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3 years ago

Describe how the design of a vacuum flask keeps the liquid inside hot ?

2 answers:

4 0

The gap between the two flasks is partially evacuated of air, creating a near-vacuum which significantly reduces heat transfer by conduction or convection. Vacuum flasks are used domestically to keep beverageshot or cold.

JulijaS [17]3 years ago

3 0

The gap between the two flasks is partially evacuated of air creating a near vacuum which significantly reduces heat transfer by conduction or convection

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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

Q.Solve the following circuit find total resistance RT. Also find value of voltage across resister RC.

vagabundo [1.1K]

Answer:

R_total = 14.57 Ω , V_C = 1.176 V

Explanation:

To solve this circuit we are going to find the equivalent resistance of each branch, let's remember

* Serial resistance

$R_{eq}$ = ∑ $R_{i}$

* For resistance in parallel

1 / R_{eq} = ∑ 1/R_{i}

We solve the two branches of the wheatstone bridge

Series resistors

Branch B

R_B = Rb + R4

R_B = 2 + 18

R_B = 20 Ω

Branch C

R_C5 = Rc + R5

R_C5 = 3 + 12

R_C5 = 15 Ω

Resistance in parallel R_B and R_C5

1 / R_BC = 1 / R_B + 1 / R_C5

1 / R_BC = 1/20 + 1/15 = 0.116666

R_BC = 8.57 Ω

Now we have a single branch, we solve the series resistance

R_total = R_A + R_BC

R_total = 6 + 8.57

R_total = 14.57 Ω

b) they ask us for the voltage in the resistance R_C

Let's remember that the voltage in a series circuit is the sum of the voltages

10 = V_a + V_BC

10 = i R_a + i R_BC = i (R_a + R_BC)

i = 10 / (R_a + R_BC)

i = 10 / (14.57)

i = 0.6863 A

The current in the series circuit is constant

V_BC = i R_BC

V_BC = 0.6863 8.57

V_BC = 5.8819 V

This voltage is divided in the bridge, for the two branches in parallel it is the same, but the resistance is different in each branch.

Branch C

V_BC = i R_C5

i = V_BC / R_C5

i = 5.8819 / 15

i = 0.39213 A

In this branch we have two resistors in series, let's remember that the current in a series circuit is constant

V_C = i R_C

V_C = 0.39213 3

V_C = 1.176 V

3 0

3 years ago

How many digits are in front of the decimal in scientific notation?

nikklg [1K]

Answer:

Explanation:

How many digits are in front of the decimal in scientific notation?

7 0

3 years ago

Importance of electric circuit

Feliz [49]

Answer:

It is important because it carries useful energy through your house that you can use for a variety of tasks.

Explanation:

Hope this helped !

8 0

3 years ago

What is the peak emf generated by a 0.250 m radius, 500-turn coil is rotated one-fourth of a revolution in 4.17 ms, originally h

zysi [14]

Complete question:

What is the peak emf generated by a 0.250 m radius, 500-turn coil is rotated one-fourth of a revolution in 4.17 ms, originally having its plane perpendicular to a uniform magnetic field 0.425 T. (This is 60 rev/s.)

Answer:

The peak emf generated by the coil is 15.721 kV

Explanation:

Given;

Radius of coil, r = 0.250 m

Number of turns, N = 500-turn

time of revolution, t = 4.17 ms = 4.17 x 10⁻³ s

magnetic field strength, B = 0.425 T

Induced peak emf = NABω

where;

A is the area of the coil

A = πr²

ω is angular velocity

ω = π/2t = (π) /(2 x 4.17 x 10⁻³) = 376.738 rad/s = 60 rev/s

Induced peak emf = NABω

= 500 x (π x 0.25²) x 0.425 x 376.738

= 15721.16 V

= 15.721 kV

Therefore, the peak emf generated by the coil is 15.721 kV

5 0

3 years ago