Answer:
0.4455 m
Explanation:
g = Acceleration due to gravity = 9.81 m/s²
Total mass is
[tex]m=62\times 25+15\times 3+5\times 3+25\\\Rightarrow m=1635\ kg[/tex]
Here the spring constant is not given so let us assume it as [tex]k=36000\ N/m[/tex]
Here, the forces are balanced
[tex]mg=kx\\\Rightarrow 1635\times 9.81=36000\times x\\\Rightarrow x=\dfrac{1635\times 9.81}{36000}\\\Rightarrow x=0.4455\ m[/tex]
The springs are compressed by 0.4455 m
Through what potential difference would you need to accelerate an alpha particle, starting from rest, so that it will just reach the surface of a 15-fmfm-diameter 238U238U nucleus
Answer:
[tex]\Delta V = 1.8 \times 10^7 V[/tex]
Explanation:
GIVEN
diameter = 15 fm =[tex]15 \times 10^{-15}[/tex]m
we use here energy conservation
[tex]K_{i}+U_{i} =K_{f}+U_{f}[/tex]
there will be some initial kinetic energy but after collision kinetic energy will zero
[tex]K_{i} + 0 = 0 + \frac{1}{4 \pi \epsilon _{0}} \frac{(2e)(92e)}{7.5 \times 10^{-15}}[/tex]
on solving these equations we get kinetic energy initial
[tex]KE_{i} = 5.65\times 10 ^{-12} \times \frac {1 eV}{1.6 \times 10^{-19}}[/tex]
[tex]KE_{i} = 35.33[/tex] J ..............(i)
That is, the alpha particle must be fired with 35.33 MeV of kinetic energy. An alpha particle with charge q = 2 e
and gains kinetic energy K =e∆V ..........(ii)
by accelerating through a potential difference ∆V
Thus the alpha particle will
just reach the [tex]{238}_U[/tex] nucleus after being accelerated through a potential difference ∆V
equating (i) and second equation we get
e∆V = 35.33 Me V
[tex]\Delta V = \frac{35.33}{2} MV\\\Delta V = 1.8 \times 10^7 V[/tex]
Unpolarized light with intensity I0I0I_0 is incident on an ideal polarizing filter. The emerging light strikes a second ideal polarizing filter whose axis is at 40.0∘∘ to that of the first. Determine the intensity of the beam after it has passed through the second polarizer. g
Answer:
[tex]0.293I_0[/tex]
Explanation:
When the unpolarized light passes through the first polarizer, only the component of the light parallel to the axis of the polarizer passes through.
Therefore, after the first polarizer, the intensity of light passing through it is halved, so the intensity after the first polarizer is:
[tex]I_1=\frac{I_0}{2}[/tex]
Then, the light passes through the second polarizer. In this case, the intensity of the light passing through the 2nd polarizer is given by Malus' law:
[tex]I_2=I_1 cos^2 \theta[/tex]
where
[tex]\theta[/tex] is the angle between the axes of the two polarizer
Here we have
[tex]\theta=40^{\circ}[/tex]
So the intensity after the 2nd polarizer is
[tex]I_2=I_1 (cos 40^{\circ})^2=0.587I_1[/tex]
And substituting the expression for I1, we find:
[tex]I_2=0.587 (\frac{I_0}{2})=0.293I_0[/tex]
Approximately 80% of the energy used by the body must be dissipated thermally. The mechanisms available to eliminate this energy are radiation, evaporation of sweat, evaporation from the lungs, conduction, and convection. In this question, we will focus on the evaporation of sweat alone, although all of these mechanisms are needed to survive. The latent heat of vaporization of sweat at body temperature (37 °C) is 2.42 x 10^6 J/kg and the specific heat of a body is approximately 3500 J/(kg*°C).
(A) To cool the body of a jogger of mass 90 kg by 1.8°C , how much sweat has to evaporate?
O 130 g
O 230 g
O 23 g
O 13 g
Answer:
the correct answer is c) 23 g
Explanation:
The heat lost by the runner has two parts: the heat absorbed by sweat in evaporation and the heat given off by the body
Q_lost = - Q_absorbed
The latent heat is
Q_absorbed = m L
The heat given by the body
Q_lost = M [tex]c_{e}[/tex] ΔT
where m is the mass of sweat and M is the mass of the body
m L = M c_{e} ΔT
m = M c_{e} ΔT / L
let's replace
m = 90 3.500 1.8 / 2.42 10⁶
m = 0.2343 kg
reduced to grams
m = 0.2342 kg (1000g / 1kg)
m = 23.42 g
the correct answer is c) 23 g
230 g should have to evaporate.
Given that,
The latent heat of vaporization of sweat at body temperature (37 °C) is 2.42 x 10^6 J/kg and the specific heat of a body is approximately 3500 J/(kg*°C).The calculation is as follows:
[tex]= \frac{(90kg)(3500J/kg^{\circ})(1.8^{\circ}C)}{2.42\times 10^6J/kg}[/tex]
= 0.23 kg
= 230g
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Three point charges lie in a straight line along the y-axis. A charge of q1 = − 9.0 μC is at y = 6.0 m, and a charge of q2 = −8.0 μC is at y = − 4.0m. The net electric force on the third point charge is zero. Where is this charge located?
The third point charge must be located at the point where the net electric force on it is zero.
How can the charge be located?The net electric force on a point charge is the vector sum of the electric forces exerted on it by all the other point charges. The electric force exerted by a point charge q1 on a point charge q2 is given by Coulomb's law:
F = k * |q1| * |q2| / r^2
where k is Coulomb's constant, |q1| and |q2| are the magnitudes of the charges, and r is the distance between the charges.
The direction of force will be opposite for both charges, and since net force is zero, the force due to both charges will be equal and opposite.
So, |q1| * (1/r1^2) = |q2| * (1/r2^2)
Where r1 is the distance between point charge 1 and point charge 3, r2 is the distance between point charge 2 and point charge 3.
Solving for position of point charge 3, we can say that it is located at y = 0.
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To find the position where a third charge experiences zero net force due to two other charges along the y-axis, we must consider the magnitudes and signs of all charges and use Coulomb's Law. The third charge will be or negative and situated closer to the smaller magnitude charge to balance the forces. The exact position, however, cannot be determined without the magnitude of the third charge.
Explanation:The question pertains to the concept of electric forces and charge equilibrium in Physics. We are asked to find the position where a third charge would experience a net electric force of zero due to the presence of two other point charges arranged along the y-axis. The charges are q1 = − 9.0 μC at y = 6.0 m and q2 = − 8.0 μC at y = − 4.0 m.
Let's denote the position of the third charge as y3 where the net force is zero. To ensure a net force of zero on the third charge, the electric force due to q1 must be equal in magnitude and opposite in direction to the force due to q2. Since both q1 and q2 are negative, the third charge must also be negative (so it will be repelled by both). The third charge must be placed between q1 and q2.
Likewise, since q1 is larger in magnitude than q2, the third charge must be placed closer to q2 to balance the forces (since electric force is inversely related to the square of the distance). Mathematically, we can set up the equation where the magnitude of the forces due to each charge on the third charge is equal. This investigation involves Coulomb's Law (F = k · |q1 · q3| / r^2), where F is the force between charges, k is Coulomb's constant, q3 is the third charge, and r is the distance between the charges involved.
Through setting up the equations and solving for y3, we can determine the precise location. Unfortunately, the problem does not provide the magnitude of the third charge, so while we can describe the process, we can't compute a specific answer without this information. In a real scenario though, any negative third charge would suffice, and its magnitude would not affect the balance point.
A barge floating in fresh water (p = 1000 kg/m3) is shaped like a hollow rectangular prism with base area A = 550 m2 and height H = 2.0 m. When empty the
bottom of the barge is located H0 = 0.45 m below the surface of the water. When fully
loaded with coal the bottom of the barge is located H; = 1.05 m below the surface. Part (a) Find the mass of the coal in kilograms.
Numeric : A numeric value is expected and not an expression.
m1 =
Part (b) How far would the barge be submerged (in meters) if m; = 450000 kg of coal had been placed on the empty barge? Numeric : A numeric value is expected and not an expression.
Answer:
a) [tex]\Delta m = 330000\,kg[/tex], b) [tex]h = 1.268\,m[/tex]
Explanation:
a) According the Archimedes' Principle, the buoyancy force is equal to the displaced weight of surrounding liquid. The mass of the coal in the barge is:
[tex]\Delta m \cdot g = \rho_{w}\cdot g \cdot \Delta V[/tex]
[tex]\Delta m = \rho_{w}\cdot \Delta V[/tex]
[tex]\Delta m = \left(1000\,\frac{kg}{m^{3}} \right)\cdot (550\,m^{2})\cdot (1.05\,m-0.45\,m)[/tex]
[tex]\Delta m = 330000\,kg[/tex]
b) The submersion height is found by using the equation derived previously:
[tex]\Delta m = \rho_{w}\cdot \Delta V[/tex]
[tex]450000\,kg = \left(1000\,\frac{kg}{m^{3}}\right)\cdot (550\,m^{2})\cdot (h-0.45\,m)[/tex]
The final submersion height is:
[tex]h = 1.268\,m[/tex]
Answer:
a) m = 330000 kg = 330 tons
b) H3 = 1.268 meters
Explanation:
Given:-
- The density of fresh-water, ρ = 1000 kg/m^3
- The base area of the rectangular prism boat, A = 550 m^2
- The height of the boat, H = 2.0 m ( empty )
- The bottom of boat barge is H1 = 0.45 m of the total height H under water. ( empty )
- The bottom of boat barge is H2 = 1.05 m of the total height H under water
Find:-
a) Find the mass of the coal in kilograms.
b) How far would the barge be submerged (in meters) if m; = 450000 kg of coal had been placed on the empty barge?
Solution:-
- We will consider the boat as our system with mass ( M ). The weight of the boat "Wb" acts downward while there is an upward force exerted by the body of water ( Volume ) displaced by the boat called buoyant force (Fb):
- We will apply the Newton's equilibrium condition on the boat:
Fnet = 0
Fb - Wb = 0
Fb = Wb
Where, the buoyant force (Fb) is proportional to the volume of fluid displaced ( V1 ). The expression of buoyant force (Fb) is given as:
Fb = ρ*V1*g
Where,
V1 : Volume displaced when the boat is empty and the barge of the boat is H1 = 0.45 m under the water:
V1 = A*H1
Hence,
Fb = ρ*A*g*H1
Therefore, the equilibrium equation becomes:
ρ*A*g*H1 = M*g
M = ρ*A*H1
- Similarly, apply the Newton's equilibrium condition on the boat + coal:
Fnet = 0
Fb - Wb - Wc = 0
Fb = Wb + Wc
Where, the buoyant force (Fb) is proportional to the volume of fluid displaced ( V2 ). The expression of buoyant force (Fb) is given as:
Fb = ρ*V2*g
Where,
V2 : Volume displaced when the boat is filled with coal and the barge of the boat is H2 = 1.05 m under the water:
V2 = A*H2
Hence,
Fb = ρ*A*g*H2
Therefore, the equilibrium equation becomes:
ρ*A*g*H2 = g*( M + m )
m+M = ρ*A*H2
m = ρ*A*H2 - ρ*A*H1
m = ρ*A*( H2 - H1 )
m = 1000*550*(1.05-0.45)
m = 330000 kg = 330 tons
- We will set the new depth of barge under water as H3, if we were to add a mass of coal m = 450,000 kg then what would be the new depth of coal H3.
- We will use the previously derived result:
m = ρ*A*( H3 - H1 )
H3 = m/ρ*A + H1
H3 = (450000 / 1000*550) + 0.45
H3 = 1.268 m
The primary coil of a transformer has N1 = 275 turns, and its secondary coil has N2 = 2,200 turns. If the input voltage across the primary coil is Δv = (160 V)sin ωt, what rms voltage is developed across the secondary coil?
Answer:
Secondary voltage of transformer is 905.23 volt
Explanation:
It is given number of turns in primary of transformer [tex]N_1=275[/tex]
Number of turns in secondary [tex]N_2=2200[/tex]
Input voltage equation of the transformer
[tex]\Delta v=160sin\omega t[/tex]
Here [tex]v_{max}=160volt[/tex]
[tex]v_{rms}=\frac{160}{\sqrt{2}}=113.15volt[/tex]
For transformer we know that
[tex]\frac{V_1}{V_2}=\frac{N_1}{N_2}[/tex]
[tex]\frac{113.15}{V_2}=\frac{275}{2200}[/tex]
[tex]V_2=905.23Volt[/tex]
Therefore secondary voltage of transformer is 905.23 volt
Unpolarized light with an average intensity of 845 W/m2 moves along the x-axis when it enters a Polarizer A with a vertical transmission axis (along the y-axis). The transmitted light then enters a second polarizer, B at an angle in the y-z plane . The light that exits the second polarizer is found to have an average intensity of 275 W/m2. What is the orientation angle of the second polarizer (B) relative to the first one (A)
Answer:
θ = 36.2º
Explanation:
When light passes through a polarizer it becomes polarized and if it then passes through a second polarizer, it must comply with Malus's law
I = I₀ cos² tea
The non-polarized light between the first polarized of this leaves half the intensity, with vertical polarization
I₁ = I₀ / 2
I₁ = 845/2
I₁ = 422.5 W / m²
In this case, the incident light in the second polarizer has an intensity of I₁ = 422.5 W / m² and the light that passes through the polarizer has a value of
I = 275 W / m ²
Cos² θ = I / I₁
Cos θ = √ I / I₁
Cos θ = √ (275 / 422.5)
Cos θ = 0.80678
θ = cos⁻¹ 0.80678
θ = 36.2º
This is the angle between the two polarizers
A ray of laser light travels through air and enters an unknown material. The laser enters the material at an angle of 36 degrees to the normal. The refracted angle is 27.5 degrees. If the index of refraction of air is n = 1.00, what is the index of refraction of the unknown material? g
Answer:
n = 1.27
Explanation:
just took test :)
Answer:
N= 1.27
Explanation:
A bullet with a mass of 4.5 g is moving with a speed of 300 m/s (with respect to the ground) when it collides with a rod with a mass of 3 kg and a length of L = 0.25 m. The rod is initially at rest, in a vertical position, and pivots about an axis going through its center of mass which is located exactly halfway along the rod. The bullet imbeds itself in the rod at a distance L/4 from the pivot point. As a result, the bullet-rod system starts rotating together. What is the magnitude of the angular velocity (in rad/s) of the rotation (with respect to the ground) immediately after the collision? You may treat the bullet as a point particle.
The magnitude of the angular velocity immediately after the collision is
[tex]\( 14.4 \, \text{rad/s} \)[/tex].
To find the angular velocity immediately after the collision, we can apply the principle of conservation of angular momentum. The angular momentum before the collision equals the angular momentum after the collision.
The angular momentum before the collision is zero since the rod is at rest. After the collision, the bullet-rod system rotates together. We can calculate the moment of inertia [tex]\( I \)[/tex] of the system and then use the equation [tex]\( L = I\omega \)[/tex], where [tex]\( L \)[/tex] is the angular momentum and [tex]\( \omega \)[/tex] is the angular velocity.
First, calculate the moment of inertia of the system:
[tex]\[ I = \frac{1}{3}mL^2 + md^2 \][/tex]
[tex]\[ I = \frac{1}{3}(3)(0.25)^2 + (0.0045)(0.25/4)^2 \][/tex]
[tex]\[ I = 0.015625 \, \text{kg m}^2 \][/tex]
Now, apply conservation of angular momentum:
[tex]\[ I\omega = mvd \][/tex]
[tex]\[ \omega = \frac{mvd}{I} \][/tex]
[tex]\[ \omega = \frac{(0.0045)(300)(0.25/4)}{0.015625} \][/tex]
[tex]\[ \omega = 14.4 \, \text{rad/s} \][/tex]
So, the magnitude of the angular velocity immediately after the collision is [tex]\( 14.4 \, \text{rad/s} \)[/tex].
We have two solenoids: solenoid 2 has twice the diameter, half the length, and twice as many turns as solenoid 1. The current in solenoid 2 is three times that in solenoid 1. How does the field B2 at the center of solenoid 2 compare to B1 at the center of solenoid 1?
Answer:
the field at the center of solenoid 2 is 12 times the field at the center of solenoid 1.
Explanation:
Recall that the field inside a solenoid of length L, N turns, and a circulating current I, is given by the formula:
[tex]B=\mu_0\, \frac{N}{L} I[/tex]
Then, if we assign the subindex "1" to the quantities that define the magnetic field ([tex]B_1[/tex]) inside solenoid 1, we have:
[tex]B_1=\mu_0\, \frac{N_1}{L_1} I_1[/tex]
notice that there is no dependence on the diameter of the solenoid for this formula.
Now, if we write a similar formula for solenoid 2, given that it has :
1) half the length of solenoid 1 . Then [tex]L_2=L_1/2[/tex]
2) twice as many turns as solenoid 1. Then [tex]N_2=2\,N_1[/tex]
3) three times the current of solenoid 1. Then [tex]I_2=3\,I_1[/tex]
we obtain:
[tex]B_2=\mu_0\, \frac{N_2}{L_2} I_2\\B_2=\mu_0\, \frac{2\,N_1}{L_1/2} 3\,I_1\\B_2=\mu_0\, 12\,\frac{N_1}{L_1} I_1\\B_2=12\,B_1[/tex]
The magnetic field at the center of solenoid 2 (B2) will be twelve times larger than the magnetic field at the center of solenoid 1 (B1), due to having four times the number of turns per unit length and three times the current.
Explanation:The magnetic field inside a solenoid is given by the formula B = µnI, where B is the magnetic field, µ (mu) is the magnetic permeability of the medium, n is the number of turns per unit length, and I is the current through the solenoid. Given solenoid 2 has twice the diameter of solenoid 1, half the length, and twice as many turns, with the current being three times that of solenoid 1, several factors will influence the magnetic field in solenoid 2 (B2) compared to solenoid 1 (B1).
The number of turns per unit length for solenoid 2 is four times that of solenoid 1, since it has twice as many turns and half the length. Additionally, the current in solenoid 2 is three times that in solenoid 1. Therefore, B2 will be twelve times larger than B1, since the magnetic field inside a solenoid is directly proportional to both the number of turns per unit length of the solenoid and the current through it (B2 = 12 * B1).