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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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The compressor on an air conditioner draws 35.0 A when it starts up. If the start-up time is 0.59 s, how much charge passes a cr
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6 0
3 years ago
(15 Points)
oksian1 [2.3K]

The vertical weight carried by the builder at the rear end is F = 308.1 N

<h3>Calculations and Parameters</h3>

Given that:

The weight is carried up along the plane in rotational equilibrium condition

The torque equilibrium condition can be used to solve

We can note that the torque due to the force of the rear person about the position of the front person = Torque due to the weight of the block about the position of the front person

This would lead to:

F(W*cosθ) = mgsinθ(L/2) + mgcosθ(W/2)

F(1cos20)= 197/2(3.10sin20 + 2 cos 20)

Fcos20= 289.55

F= 308.1N

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5 0
2 years ago
The 1.53-kg uniform slender bar rotates freely about a horizontal axis through O. The system is released from rest when it is in
OlgaM077 [116]

Answer:

The spring constant = 104.82 N/m

The angular velocity of the bar when θ = 32° is 1.70 rad/s

Explanation:

From the diagram attached below; we use the conservation of energy to determine the spring constant by using to formula:

T_1+V_1=T_2+V_2

0+0 = \frac{1}{2} k \delta^2 - \frac{mg (a+b) sin \ \theta }{2}  \\ \\ k \delta^2 = mg (a+b) sin \ \theta \\ \\ k = \frac{mg(a+b) sin \ \theta }{\delta^2}

Also;

\delta = \sqrt{h^2 +a^2 +2ah sin \ \theta} - \sqrt{h^2 +a^2}

Thus;

k = \frac{mg(a+b) sin \ \theta }{( \sqrt{h^2 +a^2 +2ah sin \ \theta} - \sqrt{h^2 +a^2})^2}

where;

\delta = deflection in the spring

k = spring constant

b = remaining length in the rod

m = mass of the slender bar

g = acceleration due to gravity

k = \frac{(1.53*9.8)(0.6+0.2) sin \ 64 }{( \sqrt{0.6^2 +0.6^2 +2*0.6*0.6 sin \ 64} - \sqrt{0.6^2 +0.6^2})^2}

k = 104.82\ \  N/m

Thus; the spring constant = 104.82 N/m

b

The angular velocity can be calculated by also using the conservation of energy;

T_1+V_1 = T_3 +V_3  \\ \\ 0+0 = \frac{1}{2}I_o \omega_3^2+\frac{1}{2}k \delta^2 - \frac{mg(a+b)sin \theta }{2} \\ \\ \frac{1}{2} \frac{m(a+b)^2}{3}  \omega_3^2 +  \frac{1}{2} k \delta^2 - \frac{mg(a+b)sin \ \theta }{2} =0

\frac{m(a+b)^2}{3} \omega_3^2  + k(\sqrt{h^2+a^2+2ah sin \theta } - \sqrt{h^2+a^2})^2 - mg(a+b)sin \theta = 0

\frac{1.53(0.6+0.6)^2}{3} \omega_3^2  + 104.82(\sqrt{0.6^2+0.6^2+2(0.6*0.6) sin 32 } - \sqrt{0.6^2+0.6^2})^2 - (1.53*9.81)(0.6+0.2)sin \ 32 = 0

0.7344 \omega_3^2 = 2.128

\omega _3 = \sqrt{\frac{2.128}{0.7344} }

\omega _3 =1.70 \ rad/s

Thus, the angular velocity of the bar when θ = 32° is 1.70 rad/s

7 0
3 years ago
At what speed does a 1,248 kg compact car have the same kinetic energy as a 18,777 kg truck going 26 km/h?
notka56 [123]

The speed of car is 100.8km/h

KE =  \frac{1}{2} m(v \: truck ) {}^{2}

= 0.5 \times 18777 \times  \frac{26}{3.6} \times  \frac{26}{3.6}

= 489708.79j

=  \frac{1}{2}  \times 1248 \times( v \: car) {}^{2}

= 624v {}^{2}

so \: v {}^{2}  =  \frac{489708.7}{624}

= 784

v =  \sqrt{784}

v = 28m/s

v car= 28×3.6

=100.8km/h

Hence, the speed of the car is 100.8km/h

learn more about speed from here:

brainly.com/question/28326855

#SPJ4

3 0
1 year ago
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