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wolverine [178]

3 years ago

7

Physical quantity are measurable quantities

1 answer:

Vilka [71]3 years ago

7 0

Answer:

yeah physical quantities are the quantities which can be meaured

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e force acting between two charged particles A and B is 5.2 × 10-5 newtons. Charges A and B are 2.4 × 10-2 meters apart. If the

anastassius [24]

The force acting between the particles is

$F=k \frac{Q_{1}Q_{2}}{r^2}$
Then
$Q_{2}= \frac{5.2 \times 10^-^5 \times 0.024^2}{ 9.0 \times 10^9 7.2 \times 10^-^8} =4.622 \times 10^-^1^1C$

7 0

3 years ago

Read 2 more answers

How long was a 60 W light bulb turned on if it used a total of 580 J of energy?

icang [17]

Here's the tool you need. You can't answer the question without this:

   "1 watt"
means
   "1 joule of energy, generated, used, or moved, every second".

So 60 watts = 60 joules per second

   Total energy generated,
      used, or moved = (power) x (time).

   580 joules = (60 watts) x (time)

Divide each side
by (60 watts): Time = (580 joules) / (60 joules/sec)

   = (9 and 2/3) seconds .

7 0

3 years ago

A cylindrical capacitor has an inner conductor of radius 2.7 mmmm and an outer conductor of radius 3.1 mmmm. The two conductors

Mars2501 [29]

Answer:

(A) Capacitance per unit length = $4.02 \times 10^{-10}$

(B) The magnitude of charge on both conductor is $Q = 4.22 \times 10^{-19}$ C and the sign of charge on inner conductor is $+Q$ and the sign on outer conductor is $-Q$

Explanation:

Given :

Radius of inner part of conductor $(R_{1})$ = $2.7 \times 10^{-3}$ m

Radius of outer part of conductor $(R_{2})$ = $3.1 \times 10^{-3}$ m

The length of the capacitor $(l)$ = $3 \times 10^{-3}$ m

(A)

Capacitance is purely geometrical property. It depends only on length, radius of conductor.

From the formula of cylindrical capacitor,

$C = \frac{2\pi\epsilon_{o} l }{ln\frac{R_{2} }{R_{1} } }$

Where, $\epsilon_{o} = 8.85 \times 10^{-12}$

But we need capacitance per unit length so,

$\frac{C}{l} = \frac{2\pi\epsilon_{o} }{ln\frac{R_{2} }{R_{1} } }$

capacitance per unit length = $\frac{6.28 \times 8.85 \times 10^{-12} }{ln(1.148)} = 4.02 \times 10^{-10}$

(B)

The charge on both conductors is given by,

$Q = C \Delta V$

Where, C = capacitance of cylindrical capacitor and value of $C = 12.06 \times 10^{-13}$ F, $\Delta V = 350 \times 10^{-3}$ V

∴ $Q = 4.22 \times 10^{-19}$ C

The magnitude of charge on both conductor is same as above but the sign of charge is different.

Charge on inner conductor is $+Q$ and Charge on outer conductor is $-Q$ .

8 0

3 years ago

aleksandrvk [35]

ribosomes

an animal cell with a nucleus

animal

7 0

3 years ago

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Question 2 Ammonium phosphate NH43PO4 is an important ingredient in many fertilizers. It can be made by reacting phosphoric acid

Mama L [17]

Answer:

The answer is 12.27 g (NH4)3PO4

Explanation:

Step 1: balance the chemical equation:

H3PO4 + 3NH3 → (NH4)3PO4

step 2: the molar masses of each of the reagents and products are calculated using the periodic table:

H3PO4:

3 atoms of H: 3x1=3 g/mol

1 atom of P: 1x30.97=30.97 g/mol

4 atoms of O: 4x16=64 g/mol

Molar mass=3+30.97+64=97.97 g/mol

3NH3:

3 atoms of N: 3x14=42 g/mol

9 atoms of H: 9x1=9 g/mol

Molar mass=42+9=51 g/mol

(NH4)3PO4:

3 atoms of N: 3x14=42 g/mol

12 atoms of H: 12x1=12 g/mol

1 atom of P: 1x30.97=30.97 g/mol

4 atoms of O: 4x16=64 g/mol

Molar mass=42+12+30.97+64=148.97 g/mol

we make a rule of three to calculate the amount of ammonium phosphate:

51 g NH3------------------148.97 g (NH4)3PO4

4.2 g NH3-----------------x g (NH4)3PO4

Clearing the x, we have:

$x g (NH4)3PO4 = \frac{(4.2 g NH3)x(148.97 g (NH4)3PO4)}{51 g NH3}=12.27 g (NH4)3PO4$

7 0

3 years ago