PHYSICS COLLEGE

a. of water a. of bread a. of soap a. of juice a. of packet a. of sand option_piece packet grain bottle cake slice bale tank pane note tube bar can sack pile sheaf​

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Answer 1
Answer:

Answer:

is it a poem or recipe :(

don't know

Related Questions

What happens at night- describing air circulation
Two large parallel metal plates are 1.6 cm apart and have charges of equal magnitude but opposite signs on their facing surfaces. Take the potential of the negative plate to be zero. If the potential halfway between the plates is then +3.8 V, what is the electric field in the region between the plates?
At what partial pressure are argon atoms expected to have a free travel of approximately 5 µm, if the gas is at a temperature of 400 K? The cross section of collision, σ, or Argon is 0.28 nm2Ar molar mass is 39.9 g/mole
The ____ contains the highest concentration of ozone
Optical astronomers need a clear, dark sky to collect good data. Explain why radio astronomers have no problem observing in the UK where it is often very cloudy.

A research Van de Graaff generator has a 2.00-m diameter metal sphere with a charge of 5.00 mC on it. (a) What is the potential near its surface?
(b) At what distance from its center is the potential 1.00 MV?
(c) An oxygen atom with three missing electrons is released near the Van de Graaff generator. What is its energy in MeV when the atom is at the distance found in part b?

Answers

Answer:

a)V=49.5MV

b)r=49.5m

c)\Delta U=3*(49.5-1)=145.5 MeV  

Explanation:

a) The potential equation is given by:

V=k(Q)/(r)

k is the electrostatic constant (k=9.9*10^(9)Nm^(2)/C^(2))

Q is the charge Q = 5mC

r is the radius of the sphere r = 1 m

V=9.9*10^(9)(5*10^(-3))/(1)=49.5MV

b) We solve it using the same equation.

Here we need to find r:

r=k(Q)/(V)

r=9.9*10^(9)(5*10^(-3))/(1*10^(6))

r=49.5m

c) The relation between difference potential and electrical energy is:

\Delta U=\Delta Vq

here q is 3e becuase oxygen atom has three missing electrons

Therefore:

\Delta U=3*(49.5-1)=145.5 MeV  

I hope it heps you!

A ball with radius .15 m is rolling with an angular velocity of 6.5 rad/s. (a) What linear distance will the ball roll in 5.0 seconds? The ball then slows down with a linear acceleration of -1.2 rad/s2. (B) How fast will it be rolling rad/s after .65 second? (C) If a piece of gum is stuck on the outer part of the ball, what is its linear speed? (4.9, 5.72, .858)

Answers

Answer:

(a) 4.875 m

(b) 5.72 rad/s

(c) 0.858 m/s

Explanation:

(a) Assuming constant angular speed, the angular distance the ball would have traveled after 5s at the rate of 6.5 rad/s is

6.5 * 5 = 32.5 rad

With radius of 0.15m, the linear distance it would have traveled is

32.5 * 0.15 = 4.875 m

(b)The angular velocity of the ball after 0.65s when subjected to an angular acceleration of -1.2 rad/s is

6.5 - 1.2*0.65 = 5.72 rad/s

(c)The linear speed of the ball is the product of the angular speed and radius 0.15 m

5.72 * 0.15 = 0.858 m/s

In an evironmental system of subsystem, the mass balance equation is:__________.

Answers

Answer:

Explanation:

The mass balance is an application of conservation of mass, to the analysis of physical system. This is given in an equation form as

Input = Output + Accumulation

The conservation law that is used in this analysis of the system actually depends on the context of the problem. Nevertheless, they all revolve around conservation of mass. By conservation of mass, I mean that the fact that matter cannot disappear or be created spontaneously.

Neptunium. In the fall of 2002, scientists at Los Alamos National Laboratory determined that the critical mass of neptunium-237 is about 60 kg. The critical mass of a fissionable material is the minimum amount that must be brought together to start a nuclear chain reaction. Neptunium-237 has a density of 19.5 g/cm3. What would be the radius of a sphere of this material that has a critical mass?

Answers

To solve this problem it is necessary to apply the concepts related to density, such as the relationship between density and Volume.

The volume of a sphere can be expressed as

V = (4)/(3) \pi r^3

Here r is the radius of the sphere and V is the volume of Sphere

Using the expression of the density we know that

\rho = (m)/(V) \rightarrow V = (m)/(\rho)

The density is given as

\rho = (19.5g/cm^3)((10^3kg/m^3)/(1g/cm^3))

\rho = 19.5*10^3kg/m^3

Now replacing the mass given and the actual density we have that the volume is

V = (60kg)/(19.5*10^3kg/m^3 )

V = 3.0769*10^(-3) m ^3

The radius then is,

V = (4)/(3) \pi r^3

r = \sqrt[3]{(3V)/(4\pi)}

Replacing,

r = \sqrt[3]{(3(3.0769*10^(-3)))/(4\pi)}

The radius of a sphere made of this material that has a critical mass is 9.02 cm.

Air contains 78.08% nitrogen, 20.095% oxygen, and 0.93% argon. calculate the partial pressure of oxygen if the total pressure of the air sample is 1.7 atm.a.

Answers

partial pressure in a mixture of two or more gases will be given by formula

P_(partial) = mole fraction of gas * total pressure

now here mole fraction is same as percentage of gas in the mixture

Now mole fraction of oxygen is 0.20095 (20.095%)

now here pressure of oxygen in the mixture is given as

P_(o_2) = 0.20095 * P_(total)

P_(o_2) = 0.20095 * 1.7

P_(o_2) = 0.342 atm

so pressure due to oxygen in the mixture will be 0.342 atm

Answer:

20.095

Explanation:

Tonya picks up a leaf from the ground and holds it at arm’s length. She lets go, and the leaf falls to the ground. What force pulled the leaf to the ground?

Answers

Answer:

Gravity

Explanation:

Gravity is constantly pulling objects downward. Without it, everything would float out into space.

I hope this answer helps :)

Answer:

The answer for the given question above would be option C. GRAVITATIONAL FORCE. Based on the given scenario above of a leaf that falls to the ground when Tonya let it go, the force that pulled the leaf to the ground is the gravitational force. This kind of force is a force that attracts any object with mass.

Hope this helps!!!
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. Using your knowledge of circular (centripetal) motion, derive an equation for the radius r of the circular path that electrons follow in terms of the magnetic field B, the electrons' velocity v, charge e, and mass m. You may assume that the electrons move at right angles to the magnetic field.2. Recall from electrostatics, that an electron obtains kinetic energy when accelerated across a potential difference V. Since we can directly measure the accelerating voltage V in this expierment, but not the electrons' velocity v, replace velocity in your previous equation with an expression containing voltage. The electron starts at rest. Now solve this equation for e/m.You should obtain e/m = 2V/(B^2)(r^2)3. The magnetic field on the axis of a circular current loop a distance z away is given byB = mu I R^2 / 2(R^2 + z^2)^ (3/2)where R is the radius of the loops and I is the current. Using this result , calculate the magnetic field at the midpoint along the axis between the centers of the two current loops that make up the Helmholtz coils, in terms of their number of turns N, current I, and raidus R.Helmholtz coils are separated by a distance equal to their raidus R. You should obtain:|B| = (4/5)^(3/2) *mu *NI/R = 9.0 x 10^-7 NI/Rwhere B is magnetic field in tesla, I is in current in amps, N is number of turns in each coil, and R is the radius of the coils in meters
The cart is initially at rest. ForceF is applied to the cart for time ? t, after which the car has speed v. Suppose the same force is applied for the same time to a second cart with twice the mass. Friction is negligible. Afterward, the second cart's speed will bea) 2vb) 1/2vc) vd) 4ve) 1/4v