I need help for my science class

Is it true weather refers to the conditions of the atmosphere at a given time and place

damaskus [11]

yes that is true because climate is over a period of time

8 0

4 years ago

What what is the change in internal energy if 500 joule of heat is added to a system and 125 joule of work are done on a system

eduard

Answer:

DU = 375 Joules

Explanation:

Given the following data;

Quantity of heat = 500 Joules

Work done = 125 Joules

To find the change in internal energy;

Mathematically, the change in internal energy of a system is given by the formula;

DU = Q - W

Where;

DU is the change in internal energy.
Q is the quantity of energy.
W is the work done.

Substituting into the formula, we have;

DU = 500 - 125

<em>DU = 375 Joules</em>

8 0

3 years ago

Two metal spheres of different sizes are charged such that the electric potential is the same at the surface of each. Sphere A h

vodka [1.7K]

Answer:

Explanation:

Given

Radius of A is twice of B i.e.

$R_A=2R_B$

Also Potential of both sphere is same

$V_A=V_B$

$V=\frac{kQ}{R}$

thus

$k\frac{Q_A}{R_A}=K\frac{Q_B}{R_B}$

$\frac{Q_A}{Q_B}=\frac{R_A}{R_B}$

$\frac{Q_A}{Q_B}=\frac{2}{1}=2$

$\frac{Q_B}{Q_A}=\frac{1}{2}$

(b)Ratio of $\frac{E_B}{E_A}$

Electric Field is given by $E=\frac{kQ}{R^2}$

thus $E_A=\frac{kQ_A}{R_A^2}$ ----1

$E_B=\frac{kQ_B}{R_B^2}$ ----2

Divide 2 by 1

$\frac{E_B}{E_A}=\frac{Q_B}{R_B^2}\times \frac{R_A^2}{Q_A}$

$\frac{E_B}{E_A}=\frac{1}{2}\times 4=2$

8 0

3 years ago

Q|C A string with linear density 0.500 g/m is held under tension 20.0 N. As a transverse sinusoidal wave propagates on the strin

julia-pushkina [17]

A string with linear density 0.500 g/m.

Tension 20.0 N.

The maximum speed $v_{y, max}$

The energy contained in a section of string 3.00 m long as a function of $v_{y, max}$ .

We are given following data for string with linear density held under tension :

μ = 0.5 $\frac{g}{m}$

= 0.5 x 10⁻³ $\frac{kg}{m}$

T = 20 N

If string is L = 3m long, total energy as a function of $v_{y, max}$ is given by:

E = 1/2 x μ x L x ω² x A²

= 1/2 x μ x L x $v^{2} _{y, max}$

= 7.5 x 10⁻⁴ $v^{2} _{y, max}$

So, The total energy as a function of $v^{2} _{y, max}$ = 7.5 x 10⁻⁴ $v^{2} _{y, max}$

Learn more about linear density problem here:

brainly.com/question/17190616

#SPJ4

3 0

2 years ago

What is the moment of inertia of the object starting from rest if it has a final velocity of 5.9 m/s? Express the moment of iner

Bogdan [553]

Answer:

The moment of inertia is I = 0.126*R^2*M

Explanation:

We can calculate the moment of inertia of an object that starts from rest and has a final velocity using the energy conservation equation, as follows:

Ek1 + Ep1 = Ek2 + Ep2, where

Ek1 = kinetic energy of the object before to roll down

Ep1 = potential energy of the object

Ek2 = kinetic energy when the object comes down

Ep2 = potential energy of the object at the bottom

We have the follow:

Ek1 = 0

Ep1 = M*g*h

Ek2 = ((I*w)/2) + ((M*v^2)/2)

Ep2 = 0

Replacing values:

0 + M*g*h = ((I*w)/2) + ((M*v^2)/2) + 0

where:

M = mass of the object

g = gravitational acceleration

I = moment of the inertia

w = angular velocity = v/R

h = height

M*g*h = ((1/2) * I * (v^2/R^2)) + ((M*v^2)/2)

M*9.8*2 = (I*(5.9^2)/(2*R^2)) + ((5.9^2 * M)/2)

19.6 * M = ((17.4*I)/R^2) + 17.4*M

Clearing I, we have:

I = 0.126*R^2*M

5 0

3 years ago