Answer:
The terminal velocity at sea level is 7.99 m/s
The terminal velocity at an altitude of 5000 m is 10.298 m/s
Explanation:
mass of sphere m = 10 kg
radius of sphere r = 0.5 m
air density at sea level p = 1.22 kg/m^3
drag coefficient Cd = 0.8
terminal velocity = ?
Area of the sphere A = [tex]4\pi r^{2}[/tex] = 4 x 3.142 x [tex]0.5^{2}[/tex] = 3.142 m^2
terminal velocity is gotten from the relationship
[tex]Vt = \sqrt{\frac{2mg}{pACd} }[/tex]
where g = acceleration due to gravity = 9.81 m/s^2
imputing values into the equation
[tex]Vt = \sqrt{\frac{2*10*9.81}{1.22*3.142*0.8} }[/tex] = 7.99 m/s
If at an altitude of 5000 m where air density = 0.736 kg/m^3, then we replace value of air density in the relationship as 0.736 kg/m^3
[tex]Vt = \sqrt{\frac{2mg}{pACd} }[/tex]
[tex]Vt = \sqrt{\frac{2*10*9.81}{0.736*3.142*0.8} }[/tex] = 10.298 m/s
The value of terminal velocity of object is 7.99 m/s and the value of terminal velocity at an altitude of 5000 m is 10.30 m/s.
Given data:
The mass of sphere is, m = 10 kg.
The radius of sphere is, r = 0.5 m.
The density of air is, [tex]\rho = 1.22 \;\rm kg/m^{3}[/tex].
The drag coefficient of object is, [tex]C_{d}=0.8[/tex].
The altitude is, h = 5000 m.
The density of air at altitude is, [tex]\rho' =0.736 \;\rm kg/m^{3}[/tex].
The mathematical expression for the terminal velocity of an object is,
[tex]v_{t}=\sqrt\dfrac{2mg}{\rho \times A \times C_{d}}[/tex]
here,
g is the gravitational acceleration.
A is the area of sphere.
Solving as,
[tex]v_{t}=\sqrt{\dfrac{2 \times 10 \times 9.8}{1.22 \times (4 \pi r^{2}) \times C_{d}}}\\\\\\v_{t}=\sqrt{\dfrac{2 \times 10 \times 9.8}{1.22 \times (4 \pi \times 0.5^{2}) \times 0.8}}\\\\\\\v_{t}=7.99 \;\rm m/s[/tex]
Now, the terminal velocity at the altitude of 5000 m is given as,
[tex]v_{t}'=\sqrt\dfrac{2mg}{\rho' \times A \times C_{d}}[/tex]
Solving as,
[tex]v_{t}'=\sqrt{\dfrac{2 \times 10 \times 9.8}{0.736 \times (4 \pi \times 0.5^{2}) \times 0.8}}\\\\\\v_{t}'=10.30 \;\rm m/s[/tex]
Thus, we can conclude that the value of terminal velocity of object is 7.99 m/s and the value of terminal velocity at an altitude of 5000 m is 10.30 m/s.
Learn more about the terminal velocity here:
https://brainly.com/question/2654450
1. In a Millikan type experiment, two horizontal plates are 2.5 cm apart. A latex sphere of
mass 1.5 x 10-15 kg remains stationary when the potential difference between the
plates is 460 V, with the upper plate positive. [2+2+2+2 = 8 marks]
a. Is the sphere charged negatively or positively?
b. What is the magnitude of the electric field intensity between the plates?
C. Calculate the magnitude of the charge on the latex sphere.
d. How many excess or deficit electrons does the sphere have?
Answer:
Explanation:
a. Is the sphere charged negatively or positively?
The sphere us negatively charged. In a Millikan type experiment, there will be two forces that will be acting on the sphere which are the electric force which acts upward and also the gravity which acts downward.
Because the upper plate is positively charged, there'll what an attractive curve with an upward direction which will be felt by the negatively charged sphere.
b. What is the magnitude of the electric field intensity between the plates?
The magnitude of the electric field intensity between the plates is 18400v/m.
C. Calculate the magnitude of the charge on the latex sphere.
The magnitude of the charge on the latex sphere hae been solved and attached
d. How many excess or deficit electrons does the sphere have?
There are 5 excess electrons that the sphere has.
Check the attachment for further explanation.
Monochromatic coherent light shines through a pair of slits. If the wavelength of the light is decreased, which of the following statements are true of the resulting interference pattern? (There could be more than one correct choice.)
a. The distance between the maxima decreases.
b. The distance between the minima decreases.
c. The distance between the maxima stays the same.
d. The distance between the minima increases.
e. The distance between the minima stays the same.
Answer:
he correct answers are a, b
Explanation:
In the two-slit interference phenomenon, the expression for interference is
d sin θ= m λ constructive interference
d sin θ = (m + ½) λ destructive interference
in general this phenomenon occurs for small angles, for which we can write
tanθ = y / L
tan te = sin tea / cos tea = sin tea
sin θ = y / La
un
derestimate the first two equations.
Let's do the calculation for constructive interference
d y / L = m λ
the distance between maximum clos is and
y = (me / d) λ
this is the position of each maximum, the distance between two consecutive maximums
y₂-y₁ = (L 2/d) λ - (L 1 / d) λ₁ y₂ -y₁ = L / d λ
examining this equation if the wavelength decreases the value of y also decreases
the same calculation for destructive interference
d y / L = (m + ½) κ
y = [(m + ½) L / d] λ
again when it decreases the decrease the distance
the correct answers are a, b
Using energy considerations and assuming negligible air resistance, show that a rock thrown from a bridge 25.0 m above water with an initial speed of 20.0 m/s strikes the water with a final speed of what, independent of the direction thrown.
Complete question is;
Using energy considerations and assuming negligible air resistance, show that a rock thrown from a bridge 25.0 m above water with an initial speed of 20.0 m/s strikes the water with a final speed of 31.1 m/s, independent of the direction thrown
Answer:
It is proved that the final speed is truly 31.1 m/s
Explanation:
From energy - conservation principle;
E_i = Initial potential energy + Initial Kinetic Energy
Or
E_i = U_i + K_i
Similarly, for final energy
E_f = U_f + K_f
So, expressing the formulas for potential and kinetic energies, we now have;
E_i = (m × g × y_i) + (½ × m × v_i²)
Similarly,
E_f = (m × g × y_f) + (½ × m × v_f²)
We are given;
y_i = 25 m
y_f = 0 m
v_i = 20 m/s
v_f = 31.1 m/s
So, plugging in relevant values;
E_i = m((9.8 × 25) + (½ × 20²))
E_i = 485m
Similarly;
E_f = m((9.8 × 0) + (½ × v_f²)
E_f ≈ ½m•v_f²
From energy conservation principle, E_i = E_f.
Thus;
485m = ½m•v_f²
m will cancel out to give;
½v_f² = 485
v_f² = 485 × 2
v_f² = 970
v_f = √970
v_f ≈ 31.1 m/s
A package is dropped from a helicopter moving upward at 1.5 m/s. If it takes 16.0 s before the package strikes the ground, how high above the ground was the package when it was released if air resistance is negligible?
Two beams of coherent light start out at the same point in phase and travel different paths to arrive at point P. If the maximum destructive interference is to occur at point P, the two beams must travel paths that differ by
Answer:
the two beams must travel paths that differ by one-half of a wavelength.
An 88.0 kg spacewalking astronaut pushes off a 645 kg satellite, exerting a 110 N force for the 0.450 s it takes him to straighten his arms. How far apart are the astronaut and the satellite after 1.40 min?
Answer:
The astronaut and the satellite are 53.718 m apart.
Explanation:
Given;
mass of spacewalking astronaut, = 88 kg
mass of satellite, = 645 kg
force exerts by the satellite, F = 110N
time for this action, t = 0.45 s
Determine the acceleration of the satellite after the push
F = ma
a = F / m
a = 110 / 645
a = 0.171 m/s²
Determine the final velocity of the satellite;
v = u + at
where;
u is the initial velocity of the satellite = 0
v = 0 + 0.171 x 0.45
v = 0.077 m/s
Determine the displacement of the satellite after 1.4 m
d₁ = vt
d₁ = 0.077 x (1.4 x 60)
d₁ = 6.468 m
According to Newton's third law of motion, action and reaction are equal and opposite;
Determine the backward acceleration of the astronaut after the push;
F = ma
a = F / m
a = 110 / 88
a = 1.25 m/s²
Determine the final velocity of the astronaut
v = u + at
The initial velocity of the astronaut = 0
v = 1.25 x 0.45
v = 0.5625 m/s
Determine the displacement of the astronaut after 1.4 min
d₂ = vt
d₂ = 0.5625 x (1.4 x 60)
d₂ = 47.25 m
Finally, determine the total separation between the astronaut and the satellite;
total separation = d₁ + d₂
total separation = 6.468 m + 47.25 m
total separation = 53.718 m
Therefore, the astronaut and the satellite are 53.718 m apart.
A gun has a muzzle speed of 90 meters per second. What angle of elevation should be used to hit an object 150 meters away? Neglect air resistance and use g=9.8m/sec2 as the acceleration of gravity.
Answer:
θ₀ = 84.78° (OR) 5.22°
Explanation:
This situation can be treated as projectile motion. The parameters of this projectile motion are:
R = Range of Projectile = 150 m
V₀ = Launch Speed of Projectile = 90 m/s
g = 9.8 m/s²
θ₀ = Launch angle (OR) Angle of Elevation = ?
The formula for range of a projectile is given as:
R = V₀² Sin 2θ₀/g
Sin 2θ₀ = Rg/V₀²
Sin 2θ₀ = (150 m)(9.8 m/s²)/(90 m/s)²
2θ₀ = Sin⁻¹ (0.18)
θ₀ = 10.45°/2
θ₀ = 5.22°
Also, we know that for the same launch velocity the range will be same for complementary angles. Therefore, another possible value of angle is:
θ₀ = 90° - 5.22°
θ₀ = 84.78°
How does an atom of rubidium-85 become a rubidium ion with a +1 charge?
Answer:
C. The atom loses 1 electron to have a total of 36.
Explanation:
Cations have a positive charge. Cations lose electrons.
The number of electrons in a Rubidium atom is 37. If the atom loses 1 electron, then it has 36 left.
A hard drive disk rotates at 7200 rpm. The disk has a diameter of 5.1 in (13 cm). What is the speed of a point 6.0 cm from the center axle
Answer:
The speed will be "3.4×10⁴ m/s²".
Explanation:
The given values are:
Angular speed,
w = 7200 rpm
i.e.,
= [tex]7200 \times \frac{2 \pi}{60}[/tex]
= [tex]753.6 \ rad/s[/tex]
Speed from the center,
r = 6.0 cm
As we know,
⇒ Linear speed, [tex]v=wr[/tex]
On putting the estimated values, we get
[tex]=753.6\times 0.06[/tex]
[tex]=45.216 \ m[/tex]
Now,
Acceleration on disk will be:
⇒ [tex]a=\frac{v^2}{r}[/tex]
[tex]=34074 \ m/s^2[/tex]
[tex]=3.4\times 10^4 \ m/s^2[/tex]
An erect object is placed on the central axis of a thin lens, further from the lens than the magnitude of its focal length. The magnification is +0.4. This means:
A) the image is real and erect and the lens is a converging lens
B) the image is real and inverted and the lens is a converging lens
C) the image is virtual and erect, and the lens is a diverging lens
D) the image is virtual and erect, and the lens is a converging lens
E) the image is virtual and inverted and the lens is a diverging lens
Answer:
the image is virtual and erect and the lens divergent; therefore the correct answer is C
Explanation:
In a thin lens the magnification given by
m = h '/ h = - q / p
where h ’is the height of the image, h is the height of the object, q is the distance to the image and p is the distance to the object.
It indicates that the object is straight and is placed at a distance p> f
analyze the situation tells us that the magnification is positive so the distance to the image must be negative, that is, that the image is on the same side as the object.
Consequently the lens must be divergent
The magnification value is
0.4 = h ’/ h
h ’= 0.4 h
therefore the erect images
therefore the image is virtual and erect and the lens divergent; therefore the correct answer is C
A simple pendulum of length 1.62 m has a mass of 117 g attached. It is drawn back 38.0 degrees and then released. What is the maximum speed of the mass
Answer:
The maximum speed of the mass is 4.437 m/s.
Explanation:
Given;
length of pendulum, L = 1.62 m
attached mass, m = 117 g
angle of inclination, θ = 38°
This mass was raised to a height of
h = 1.62 - cos38° = 1.0043 m
Apply the principle of conservation of mechanical energy
PE = KE
mgh = ¹/₂mv²
v = √(2gh)
v = √(2 * 9.8 * 1.0043)
v = 4.437 m/s.
Therefore, the maximum speed of the mass is 4.437 m/s.
What is the transmitted intensity of light if an additional polarizer is added perpendicular to the first polarizer in the setup described in Question 3
Answer:
3) Transmitted intensity of light if unpolarized light passes through a single polarizing filter = 40 W/m²
- Transmitted intensity of light if an additional polarizer is added perpendicular to the first polarizer in the setup described = 7.5 W/m²
Explanation:
Complete Question
3) What is the transmitted intensity of light if unpolarized light passes through a single polarizing filter and the initial intensity is 80 W/m²?
- What is the transmitted intensity of light if an additional polarizer is added perpendicular to the first polarizer in the setup described in Question 3 (the setup)? Show all work in your answer.
The image of this setup attached to this question as obtained from online is attached to this solution.
Solution
3) When unpolarized light passes through a single polarizer, the intensity of the light is cut in half.
Hence, if the initial intensity of unpolarized light is I₀ = 80 W/m²
The intensity of the light rays thay pass through the first single polarizer = I₁ = (I₀/2) = (80/2) = 40 W/m²
- According to Malus' law, the intensity of transmitted light through a polarizer is related to the intensity of the incident light and the angle at which the polarizer is placed with respect to the major axis of the polarizer before the current polarizer of concern.
I₂ = I₁ cos² θ
where
I₂ = intensity of light that passes through the second polarizer = ?
I₁ = Intensity of light from the first polarizer that is incident upon the second polarizer = 40 W/m²
θ = angle between the major axis of the first and second polarizer = 30°
I₂ = 40 (cos² 30°) = 40 (0.8660)² = 30 W/m²
In the same vein, the intensity of light that passes through the third/additional polarizer is related to the intensity of light that passes through the second polarizer and is incident upon this third/additional polarizer through
I₃ = I₂ cos² θ
I₃ = intensity of light that passes through the third/additional polarizer = ?
I₂ = Intensity of light from the second polarizer that is incident upon the third/additional polarizer = 30 W/m²
θ = angle between the major axis of the second and third/additional polarizer = 60° (although, it is 90° with respect to the first polarizer, it is the angle it makes with the major axis of the second polarizer, 60°, that matters)
I₃ = 30 (cos² 60°) = 30 (0.5)² = 7.5 W/m²
Hope this Helps!!!
The magnitude of the magnetic flux through the surface of a circular plate is 6.80 10-5 T · m2 when it is placed in a region of uniform magnetic field that is oriented at 43.0° to the vertical. The radius of the plate is 8.50 cm. Determine the strength of the magnetic field. mT A circular plate of radius r is lying flat. A field of arrows labeled vector B rising up and to the right pass through the plate.
Answer:
B = 4.1*10^-3 T = 4.1mT
Explanation:
In order to calculate the strength of the magnetic field, you use the following formula for the magnetic flux trough a surface:
[tex]\Phi_B=S\cdot B=SBcos\alpha[/tex] (1)
ФB: magnetic flux trough the circular surface = 6.80*10^-5 T.m^2
S: surface area of the circular plate = π.r^2
r: radius of the circular plate = 8.50cm = 0.085m
B: magnitude of the magnetic field = ?
α: angle between the direction of the magnetic field and the normal to the surface area of the circular plate = 43.0°
You solve the equation (1) for B, and replace the values of the other parameters:
[tex]B=\frac{\Phi_B}{Scos\alpha}=\frac{6.80*10^{-5}T.m^2}{(\pi (0.085m)^2)cos(43.0\°)}\\\\B=4.1*10^{-3}T=4.1mT[/tex]
The strength of the magntetic field is 4.1mT
How far apart (in mm) must two point charges of 90.0 nC (typical of static electricity) be to have a force of 3.80 N between them
Answer:
The distance between the two charges is =4.4mm
Please Help!!!! I WILL GIVE BRAINLIEST!!!!!!!!!!!!!
Upon using Thomas Young’s double-slit experiment to obtain measurements, the following data were obtained. Use these data to determine the wavelength of light being used to create the interference pattern. Do this using three different methods.
The angle to the eighth maximum is 1.12°.
The distance from the slits to the screen is 302.0 cm.
The distance from the central maximum to the fifth minimum is 3.33 cm.
The distance between the slits is 0.000250 m.
The 3 equations I used were 1). d sin θ_m =(m)λ 2). delta x =λL/d and 3.) d(x_n)/L=(n-1/2)λ
but all my answers are different.
DID I DO SOMETHING WRONG!!!!!!!
Given info
d = 0.000250 meters = distance between slits
L = 302 cm = 0.302 meters = distance from slits to screen
[tex]\theta_8 = 1.12^{\circ}[/tex] = angle to 8th max (note how m = 8 since we're comparing this to the form [tex]\theta_m[/tex])
[tex]x_n = x_5 = 3.33 \text{ cm} = 0.0333 \text{ meters}[/tex] (n = 5 as we're dealing with the 5th minimum )
---------------
Method 1
[tex]d\sin(\theta_m) = m\lambda\\\\0.000250\sin(\theta_8) = 8\lambda\\\\8\lambda = 0.000250\sin(1.12^{\circ})\\\\\lambda = \frac{0.000250\sin(1.12^{\circ})}{8}\\\\\lambda \approx 0.000 000 61082633\\\\\lambda \approx 6.1082633 \times 10^{-7} \text{meters}\\\\ \lambda \approx 6.11 \times 10^{-7} \text{ meters}\\\\ \lambda \approx 611 \text{ nm}[/tex]
Make sure your calculator is in degree mode.
-----------------
Method 2
[tex]\Delta x = \frac{\lambda*L*m}{d}\\\\L*\tan(\theta_m) = \frac{\lambda*L*m}{d}\\\\\tan(\theta_m) = \frac{\lambda*m}{d}\\\\\tan(\theta_8) = \frac{\lambda*8}{0.000250}\\\\\tan(1.12^{\circ}) = \frac{\lambda*8}{0.000250}\\\\\lambda = \frac{1}{8}*0.000250*\tan(1.12^{\circ})\\\\\lambda \approx 0.00000061094306 \text{ meters}\\\\\lambda \approx 6.1094306 \times 10^{-7} \text{ meters}\\\\\lambda \approx 611 \text{ nm}\\\\[/tex]
-----------------
Method 3
[tex]\frac{d*x_n}{L} = \left(n-\frac{1}{2}\right)\lambda\\\\\frac{0.000250*3.33}{302.0} = \left(5-\frac{1}{2}\right)\lambda\\\\0.00000275662251 \approx \frac{9}{2}\lambda\\\\\frac{9}{2}\lambda \approx 0.00000275662251\\\\\lambda \approx \frac{2}{9}*0.00000275662251\\\\\lambda \approx 0.00000061258279 \text{ meters}\\\\\lambda \approx 6.1258279 \times 10^{-7} \text{ meters}\\\\\lambda \approx 6.13 \times 10^{-7} \text{ meters}\\\\\lambda \approx 613 \text{ nm}\\\\[/tex]
There is a slight discrepancy (the first two results were 611 nm while this is roughly 613 nm) which could be a result of rounding error, but I'm not entirely sure.
Two identical small charged spheres are a certain distance apart, and each one initially experiences an electrostatic force of magnitude F due to the other. With time, charge gradually leaks off of both spheres. When each of the spheres has lost half its initial charge, the magnitude of the electrostatic force will be
Answer:
F' = F/4
Thus, the magnitude of electrostatic force will become one-fourth.
Explanation:
The magnitude of force applied by each charge on one another can be given by Coulomb's Law:
F = kq₁q₂/r² -------------- equation 1
where,
F = Force applied by charges
k = Coulomb's Constant
q₁ = magnitude of first charge
q₂ = magnitude of 2nd charge
r = distance between the charges
Now, in the final state the charges on both spheres are halved. Therefore,
q₁' = q₁/2
q₂' = q₂/2
Hence, the new force will be:
F' = kq₁'q₂'/r²
F' = k(q₁/2)(q₂/2)/r²
F' = (kq₁q₂/r²)(1/4)
using equation 1:
F' = F/4
Thus, the magnitude of electrostatic force will become one-fourth.
The magnitude of the electrostatic force will be F' = F/4
The magnitude of the electrostatic force:Here we used Coulomb's Law:
F = kq₁q₂/r² -------------- equation 1
Here
F = Force applied by charges
k = Coulomb's Constant
q₁ = magnitude of first charge
q₂ = magnitude of 2nd charge
r = distance between the charges
Now
q₁' = q₁/2
q₂' = q₂/2
So, the new force should be
F' = kq₁'q₂'/r²
F' = k(q₁/2)(q₂/2)/r²
F' = (kq₁q₂/r²)(1/4)
So,
F' = F/4
Learn more about force here: https://brainly.com/question/14282312
a car slows down from - 27.7 m/s to -10.9 m/s while undergoing a displacement of -105 m .what is its acceleration?
Answer:
3.09 m/s²
Explanation:
Given:
Δx = -105 m
v₀ = -27.7 m/s
v = -10.9 m/s
Find: a
v² = v₀² + 2aΔx
(-10.9 m/s)² = (-27.7 m/s)² + 2a (-105 m)
a = 3.09 m/s²
A projectile is launched with V0 = 7.6 m/s and initial angle = 1.27 radians above the horizontal. What is the initial horizontal component of the projectile velocity in miles per hour?
Answer:
The horizontal component is [tex]v_h = 1.7096 \ m/s[/tex]
Explanation:
A diagram illustrating the projection is shown on the first uploaded image (from IB Maths Resources from British international school Phuket )
From the question we are told that
The initial velocity is [tex]v_o = 7.6 \ m/s[/tex]
The angle of projection is [tex]\theta = 1.27 \ rad = 72.77^o[/tex]
The horizontal component of this projectile velocity is mathematically represented as
[tex]v_h = v_o * cos (\theta )[/tex]
substituting values
[tex]v_h = 7.6 * cos (72.77 )[/tex]
[tex]v_h = 1.7096 \ m/s[/tex]
Two 10-cm-diameter charged rings face each other, 25 cm apart. The left ring is charged to ? 25 nC and the right ring is charged to + 25 nC.A) What is the magnitude of the electric field E? at the midpoint between the two rings?B) What is the direction of the electric field E? at the midpoint between the two rings?C) What is the magnitude of the force on a proton at the midpoint?D) What is the direction of the force F? on a proton at the midpoint?
Answer:
A) E = 0N/C
B) 0i + 0^^j
C) F = 0N
D) 0^i + 0^j
Explanation:
You assume that the rings are in the zy plane but in different positions.
Furthermore, you can consider that the origin of coordinates is at the midway between the rings.
A) In order to calculate the magnitude of the electric field at the middle of the rings, you take into account that the electric field produced by each ring at the origin is opposite to each other and parallel to the x axis.
You use the following formula for the electric field produced by a charge ring at a perpendicular distance of r:
[tex]E=k\frac{rQ}{(r+R^2)^{3/2}}[/tex] (1)
k: Coulomb's constant = 8.98*10^9Nm^2/C
Q: charge of the ring
r: perpendicular distance to the center of the ring
R: radius of the ring
You use the equation (1) to calculate the net electric field at the midpoint between the rings:
[tex]E=k\frac{rQ}{(r^2+R^2)^{3/2}}-k\frac{rQ}{(r^2+R^2)^{3/2}}=0\frac{N}{C}[/tex]
The electric field produced by each ring has the same magnitude but opposite direction, then, the net electric field is zero.
B) The direction of the electric field is 0^i + 0^j
C) The magnitude of the force on a proton at the midpoint between the rings is:
[tex]F=qE=q(0N/C)=0N[/tex]
D) The direction of the force is 0^i + 0^j
Part A: The magnitude of the electric field generated at the midpoint between two rings is equal that is 0 N/C.
Part B: The direction of the electric field at the midpoint is opposite.
Part C: The magnitude of the force generated on a proton at the midpoint between two rings is equal that is 0 N.
Part D: The direction of the force on a proton at the midpoint is opposite.
Electric fieldAn electric field is defined as the region that surrounds electrically charged particles and exerts a force on all other charged particles within the region, either attracting or repelling them.
Given that diameter of the ring is 10 m and they are 25 m apart from each other. The charge on the left ring is -25nC and on the right ring is 25nC. The electric field can be given as below.
[tex]E = \dfrac {kQ}{(r+R)^2}\\ [/tex]
Where Q is the charge, r is the radius of the ring, R is the mid-point distance and k is the constant.
Part A
The electric field at the mid-point will be the sum of the electric field generated by both the rings. Substituting the values in the above equation,
[tex]E = \dfrac {8.9\times 10^9\times 25}{(10 +12.5)^2}+\dfrac {8.9\times 10^9\times (-25)}{(10 +12.5)^2}[/tex]
[tex]E = \dfrac {222.5\times 10^9}{506.25} - \dfrac {222.5\times 10^9}{506.25}\\ [/tex]
[tex]E = 0\;\rm N/C[/tex]
Hence we can conclude that both the rings generate the electric field with the same magnitude but they are opposite in direction.
Part B
The electric field at the mid-point is 0 N/C. In the vector form, the electric field can be given as below.
[tex]E = 0i+0j[/tex]
The vector form shows that the electric field at the mid-point between the two rings has the same magnitude but is opposite in direction.
Part C
The force can be given as below.
[tex]F = qE[/tex]
[tex]F = 0 \;\rm N[/tex]
If the electric field at the mid-point is zero, then the force at the mid-point will be zero.
Part D
The vector form of the force at the midpoint is given below.
[tex]F = 0i+0j[/tex]
Hence we can conclude that at the midpoint of two rings, the electric field generates an equal force on the proton but in opposite direction. Hence the net force will be zero.
To know more about the electric field, follow the link given below.
https://brainly.com/question/4440057.
A very large sheet of a conductor carries a uniform charge density of on its surfaces. What is the electric field strength 3.00 mm outside the surface of the conductor?
Complete Question
A very large sheet of a conductor carries a uniform charge density of [tex]4.00\ pC/mm^2[/tex] on its surfaces. What is the electric field strength 3.00 mm outside the surface of the conductor?
Answer:
The electric field is [tex]E = 4.5198 *10^{5} \ N/C[/tex]
Explanation:
From the question we are told that
The charge density is [tex]\sigma = 4.00pC /mm^2 = 4.00 * 10^{-12 } * 10^{6} = 4.00 *10^{-6}C/m[/tex]
The position outside the surface is [tex]a = 3.00 \ mm = 0.003 \ m[/tex]
Generally the electric field is mathematically represented as
[tex]E = \frac{\sigma}{\epsilon _o }[/tex]
Where [tex]\epsilon_o[/tex] is the permitivity of free space with values [tex]\epsilon _o = 8.85 *10^{-12} F/m[/tex]
substituting values
[tex]E = \frac{4.0*10^{-6}}{8.85 *10^{-12} }[/tex]
[tex]E = 4.5198 *10^{5} \ N/C[/tex]
A commuter train passes a passenger platform at a constant speed of 40.0 m/s. The train horn is sounded at its characteristic frequency of 320 Hz.
(a) What overall change in requency is detected by a person on the platform as the train moves from approaching to receding?
(b) What wavelength is detected by a person on the platform as the train approaches?
Answer:
a) -75 Hz
b) 0.11 [tex]m^{-1}[/tex]
Explanation:
a) Let us first find the frequency detected by the person on the platform.
We have to find the frequency observed by the person when the train was approaching and when the train was receding.
When the train was approaching:
[tex]f_o = \frac{v}{v - v_s} f_s[/tex]
where fo = frequency observed
fs = frequency from the source = 320 Hz
v = speed of sound = 343 m/s
vs = speed of the train = 40 m/s
Therefore:
[tex]f_o = \frac{343}{343 - 40} * 320\\\\f_o = \frac{343}{303} * 320\\\\f_o = 362 Hz[/tex]
The person on the platform heard the sound at a frequency of 362 Hz when the train was approaching.
When the train was receding:
[tex]f_o = \frac{v}{v + v_s} f_s[/tex]
[tex]f_o = \frac{343}{343 + 40} * 320\\\\f_o = \frac{343}{383} * 320\\\\f_o = 287 Hz[/tex]
The person on the platform heard the sound at a frequency of 287 Hz when the train was receding.
Therefore, the frequency change is given as:
Δf = 287 - 362 = -75 Hz
b) We can find the wavelength detected by the person on the platform as the train approaches by using the formula for speed:
[tex]v = \lambda f[/tex]
where λ = wavelength
f = frequency of the train as it approaches = 362 Hz
v = speed of train = 40 m/s
Therefore, the wavelength detected is:
40 = λ * 362
λ = 40 / 362 = 0.11 [tex]m^{-1}[/tex]
mention two similarities of citizen and aliens
Answer:
The main points of difference between a citizen and alien are: (a) A citizen is a permanent resident of a state, while an alien is a temporary resident, who comes for a specific duration of time as a tourist or on diplomatic assignment. ... Aliens do not possess such rights in the state where they reside temporarily
Explanation:
A 5.0-Ω resistor and a 9.0-Ω resistor are connected in parallel. A 4.0-Ω resistor is then connected in series with this parallel combination. An ideal 6.0-V battery is then connected across the series-parallel combination of the three resistors. What is the current through (a) the 4.0-Ω resistor? (b) the 5.0-Ω resistor? (c) the 9.0-Ω resistor?
Answer:
Explanation:
The current through the resistor is 0.83 A
.
Part b
The current through resistor is 0.53 A
.
Part c
The current through resistor is 0.30 A
A commercial diffraction grating has 500 lines per mm. Part A When a student shines a 480 nm laser through this grating, how many bright spots could be seen on a screen behind the grating
Answer:
The number of bright spot is m =4
Explanation:
From the question we are told that
The number of lines is [tex]s = 500 \ lines / mm = 500 \ lines / 10^{-3} m[/tex]
The wavelength of the laser is [tex]\lambda = 480 nm = 480 *10^{-9} \ m[/tex]
Now the the slit is mathematically evaluated as
[tex]d = \frac{1}{s} = \frac{1}{500} * 10^{-3} \ m[/tex]
Generally the diffraction grating is mathematically represented as
[tex]dsin\theta = m \lambda[/tex]
Here m is the order of fringes (bright fringes) and at maximum m [tex]\theta = 90^o[/tex]
So
[tex]\frac{1}{500} * sin (90) = m * (480 *10^{-3})[/tex]
=> [tex]m = 4[/tex]
This implies that the number of bright spot is m =4
1. The uniform purely axial magnetic induction required by the experiment in a volume large enough to accommodate the Lorentz Tube is produced by the Helmholtz Coils. What is the magnetic induction due to a coil current 1.5 Ampere
Complete Question
The uniform purely axial magnetic induction required by the experiment in a volume large enough to accommodate the Lorentz Tube is produced by the Helmholtz Coils. What is the magnetic induction due to a coil current 1.5 Ampere? Convert the result in the still popular non-SI unit Gauss (1 Tesla = 10^4 Gauss).
B = N*mue*I/(2*r)
# of loops = 140
radius of the coil = 0.14m
Answer:
The magnetic induction is [tex]B = 2.639 \ Gauss[/tex]
Explanation:
From the question we are told that
The coil current is [tex]I = 1.5 \ A[/tex]
The number of loops is [tex]N = 140[/tex]
The magnetic field due to the current is mathematically represented as
[tex]B = \mu_o * N * I[/tex]
[tex]\mu_o[/tex] is the permeability of free space with value [tex]\mu_o = 4\pi * 10^{-7} N/A^2[/tex]
substituting value
[tex]B = 4\pi * 10^{-7} * 140 * 1.5[/tex]
[tex]B = 2.639*19^{-4} \ T[/tex]
From question
(1 Tesla = [tex]10^4 \ Gauss[/tex]).
=> [tex]B = 2.693 *10^{-4} *10^4 = 2.63 \ Gauss[/tex]
=> [tex]B = 2.639 \ Gauss[/tex]
You’re driving down the highway late one night at 20 m/s when a deer steps onto the road 35 m in front of you. Your reaction time before stepping on the brakes is 0.50 s, and the maximum deceleration of your car is 10 m/s2.
a. How much distance is between you and the deer when you come to a stop?
b. What is the maximum speed you could have and still not hit the deer?
Answer:
(a) Distance between deer and car = 5 m
(b) Vmax = 21.92 m/s
Explanation:
a.
First we calculate distance covered during response time:
s₁ = vt --------- equation 1
where,
s₁ = distance covered during response time = ?
v = speed of car = 20 m/s
t = response time = 0.5 s
Therefore,
s₁ = (20 m/s)(0.5 s)
s₁ = 10 m
Now, we calculate the distance covered by the car during deceleration. Using 3rd equation of motion:
2as₂ = Vf² - Vi²
s₂ = (Vf² - Vi²)/2a ------ eqation 2
where,
a = deceleration = - 10 m/s²
s₂ = Distance covered during deceleration = ?
Vf = Final Velocity = 0 m/s (since car finally stops)
Vi = Initial Velocity = 20 m/s
Therefore,
s₂ = [(0 m/s)² - (20 m/s)²]/2(-10 m/s²)
s₂ = (400 m²/s²)/(20 m/s²)
s₂ = 20 m
thus, the total distance covered by the car before coming to rest is given as:
s = s₁ + s₂
s = 10 m + 20 m
s = 30 m
Now, the distance between deer and car, when it comes to rest, can be calculated as:
Distance between deer and car = 35 m - s = 35 m - 30 m
Distance between deer and car = 5 m
b.
Since, the distance covered by the car in total must be equal to 35 m at maximum. Therefore,
s₁ + s₂ = 35 m
using equation 1 and equation 2 from previous part:
Vi t + (Vf² - Vi²)/2a = 35 m
Vi(0.5 s) + [(0 m/s)² - Vi²]/2(-10 m/s²) = 35 m
0.5 Vi + 0.05 Vi² = 35
0.05 Vi² + 0.5 Vi - 35 = 0
solving this quadratic equation, we get:
Vi = - 31.92 m/s (OR) Vi = 21.92 m/s
For maximum velocity:
Vmax = 21.92 m/s
In the child's game of tetherball, a rope attached to the top of a tall pole is tied to a ball. Players hit the ball in opposite directions in an attempt to wrap the ball and rope around the pole. Assume the rope has negligible mass and that resistive forces, such as air resistance and friction, can be neglected. As the ball wraps around the pole between hits, how does the angular speed of the ball change
Answer:
The angular speed of the ball will increase
Explanation:
the angular speed of the ball will increase because the force of hit by the players will sum up in opposite direction to increase the angular speed
A 30 L electrical radiator containing heating oil is placed in a 50 m3room. Both the roomand the oil in the radiator are initially at 10◦C. The radiator with a rating of 1.8 kW is nowturned on. At the same time, heat is lost from the room at an average rate of 0.35 kJ/s.After some time, the average temperature is measured to be 20◦C for the air in the room,and 50◦C for the oil in the radiator. Taking the density and the specific heat of the oil to be950 kg/m3and 2.2 kJ/kg◦C, respectively, determine how long the heater is kept on. Assumethe room is well sealed so that there are no air leaks.
Answer:
Explanation:
Heat absorbed by oil
= mass x specific heat x rise in temperature
= 30 x 10⁻³ x 950 x 2.2 x 10³ x ( 50-10 )
= 25.08 x 10⁵ J
Heat absorbed by air
= 50 x 1.2 x 1.0054 x 10³ x ( 20-10 )
= 6.03 x 10⁵ J
Total heat absorbed = 31.11 x 10⁵ J
If time required = t
heat lost from room
= .35 x 10³ t
Total heat generated in time t
= 1.8 x 10³ t
Heat generated = heat used
1.8 x 10³ t = .35 x 10³ t + 31.11 x 10⁵
1.45 x 10³ t = 31.11 x 10⁵
t = 31.11 x 10⁵ / 1.45 x 10³
t = 2145.5 s
Which best describes the relationship between heat, internal energy, and thermal energy?
Internal energy is heat that flows, and heat is the part of thermal energy that can be transferred.
Internal energy is thermal energy that flows, and thermal energy is the part of heat that can be transferred.
Thermal energy is heat that flows, and heat is the part of internal energy that can be transferred.
Heat is thermal energy that flows, and thermal energy is the part of internal energy that can be transferred.
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Answer:
I think it is the 4th answer choice
Explanation:
Heat is thermal energy that flows in the direction of high temp to low temp, and internal energy is the "energy contained in a system", and thermal energy is a part of that.
Five identical cylinders are each acted on by forces of equal magnitude. Which force exerts the biggest torque about the central axes of the cylinders
Answer:
From the image, the force as shown in option A will exert the biggest torque on the cylinder about its central axes.
Explanation:
The image is shown below.
Torque is the product of a force about the center of rotation of a body, and the radius through which the force acts. For a given case such as this, in which the cylinders are identical, and the forces are of equal magnitude, the torque at the maximum radius away from the center will exert the maximum torque. Also, the direction of the force also matters. To generate the maximum torque, the force must be directed tangentially away from the circle formed by the radius through which the force acts away from the center. Option A satisfies both condition and hence will exert the most torque on the cylinder.