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Dimas [21]
3 years ago
6

A sculptor has asked you to help electroplate gold onto a brass statue. You know that the charge carriers in the ionic solution

are singly ionized gold ions, and you've calculated that you must deposit 0.560g of gold to reach the necessary thickness.
How much current do you need, in mA, to plate the statue in 4.70hours ?
Physics
1 answer:
scZoUnD [109]3 years ago
3 0

Answer:

ampere = 15.95 mA

Explanation:

given data

deposit gold = 0.560 g

time = 4.70 hour  = 16920 s

solution

we get here moles of gold that is

moles of gold = 0.560 g ÷ 196.967 g/mol

moles of gold = 0.0028

moles of gold  =  0.0028 ( 1 Faraday / 1 mol )

moles of gold  = 0.0028 Faradays

moles of gold  = 0.0028 Faradays  × 96500 Coulom/ faraday

moles of gold  = 270.2 coulombs  

and we get here ampere that is

ampere = \frac{270.2}{16920}

ampere = 0.01596 A

ampere = 15.95 mA

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Please help.. I'm desperate.
rosijanka [135]

Answer:

Newton's Second Law tells us that the more mass an object has, the more force is needed to move it. A larger rocket will need stronger forces (eg. more fuel) to make it accelerate. The space shuttles required seven pounds of fuel for every pound of payload they carry.

Explanation:

7 0
2 years ago
A 25,000 kg traveling east collides with a 2,000 kg truck standing still on the tracks. After the collision the train and truck
Elis [28]

Answer:

24.084 m/s

Explanation:

From the law of conservation of linear momentum

Total momentum before collision equals to the total momentum after collision

Since momentum=mv where m is mass and v is velocity

M_{truck}V_{truck}=V_{common}*(M_{truck} +M_{standing}) where M_{truck} is the mass of the truck, V_{truck} is velocity of the truck, V_{common} is the common velocity of moving and standing truck after collision and M_{standing} is the mass of the standing truck

Making V_{truck} the subject we obtain

V_{truck}=\frac { V_{common}*(M_{truck} +M_{standing})}{M_{truck}}

Substituting M_{truck} as 25000 Kg, V_{common} as 22.3 m/s, M_{standing} as 2000 Kg we obtain

V_{truck}=\frac { 22.3 m/s *(25000 Kg +2000 Kg)}{25000}= 24.084 m/s

Therefore, assuming no friction and considering that after collision they still move eastwards hence common velocity and initial truck velocities are positive

The truck was moving at 24.084 m/s

3 0
3 years ago
Block on inclined plane experience a force due to gravity of 300N straight down. If the slope is inclined at 67.8°to the horizon
Tems11 [23]

Answer:

The component of the force due to gravity perpendicular and parallel to the slope is  113.4 N and 277.8 N respectively.

Explanation:

Force is any cause capable of modifying the state of motion or rest of a body or of producing a deformation in it. Any force can be decomposed into two vectors, so that the sum of both vectors matches the vector before decomposing. The decomposition of a force into its components can be done in any direction.

Taking into account the simple trigonometric relations, such as sine, cosine and tangent, the value of their components and the value of the angle of application, then the parallel and perpendicular components will be:

  • Fparallel = F*sinα =300 N*sin 67.8° =300 N*0.926⇒ Fparallel =277.8 N
  • Fperpendicular = F*cosα =  300 N*cos 67.8° = 300 N*0.378 ⇒ Fperpendicular= 113.4 N

<u><em>The component of the force due to gravity perpendicular and parallel to the slope is  113.4 N and 277.8 N respectively.</em></u>

6 0
3 years ago
A bullet B of mass mB traveling with a speed v0 = 1400 m/s ricochets off a fixed steel plate A of mass mA. Let mA ≫ mB so that i
Greeley [361]

Answer:

Rebounce angle is 345°

Rebounce speed is 989.95m/s

Explanation:

Calculate the x  component of the velocity of the bullet before impact by using the following relation:

Vbx= Vb Cos thetha

Here,  is the initial velocity of the bullet, Vo = 1400m/s and is the incidence angle of the bullet.= theta = 15°

Substituting

Vbx = Cos15 ×1400 = 1352.30m/s

Calculate the y component using the relation:

Vby = Vo Sin theta

Vby = sin 15° × 1400

Vby = 362.35m/s

The rebounce angle = 360 - incidence angle

Rebounce angle =( 360 - 15)° = 345°

The rebound speed V' = Vby - Vbx

V' = (1352.30 - 362.35)m/s

V' = 989.95 m/s

5 0
3 years ago
What will happen to this current if a magnet is brought near the cord? A. It will exert a force on the voltage. B. The electric
Andreyy89

Answer:

The correct answer is D)

Explanation:

When an electric magnet is brought near a cord with an electric current, the cord will most likely deflect away from the magnet because electric fields flowing through a wire generates its own magnetic field.

Cheers!

8 0
3 years ago
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