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3 years ago

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The thermal energy used by heater in 3 minutes is used to melt wax.Melting point of solid wax is 60.Specific heat of wax is 220j

/g.Heater supplies 7700j of energy,how much wax melts?

1 answer:

Romashka-Z-Leto [24]3 years ago

5 0

Answer:

m = 35 g

Explanation:

The specific heat of a material can be calculated by the following formula:

$C = \frac{Q}{m}\\$

where,

C = Specific Heat of Wax = 220 J/g

Q = Amount of Heat Supplied by the Heater = 7700 J

m = mass of wax melted = ?

Therefore,

$220\ J/g = \frac{7700\ J}{m}\\\\m = \frac{7700\ J}{220\ J/g}\\$

<u>m = 35 g </u>

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A 57 kg wagon is pulled with a constant net force of 38 N. Calculate the acceleration of the wagon. F = ma

Dahasolnce [82]

Answer:

<h2>0.67 m/s²</h2>

Explanation:

The acceleration of an object given it's mass and the force acting on it can be found by using the formula

$a = \frac{f}{m} \\$

f is the force

m is the mass

From the question we have

$a = \frac{38}{57} = \frac{2}{3} \\ = 0.666666...$

We have the final answer as

<h3>0.67 m/s²</h3>

Hope this helps you

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2 years ago

A solid conducting sphere of radius 2 cm has a charge of 8microCoulomb. A conducting spherical shell of inner radius 4 cm andout

nika2105 [10]

Answer:

C) 7.35*10⁶ N/C radially outward

Explanation:

If we apply the Gauss'law, to a spherical gaussian surface with radius r=7 cm, due to the symmetry, the electric field must be normal to the surface, and equal at all points along it.
So, we can write the following equation:

$E*A = \frac{Q_{enc} }{\epsilon_{0}} (1)$

As the electric field must be zero inside the conducting spherical shell, this means that the charge enclosed by a spherical gaussian surface of a radius between 4 and 5 cm, must be zero too.
So, the +8 μC charge of the solid conducting sphere of radius 2cm, must be compensated by an equal and opposite charge on the inner surface of the conducting shell of total charge -4 μC.
So, on the outer surface of the shell there must be a charge that be the difference between them:

$Q_{enc} = - 4e-6 C - (-8e-6 C) = + 4 e-6 C$

Replacing in (1) A = 4*π*ε₀, and Qenc = +4 μC, we can find the value of E, as follows:

$E = \frac{1}{4*\pi*\epsilon_{0} } *\frac{Q_{enc} }{r^{2} } = \frac{9e9 N*m2/C2*4e-6C}{(0.07m)^{2} } = 7.35e6 N/C$

As the charge that produces this electric field is positive, and the electric field has the same direction as the one taken by a positive test charge under the influence of this field, the direction of the field is radially outward, away from the positive charge.

6 0

3 years ago

otez555 [7]

A. Impulse is simply the product of Force and time. Therefore,

I = F * t ---> 1

where I is impulse, F is force, t is time

However another formula for solving impulse is:

I = m vf – m vi ---> 2

where m is mass, vf is final velocity and vi is initial velocity

Therefore using equation 2 to solve for impulse I:

I = 2000kg (0) – 2000kg (77 m/s)
I = -154,000 kg m/s

B. By conservation of momentum, we also know that Impulse is conserved. That means that increasing the time by a factor of 3 would still result in an impuse of -154,000 kg m/s. So,

I = F’ * (3 t) = -154,000 kg m/s

Since t is multiplied by 3, therefore this only means that Force is decreased by a factor of 3 to keep the impulse constant, therefore:

(F/3) (3t) = -154,000 kg m/s

Summary of Answers:

A. I = -154,000 kg m/s

B. Force is decreased by factor of 3

8 0

3 years ago

What is center of gravity and how does it affect rockets?

Bezzdna [24]

Answer:

Rocket Center of Gravity. As a rocket flies through the air, it both translates and rotates. The rotation occurs about a point called the center of gravity. The center of gravity is the average location of the weight of the rocket.

4 0

3 years ago

Read 2 more answers

A multimeter in an RL circuit records an rms current of 0.600 A and a 50.0-Hz rms generator voltage of 110 V. A wattmeter shows

jeka57 [31]

Answer:

(a) The impedance in the circuit is $Z=183.33\Omega$ .

(b)The resistance is $R=38.89\Omega$ .

(c) The inuctance is 0.57 H.

Explanation:

(a)

The expression for the impedance is as follows:

$Z=\frac{V_rms}{I_rms}$

Here, $V_rms$ is the rms voltage and $I_rms$ is the rms current.

Put $V_rms=110 V$ and $I_rms=0.600 A$ .

$Z=\frac{110}{0.600}$

$Z=183.33\Omega$

Therefore, the impedance in the circuit is $Z=183.33\Omega$ .

(b)

The expression for the average power is as follows;

$P_{a}=I_{rms}^{2}R$

Here, $P_{a}$ is the average power and R is the resistance.

Calculate the resistance by rearranging the above expression.

$R=\frac{P_{a}}{I_{rms}^{2}}$

Put $P_{a}=14W$ and

$R=\frac{14}{{0.600}^{2}}$

$R=38.89\Omega$

Therefore, the resistance is $R=38.89\Omega$ .

(c)

The expression for the impedance is as follows;

$Z^{2}=R^{2}+X_{L}^{2}$

Here, $X_{L}$ is the inductive reactance.

Put $Z=183.33\Omega$ and $R=38.89\Omega$ .

$(183.33)^{2}=(38.89)^{2}+X_{L}^{2}$

$X_{L}=179.16\Omega$

The expression for the inductive reactance in terms of frequency is as follows;

$X_{L}=2\pi fL$

Here, L is the inductance.

Calculate the inductance by rearranging the above expression.

$L=\frac{X_{L}}{2\pi f}$

Put $X_{L}=179.16\Omega$ and f=50Hz.

$L=\frac{179.16}{2\pi (50)}$

L=0.57 H

Therefore, the inuctance is 0.57 H.

4 0

3 years ago