What does it mean for a chemical equation to be balanced?

Please help me on 3,4 and 6. Thanks

rusak2 [61]

3. The second piston will experience the same force as compared with the first. This is because since the pressure is the same everywhere inside the fluid system, the force is proportional to the surface area. We are told that both the first and the second piston have the same surface area, therefore, they will both experience the same force/pressure.

4. The situation is much the same as number 3 above, with the exception that the second piston is twenty times larger than the first. Again, since the pressure is the same everywhere inside the fluid system, the force is proportional to the surface area. We are told that the second piston is 20 times larger than the first, therefore, the larger piston will experience 20 times larger the force of the small one.

6. The answer is TRUE. The hydraulic braking system of most cars makes use of a vacuum servo (or booster), which is located between the brake pedal and the master cylinder piston. This vacuum servo amplifies the force applied from the brake pedal.

3 0

4 years ago

What conclusions can you draw about the connection between the Sun’s altitude and the latitude of the observer on seasonal chang

Cerrena [4.2K]

Answer:

Explanation:

Altitude of the Sun and the latitude position on the earth play an important role in the season change on the earth.

When the altitude of the sun is high then the average temperature of the earth is higher because the luminous intensity of the sun rays is higher due to the focusing of high energy sun rays over a small area.

But when the sun is at higher altitudes we receive less denser rays of the sun and hence we have less heat on the earth on an average.

But despite of the altitude some places on the earth have distinct temperature than the other place at the same time of the year. This is due to their latitudinal location. The places near the equator are warmer most of the times throughout the year because they receive the most direct rays while the poles receive slanting rays and hence are colder even in summer when the earth is at lower altitudes.

6 0

3 years ago

1) You slam on the brakes of your car in a panic, and skid a certain distance on a straight level road. If you had been travelin

aleksandr82 [10.1K]

Answer:

d = 4 d₀o

Explanation:

We can solve this exercise using the relationship between work and the variation of kinetic energy

W = ΔK

In that case as the car stops v_f = 0

the work is

W = -fr d

we substitute

- fr d₀ = 0 - ½ m v₀²

d₀ = ½ m v₀² / fr

now they indicate that the vehicle is coming at twice the speed

v = 2 v₀

using the same expressions we find

d = ½ m (2v₀)² / fr

d = 4 (½ m v₀² / fr)

d = 4 d₀o

3 0

3 years ago

Violet light of wavelength 405 nm ejects electrons with a maximum kinetic energy of 0.890 eV from a certain metal. What is the b

AleksandrR [38]

Answer:

Ф = 2.179 eV

Explanation:

This exercise has electrons ejected from a metal, which is why it is an exercise on the photoelectric effect, which is explained assuming the existence of energy quanta called photons that behave like particles.

E = K + Ф

the energy of the photons is given by the Planck relation

E = h f

we substitute

h f = K + Ф

Ф= hf - K

the speed of light is related to wavelength and frequency

c = λ f

f = c /λ

Φ = $\frac{hc}{\lambda } - K$

let's reduce the energy to the SI system

K = 0.890 eV (1.6 10⁻¹⁹ J / 1eV) = 1.424 10⁻¹⁹ J

calculate

Ф = 6.63 10⁻³⁴ 3 10⁸/405 10⁻⁹ -1.424 10⁻¹⁹

Ф = 4.911 10⁻¹⁹ - 1.424 10⁻¹⁹

Ф = 3.4571 10⁻¹⁹ J

we reduce to eV

Ф = 3.4871 10⁻¹⁹ J (1 eV / 1.6 10⁻¹⁹ J)

Ф = 2.179 eV

4 0

3 years ago

A small mailbag is released from a helicopter that is descending steadily at 3 m/s.

mario62 [17]

Answer:

a) Speed of mailbag after 3 seconds = 32.4 m/s

b) Package is 44.1 meter below helicopter

c) If the helicopter was rising steadily at 3.00 m/s

Speed of mailbag after 3 seconds = 26.4 m/s

Package is 44.1 meter below helicopter

Explanation:

a) We have equation of motion, v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration and t is the time taken.

Initial velocity = 3 m/s, acceleration = 9.8 $m/s^2$ and time = 3 seconds.

v = 3+9.8*3 = 32.4 m/s

Speed of mailbag after 3 seconds = 32.4 m/s

b) We have equation of motion , $s= ut+\frac{1}{2} at^2$ , s is the displacement, u is the initial velocity, a is the acceleration and t is the time.

Velocity of helicopter = 3 m/s, time taken = 3 seconds, acceleration = 0 $m/s^2$ .

$s= 3*3+\frac{1}{2} *0*3^2\\ \\ s=9m$

Distance traveled by helicopter = 9 meter.

Velocity of package = 3 m/s, time taken = 3 seconds, acceleration = 9.8 $m/s^2$ .

$s= 3*3+\frac{1}{2} *9.8*3^2\\ \\ s= 53.1m$

Distance traveled by package = 53.1 meter.

So package is (53.1-9)meter below helicopter = 44.1 m

c) Initial velocity = -3 m/s, acceleration = 9.8 $m/s^2$ and time = 3 seconds.

v = -3+9.8*3 = 26.4 m/s

Speed of mailbag after 3 seconds = 26.4 m/s

Velocity of helicopter = -3 m/s, time taken = 3 seconds, acceleration = 0 $m/s^2$ .

$s= -3*3+\frac{1}{2} *0*3^2\\ \\ s=-9m$

Distance traveled by helicopter = 9 meter.

Velocity of package = -3 m/s, time taken = 3 seconds, acceleration = 9.8 $m/s^2$ .

$s= -3*3+\frac{1}{2} *9.8*3^2\\ \\ s= 35.1m$

Distance traveled by package = 35.1 meter.

So package is (35.1+9)meter below helicopter = 44.1 m

4 0

3 years ago