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alina1380 [7]
3 years ago
5

7. Copper can be coated on the surface of iron but not on silver? why​

Physics
1 answer:
Lisa [10]3 years ago
4 0

Answer:

Silver has greater electronegativity (pull on electrons) than copper, so it will be reduced rather than oxidized. Silver ions will plate out on copper metal. Sir the coating of iron with copper surface can be referred as IRON-COPPER PLATING.

Explanation:

Iron is used for the electroplating of so copper because iron falls above copper in the electrochemical series whereas silver will fall off below the copper in electrochemical series that makes it not reliable for the electroplating purpose

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A tuning fork with a frequency of 335 Hz and a tuning fork of unknown frequency produce beats with a frequency of 5.3 when struc
Ganezh [65]

Answer:

Explanation:

Unknown fork frequency is either

335 + 5.3 = 340.3 Hz

or

335 - 5.3 = 329.7 Hz

After we modify the known fork, the unknown fork frequency equation becomes either

(335 - x) + 8 = 340.3

(335 - x)  = 332.3

x = 2.7 Hz

or

(335 - x) + 8 = 329.7

(335 - x) = 321.7

x = 13.3 Hz

IF the unknown fork frequency was 340.3 Hz,

THEN the 335 Hz fork was detuned to 335 - 2.7 = 332.3 Hz

IF the unknown fork frequency was 329.7 Hz,

THEN the 335 Hz fork was detuned to 335 - 13.3 = 321.7 Hz

3 0
2 years ago
A ball is thrown straight up into the air and passes a window 3 seconds after being released. It passes the same window on its w
melomori [17]

The  initial velocity of the ball is 55.125 m/s.

<h3>Initial velocity of the ball</h3>

The initial velocity of the ball is calculated as follows;

During upward motion

h = vi - ¹/₂gt²

h = vi - 0.5(9.8)(3²)

h = vi - 44.1 ----------------- (1)

During downward motion

h = vi + ¹/₂gt²

h = 0 + 0.5(9.8)(1.5)²

h = 11.025 ----------- (2)

solve (1) and (2) together, to determine the initial velocity of the ball

11.025 = vi - 44.1

vi = 11.025 + 44.1

vi = 55.125 m/s

Thus, the  initial velocity of the ball is 55.125 m/s.

Learn more about initial velocity here: brainly.com/question/19365526

#SPJ1

8 0
1 year ago
The average speed between earth and the sun is 1.50 x10^8 km. Calculate the average speed of the Earth in its orbit in kilometer
cluponka [151]

Answer:

The average speed of the earth in its orbit is 29.86km/s

Explanation:

The average distance between the Earth and the Sun is 1.50x10^{8} km.

The average speed of the earth in its orbit can be found by the next equation :

v = \frac{2 \pi r}{T}  (1)

Where r is the radius and T is the period.

In this case, the orbit of the Earth can be considered as a circle

(r = 1.50x10^{8}km) instead of an ellipse.

It takes 1 year to the Earth to make one revolution around the Sun. Therefore, its period will be 365.25 days.

Notice that to express the period in terms of seconds, the following is needed:

365.25d . \frac{86400s}{1d} ⇒ 31557600s

Then, equation 1 can be used:

v = \frac{2 \pi (1.50x10^{8}km)}{31557600s}

v = 29.86km/s

7 0
3 years ago
Read 2 more answers
What component of Star Wars is possible
kolezko [41]
Comparing to current technology, here are some component of star wars that is  possible :

- The drones
Remember the flying drones that could shoot stuff in star wars ? we already made that

- Prosthetic limb
Luke's bionic hand in star wars ? we also already made that

- The probe droids
We used a similar thing when our scientist done an expedition to the moon
7 0
2 years ago
Read 2 more answers
A body of mass 2 kg is moving in the positive X-Direction with a speed of 4 m/s collides head on with an another body of mass 3
Inga [223]
m_1=2 \\ m_2=3 \\ v_1=4 \\ v_2=1 \\ v\text{ =speed after collision (to be determined)}.

The momentul of the system preserves:

m_1v_1-m_2v_2=(m_1+m_2)v \ \ \ \ \ \Rightarrow \ \ \ \ \ v=\dfrac{m_1v_1-m_2v_2}{m_1+m_2}.

Ok, we found the speed after the collision.
Now, because the impact is plastic, it produces heat, sound energy and who knows what other forms of energy. We denote all this wasted energy with E.

Now, we write the energy conservation law:

\dfrac{m_1v_1^2}{2}+\dfrac{m_2v^2_2}{2}=\dfrac{(m_1+m_2)v^2}{2}+E

From the above equation, you find E,  and then conclude that the sound energy can certainly not be greater than this.
8 0
3 years ago
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