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Readme [11.4K]
3 years ago
8

An electroplating solution is made up of nickel(II) sulfate. How much time would it take to deposit 0.500 g of metallic nickel o

n a custom car part using a current of 3.00 A
Physics
1 answer:
ELEN [110]3 years ago
6 0

Answer:

1.52 hour

Explanation:

M = 0.5 g, I = 3 A

Electrochemical equivalent of nickel

Z = 3.04 × 10^(-4) g/C

By use of Faraday's laws of electrolysis

M = Z I t

t = M / Z I

t = 0.5 / (3.04 × 10^-4 × 3)

t = 5482.45 second = 1.52 hour

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A construction crane lowers a beam into place at constant speed. Consider the work Wg done by gravity on the beam and the work W
aalyn [17]

Answer:

Wg is positive and WT negative.

(Letters in options are all wrongly written).

Explanation:

Remember that the work of a force is the internal product between the force and the displacement W=F.d.

Since the displacement is downwards like the weight, the work done by gravity is positive, while the work done by the tension is negative since it points upwards.

5 0
3 years ago
Which change would create light?
hichkok12 [17]

Answer:

Im not 100% sure but i think the answer is A. An electron in an atom jumping from a lower energy state to a higher one.

Explanation:

lmk if its wrong

6 0
3 years ago
The 500 series bullet train travels between Tokyo and Hakata, Japan. If it is heading north with a momentum of 13,194,098.64 kg
Hitman42 [59]

Answer:

The mass of the 500 series bullet train is 181.4 tonnes.

Explanation:

The momentum is given by:

p = mv

Where:

m: is the mass =?

v: is the velocity = 261.8 km/h

p: is the momentum = 13194098.64 kg*m/s

By solving the above equation for "m" we have:

m = \frac{p}{v} = \frac{13,194,098.64 kg*m/s}{261.8 km/h*\frac{1000 m}{1 km}*\frac{1 h}{3600 s}} = 181.4 tonnes

Therefore, the mass of the 500 series bullet train is 181.4 tonnes.

I hope it helps you!                

5 0
3 years ago
A cube has a drag coefficient of 0.8. what would be the terminal velocity of a sugar cube 1 cm on a side in air (ρ = 1.2 kg/m3)?
kramer
To answer this question, we should know the formula for the terminal velocity. The formula is written below:

v = √(2mg/ρAC)
where
m is the mass
g is 9.81 m/s²
ρ is density
A is area
C is the drag coefficient

Let's determine the mass, m, to be density*volume.
Volume = s³ = (1 cm*1 m/100 cm)³ = 10⁻⁶ m³
m = (1.6×10³ kg/m³)(10⁻⁶ m³) = 1.6×10⁻³ kg
A = (1 cm * 1 m/100 cm)² = 10⁻⁴ m²

v = √(2*1.6×10⁻³ kg*9.81 m/s²/1.6×10³ kg/m³*10⁻⁴ m²*0.8)
<em>v = 0.495 m/s</em>
7 0
3 years ago
A ball with 100 J of PE is released from a height of 10 m. What will be the KE of the ball at 5
harkovskaia [24]

Answer:

The kinetic energy is: 50[J]

Explanation:

The ball is having a potential energy of 100 [J], therefore

PE = [J]

The elevation is 10 [m], and at this point the ball is having only potential energy, the kinetic energy is zero.

E_{p} =m*g*h\\where:\\g= gravity[m/s^{2} ]\\m = mass [kg]\\m= \frac{E_{p} }{g*h}\\ m= \frac{100}{9.81*10}\\\\m= 1.01[kg]\\\\

In the moment when the ball starts to fall, it will lose potential energy and the potential energy will be transforme in kinetic energy.

When the elevation is 5 [m], we have a potential energy of

P_{e} =m*g*h\\P_{e} =1.01*9.81*5\\\\P_{e} = 50 [J]\\

This energy is equal to the kinetic energy, therefore

Ke= 50 [J]

8 0
3 years ago
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