What is the advantage of having two eyes instead of one eye in humans?

A platinum ball weighing 100 g is removed from a furnace and dropped into 400 g of water at 0 degree C. If the equilibrium tempe

kakasveta [241]

Answer:

T = 1010 degree Celsius

Explanation:

mass of ball (Mb) = 100 g

mass of water (Mw) = 400 g

temp of water = 0 degree

specific heat of platinum (C) = 0.04 cal/g degree celsius

we can calculate the temperature of the furnace from the equation before

Mb x C x (temp of furnace (T) - equilibrium temp) = Mw x (equilibrium temp - temp of furnace)

100 x 0.04 x ( T - 10) = 400 x (10 - 0)

4 (T - 10) = 4000

T - 10 = 1000

T = 1010 degree Celsius

3 0

3 years ago

Applying the Law of Conservation of Energy. If a car was released down the track from a height what happens to the potential ene

erastova [34]

Answer:

According to the law of conservation of energy, energy cannot be created or destroyed, although it can be changed from one form to another. KE + PE = constant. A simple example involves a stationary car at the top of a hill. As the car coasts down the hill, it moves faster and so it’s kinetic energy increases and it’s potential energy decreases. On the way back up the hill, the car converts kinetic energy to potential energy. In the absence of friction, the car should end up at the same height as it started.

This law had to be combined with the law of conservation of mass when it was determined that mass can be inter-converted with energy.

One can also imagine the energy transformation in a pendulum. When the ball is at the top of its swing, all of the pendulum’s energy is potential energy. When the ball is at the bottom of its swing, all of the pendulum’s energy is kinetic energy. The total energy of the ball stays the same but is continuously exchanged between kinetic and potential forms

4 0

2 years ago

The magnitude of Earth’s magnetic field is about 0.5 gauss near Earth’s surface. What’s the maximum possible magnetic force on a

vampirchik [111]

Answer:

$F = 1.5 \times 10^{-16} N$

this force is $1.68 \times 10^{13}$ times more than the gravitational force

Explanation:

Kinetic Energy of the electron is given as

$KE = 1 keV$

$KE = 1 \times 10^3 (1.6 \times 10^{-19}) J$

$KE = 1.6 \times 10^{-16} J$

now the speed of electron is given as

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

now we have

$v = \sqrt{\frac{2 KE}{m}}$

$v = 1.87 \times 10^7 m/s$

now the maximum force due to magnetic field is given as

$F = qvB$

$F = (1.6\times 10^{-19})(1.87 \times 10^7)(0.5 \times 10^{-4})$

$F = 1.5 \times 10^{-16} N$

Now if this force is compared by the gravitational force on the electron then it is

$\frac{F}{F_g} = \frac{1.5 \times 10^{-16}}{9.1 \times 10^{-31} (9.8)}$

$\frac{F}{F_g} = 1.68 \times 10^{13}$

so this force is $1.68 \times 10^{13}$ times more than the gravitational force

4 0

3 years ago

If 745-nm and 660-nm light passes through two slits 0.54 mm apart, how far apart are the second-order fringes for these two wave

kotegsom [21]

Answer:

0.82 mm

Explanation:

The formula for calculation an $n^{th}$ bright fringe from the central maxima is given as:

$y_n=\frac{n \lambda D}{d}$

so for the distance of the second-order fringe when wavelength $\lambda_1$ = 745-nm can be calculated as:

$y_2 = \frac{n \lambda_1 D}{d}$

where;

n = 2

$\lambda_1$ = 745-nm

D = 1.0 m

d = 0.54 mm

substituting the parameters in the above equation; we have:

$y_2 = \frac{2(745nm*\frac{10^{-9m}}{1.0nm}(1.0m) }{0.54 (\frac{10^{-3m}} {1.0mm})}$

$y_2$ = 0.00276 m

$y_2$ = 2.76 × 10 ⁻³ m

The distance of the second order fringe when the wavelength $\lambda_2$ = 660-nm is as follows:

$y^'}_2 = \frac{2(660nm*\frac{10^{-9m}}{1.0nm}(1.0m) }{0.54 (\frac{10^{-3m}} {1.0mm})}$

$y^'}_2$ = 1.94 × 10 ⁻³ m

So, the distance apart the two fringe can now be calculated as:

$\delta y = y_2-y^{'}_2$

$\delta y$ = 2.76 × 10 ⁻³ m - 1.94 × 10 ⁻³ m

$\delta y$ = 10 ⁻³ (2.76 - 1.94)

$\delta y$ = 10 ⁻³ (0.82)

$\delta y$ = 0.82 × 10 ⁻³ m

$\delta y$ = 0.82 × 10 ⁻³ m $(\frac{1.0mm}{10^{-3}m} )$

$\delta y$ = 0.82 mm

Thus, the distance apart the second-order fringes for these two wavelengths = 0.82 mm

6 0

3 years ago

Two 10 kg pucks head straight towards each other with velocities of 10 m/s and -20 m/s. They collide and stick together. Calcula

RUDIKE [14]

The final velocity of the two pucks is -5 m/s

Explanation:

We can solve the problem by using the law of conservation of momentum.

In fact, in absence of external force, the total momentum of the two pucks before and after the collision must be conserved - so we can write:

$m_1 u_1 + m_2 u_2 = (m_1 +m_2)v$

where

$m_1 = m_2 = m = 10 kg$ is the mass of each puck

$u_1 = 10 m/s$ is the initial velocity of the 1st puck

$u_2 = -20 m/s$ is the initial velocity of the 2nd puck

v is the final velocity of the two pucks sticking together

Re-arranging the equation and solving for v, we find:

$mu_1 + mu_2 = (m+m)v\\u_1 + u_2 = 2v\\v=\frac{u_1+u_2}{2}=\frac{10-20}{2}=-5 m/s$

Learn more about momentum:

brainly.com/question/7973509

brainly.com/question/6573742

brainly.com/question/2370982

brainly.com/question/9484203

#LearnwithBrainly

8 0

2 years ago