Answer:
a = 0.1962 m/s^2
Explanation:
The magnitude of kinetic friction exerted is given by
[tex]F_k=\mu_kN[/tex]
Where, μ_k= coefficient of kinetic friction= 0.02 and N = reaction force = mg
Where m= mass = 30 Kg and, g is acceleration due to gravity =9.81 m/s^2
F_k=0.02×30×9.81 =5.886 N
Now, since, there is no applied force this kinetic friction force will cause acceleration of the child
⇒ ma = F_k
here, a is the acceleration
⇒30a = 5.886
⇒ a = 0.1962 m/s^2
A physics student mounts two thin lenses along a single optical axis (the lenses are at right angles to the line connecting them, and they appear concentric when viewed from either end). The lenses are identical, each with a positive (converging) focal length of 14.8 cm. They are separated by a distance of 39.4 cm. Lens 1 is to the left of Lens 2.
Required:
a. What is the final image's distance (in cm) from Lens 2?
b. Where is the final image located?
c. What is the overall magnification of the lens pair, considered as a single optical instrument?
Answer:
A) q₂ = 75.98 cm, B) q₂' = 115.38 cm, C)
Explanation:
A) This is an exercise in geometric optics, as the two lenses are separated by a greater distance than their focal lengths from each lens, they must be worked as independent lenses.
Lens 1. More to the left
let's use the constructor equation
[tex]\frac{1}{f} = \frac{1}{p} + \frac{1}{q}[/tex]
where f is the focal length, p and q are the distance to the object and the image, respectively,
We must assume a distance to the object to perform the calculation, suppose that the object is 50 cm from lens 1 that is further to the left of the system.
[tex]\frac{1}{q_1} = \frac{1}{f} - \frac{1}{p}[/tex]
[tex]\frac{1}{q_1} = \frac{1}{14.8} - \frac{1}{50}[/tex]
1 / q₁ = 0.04756
q₁ = 21.0227 cm
this image is the object for the second lens that has f₂ = 14.8 cm
the distance must be measured from the second lens
p₂ = 39.4 -q₁
p₂ = 39.4 -21.0227
p₂ = 18.38 cm
let's use the constructor equation
1 / q₂ = 1 / f - 1 / p2
[tex]\frac{1}{q_2} = \frac{1}{14.8} - \frac{1}{18.38}[/tex]
[tex]\frac{1}{q_2}[/tex] = 0.01316
q₂ = 75.98 cm
measured from the second lens
B) the position of the final image with respect to the first lens is
q₂’= q₂ + 39.4
q₂'= 75.98 +39.4
q₂' = 115.38 cm
C) the magnification of a lens is
m = - q / p
in this case the image measured from lens 2 is q2 = 75.98 cm
the distance to the object from the first lens is p1 = 50cm
m = - 75.98 / 50
m = -1.5 X
the negative sign indicates that the image is inverted
When an object levitates, the magnetic force causes the object to repel. Without this magnetic force, ________ would pull the object down.
Answer:
gravity
Explanation:
I don't know what the explanations would be
Which element would have properties most like helium (He)?
A. Ar
B. Hg
C. H
D. O
Answer: A. Ar
Explanation: not anything else besides Ar
A 10- kg ball starting from rest rolls down a 5 m tall smooth hill from one person to another person who is standing at the bottom of the hill with a big spring whose constant is 100 N/m. How far does the spring compress in order to stop the ball
Answer: 3.13 m
Explanation:
Given
mas of the ball is m=10 kg
The ball rolls down a vertical distance of 5 m
Spring constant of spring is [tex]k=100\ N/m[/tex]
Here, the potential energy of the ball converted into kinetic energy which in turn converts into elastic potential energy
[tex]\Rightarrow mgh=\frac{1}{2}kx^2\quad [\text{x=compression in the spring}]\\\\\Rightarrow 10\times 9.8\times 5=\frac{1}{2}\cdot 100\cdot x^2\\\Rightarrow x=\sqrt{9.8}\\\Rightarrow x=3.13\ m[/tex]
Thus, the spring compresses by 3.13 m.
Suppose that an electron and a positron collide head-on. Both have kinetic energy of 1.20 MeV and rest energy of 0.511 MeV. They produce two photons, which by conservation of momentum must have equal energy and move in opposite directions. What is the energy Ephoton of one of these photons
Answer:
E = 1.711 MeV
Explanation:
From the law of the conservation of energy:
[tex]K.E_{e}+K.E_p + E_{e}+E_{p} = 2 E[/tex]
where,
[tex]K.E_e=K.Ep=[/tex] the kinetic energy of positron and electron = 1.2 MeV
[tex]E_e=E_p =[/tex] Rest energy of the electron and the positron = 0.511 MeV
E = Energy of Photon = ?
Therefore,
[tex]1.2\ MeV + 1.2\ MeV + 0.511\ MeV + 0.511\ MeV = 2E\\\\E = \frac{3.422\ MeV}{2}\\\\[/tex]
E = 1.711 MeV
A 7.5 cm tall Aragorn action figure is placed 12.4 cm in front of a double convex lens. If the image of Aragorn is located 7.5 from the lens, what is the focal length of the lens?
Answer:
f = 4.67 cm
Explanation:
Here, we can use the thin lens formula, as follows:
[tex]\frac{1}{f}= \frac{1}{p}+ \frac{1}{q}\\\\[/tex]
where,
f = focal length of lens = ?
p = distance of object from lens = 12.4 cm
q = distance of image from lens = 7.5 cm
Therefore,
[tex]\frac{1}{f} =\frac{1}{12.4\ cm} +\frac{1}{7.5\ cm}\\\\\frac{1}{f} = \frac{1}{4.67\ cm}[/tex]
f = 4.67 cm
he electric field is dependent on the distance between the plates. false: The voltage of a connected charged capacitor decreases when the plate area is increased. false: The voltage of a disconnected charged capacitor increases when the plates are brought closer together.
Answer:
False.
Explanation:
The voltage of a disconnected charged capacitor decreases when the plates are brought closer together because the capacitance is inversely proportional to the area. If the area between plates decreases, its capacitance increases and vice versa. There is direct relationship between voltage of a disconnected charged capacitor and plates. If the distance between plates decreases, the voltage of a disconnected charged capacitor is also decreases while on the other hand, if the distance between plates increases, the voltage of a disconnected charged capacitor is also increases.
You are studying a population of flowering plants for several years. When you present your research findings you make the statement that, "Increased allocation of resources to reproduction relative to growth diminished future fecundity." Which of the following graph descriptions could accurately present your data?
a) With seeds in the current year on the y-axis and seeds in the previous year on the x-axis, you would see a line that increased from left to right
b) With survivorship on the y-axis and number of seeds produced on the x-axis, you would see a line that decreased left to right.
c) With leaf area on the y-axis and number of seeds produced on the x-axis, you would see a line that increased left to right
d) With survivorship on the y-axis and number of seeds produced on the x-axis, you would see a line that increased left to right.
e) With seeds in the current year on the y-axis and seeds in the previous year on the x-axis, you would see a line that decreased from left to right
Answer:
Option A
Explanation:
The graph for this problem must depict the following ""Increased allocation of resources to reproduction relative to growth diminished future fecundity."
Hence, the survivor ship must be on the Y axis and the resources on the X axis.
Here the resources include the number of seeds produced.
hence, the higher is the number of seeds (resource), the lower is the survivorship (future fecundity)
Hence, option A is correct
two spheres of radii 5cm and 3cm are given charges on risk volume and 50 calling respectively and then connected by a wire calculate the loss of energy after connection
Answer:
Solution given:
Radius of small sphere[r]=5cm=0.05m
Radius of large sphere[R]=10cm=0.1m
capacitance of small sphere[c]=4πε0r
=[tex]4π*8.85×10^{-12}×0.05=5.56*10^{-12}F[/tex]
Charge for small sphere[Q1]=100C
Charge for small sphere[Q2]=50C
Potential difference [V1]=[tex] \frac{charge}{capacitance}=\frac{100}{5.56*10^{-12}}=1.8×10^{13}[/tex]V
.
again
capacitance of small sphere[C]=4πε0R
=[tex]4π*8.85×10^{-12}×0.1=1.11*10^{-11}F[/tex]
Potential difference [V2]=[tex] \frac{charge}{capacitance}=\frac{50}{1.11*10^{-11}}=4.5×10^{12}[/tex]V
Now
Loss of energy:
[tex] \frac{cC(V1-V2)^{2}}{2(c+C)}[/tex]
=[tex] \frac{5.56*10^{-12}*1.11*10^{-11}(1.8*10^{13}-4.5*10^{12})^{2}}{2(5.56*10^{-12}+1.11*10^{-11})}[/tex]
=25Joule
What are the biotic factors in this image?
Someone help thank you!!
A spinning disc with a mass of 2.5kg and a radius of 0.80m is rotating with an angular velocity of 1.5 rad/s. A ball of clay with unknown mass is dropped onto the disk and sticks to the very edge causing the angular velocity of the disk to slow to 1.13 rad/s. What is the mass of the ball of clay
Answer:
M = 1.90 Kg
Explanation:
Given data: mass = 2.5 Kg
radius R = 0.8 m
angular velocity ω = 1.5 rad/s
Angular momentum L =0.5×Iω^2
Where, I is the moment of inertia of the spinning disc.
I = 0.5MR^2
I = 0.5×2.5×0.8^2
I = 0.8 Kg/m^2
Then L = 0.5×0.8×1.5^2 = 0.8×2.25 = 0.9 Kg-m^2/sec
Let unknown mass be M
New mass of disc = (2.5+M) Kg, R = 0.8 m
New I = 0.5(2.5+M)(0.8)^2
Since, angular momentum is conserved
Angular momentum before = angular momentum after
0.5×0.5(2.5+M)(0.8)^2×(1.13)^2 = 0.9
Solving for M we get
0.204304(2.5+M)=0.9
M = 1.90 Kg
Please help ASAP with questions
Besides ethical considerations, what is another reason why Milgram’s experiment may be difficult to duplicate?
A 70 kg human body typically contains 140 g of potassium. Potassium has a chemical atomic mass of 39.1 u and has three naturally occurring isotopes. One of those isotopes, 40K (potassium), is radioactive with a half-life of 1.3 billion years and a natural abundance of 0.012 %. Each 40K (potassium) decay deposits, on average, 1.0 MeV of energy into the body.
What yearly dose in Gy does the typical person receive from the decay of 40K (potassium) in the body? Express your answer using two significant figures.
Answer:
Gy = 3.14x10⁻⁴ Gy
Explanation:
To get the dose in Gy we need to use the following expression:
Gy = E / m (1)
Where:
Gy: dose
E: energy absorbed per atom
m: mass of the human body.
We don't have the energy per atom, but we can calculate that by following the next procedure.
First, let's determine the number of atoms of potassium in our body. For that we need to determine the moles in the 140 g of potassium, with the molecular mass and then, use the avogadro's number:
moles = m/MM
moles = 140 / 39.1 = 3.58 moles
N° atoms = 3.58 * 6.02x10²³ atoms = 2.16x10²⁴ atoms of K.
The abundance of the ⁴⁰K is 0.012% so the atoms of this isotope would be:
N = 2.16x10²⁴ * (0.012/100) = 2.59x10²⁰ atoms of ⁴⁰K.
With this number, and the half life rate, we can determine the number of decay atoms in a year (λ) using the following expression:
λ = ln2 / t(1/2)
λ = ln2 / 1.3x10⁹ = 5.33x10⁻¹⁰ year⁻¹
This number, multiplied by the number of atoms:
R = 5.33x10⁻¹⁰ * 2.59x10²⁰ = 1.38x10¹¹ atoms/year
Now, each atom of K gives an average energy of 1 MeV, so with the atoms we have:
E = 1.38x10¹¹ * 1x10⁶ eV = 1.38x10¹⁷ eV
This value can be expressed in Joules so:
E = 1.38x10¹⁷ eV * (1 J / 6.24x10¹⁸ eV) = 0.022 J
Finally, we can use (1) to get the dose in Gy:
Gy = 0.022 / 70
Gy = 3.14x10⁻⁴ GyHope this helps
What is Hooke's law? what is meant by elastic limit?
please answer me
Answer:
Explanation:
The law is about how object will be deformed when certain amount force is applied. Elastic limit is a point where object will not return its previous shape after being deformed
In any given wave, when the frequency of the wave doubles (f = 2f), which of the following other changes would also take place?
A The wavelength would double (λ = 2λ).
B The velocity would double (v = 2v).
C The wavelength would be half (λ = λ/2).
D The velocity would be half (v = v/2).
Answer:
The correct answer is -
B. The velocity would double (v = 2v).
C. The wavelength would be half (λ = λ/2).
Explanation:
A wave has a speed or velocity that is related to the wavelength of the wave and the frequency of the wave and this relationship can be represented by the following equation-
Wave velocity V = Wavelength (λ) * Frequency (f)
Frequency (f) = Velocity (V) / Wavelength(λ).
The frequency and wavelength are inversely proportional and frequency and velocity are directly proportional to each other.
So, if f = 2f then,
putting value in the formula,
2f = 2v/λ, which means, f = 2v and f = λ/2
when the frequency is doubled, the wavelength will be halved and velocity will be doubled.
The tension in a pulley belt is 31 N when stationary. Calculate power in watts transmitted when the belt is on the point of slipping on the smaller wheel. the wheel is 379 mm diameter and the coefficient of friction is 0.3. The angle of lap is 1610. The wheel speed is 1,547 rev/min.
Answer:
P = 756.84 Watts
Explanation:
As the tension is stationary or innitial, T₀ = 31 N, the mean would be:
T₁ + T₂ / 2 = T₀ (1)
T₁ + T₂ = 2 * 31 = 62 N
Now, with the following expression we can determine the linear speed:
V = πWD (2)
W: angular speed of the wheel (rev/s)
D: diameter of the wheel (in meters)
W = 1547 rev/min * (1 min/60 s) = 25.78 rev/s
V = π * 25.78 * 0.379 = 30.695 m/s
We also know that:
T₁ / T₂ = exp (μθ)
T₁ = T₂ exp(μθ) (3)
We already have those values so replacing:
T₁ = T₂ exp(0.3 * 161 * π/180)
T₁ = 2.32T₂ (4)
We can now replace (4) in (1) like this:
T₁ + T₂ = 62 N
2.32T₂ + T₂ = 62
3.32T₂ = 62
T₂ = 18.67 N
Which means that T₁:
T₁ = 2.32(18.67)
T₁ = 43.33 N
Finally, the power can be determined using the following expression:
P = (T₁ - T₂)V (5) Replacing we have:
P = (43.33 - 18.67)*30.695
P = 756.84 WattsHope this helps