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4 years ago

HELP PLEASE

Physics

2 answers:

Artemon [7]4 years ago

5 0

Explanation:

Electromagnetic induction: It is the phenomenon of the production of electromagnetic force across the conductor by changing the magnetic field.

The current can be induced by moving the conductor in the magnetic field. The current can also be induced by changing the magnetic field across the conductor.

The induced current can be increased in the coil by the following ways:

By increasing the strength of the magnet.

By increasing the speed of the magnet through the coil.

By adding more loops of wire to the coil.

By increasing the magnetic flux.

Therefore, "by moving the magnet away from the coil" will not increase the electric current induced in a wire. It will decrease the electric current induced in the wire.

Ira Lisetskai [31]4 years ago

3 0

The answer is magnet away from the coil

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A 0.468 g sample of pentane, C 5H 12, was burned in a bomb calorimeter. The temperature of the calorimeter and 1.00 kg of water

Fittoniya [83]

Answer:

$Q_{lost}$ per mole of pentane = 3157.53 kJ/mol

Explanation:

Given:

Mass of pentane, m = 0.468 gram

Molar mass of pentane, M = 72.15

Now, mol of pentane, n = mass/M = 0.468/72.15 = 0.00648 mol of C5H12

Now,

ΔT = 23.65 - 20.45 = 3.2°C

Heat capacity of the calorimeter, C = 2.21 kJ/°C

Specific heat capacity of the water, Cp = 4.184 J/g.°C

Now,

the heat gained = the heat lost

$Q_{gained} = -Q_{lost}$

also,

$Q_{gained} = Q_{water} + Q_{calorimeter}$

$Q_{water} = m\times C\times(T_f-T_i)$

$Q_{water} = 1000\times4.184\times(23.65-20.45) = 13388.8\ J$

and

$Q_{calorimeter} = C\times\Delta T = 2.21\times1000\times3.2 = 7072\ J$

Now,

$Q_{total} = 13388.8 +7072 = 20460.8\ J$

we have,

$Q_{lost} = -Q_{gain} = - 20460.8\ J$ (Here negative sign depicts the release of the heat)

$Q_{lost}$ per mole of pentane =-20460.8/(0.00648 ) = 3157.53 kJ/mol

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3 years ago

Different between pressure and force

9966 [12]

Force is mass into acceleration

and pressure is force applied per unit area.

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3 years ago

Which of the following statements describes mass defect? A. Mass defect is the energy that balances the protons and electrons in

Yanka [14]

c. mass defect is the energy that binds the protons and neutrons together in the nucleus

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3 years ago

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A) At a certain instant, a particle-like object is acted on by a force F = (3.0 N) ihat - (3.0 N) jhat + (9.0 N) khat while the

Paha777 [63]

Answer:

a) Instantaneous rate at which the force does work on the object = -6 W

b) $\texttt{Velocity of object,}\vec{v}=4\hat{j}$

Explanation:

a) Given that

$\vec{F}=3\hat{i}-3\hat{j}$

and

$\vec{v}=-2\hat{i}+4\hat{k}$

Instantaneous rate at which the force does work on the object is called power.

Power is the dot product of force and velocity.

$P=\vec{F}.\vec{v}=(3\hat{i}-3\hat{j}).(-2\hat{i}+4\hat{k})=-6W$

Instantaneous rate at which the force does work on the object = -6 W

b) Here given that $\vec{v}=-a\hat{j}$

Power = -12 W

$P=\vec{F}.\vec{v}=(3\hat{i}-3\hat{j}).a\hat{j}=-12W\\\\-3a=-12\\\\a=4\\\\\vec{v}=4\hat{j}$

$\texttt{Velocity of object,}\vec{v}=4\hat{j}$

4 0

3 years ago

A rocket accelerates upward from rest, due to the first stage, with a constant acceleration of a1 = 67 m/s2 for t1 = 39 s. The f

Igoryamba

Answer:

(a) $v_1= a_1t_1$

(b) $v_2 =a_1t_1+a_2t_2$

(c) 44133.5 m

Explanation:

<u>Given:</u>

$u$ = initial speed of the rocket in the first stage = 0 m/s
$v_1$ = final speed of the rocket in the first stage
$v_2$ = final speed of the rocket in the second stage
$t_1$ = time interval of the first stage
$t_2$ = time interval of the second stage
$s_1$ = distance traveled by the rocket in the first stage

$s_2$ = distance traveled by the rocket in the second stage
$s$ = distance traveled by the rocket in whole time interval

Part (a):

Since the rocket travels at constant acceleration.

$\therefore v_1 = u+a_1t_1\\\Rightarrow v_1 = a_1t_1$

Hence, the expression of the rocket's speed at time $t_1\ is\ v_1 = a_1t_1$ .

Part (b):

In this part also, the rocket moves with a constant acceleration motion.

$\therefore v_2 = v_1+a_2t_2\\\Rightarrow v_2 = a_1t_1+a_2t_2$

Hence, the expression of the rocket's speed in the time interval $t_2$ is $v_2 = a_1t_1+a_2t_2$ .

Part (c):

For the constant acceleration of rocket, let us first calculate the distance traveled by the rocket in both the time intervals.

$s_1 = u+\dfrac{1}{2}a_1t_1^2\\\Rightarrow s_1 = 0+\dfrac{1}{2}67\times(1)^2\\\Rightarrow s_1 =33.5\ m$

Similarly,

$s_2 = v_1t_2+\dfrac{1}{2}a_2t_2^2\\\Rightarrow s_2 = a_1t_1t_2+\dfrac{1}{2}a_2t_2^2\\\Rightarrow s_2 = 67\times1\times49+\dfrac{1}{2}\times 34\times(49)^2\\\Rightarrow s_2 =44100\ m\\\therefore s = s_1+s_2\\\Rightarrow s = 33.5\ m+44100\ m\\\Rightarrow s =44133.5\ m$

Hence, the rocket moves a total distance of 44133.5 m until the end of the second period of acceleration.

5 0

3 years ago

Read 2 more answers