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3 years ago

Pen and ink manufacturers are asked to submit their new ink formulations to what database?

Physics

1 answer:

Ilia_Sergeevich [38]3 years ago

3 0

The international ink library.

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When 525 J of heat is added to 7.6 g of metal at 26°C, the temperature increases to 80°C. What is the specific heat of the metal

Temka [501]

Ans the answer is 0.25J

Explanation

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3 years ago

If a toy car with a coin on top of it is rolling down a hill what force is keeping the coin on top of the car?

forsale [732]

The answer is gravity.

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3 years ago

When a piece of copper is taken to the moon , a change will be observed in it's ?

SpyIntel [72]

Weight. Because there is less gravity on the moon.

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3 years ago

147.5 ml of ethyl alcohol (coefficient of volume expansion = 1120 x 10-6K-1) at a temperature of 273.1 K is measured into a 150.

marishachu [46]

Answer:

288.2 K

Explanation:

$\beta$ = Volumetric expansion coefficient = $1120\times 10^{-6}\ /K$

$T_i$ = Initial temperature = 273.1 K

$T_f$ = Final temperature

$v_0$ = Original volume = 150 mL

Change in volume is given by

$\Delta v=\beta v_0(T_f-T_i)\\\Rightarrow T_f=\frac{\Delta v}{\beta v_0}+T_i\\\Rightarrow T_f=\frac{150-147.5}{1120\times 10^{-6}\times 147.5}+273.1\\\Rightarrow T_f=288.2\ K$

The temperature of the ethyl alcohol should be 288.2 K to reach 150 mL

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3 years ago

Aconstant current of 3 Afor 4 hours is required to charge an automotive battery. If the terminal voltage is V, where t is in hou

kirza4 [7]

Answer:

(a) 43.2 kC

(b) 0.012V kWh

(c) 0.108V cents

Explanation:

<u>Given:</u>

i = current flow = 3 A
t = time interval for which the current flow = $4\ h = 4\times 3600\ s = 14400\ s$
V = terminal voltage of the battery
R = rate of energy = 9 cents/kWh

<u>Assume:</u>

Q = charge transported as a result of charging
E = energy expended
C = cost of charging

Part (a):

We know that the charge flow rate is the electric current flow through a wire.

$\therefore i = \dfrac{Q}{t}\\\Rightarrow Q =it\\\Rightarrow Q = 3\times 14400\\\Rightarrow Q = 43200\ C\\\Rightarrow Q = 43.200\ kC\\$

Hence, 43.2 kC of charge is transported as a result of charging.

Part (b):

We know the electrical energy dissipated due to current flow across a voltage drop for a time interval is given by:

$E = Vit\\\Rightarrow E = V\times 3\times 4\\\Rightarrow E = 12V\ Wh\\\Rightarrow E = 0.012V\ kWh\\$

Hence, 0.012V kWh is expended in charging the battery.

Part (c):

We know that the energy cost is equal to the product of energy expended and the rate of energy.

$\therefore \textrm{Cost}=\textrm{Energy}\times \textrm{Rate}\\\Rightarrow C = ER\\\Rightarrow C = 0.012V\times 9\\\Rightarrow C =0.108V\ cents$

Hence, 0.108V cents is the charging cost of the battery.

4 0

4 years ago