PHYSICS COLLEGE

Two cylinders with the same mass density rhoC = 713 kg / m3 are floating in a container of water (with mass density rhoW = 1025 kg / m3). Cylinder #1 has a length of L1 = 20 cm and radius r1 = 5 cm. Cylinder #2 has a length of L2 = 10 cm and radius r2 = 10 cm. If h1 and h2 are the heights that these cylinders stick out above the water, what is the ratio of the height of Cylinder #2 above the water to the height of Cylinder #1 above the water (h2 / h1)? h2 / h1 =

Answers

Answer 1

Answer:

Answer:

Explanation:

Given

density of cylinder is $\rho _c=713 kg/m^3$

Length of first cylinder is $L_1=20 cm$

radius $r_1=5 cm$

For cylinder 2 $L_2=10 cm$

$r_2=10 cm$

$h_1$ and $h_2$ are the height above water

as object is floating so its weight must be balanced with buoyant force

$\rho _c(\pi )/(4)d_1^2L_1g=\rho _w(\pi )/(4)d_1^2(L_1-h_1)g----1$

For 2nd cylinder

$\rho _c(\pi )/(4)d_2^2L_2g=\rho _w(\pi )/(4)d_2^2(L_2-h_2)g----2$

Dividing 1 and 2 we get

$(L_1)/(L_2)=(L_1-h_1)/(L_2-h_2)$

$(20)/(10)=(20-h_1)/(10-h_2)$

$2h_2=h_1$

$\n\Rightarrow(h_2)/(h_1)=(1)/(2)$

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Which one of the following statements concerning the Stefan-Boltzmann equation is correct? The equation can be used to calculate the power absorbed by any surface. The equation applies only to perfect radiators. The equation applies only to perfect absorbers. The equation is valid with any temperature units. The equation describes the transport of thermal energy by conduction.

Answers

"The equation can be used to calculate the power absorbed by any surface" statement concerning the Stefan-Boltzmann equation is correct.

Answer: Option A

Explanation:

According to Stefan Boltzmann equation, the power radiated by black body radiation source is directly proportionate to the fourth power of temperature of the source. So the radiation transferred is absorbed by another surface and that absorbed power will also be equal to the fourth power of the temperature. So the equation describes the relation of net radiation loss with the change in temperature from hotter temperature to cooler temperature surface.

$P=e \sigma A\left(T^(4)-T_(c)^(4)\right)$

So this law is application for calculating power absorbed by any surface.

Blocks A (mass 2.00 kg) and B (mass 6.00 kg) move on a frictionless, horizontal surface. Initially, block B is at rest and block A is moving toward it at 2.00 m/s. The blocks are equipped with ideal spring bumpers. The collision is head-on, so all motion before and after the collision is along a straight line. (a) Find the maximum energy stored in the spring bumpers and the velocity of each block at that time. (b) Find the velocity of each block after they have moved apart.

Answers

Answer:

av=0.333m/s, U=3.3466J

$v_(A2)=-1.333m/s,\n v_(B2)=0.667m/s$

Explanation:

a. let $m_A$ be the mass of block A, and $m_B=10.0kg$ be the mass of block B. The initial velocity of A, $\rightarrow v_A_1=2.0m/s$

-The initial momentum =Final momentum since there's no external net forces.

$pA_1+pB_1=pA_2+pB_2\n\nP=mv\n\n\therefore m_Av_A_1+m_Bv_B_1=m_Av_(A2)+m_Bv_(B2)$

Relative velocity before and after collision have the same magnitude but opposite direction (for elastic collisions):

$v_A_1-v_B_1=v_(B2)-v_(A2)$

-Applying the conservation of momentum. The blocks have the same velocity after collision:

$v_(B2)=v_(A2)=v_2\n\n2* 2+10* 0=2v_2+10v_2\n\nv_2=0.3333m/s$

#Total Mechanical energy before and after the elastic collision is equal:

$K_1+U_(el,1)=K_2+U_(el,2)\n\n#Springs \ in \ equilibrium \ before \ collision\n\nU_(el,2)=K_1-K_2=0.5m_Av_A_1^2-0.5(m_A+m_B)v_2^2\n\nU_(el,2)=0.5* 2* 2^2-0.5(2+10)(0.333)^2\n\nU_(el,2)=3.3466J$

Hence, the maxumim energy stored is U=3.3466J, and the velocity=0.333m/s

b. Taking the end collision:

From a above, $m_A=2.0kg, m_B=10kg, v_A=2.0,v_B_1=0$

We plug these values in the equation:

$m_Av_A_1+m_Bv_B_1=m_Av_(A2)+m_Bv_(B2)$

$2*2+10*0=2v_A_2+10v_B_2\n\n2=v_A_2+5v_B_2\n\n#Eqtn 2:\nv_A_1-v_B_1=v_(B2)-v_(A2)\n\n2-0=v_(B2)-v_(A2)\n\n2=v_(B2)-v_(A2)\n\n#Solve \ to \ eliminate \ v_(A2)\n\n6v_(B2)=2.0\n\nv_(B2)==0.667m/s\n\n#Substitute \ to \ get \ v_(A2)\n\nv_(A2)=(4)/(6)-2=1.333m/s$

The radius of Earth is 6370 km in the Earth reference frame. The cosmic ray is moving at 0.880Co relative to Earth.a. In the reference frame of a cosmic ray how wide does Earth seem along the flight direction?
b. In the reference frame of a cosmic ray how wide does Earth seem perpendicular to the flight direction?
Express your answer with the appropriate units.

Answers

Answer:

6052114.67492 m

$12.742* 10^(6)\ m$

Explanation:

v = Velocity of cosmic ray = 0.88c

c = Speed of light = $3* 10^8\ m/s$

d = Width of Earth = Diameter of Earth = $12.742* 10^(6)\ m$

When the cosmic ray is moving towards Earth then in the frame of the cosmic ray the width of the Earth appears smaller than the original

This happens due to length contraction

Length contraction is given by

$d_e=d\sqrt{1-(v^2)/(c^2)}\n\Rightarrow d_e=12.742* 10^(6)\sqrt{1-(0.88^2c^2)/(c^2)}\n\Rightarrow d_e=6052114.67492\ m$

The Earth's width is 6052114.67492 m

Contraction only occurs in the cosmic ray's frame of reference in the direction of the ray. But in perpendicular direction the width remains unchanged.

Hence, the width is $12.742* 10^(6)\ m$

g Adjacent rows in the first part of the experiment are found to have potentials of 3.66 V and 4.22 V. If the distance between rows is found to be 0.4 cm, what is the magnitude of the electric field at the location between the rows

Answers

Answer:

$E=140V/m$

Explanation:

If the electric field is uniform, the electric field between two points at potentials $V_1$ and $V_2$ which are separated by a distance d will be given by the formula:

$E=(\Delta V)/(d)$

So in our case, we have $E=(4.22V-3.66V)/(0.004m)=140V/m$

) An electron moving along the x-axis enters a magnetic field. If the electron experiences a magnetic deflection in the -y direction, what is the direction of the magnetic field in this region

Answers

Answer:

- z direction

Explanation:

To find the direction of the magnetic field, you take into account that the magnetic force over a charge, is given by the following cross product:

$\vec{F_B}=q\vec{v}\ X\ \vec{B}$ (1)

F_B: magnetic force

q: charge of the particle

v: velocity of the charge

B: magnetic field

In this case you have that the electron is moving along x-axis. You can consider this direction as the ^i direction. The electron experiences a magnetic deflection in the -y direction, that is, in the -^j direction.

By the cross products between unit vectors, you have that:

-^j = ^i X ^k

That is, the cross product between two vectors, one in the +x direction, and another one in the +z direction, generates a vector in the -y direction. However, it is necessary to take into account that the negative charge of the electron change the sign of the result of the cross product, which demands that the second vector is in the -z direction. That is:

-^i X -k^ = ^i X ^k = - ^j

Hence, the direction of the magnetic field is in the -z direction

What force causes oppositely charged particles to attract each other? A. Magnetic force B. Compression
C. Electrical Force D. Gravity

Answers

Answer:

It is electrical force

Explanation:

i got it wrong on A P E X with magnetic hope this helps!

Answer:

The electromagnetic force.

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