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myrzilka [38]
3 years ago
10

A 30.0 g mass of iron at 24.5°C is heated to 45.0°C. The theoretical specific

Physics
1 answer:
diamong [38]3 years ago
8 0

Answer:

276.135 J

Explanation:

Given that:

mass of Fe = 30.0 g

initial temperature = 24.5°C

final temperature = 45.0°C

specific heat of Fe = 0.449 J/g°C

We can determine the thermal energy added by using the formula;

Q = mcΔT

Q = 30.0g × 0.449 J/g°C × (45.0 - 24.5)°C

Q = 276.135 J

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If a conducting loop of radius 10 cm is onboard an instrument on Jupiter at 45 degree latitude, and is rotating with a frequency
Pepsi [2]

Answer:

a)  fem = - 2.1514 10⁻⁴ V,  b) I = - 64.0 10⁻³ A, c)    P = 1.38  10⁻⁶ W

Explanation:

This exercise is about Faraday's law

         fem = - \frac{ d \Phi_B}{dt}

where the magnetic flux is

        Ф = B x A

the bold are vectors

        A = π r²

we assume that the angle between the magnetic field and the normal to the area is zero

         fem = - B π 2r dr/dt = - 2π B r v

linear and angular velocity are related

        v = w r

        w = 2π f

        v = 2π f r

we substitute

        fem = - 2π B r (2π f r)

        fem = -4π² B f r²

For the magnetic field of Jupiter we use the equatorial field B = 428 10⁻⁶T

we reduce the magnitudes to the SI system

       f = 2 rev / s (2π rad / 1 rev) = 4π Hz

we calculate

       fem = - 4π² 428 10⁻⁶ 4π 0.10²

       fem = - 16π³ 428 10⁻⁶ 0.010

       fem = - 2.1514 10⁻⁴ V

for the current let's use Ohm's law

        V = I R

        I = V / R

         I = -2.1514 10⁻⁴ / 0.00336

         I = - 64.0 10⁻³ A

Electric power is

        P = V I

        P = 2.1514 10⁻⁴ 64.0 10⁻³

        P = 1.38  10⁻⁶ W

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To do work, this truck uses energy stored in chemical
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800 J Got it right on edgenuity

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Monochromatic light of a given wavelength is incident on a metal surface. however, no photoelectrons are emitted. if electrons a
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2 years ago
A ski jumper travels down a slope and
AleksandrR [38]

Answer:

304.86 metres

Explanation:

The x and y cordinates are dcos\theta and dsin\theta respectively

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Making t the subject, t=\frac{dcos\theta}{v_{ox}}

Since y=0.5gt^2=dsin\theta, we substitute t with the above and obtain

0.5g(\frac{dcos\theta}{v_{ox}})^2=dsin\theta

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d=\frac{2v_{ox}^2sin\theta}{gcos^2\theta}

d=\frac{2*30^2sin48}{9.8cos^248}

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d=304.86m

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