Answer:
Airbags reduce chances of injury by absorbing most of the impact force from the body during a car crash
Explanation:
In a car collision, the speed of the vehicle is suddenly bought to rest. All the kinetic energy is suddenly converted into other forms of energy.
The body of the driver keeps travelling forward under his inertia force due to his mass until he is slammed against the steering wheel. The steering wheel is a very rigid component, and so when the body slams against it, the body takes the deformation, absorbing some of the energy of the moving car. This sudden impact of energy can be fatal enough to gravely injure the driver because the body does not undergo much deformation. When an airbag is used, the crash automatically triggers the release of the airbag. Instead of the body colliding against the rigid steering wheel, it is now collided against the soft air bag. The airbag is very collapsible, and some of the kinetic energy of the car on the driver is converted into the deformation energy used to deform the airbag when they collide. In the process of deformation, the time of impact is extended, reducing the force impacted on the driver, reducing the fatality of the impact.
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
A wave with a frequency of 1200 Hz propagates along a wire that is under a tension of 800 N. Its wavelength is 39.1 cm. What will be the wavelength if the tension is decreased to 600 N and the frequency is kept constant
Answer:
The wavelength will be 33.9 cm
Explanation:
Given;
frequency of the wave, F = 1200 Hz
Tension on the wire, T = 800 N
wavelength, λ = 39.1 cm
[tex]F = \frac{ \sqrt{\frac{T}{\mu} }}{\lambda}[/tex]
Where;
F is the frequency of the wave
T is tension on the string
μ is mass per unit length of the string
λ is wavelength
[tex]\sqrt{\frac{T}{\mu} } = F \lambda\\\\\frac{T}{\mu} = F^2\lambda^2\\\\\mu = \frac{T}{F^2\lambda^2} \\\\\frac{T_1}{F^2\lambda _1^2} = \frac{T_2}{F^2\lambda _2^2} \\\\\frac{T_1}{\lambda _1^2} = \frac{T_2}{\lambda _2^2}\\\\T_1 \lambda _2^2 = T_2\lambda _1^2\\\\[/tex]
when the tension is decreased to 600 N, that is T₂ = 600 N
[tex]T_1 \lambda _2^2 = T_2\lambda _1^2\\\\\lambda _2^2 = \frac{T_2\lambda _1^2}{T_1} \\\\\lambda _2 = \sqrt{\frac{T_2\lambda _1^2}{T_1}} \\\\\lambda _2 = \sqrt{\frac{600* 0.391^2}{800}}\\\\\lambda _2 = \sqrt{0.11466} \\\\\lambda _2 =0.339 \ m\\\\\lambda _2 =33.9 \ cm[/tex]
Therefore, the wavelength will be 33.9 cm
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.
Two space ships collide in deep space. Spaceship P, the projectile, has a mass of 4M,
while the target spaceship T has a mass of M. Spaceship T is initially at rest and the
collision is elastic. If the final velocity of Tis 8.1 m/s, what was the initial velocity of
P?
Answer:
The initial velocity of spaceship P was u₁ = 5.06 m/s
Explanation:
In an elastic collision between two bodies the expression for the final velocity of the second body is given as follows:
[tex]V_{2} = \frac{(m_{2}-m_{1}) }{(m_{1}+m_{2})}u_{2} + \frac{2m_{1} }{(m_{1}+m_{2})}u_{1}[/tex]
Here, subscript 1 is used for spaceship P and subscript 2 is used for spaceship T. In this equation:
V₂ = Final Speed of Spaceship T = 8.1 m/s
m₁ = mass of spaceship P = 4 M
m₂ = mass of spaceship T = M
u₁ = Initial Speed of Spaceship P = ?
u₂ = Initial Speed of Spaceship T = 0 m/s
Using these values in the given equation, we get:
[tex]8.1 m/s = \frac{M-4M }{4M+M}(0 m/s) + \frac{2(4M) }{4M+M}u_{1}[/tex]
8.1 m/s = (8 M/5 M)u₁
u₁ = (5/8)(8.1 m/s)
u₁ = 5.06 m/s
Objects A and B are both positively charged. Both have a mass of 900 g, but A has twice the charge of B. When A and B are placed 30.0 cm apart, B experiences an electric force of 0.870 N.
How large is the force on A?
What is the charge on qA and qB?
If the objects are released, what is the initial acceleration of A?
Answer:
- Force on A = 0.870N
- charge of the object B = q = 2.1 μC
charge of the object A = 2q = 4.2 μC
- a = 0.966 m/s^2
Explanation:
- In order to determine the force on the object A, you take into account the third Newton law, which states that the force experienced by A has the same magnitude of the force experienced by B, but with an opposite direction.
Then, the force on A is 0.870N
- In order to calculate the charge of both objects, you use the following formula:
[tex]F_e=k\frac{q_Aq_B}{r^2}[/tex] (1)
k: Coulomb's constant = 8.98*10^9 Nm^2/C^2
r: distance between the objects = 30.0cm = 0.30m
A has twice the charge of B. If the charge of B is qB=q, then the charge of A is qA=2qB = 2q.
You replace the expression for qA and qB into the equation (1), solve for q, and replace the values of the parameters.
[tex]F_e=k\frac{(2q)(q)}{r^2}=2k\frac{q^2}{r^2}\\\\q=\sqrt{\frac{r^2Fe}{2k}}\\\\q=\sqrt{\frac{(0.30m)^2(0.870N)}{2(8.98*10^9Nm^2/C^2)}}=2.1*10^{-6}C\\\\q=2.1\mu C[/tex]
Then, you have:
charge of the object B = q = 2.1 μC
charge of the object A = 2q = 4.2 μC
- In order to calculate the acceleration of A, you use the second Newton law with the electric force, as follow:
[tex]F_e=ma\\\\a=\frac{F_e}{m}[/tex]
m: mass of the object A = 900g = 0.900kg
[tex]a=\frac{0.870N}{0.900kg}=0.966\frac{m}{s^2}[/tex]
The acceleration of A is 0.966m/s^2
A 3-liter container has a pressure of 4 atmospheres. The container is sent underground, with resulting compression into 2 L. Applying Boyle's Law, what will the new pressure be? choices: 2.3 atm 8 atm 6 atm 1.5 atm
Answer:
6 atm
Explanation:
PV = PV
(4 atm) (3 L) = P (2 L)
P = 6 atm
how do a proton and neutron compare?
Answer:
c.they have opposite charges.
Explanation:
because the protons have a positive charge and the neutrons have no charge.
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
An ice skater is in a fast spin with her arms held tightly to her body. When she extends her arms, which of the following statements in NOT true?
A. Het total angular momentum has decreased
B. She increases her moment of inertia
C. She decreases her angular speed
D. Her moment of inertia changes
Answer:
A. Her total angular momentum has decreased
Explanation:
Total angular momentum is the product of her moment of inertia and angular velocity. In this scenario it doesn’t decrease but rather remains constant as the movement of the arms doesn’t have any effect on the total angular momentum.
The movement of the arm under certain conditions however has varying effects and changes on parameters such as the moment of inertia and the angular speed.
An asteroid that has an orbit with a semi-major axis of 4 AU will have an orbital period of about ______ years.
Answer:
16 years.
Explanation:
Using Kepler's third Law.
P2=D^3
P=√d^3
Where P is the orbital period and d is the distance from the sun.
From the question the semi major axis of the asteroid is 4 AU= distance. The distance is always express in astronomical units.
P=?
P= √4^3
P= √256
P= 16 years.
Orbital period is 16 years.
An ice skater spinning with outstretched arms has an angular speed of 5.0 rad/s . She tucks in her arms, decreasing her moment of inertia by 11 % . By what factor does the skater's kinetic energy change? (Neglect any frictional effects.)
Answer:
K_{f} / K₀ =1.12
Explanation:
This problem must work using the conservation of angular momentum (L), so that the moment is conserved in the system all the forces must be internal and therefore the torque is internal and the moment is conserved.
Initial moment. With arms outstretched
L₀ = I₀ w₀
the wo value is 5.0 rad / s
final moment. After he shrugs his arms
[tex]L_{f}[/tex] = I_{f} w_{f}
indicate that the moment of inertia decreases by 11%
I_{f} = I₀ - 0.11 I₀ = 0.89 I₀
L_{f} = L₀
I_{f} w_{f} = I₀ w₀
w_{f} = I₀ /I_{f} w₀
let's calculate
w_{f} = I₀ / 0.89 I₀ 5.0
w_{f} = 5.62 rad / s
Having these values we can calculate the change in kinetic energy
[tex]K_{f}[/tex] / K₀ = ½ I_{f} w_{f}² (½ I₀ w₀²)
K_{f} / K₀ = 0.89 I₀ / I₀ (5.62 / 5)²
K_{f} / K₀ =1.12
A very long, solid cylinder with radius R has positive charge uniformly distributed throughout it, with charge per unit volume \rhorho.
(a) Derive the expression for the electric field inside the volume at a distance r from the axis of the cylinder in terms of the charge density \rhorho.
(b) What is the electric field at a point outside the volume in terms of the charge per unit length \lambdaλ in the cylinder?
(c) Compare the answers to parts (a) and (b) for r = R.
(d) Graph the electric-field magnitude as a function of r from r = 0 to r = 3R.
Answer:
the answers are provided in the attachments below
Explanation:
Gauss law state that the net electric field coming out of a closed surface is directly proportional to the charge enclosed inside the closed surface
Applying Gauss law to the long solid cylinder
A) E ( electric field ) = p*r / 2 * [tex]e_{0}[/tex]
B) E = 2K λ / r
C) Answers from parts a and b are the same
D) attached below
Applying Gauss's law which states that the net electric field in an enclosed surface is directly ∝ to the charge found in the enclosed surface.
A ) The expression for the electric field inside the volume at a distance r
Gauss law : E. A = [tex]\frac{q}{e_{0} }[/tex] ----- ( 1 )
where : A = surface area = 2πrL , q = p(πr²L)
back to equation ( 1 )
E ( electric field ) = p*r / 2 * [tex]e_{0}[/tex]
B) Electric field at point Outside the volume in terms of charge per unit length λ
Given that: linear charge density = area * volume charge density
λ = πR²P
from Gauss's law : E ( 2πrL) = [tex]\frac{q}{e_{0} }[/tex]
∴ E = [tex]\frac{\pi R^{2}P }{2e_{0}r\pi }[/tex] ----- ( 2 )
where : πR²P = λ
Back to equation ( 2 )
E = λ / 2e₀π*r where : k = 1 / 4πe₀
∴ The electric field ( E ) at point outside the volume in terms of charge per unit Length λ
E = 2K λ / r
C) Comparing answers A and B
Answers to part A and B are similar
Hence we can conclude that Applying Gauss law to the long solid cylinder
E ( electric field ) = p*r / 2 * [tex]e_{0}[/tex], E = 2K λ / r also Answers from parts a and b are the same.
Learn more about Gauss's Law : https://brainly.com/question/15175106
A 150m race is run on a 300m circular track of circumference. Runners start running from the north and turn west until reaching the south. What is the magnitude of the displacement made by the runners?
Answer:
95.5 m
Explanation:
The displacement is the position of the ending point relative to the starting point.
In this case, the magnitude of the displacement is the diameter of the circular track.
d = 300 m / π
d ≈ 95.5 m
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°
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!!!
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.
The potential energy function
U(x,y)=A[(1/x2) + (1/y2)] describes a conservative force, where A>0.
Derive an expression for the force in terms of unit vectors i and j.
Answer:
[tex]F=-2A[\frac{1}{x^3}\hat{i}+\frac{1}{y^3}\hat{j}][/tex]
Explanation:
You have the following potential energy function:
[tex]U(x,y)=A[\frac{1}{x^2}+\frac{1}{y^2}}][/tex] (1)
A > 0 constant
In order to find the force in terms of the unit vectors, you use the gradient of the potential function:
[tex]\vec{F}=\bigtriangledown U(x,y)=\frac{\partial}{\partial x}U\hat{i}+\frac{\partial}{\partial y}U\hat{j}[/tex] (2)
Then, you replace the expression (1) into the expression (2) and calculate the partial derivatives:
[tex]\vec{F}=A\frac{\partial}{\partial x}[\frac{1}{x^2}+\frac{1}{y^2}]} \hat{i}+A\frac{\partial}{\partial x}[\frac{1}{x^2}+\frac{1}{y^2}]\hat{j}\\\\\vec{F}=A(-2x^{-3})\hat{i}+A(-2y^{-3})\hat{j}\\\\F=-2A[\frac{1}{x^3}\hat{i}+\frac{1}{y^3}\hat{j}][/tex](3)
The result obtained in (3) is the force expressed in terms of the unit vectors, for the potential energy function U(x,y).
The interference of two sound waves of similar amplitude but slightly different frequencies produces a loud-soft-loud oscillation we call __________.
a. the Doppler effect
b. vibrato
c. constructive and destructive interference
d. beats
Answer:
the correct answer is d Beats
Explanation:
when two sound waves interfere time has different frequencies, the result is the sum of the waves is
y = 2A cos 2π (f₁-f₂)/2 cos 2π (f₁ + f₂)/2
where in this expression the first part represents the envelope and the second part represents the pulse or beatings of the wave.
When examining the correct answer is d Beats
The objective lens of a microscope has a focal length of 5.5mm. Part A What eyepiece focal length will give the microscope an overall angular magnification of 300
Complete Question
The distance between the objective and eyepiece lenses in a microscope is 19 cm . The objective lens has a focal length of 5.5 mm .
What eyepiece focal length will give the microscope an overall angular magnification of 300?
Answer:
The eyepiece focal length is [tex]f_e = 0.027 \ m[/tex]
Explanation:
From the question we are told that
The focal length is [tex]f_o = 5.5 \ mm = -0.0055 \ m[/tex]
This negative sign shows the the microscope is diverging light
The angular magnification is [tex]m = 300[/tex]
The distance between the objective and the eyepieces lenses is [tex]Z = 19 \ cm = 0.19 \ m[/tex]
Generally the magnification is mathematically represented as
[tex]m = [\frac{Z - f_e }{f_e}] [\frac{0.25}{f_0} ][/tex]
Where [tex]f_e[/tex] is the eyepiece focal length of the microscope
Now making [tex]f_e[/tex] the subject of the formula
[tex]f_e = \frac{Z}{1 - [\frac{M * f_o }{0.25}] }[/tex]
substituting values
[tex]f_e = \frac{ 0.19 }{1 - [\frac{300 * -0.0055 }{0.25}] }[/tex]
[tex]f_e = 0.027 \ m[/tex]
When looking at the chemical symbol, the charge of the ion is displayed as the
-superscript
-subscript
-coefficient
-product
Answer:
superscript
Explanation:
When looking at the chemical symbol, the charge of the ion is displayed as the Superscript. This is because the charge of ions is usually written up on the chemical symbol while the atom/molecule is usually written down the chemical symbol. The superscript refers to what is written up on the formula while the subscript is written down on the formula.
An example is H2O . The 2 present represents two molecule of oxygen and its written as the subscript while Fe2+ in which the 2+ is written up is known as the superscript.
Answer:
superscript
Explanation:
A charge of 87.6 pC is uniformly distributed on the surface of a thin sheet of insulating material that has a total area of 65.2 cm^2. A Gaussian surface encloses a portion of the sheet of charge. If the flux through the Gaussian surface is 9.20 N⋅m^2/C, what area of the sheet is enclosed by the Gaussian surface?
Answer:
60.8 cm²
Explanation:
The charge density, σ on the surface is σ = Q/A where q = charge = 87.6 pC = 87.6 × 10⁻¹² C and A = area = 65.2 cm² = 65.2 × 10⁻⁴ m².
σ = Q/A = 87.6 × 10⁻¹² C/65.2 × 10⁻⁴ m² = 1.34 × 10⁻⁸ C/m²
Now, the charge through the Gaussian surface is q = σA' where A' is the charge in the Gaussian surface.
Since the flux, Ф = 9.20 Nm²/C and Ф = q/ε₀ for a closed Gaussian surface
So, q = ε₀Ф = σA'
ε₀Ф = σA'
making A' the area of the Gaussian surface the subject of the formula, we have
A' = ε₀Ф/σ
A' = 8.854 × 10⁻¹² F/m × 9.20 Nm²/C ÷ 1.34 × 10⁻⁸ C/m²
A' = 81.4568/1.34 × 10⁻⁴ m²
A' = 60.79 × 10⁻⁴ m²
A' ≅ 60.8 cm²
The flux through the Gaussian surface is 9.20 N⋅m^2/C then the surface area of the Gaussian Sheet is 60.76 square cm.
Charge and Charge DensityA certain amount of electrons in excess or defect is called a charge. Charge density is the amount of charge distributed over per unit of volume.
Given that, for a thin sheet of insulating material, the charge Q is 87.6 pC and surface area A is 65.2 square cm. Then the charge density for a thin sheet is given below.
[tex]\sigma = \dfrac {Q}{A}[/tex]
[tex]\sigma = \dfrac {87.6\times 10^{-12}}{65;.2\times 10^{-4}}[/tex]
[tex]\sigma = 1.34\times 10^{-8} \;\rm C/m^2[/tex]
Thus the charge density for a thin sheet of insulating material is [tex]1.34\times 10^{-8} \;\rm C/m^2[/tex].
Now, the flux through the Gaussian surface is 9.20 N⋅m^2/C. The charge over the Gaussian Surface is given as below.
[tex]Q' = \sigma A'[/tex]
Where Q' is the charge at the Gaussian Surface, A' is the surface area of the Gaussian surface and [tex]\sigma[/tex] is the charge density.
For the closed Gaussian Surface, Flux is given below.
[tex]\phi = \dfrac {Q'}{\epsilon_\circ}[/tex]
Hence the charge can be written as,
[tex]Q' = \phi\epsilon_\circ[/tex]
So the charge can be given as below.
[tex]Q' = \phi\epsilon_\circ = \sigma A'[/tex]
Then the surface area of the Gaussian surface is given below.
[tex]A' = \dfrac {\phi\epsilon_\circ}{\sigma}[/tex]
Substituting the values in the above equation,
[tex]A' = \dfrac {9.20 \times 8.85\times 10^{-12}}{1.38\times 10^{-8}}[/tex]
[tex]A' =0.006076\;\rm m^2[/tex]
[tex]A' = 60.76 \;\rm cm^2[/tex]
Hence we can conclude that the area of the Gaussian Surface is 60.76 square cm.
To know more about the charge and charge density, follow the link given below.
https://brainly.com/question/8532098.
The center of gravity of an ax is on the centerline of the handle, close to the head. Assume you saw across the handle through the center of gravity and weigh the two parts. What will you discover?
Answer:
I believe it is they will weigh the same
Explanation:
Center of gravity is the axis on which the mass rotates evenly if I remember correctly from AP Physics
The head side is heavier than the handle side. - this will be discovered.
What is center of gravity of a object?Theoretically, the body's center of gravity is where all of the weight is believed to be concentrated. Knowing the centre of gravity is crucial because it may be used to forecast how a moving object will behave when subjected to the effects of gravity. In designing immobile constructions like buildings and bridges, it is also helpful.
We know that center of gravity is close to some particular point refers the mass of the point is greater then others. It is given that: The center of gravity of an ax is on the centerline of the handle, close to the head.
So, we can conclude that the head side of the ax is heavier than the handle side of it.
Learn more about center of gravity here:
https://brainly.com/question/17409320
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g At some point the road makes a right turn with a radius of 117 m. If the posted speed limit along this part of the highway is 25.1 m/s, how much should Raquel bank the turn so that a vehicle traveling at the posted speed limit can make the turn without relying on the frictional force between the tires and the road
Answer:
Ф = 28.9°
Explanation:
given:
radius (r) = 117m
velocity (v) = 25.1 m/s
required: angle Ф
Ф = inv tan (v² / (r * g)) we know that g = 9.8
Ф = inv tan (25.1² / (117 * 9.8))
Ф = 28.9°
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
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
A force of 44 N will stretch a rubber band 88 cm (0.080.08 m). Assuming that Hooke's law applies, how far will aa 11-N force stretch the rubber band? How much work does it take to stretch the rubber band this far?
Answer:
The rubber band will be stretched 0.02 m.
The work done in stretching is 0.11 J.
Explanation:
Force 1 = 44 N
extension of rubber band = 0.080 m
Force 2 = 11 N
extension = ?
According to Hooke's Law, force applied is proportional to the extension provided elastic limit is not extended.
F = ke
where k = constant of elasticity
e = extension of the material
F = force applied.
For the first case,
44 = 0.080K
K = 44/0.080 = 550 N/m
For the second situation involving the same rubber band
Force = 11 N
e = 550 N/m
11 = 550e
extension e = 11/550 = 0.02 m
The work done to stretch the rubber band this far is equal to the potential energy stored within the rubber due to the stretch. This is in line with energy conservation.
potential energy stored = [tex]\frac{1}{2}ke^{2}[/tex]
==> [tex]\frac{1}{2}* 550* 0.02^{2}[/tex] = 0.11 J
The rock and meterstick balance at the 25-cm mark, as shown in the sketch. The meterstick has a mass of 1 kg. What must be the mass of the rock? (Show work).
Answer:
1 kgExplanation:
Check the diagram attached below for the diagram.
Let the weight of the rock be W and the mass of the meter stick be M. Note that the mass of the meter stick will be placed at the middle of the meter stick i.e at the 50cm mark
Using the principle of moment to calculate the weight of the rock. It states that the sum of clockwise moments is equal to the sum of anti clockwise moment.
Moment = Force * perpendicular distance
The meterstick acts in the clockwise direction while the rock acys in the anti clockwise direction
Clockwise moment = 1kg * 25 = 25kg/cm
Anticlockwise moment = W * 25cm = 25W kg/cm
Equating both moments of forces
25W = 25
W = 25/23
W = 1 kg
The mass of the rock is also 1 kg
the density of gold is 19 300kg/m^3. what is the mass of gold cube with the length 0.2015m?
Answer:
The mass is [tex]157.87m^3[/tex]Explanation:
Given data
length of cube= 0.2015 m
density = 19300 kg/m^3.
But the volume of cube is given as [tex]l*l*l= l^3[/tex]
[tex]volume -of- cube= 0.2015*0.2015*0.2015= 0.00818 m^3[/tex]
The density is expressed as = mass/volume
[tex]mass=19300*0.00818= 157.87m^3[/tex]
Pulling out of a dive, the pilot of an airplane guides his plane into a vertical circle with a radius of 600 m. At the bottom of the dive, the speed of the airplane is 150 m/s. What is the apparent weight of the 70.0-kg pilot at that point?
Answer:
The apparent weight of the pilot is 3311 N
Explanation:
Given;
radius of the vertical circle, r = 600 m
speed of the plane, v = 150 m/s
mass of the pilot, m = 70 kg
Weight of the pilot due to his circular motion;
[tex]W= F_v\\\\F_v = \frac{mv^2}{r} \\\\F_v = \frac{70*150^2}{600} \\\\F_v = 2625 \ N[/tex]
Real weight of the pilot;
[tex]W_R = mg\\\\W_R = 70 *9.8\\\\W_R = 686 \ N[/tex]
Apparent weight - Real weight of pilot = weight due to centripetal force
[tex]F_N - mg = \frac{mv^2}{r} \\\\F_N = \frac{mv^2}{r} + mg\\\\F_N = 2625 \ N + 686 \ N\\\\F_N = 3311\ N[/tex]
Therefore, the apparent weight of the pilot is 3311 N
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.