Which element would most likely have a positive electron affinity?

A positively charged particle travels horizontally northward and enters a region where a field may exist. This region may contai

nignag [31]

Answer:

Only magnetic field present

Explanation:

Since, the positively charged particle does not speed up or slow down, but it does deflect in the downward direction. This means only magnetic field is present.

This is because electric field changes the velocity and magnetic field changes the direction.

The magnetic force F is given by

F = qvBsinθ

Where, q = charge magnitude, v = velocity of charge, B = strength of magnetic field. and θ =the angle between the directions of v and B.

4 0

3 years ago

Refer to the image. Calculate the magnitude of change in velocity of the cue ball.

valkas [14]

Answer:

If you add the 2 velocity vectors you get the velocity of the third vector which is the (change in) velocity of the cue ball after striking the cushion

V * 2 * cos 45 = 4 * cos 45 = 2.83 m/s

6 0

2 years ago

What must be the distance (in meters) between a point charge q1 = 16 μC and a point charge q2 = 32 μC for the electrostatic forc

adoni [48]

Answer:

d = 0.71 meters

Explanation:

It is given that,

Charge 1, $q_1=16\ \mu C=16\times 10^{-6}\ C$

Charge 2, $q_2=32\ \mu C=32\times 10^{-6}\ C$

Electrostatic force between charges, F = 9 N

Let d is the distance between the charges. The electrostatic force between the charges is given by the product of charges and divided by square of distance between them. Mathematically, it is given by :

$F=k\dfrac{q_1q_2}{d^2}$

$d=\sqrt{\dfrac{kq_1q_2}{F}}$

$d=\sqrt{\dfrac{9\times 10^9\times 16\times 10^{-6}\times 32\times 10^{-6}}{9}}$

d = 0.71 meters

So, the distance between the charges is 0.71 meters. Hence, this is the required solution.

6 0

4 years ago

A sample of 20.0 moles of a monatomic ideal gas (γ = 1.67) undergoes an adiabatic process. The initial pressure is and the initi

Alexeev081 [22]

There are some missing data in the text of the exercise. Here the complete text:
"A sample of 20.0 moles of a monatomic ideal gas (γ = 1.67) undergoes an adiabatic process. The initial pressure is 400kPa and the initial temperature is 450K. The final temperature of the gas is 320K. What is the final volume of the gas? Let the ideal-gas constant R = 8.314 J/(mol • K). "

Solution:

First, we can find the initial volume of the gas, by using the ideal gas law:
 $pV=nRT$
where
p is the pressure
V the volume
n the number of moles
R the gas constant
T the absolute temperature

Using the initial data of the gas, we can find its initial volume:
 $V_i = \frac{nRT_i}{p_i} = \frac{(20.0 mol)(8.31 J/molK)(450 K)}{4 \cdot 10^5 Pa} =0.187 m^3$

Then the gas undergoes an adiabatic process. For an adiabatic transformation, the following relationship between volume and temperature can be used:
 $TV^{\gamma-1} = cost.$
where $\gamma=1.67$ for a monoatomic gas as in this exercise. The previous relationship can be also written as
$T_i V_i^{\gamma-1} = T_f V_f^{\gamma-1}$
where i labels the initial conditions and f the final conditions. Re-arranging the equation and using the data of the problem, we can find the final volume of the gas:
$V_f = V_i \sqrt[\gamma-1]{ \frac{T_i}{T_f} }=(0.187 m^3) \sqrt[0.67]{ \frac{450 K}{320 K} }=0.310 m^3 = 310 L$
So, the final volume of the gas is 310 L.

5 0

3 years ago

Plssss I need help——— Balloon A is (A. Negatively, B. Not, C. Positively) charged, and balloon C is (A. Negatively, B. Not, C. P

makvit [3.9K]

Answer:

B. Not

C. Positively

A. Attraction

8 0

3 years ago