car has a velocity of 50 miles/hr. the direction in which the car moves must be included to fully describe the velocity of the car. The correct way would be the car has a velocity of 50 miles/hour East.
ACCELERATIONIn physics, acceleration is defined as the rate of change of velocity. By altering an object’s speed or direction which changes its velocity hence its acceleration. Just like velocity, acceleration is a vector quantity. The SI unit of acceleration is meter per second squared (m/s^2)
Determine the energy required to accelerate an electron between each of the following speeds. (a) 0.500c to 0.900c MeV (b) 0.900c to 0.942c MeV
Answer:
The energy required to accelerate an electron is 0.582 Mev and 0.350 Mev.
Explanation:
We know that,
Mass of electron [tex]m_{e}=9.11\times10^{-31}\ kg[/tex]
Rest mass energy for electron = 0.511 Mev
(a). The energy required to accelerate an electron from 0.500c to 0.900c Mev
Using formula of rest,
[tex]E=\dfrac{E_{0}}{\sqrt{1-\dfrac{v_{f}^2}{c^2}}}-\dfrac{E_{0}}{\sqrt{1-\dfrac{v_{i}^2}{c^2}}}[/tex]
[tex]E=\dfrac{0.511}{\sqrt{1-\dfrac{(0.900c)^2}{c^2}}}-\dfrac{0.511}{\sqrt{1-\dfrac{(0.500c)^2}{c^2}}}[/tex]
[tex]E=0.582\ Mev[/tex]
(b). The energy required to accelerate an electron from 0.900c to 0.942c Mev
Using formula of rest,
[tex]E=\dfrac{E_{0}}{\sqrt{1-\dfrac{v_{f}^2}{c^2}}}-\dfrac{E_{0}}{\sqrt{1-\dfrac{v_{i}^2}{c^2}}}[/tex]
[tex]E=\dfrac{0.511}{\sqrt{1-\dfrac{(0.942c)^2}{c^2}}}-\dfrac{0.511}{\sqrt{1-\dfrac{(0.900c)^2}{c^2}}}[/tex]
[tex]E=0.350\ Mev[/tex]
Hence, The energy required to accelerate an electron is 0.582 Mev and 0.350 Mev.
A car starts from rest and speeds up at a constant rate of 2.03 m/s^2 in a straight line until it reaches a speed of 21.1 m/s. The car then slows down at half the rate that it sped up until it stops. How far (in m) does the car move from start to stop?
Answer:
The car moves from start to stop 328.88 m in total.
Explanation:
Vo= 0 m/s
V1= 21.1 m/s
V2= 0 m/s
a1= 2,03 m/s²
a2= -1.015 m/s²
Speed Up:
Speed up time:
V1= Vo + a1 * t1
t1= V1/a1
t1= 10.39 sec
total distance of speed up:
d1= Vo * t1 + (a1 * t1²)/2
d1= 109.57m
Slow Down:
V2= V1 - a2 * t2
t2= V1/a2
t2= 20.78 sec
total distance of slow down:
d2= V1 * t2 - (a2 * t2²)/2
d1= 219.31m
Total Distance:
TD= d1+d2= 109.57m + 219.31m
TD= 328.88 m
A person pushes a 16.0 kg lawn mower at constant speed with a force of 87.5 N directed along the handle, which is at an angle of 45.0 to the horizontal. Part D friction force? What force must the person exert on the lawn mower to accelerate it from rest to 1.3 m/s in 2.5 seconds, assuming the same Express your answer to three significant figures and Include the appropriate units
The person must exert a force of 8.32 N to accelerate the lawn mower from rest to 1.3 m/s in 2.5 seconds.
When an object is moving at a constant speed, the net force acting on it is zero.
In this case, the force applied by the person along the handle is balanced by the frictional force opposing the motion.
So, the frictional force [tex](\(F_{\text{friction}}\))[/tex] is equal in magnitude but opposite in direction to the force applied by the person.
Given:
- Force applied by the person = 87.5 N
- The angle of the handle = 45.0 degrees to the horizontal
To find the horizontal component of the force applied by the person, use trigonometry:
[tex]\[F_{\text{person-horizontal}} = F_{\text{person}} \cdot \cos(\theta)\][/tex]
[tex]\[F_{\text{person-horizontal}} = 87.5 \, \text{N} \cdot \cos(45.0^\circ)\][/tex]
[tex]= 87.5 \, \text{N} \cdot 0.707 = 61.29 \, \text{N}[/tex]
So, the person exerts a horizontal force of 61.29 N.
Since the lawn mower moves at a constant speed, the frictional force must be equal in magnitude and opposite in direction to this horizontal force:
[tex]\[F_{\text{friction}} = -61.29 \, \text{N}\][/tex]
To calculate this force [tex](\(F_{\text{acceleration}}\))[/tex], we can use Newton's second law:
[tex]\[F = m \cdot a\][/tex]
Given:
Mass of the lawn mower = 16.0 kg
Acceleration = [tex]\(\frac{\Delta v}{\Delta t}\)[/tex],
Now, calculate [tex]\(\Delta v\):[/tex]
[tex]\[\Delta v = 1.3 \, \text{m/s} - 0 \, \text{m/s} = 1.3 \, \text{m/s}\][/tex]
Now, using Newton's second law:
[tex]\[F_{\text{acceleration}} = m \cdot a = (16.0 \, \text{kg}) \cdot \left(\frac{1.3 \, \text{m/s}}{2.5 \, \text{s}}\right)\][/tex]
[tex]\[F_{\text{acceleration}} = 8.32 \, \text{N}\][/tex]
Thus, the person must exert a force of 8.32 N.
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To accelerate a 16.0 kg lawn mower from rest to 1.3 m/s in 2.5 seconds, a person needs to exert a force of 136 N along the handle, assuming that the same friction force is present as when the mower is pushed at constant speed.
Explanation:In order to find the force that the person must exert on the lawn mower to accelerate it from rest to 1.3 m/s in 2.5 seconds, we can use the second law of motion that states that the net force acting on an object is equal to its mass times its acceleration: F = m*a.
In this case, the acceleration would be the change in speed over time, which is (1.3 m/s - 0 m/s) / 2.5 s = 0.52 m/s². Consequently, F = (16.0 kg) * (0.52 m/s²) = 8.3 N.
It's important to mention this force is in addition to the friction force and the force to move the mower horizontally. Asume that the 87.5 N force previously exerted was just overcoming friction, then the total horizontal force required would be 87.5 N (for friction) + 8.3 N (for acceleration) = 95.8 N. However, since this force makes an angle of 45.0 with the horizontal, the person must exert a larger force along the handle, about 135.5 N. Hence, the correct answer, expressed to three significant figures, is 136 N.
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