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3 years ago

Name the effect of current in electroplating

1 answer:

4 0

<span>For this, you need a compass needle, an empty matchbox, two electric cells, some wires, a plastic bottle cap and the liquid which is to be tested. Keep the compass needle inside the empty tray of matchbox and wrap a couple of rubber bands around it. Connect the wires to the battery and insert two wires in the liquid which is kept in the plastic bottle cap. You will observe that when the current flows through the wire, there is deflection in the compass needle. This shows that the given liquid conducts electricity. In fact, magnetic compass needle can detect even feeble current. LED can also be used in place of compass needle.</span>

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If you are building a clock by using a pendulum's oscillations to keep time and you find the period is a little too small (the c

Leviafan [203]

Time period of pendulum is given by

$T = 2\pi\sqrt{\frac{L}{g}}$

Here,

L is length

g is acceleration due to gravity

Therefore we can determine that the period of the pendulum is directly proportional to the square root of the length of the rope.

$T \propto \sqrt{L}$

Therefore, it is necessary to increase the length of the rope to increase the period so that it counts the seconds more slowly. The correct answer is A.

3 0

3 years ago

A jet plane travels at 40.8 km at an average speed of 340 m/s calculate how long this journey took- giveyour answer to the neare

Tju [1.3M]

Answer:

$2\ \text{minutes}$

Explanation:

d = Distance traveled = 40.8 km

s = Speed of jet = 340 m/s

Time is given by

$t=\dfrac{d}{s}$

$\Rightarrow t=\dfrac{40.8\times 10^3}{340}$

$\Rightarrow t=120\ \text{s}$

$\Rightarrow t=\dfrac{120}{60}=2\ \text{minutes}$

The time taken to complete the journey is $2\ \text{minutes}$ .

6 0

3 years ago

A soft drink (mostly water) flows in a pipe at a beverage plant with a mass flow rate that would fill 220 0.355 - L cans per min

jekas [21]

Answer:

a) 1.301 kg/s

b) 0.001301 m³/s

c) V₁ = 6.505 m/s, V₂ = 1.626 m/s

d) 118.93 kPa

Explanation:

Given:

The number of cans = 220

The volume of can, V = 0.355 L = 0.355 × 10⁻³ m³

time = 1 minute = 60 seconds

gauge pressure at point 2, P₂ = 152 kPa

b) Thus, the volume flow rate, Q = Volume/ time

Q = (220 × 0.355 × 10⁻³)/60 = 0.001301 m³/s

a) mass flow rate = Volume flow rate × density

since it is mostly water, thus density of the drink = 1000 kg/m³

thus,

mass flow rate = 0.001301 m³/s × 1000 kg/m³ = 1.301 kg/s

c) Given:

Cross section at point 1 = 2.0 cm² = 2 × 10 ⁻⁴ m²

Cross section at point 2 = 8.0 cm² = 8 × 10 ⁻⁴ m²

also,

Q = Area × Velocity

thus, for point 1

0.001301 m³/s = 2 × 10 ⁻⁴ m² × velocity at point 1 (V₁)

V₁ = 6.505 m/s

for point 2

0.001301 m³/s = 8 × 10 ⁻⁴ m² × velocity at point 1 (V₂)

V₂ = 1.626 m/s

d) Applying the Bernoulli's theorem between the points 1 and 2 we have

$P_1+\rho gV_1 + \frac{\rho V_1^2}{2}=P_2+\rho gV_2 + \frac{\rho V_2^2}{2}$

$P_1=P_2+\rho\timesg(y_2-y_1)+\frac{\rho}{2}(V_2^2-V_1^2))$

on substituting the values in the above equation, we get

$P_1=152+1000\times 9.8(1.35)+\frac{1000}{2}(1.626^2-6.505^2))$

it is given that point 1 is above point 2 thus, y₂ -y₁ is negative

$P_1=118.93\ kPa$

thus, gauge pressure at point 1 is 118.93 kPa

8 0

3 years ago

A car starts from rest and speeds up to 30 m/s in a total of 6 s.

sveta [45]

5 m/s
30 divided by 6 is 5

4 0

2 years ago

Read 2 more answers

Determine the gravitational field 300km above the surface of the earth. How does this compare to the field on the earth's surfac

Serjik [45]

The strength of the gravitational field is given by:
$g= \frac{GM}{r^2}$
where
G is the gravitational constant
M is the Earth's mass
r is the distance measured from the centre of the planet.

In our problem, we are located at 300 km above the surface. Since the Earth radius is R=6370 km, the distance from the Earth's center is:
$r=R+h=6370 km+300 km=6670 km= 6.67 \cdot 10^{6} m$

And now we can use the previous equation to calculate the field strength at that altitude:
$g= \frac{GM}{r^2}= \frac{(6.67 \cdot 10^{-11} m^3 kg^{-1} s^{-2})(5.97 \cdot 10^{24} kg)}{(6.67 \cdot 10^6 m)^2} = 8.95 m/s^2$

And we can see this value is a bit less than the gravitational strength at the surface, which is $g_s = 9.81 m/s^2$ .

4 0

3 years ago