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3 years ago

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A 0.0414 kg ingot of metal is heated to 243◦C

1 answer:

tino4ka555 [31]3 years ago

3 0

Answer:

448 J/kg/°C

Explanation:

m₁ C₁ (T₁ − T) + m₂ C₂ (T₂ − T) = 0

(0.0414 kg) C (243°C − 20.4°C) + (0.411 kg) (4186 J/kg/°C) (18°C − 20.4°C) = 0

(9.22 kg°C) C − 4129 J = 0

C = 448 J/kg/°C

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A flat loop of wire consisting of a single turn of cross-sectional area 8.00 cm2 is perpendicular to a magnetic field that incre

Evgen [1.6K]

Complete Question

A flat loop of wire consisting of a single turn of cross-sectional area 8.00 cm2 is perpendicular to a magnetic field that increases uniformly in magnitude from 0.500 T to 1.60 T in 0.99 s. What is the resulting induced current if the loop has a resistance of $1.20 \ \Omega$

Answer:

The current is $I = 0.0007 41 \ A$

Explanation:

From the question we are told that

The area is $A = 8.00 \ cm^2 = 8.0 *10^{-4} \ m^2$

The initial magnetic field at $t_o = 0 \ seconds$ is $B_i = 0.500 \ T$

The magnetic field at $t_1 = 0.99 \ seconds$ is $B_f = 1.60 \ T$

The resistance is $R = 1.20 \ \Omega$

Generally the induced emf is mathematically represented as

$\epsilon = A * \frac{B_f - B_i }{ t_f - t_o }$

=> $\epsilon = 8.0 *10^{-4} * \frac{1.60 - 0.500 }{ 0.99- 0 }$

=> $\epsilon = 0.000889 \ V$

Generally the current induced is mathematically represented as

$I = \frac{\epsilon}{R }$

=> $I = \frac{0.000889}{ 1.20 }$

=> $I = 0.0007 41 \ A$

6 0

2 years ago

38.4 mol of krypton is in a rigid box of volume 64 cm^3 and is initially at temperature 512.88°C. The gas then undergoes isobari

kolbaska11 [484]

Answer:

Final volumen first process $V_{2} = 98,44 cm^{3}$

Final Pressure second process $P_{3} = 1,317 * 10^{10} Pa$

Explanation:

Using the Ideal Gases Law yoy have for pressure:

$P_{1} = \frac{n_{1} R T_{1} }{V_{1} }$

where:

P is the pressure, in Pa

n is the nuber of moles of gas

R is the universal gas constant: 8,314 J/mol K

T is the temperature in Kelvin

V is the volumen in cubic meters

Given that the amount of material is constant in the process:

$n_{1} = n_{2} = n$

In an isobaric process the pressure is constant so:

$P_{1} = P_{2}$

$\frac{n R T_{1} }{V_{1} } = \frac{n R T_{2} }{V_{2} }$

$\frac{T_{1} }{V_{1} } = \frac{T_{2} }{V_{2} }$

$V_{2} = \frac{T_{2} V_{1} }{T_{1} }$

Replacing : $T_{1} =786 K, T_{2} =1209 K, V_{1} = 64 cm^{3}$

$V_{2} = 98,44 cm^{3}$

Replacing on the ideal gases formula the pressure at this piont is:

$P_{2} = 3,92 * 10^{9} Pa$

For Temperature the ideal gases formula is:

$T = \frac{P V }{n R }$

For the second process you have that $T_{2} = T_{3}$ So:

$\frac{P_{2} V_{2} }{n R } = \frac{P_{3} V_{3} }{n R }$

$P_{2} V_{2} = P_{3} V_{3}$

$P_{3} = \frac{P_{2} V_{2}}{V_{3}}$

$P_{3} = 1,317 * 10^{10} Pa$

7 0

2 years ago

The wave shown below is produced in a rope.

Kay [80]

Answer:

<h3>B) A dot vector drawn up and vector drawn down.</h3>

Explanation:

Brainliest?

6 0

3 years ago

Read 2 more answers

A particle of mass m collides with a second particle of mass m. Before the collision, the first particle is moving in the x-dire

oee [108]

Answer:

a) v, v

b) 2mv^2

c) Elastic collion

Explanation:

(a) The velocity of the second particle after the collision is (v2x,v2y)=(v,−v). From momentum conservation in x-direction

Here x, y represent direction.They are not variable. 1 and 2 represent before and after.

2vm=v1xm+v2xm, we find v1x=v.

From momentum conservation in y-direction

0 =v1ym+v2ym, we findv1y=v.

(b) By energy conservation principle

Before: K=1/2m(2v)^2=2mv^2.

After: K=1/2m(v^2(1x)+v^2(1y))+12m(v22x+v22y)=2mv^2

(c) The collision is elastic

6 0

3 years ago

You throw a glob of putty straight up toward the ceiling, which is 3.50 mm above the point where the putty leaves your hand. The

LuckyWell [14K]

Answer: $4.65\ m/s$

Explanation:

Given

Distance putty has to travel is 3.5 m

The initial speed of putty is 9.50 m/s

Using equation of motion to determine the velocity of putty just before it hits ceiling

$v^2-u^2=2as$

$\Rightarrow v^2-(9.5)^2=2(-9.8)(3.5)\\\\\Rightarrow v^2=9.5^2-68.6\\\Rightarrow v=\sqrt{90.25-68.6}\\\Rightarrow v=4.65\ m/s$

So, the velocity of putty just before hitting is $4.65\ m/s$

5 0

3 years ago