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gregori [183]
3 years ago
8

What is the importance of measurement? ​

Physics
1 answer:
Kruka [31]3 years ago
8 0

Answer:

To know the time ,length area

You might be interested in
What would happen when two or more capacitors are connected in parallel across a potential difference?
Varvara68 [4.7K]

Answer:

The Current Iₜ = I₁ + I₂ + I₃

Charge Qₜ = Q₁ + Q₂ + Q₃

Potential difference Vₜ = V₁ = V₂ = V₃

The total capacitance Cₜ = C₁ + C₂ + C₃

Explanation:

According to the attached image;

For parallel arrangements of capacitors, the current flowing through each of the capacitors sums up to the total current flowing through the circuit;

Iₜ = I₁ + I₂ + I₃

Also the charge storage by each capacitor sums up to give the total charge stored;

Qₜ = Q₁ + Q₂ + Q₃

The potential difference across each of the capacitors are the same and equal to the total voltage across the circuit;

Vₜ = V₁ = V₂ = V₃

The total capacitance equals the sum of the capacitances of each of the capacitors;

Cₜ = C₁ + C₂ + C₃

6 0
3 years ago
If 190 grams of water is cooled from 42.7°C to 21.2° C how much energy was lost by the water?
yanalaym [24]

Answer:

Energy lost by the water is 17.1 x 10³ J .

Explanation:

Specific heat is defined as the amount of energy per unit mass needed to raise the temperature of the substance by a one degree Celsius.

Heat energy gain or loss by any substance is given by :

Q = m x C x ( T₁ - T₂ )     .....(1)

Here m is the mass of the substance, C is specific heat of the substance and T₁ and T₂ are the initial and final temperature of the substance.

According to the problem,

Mass of water, m = 190 gm

Specific heat of water, C = 4.186 J gm⁻¹ ⁰C⁻¹

Initial temperature, T₁ = 42.7⁰ C

Final temperature, T₂ = 21.2⁰ C  

Substitute these values in equation (1).

Heat energy loss by water = 190 x 4.186 x ( 21.2 - 42.7 )

                                            =  -17.1 x 10³ J

In the above value, negative sign denotes the loss in energy.

7 0
3 years ago
To initiate a nuclear reaction, an experimental nuclear physicist wants to shoot a proton into a 5.50-fm-diameter 12C nucleus. T
Fantom [35]

Answer:

V_1= 3.4*10^7m/s

Explanation:

From the question we are told that

Nucleus diameter d=5.50-fm

a 12C nucleus

Required kinetic energy K=2.30 MeV

Generally initial speed of proton must be determined,applying the law of conservation of energy we have

            K_2 +U_2=K_1+U_1

where

K_1 =initial kinetic energy

K_2 =final kinetic energy

U_1 =initial electric potential

U_2 =final electric potential

mathematically

   U_2 = \frac{Kq_pq_c}{r_2}

where

r_f=distance b/w charges

q_c=nucleus charge =6(1.6*10^-^1^9C)

K=constant

q_p=proton charge

Generally kinetic energy is know as

         K=\frac{1}{2}  mv^2

Therefore

         U_2 = \frac{Kq_pq_c}{r_2} + K_2=\frac{1}{2}  mv_1^2 +U_1

Generally equation for radius is d/2

Mathematically solving for radius of nucleus

         R=(\frac{5.50}{2}) (\frac{1*10^-^1^5m}{1fm})

         R=2.75*10^-^1^5m

Generally we can easily solving mathematically substitute into v_1

   q_p=6(1.6*10^-^1^9C)

   K_1=9.0*10^9 N-m^2/C^2

   U_1= 0

   R=2.75*10^-^1^5m

   K=2.30 MeV

   m= 1.67*10^-^2^7kg

   V_1= (\frac{2}{1.67*10^-^2^7kg})^1^/^2 (\frac{(9.0*10^9 N-m^2/C^2)*(6(1.6*10^-^1^9C)(1.6*10^-^1^9C)}{2.75*10^-^1^5m+2.30 MeV(\frac{1.6*10^-^1^3 J}{1 MeV}) }

    V_1= 3.4*10^7m/s

Therefore the proton must be fired out with a speed of V_1= 3.4*10^7m/s

8 0
3 years ago
How does the transfer of thermal energy occur?
natali 33 [55]
Heat<span> may be </span>transferred<span> by means of conduction, convection, or radiation. </span>
8 0
3 years ago
An initially uncharged 3.47-μF capacitor and a 6.43-kΩ resistor are connected in series to a 1.50-V battery that has negligible
harkovskaia [24]

Answer: a) io=233.28 A ( initial current); b) τ=R*C= 22.31 ms; c) 81.7 ms

Explanation:  In order to explain this problem we have to use, the formule for the variation of the current in a RC circuit:

I(t)=io*Exp(-t/τ)

and also we consider that io=V/R=(1.5/6.43*10^3)

=233.28 A

then the time constant for the RC circuit is τ=R*C=6.43*10^3*3.47*10^-6

=22.31 ms

Finally the time to reduce the current to 2.57% of its initial value is obtained from:

I(t)=io*Exp(-t/τ)  for I(t)/io=0.0257=Exp(-t/τ) then

ln(0.0257)*τ =-t

t=-ln(0.0257)*τ=81.68 ms

3 0
3 years ago
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