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3 years ago

8

How does reducing the volume of a gas affect its pressure if the temperature of the gas and the number of particles are constant

?

1 answer:

Tema [17]3 years ago

5 0

have no idea but I think the particles would be affected because they would move faster then normal and have the space to

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Resistors labeled 100 Ω have true resistances that are between 80 Ω and 120 Ω. Let X be the mass of a randomly chosen resistor.

STatiana [176]

Answer:

See explanation for step by step explanations.

Explanation:

Let X be the mass of a randomly selected resistor

The probability density function is given as

, 80<x<120

=0, otherwise

a) By using the given pdf we have to find the probability of the resistors having resistance less than 90

=

=

=0.0625

the probability of the resistors having resistance less than 90 is 0.0625

Step 2 of 4<

/p>

b) we have to find the mean resistance

Mean=

=

=

=106.67

The mean resistance is 106.67

5 0

4 years ago

At a particular instant, a moving body has a kinetic energy of 295 J and a momentum of magnitude 25.1 kg · m/s.(a)What is the sp

motikmotik

Answer:

a) 23.51 m/s

b) 1.07 kg

Explanation:

Parameters given:

Kinetic energy, K = 295 J

Momentum, p = 25.1 kgm/s

a) The kinetic energy of a body is given as:

$K = \frac{1}{2} mv^2$

where m = mass of the body and v = speed of the body

We know that momentum is given as:

p = mv

Therefore:

K = 1/2 * pv

=> v = 2K / p

v = (2 * 295) / 25.1 = 23.51 m/s

The velocity of the body at that instant is 23.51 m/s.

b) Momentum is given as:

p = mv

=> m = p / v

m = 25.1 / 23.51 = 1.07 kg

The mass of the body at that instant is 1.07 kg

5 0

3 years ago

A combination reaction is when two or more reactants __ to form one product

True [87]

Answer:

A combined reaction is when two or more reagents combine __ to form a product

Explanation:

A combination or synthesis reaction, consists of the formation of a substance from the reaction of two or more substances.

Example: A + B -> C

2Na + Cl2 ---> 2NaCl (sodium chloride formation reaction)

7 0

3 years ago

Which of the following statements are true concerning electromagnetic radiation fields?Select all that apply.a.) The electric an

ale4655 [162]

Answer:

a.) The electric and magnetic fields are in phase with each other as they propagate through space.

Explanation:

Electromagnetic wave is a transverse wave in which magnetic field and electric field both induces each other as both changes with time

Here magnetic field induces electric field and similarly magnetic field induces electric field.

As we know that this is a transverse wave so here magnetic field and electric field lies in perpendicular planes. but they both propagate in same direction in such a wave that both fields reaches their maximum position and minimum positions simultaneously

So the correct answer is

a.) The electric and magnetic fields are in phase with each other as they propagate through space.

6 0

4 years ago

Two stationary point charges of 3.00 nC and 2.00 nC are separated by a distance of 50.0 cm. An electron is released from rest at

andrew-mc [135]

Answer:

1. the electric potential energy of the electron when it is at the midpoint is - 2.9 x $10^{-17}$ J

2. the electric potential energy of the electron when it is 10.0 cm from the 3.00 nC charge is - 5.04 x $10^{-17}$ J

Explanation:

given information:

$q_{1}$ = 3 nC = 3 x $10^{-9}$ C

$q_{2}$ = 2 nC = 2 x $10^{-9}$ C

r = 50 cm = 0.5 m

the electric potential energy of the electron when it is at the midpoint

potential energy of the charge, F

F = k $\frac{q_{e}q}{r}$

where

k = constant (8.99 x $10^{9} Nm^{2} /C^{2}$ )

electron charge, $q_{e}$ = - 1.6 x $10^{-19}$ C

since it is measured at the midpoint,

r = $\frac{0.5}{2}$

= 0.25 m

thus,

F = $F_{1}+ F_{2}$

= k $\frac{q_{e} q_{1} }{r}$ + k $\frac{q_{e} q_{2} }{r}$

= $\frac{kq_{e} }{r}$ ( $q_{1} +q_{2}$ )

= (8.99 x $10^{9}$ )( - 1.6 x $10^{-19}$ )(3 x $10^{-9}$ +2 x $10^{-9}$ )/0.25

= - 2.9 x $10^{-17}$ J

the electric potential energy of the electron when it is 10.0 cm from the 3.00 nC charge

$r_{1}$ = 10 cm = 0.1 m

$r_{2}$ = 0.5 - 0.1 = 0.4 m

F = k $\frac{q_{e} q_{1} }{r}$ + k $\frac{q_{e} q_{2} }{r}$

= $kq_{e}$ ( $\frac{q_{1} }{r_{1} }$ + $\frac{q_{2} }{r_{2} }$ )

= (8.99 x $10^{9}$ )( - 1.6 x $10^{-19}$ )(3 x $10^{-9}$ /0.1+2 x $10^{-9}$ /0.4)

= - 5.04 x $10^{-17}$ J

3 0

3 years ago