________ law is called the law of partial pressure

How many joules are required to heat .250 kg of liquid water from 0 °C to 100 °C ?

dedylja [7]

Answer:

Specific heat of water is 4.186 J/g/C. The heat required to raise the temperature by

is

Here is mass of water being heated, specific heat of water and change in temperature.

Here .

Heat energy required is

Explanation:

4 0

3 years ago

What is the kinetic energy of a 120-cm thin uniform rod with a mass of 450 g that is rotating about its center at 3.60 rad/s?

goldfiish [28.3K]

Answer:

1.05 J.

Explanation:

Kinetic Energy: This is the energy possessed by a body due to its motion. The S.I unit of kinetic energy is Joules (J). The formula of kinetic energy is given as

Ek = 1/2mv²................. Equation 1

Where Ek = kinetic energy, m = mass of the uniform rod, v = liner velocity of the rod.

But,

v = αr .......................... Equation 2

Where α = angular velocity of the rod, r = radius of the circle.

Given: α = 3.6 red/s, r = 120/2 = 60 cm = 0.6 m.

Substitute into equation 2

v = 3.6(0.6)

v = 2.16 m/s.

Also given: m = 450 g = 0.45 kg.

Substitute into equation 1

Ek = 1/2(0.45)(2.16²)

Ek = 1.05 J.

4 0

3 years ago

The question is in the picture

Sedbober [7]

Answer:

e) 120m/s

Explanation:

When the ball reaches its highest point, its velocity becomes zero, meaning

$v_0-gt = 0$ .

where $v_0$ is the initial velocity.

Solving for $t$ we get

$t = \dfrac{v_0}{g}$

which is the time it takes the ball to reach the highest point.

Now, after the ball has reached its highest point, it turns around and falls downwards. After time $t_0$ since it had reached the highest point, the ball has traveled downwards and the velocity $v_f$ it has gained is

$v_f = gt_0$ ,

and we are told that this is twice the initial velocity $v_0$ ; therefore,

$v_f = 2v_0 = gt_0$

which gives

$t_0 = \dfrac{2v_0}{g}.$

Thus, the total time taken to reach velocity $2v_0$ is

$t_{tot} = t+t_0 = \dfrac{v_0}{g}+\dfrac{2v_0}{g}$

$t_{tot} = \dfrac{3v_0}{g}.$

This $t_{tot}$ , we are told, is 36 seconds; therefore,

$36= \dfrac{3v_0}{g},$

and solving for $v_0$ we get:

$v_0 = \dfrac{36g}{3}$

$v_0 = \dfrac{36s(10m/s^2)}{3}$

$\boxed{v_0 = 120m/s}$

which from the options given is choice e.

7 0

4 years ago

According to the Heisenberg uncertainty principle, if the uncertainty in the speed of an electron is 3.5 × 103 m/s, the uncertai

GREYUIT [131]

Explanation:

It is given that,

Uncertainty in the speed of an electron, $\Delta v=3.5\times 10^3\ m/s$

According to Heisenberg uncertainty principle,

$\Delta x.\Delta p=\dfrac{h}{4\pi}$

$\Delta x$ is the uncertainty in the position of an electron

Since, $\Delta p=m\Delta v$

$\Delta x=\dfrac{h}{4\pi.m \Delta v}$

$\Delta x=\dfrac{6.6\times 10^{-34}}{4\pi\times 9.1\times 10^{-31}\times 3.5\times 10^3}$

$\Delta x=1.64\times 10^{-8}\ m$

So, the uncertainty in its position is $1.64\times 10^{-8}\ m$ . Hence, this is the required solution.

6 0

3 years ago

Two balls of equal size are dropped from the same height from the roof of a building. the mass of ball a is twice that of ball b

Sindrei [870]

The mass of ball a is twice the mass of ball b:
$m_a = 2 m_b$
This means that the initial potential energy of ball a ( $U_a = m_a gh=2 m_b gh$ ) is twice the potential energy of ball b ( $U_b = m_b g h$ ):
$U_a = 2 U_b$
When the two balls reach the ground, the potential energy of each ball has converted into kinetic energy (since now their altitude is h=0), because the total mechanical energy of each ball must be conserved. Therefore:
$K_a = U_a$
$K_b = U_b$
and so the kinetic energy of ball a must be twice the kinetic energy of ball b:
$K_a = 2 K_b$

8 0

3 years ago

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