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puteri [66]
3 years ago
11

A hot cube of iron was heated up using 1500 J of thermal energy and was placed in a beaker of water. Before it was heated, the i

ron cube was 20.0 ˚C. The cube of iron raised the temperature of the water to 45.0 ˚C. If the iron cube has a mass of 133 g, what is the specific heat capacity of iron?
Physics
1 answer:
Mariana [72]3 years ago
4 0

Answer:

451.13 J/kg.°C

Explanation:

Applying,

Q = cm(t₂-t₁)............... Equation 1

Where Q = Heat, c = specific heat capacity of iron, m = mass of iron, t₂= Final temperature, t₁ = initial temperature.

Make c the subject of the equation

c = Q/m(t₂-t₁).............. Equation 2

From the question,

Given: Q = 1500 J, m = 133 g = 0.113 kg, t₁ = 20 °C, t₂ = 45 °C

Substitute these values into equation 2

c = 1500/[0.133(45-20)]

c = 1500/(0.133×25)

c = 1500/3.325

c = 451.13 J/kg.°C

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The force on the proton is 17.4 N.

<h3>What is the force on the proton?</h3>

Now we know that the proton is positively charged and that the force on the charge as it moved through the magnetic field could be given by the relation; F = qvB

Where;

F = force

q = charge

v = velocity

B = magnetic field

Having said this, we can see that;

q = 1.601019 As or C

v =  2.4105 m/s

T = 4.5 T

F =  1.601019 As * 2.4105 m/s * 4.5 T

F = 17.4 N

Learn more about magnetic force:brainly.com/question/12824331

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2 years ago
How are electromagnetic waves that are produced by oscillating charges and sound waves that are produced by oscillating tuning f
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4 years ago
Which countries that had sea ice along their coasts in September 1986 were bordered by open water in September 2017?
mrs_skeptik [129]

Answer:

Norway, Sweden, Finland and Iceland

Explanation:

Sea ice is a frozen seawater that floats on the ocean surface. It is formed between the Antarctic and Arctic hemisphere. It disappears in summer but not completely. The countries that experienced sea ice in 1986 were eight (8) in number but the countries bordered by open water were in September 2017 were Norway, Iceland, Finland and Russia.

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3 years ago
6. Velocity is a ____________. a) vector b) value c) arrow d) none of these
Olenka [21]
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4 0
3 years ago
Read 2 more answers
A soccer player takes a corner kick, lofting a stationary ball 30.0° above the horizon at 18.0 m/s. If the soccer ball has a mas
Radda [10]

Answer:

change in momentum, \Delta p=7.65 \,kg.m.s^{-1}

  • \Delta p_x= 6.6251 \,kg.m.s^{-1}
  • \Delta p_y= 3.825 \,kg.m.s^{-1}

Average Force, F=144.3396\,N

  • F_x=125.0018\,N
  • F_y=72.1698\,N

Explanation:

Given:

angle of kicking from the horizon, \theta= 30^{\circ}

velocity of the ball after being kicked, v=18 m.s^{-1}

mass of the ball, m=0.425\, kg

time of application of force, t=5.3\times 10^{-2}\,s

We know, since body is starting from the rest

\Delta p=m.v.....................(1)

\Delta p=0.425\times 18

\Delta p=7.65 \,kg.m.s^{-1}

Now the components:

\Delta p_x= 7.65\times cos 30^{\circ}

\Delta p_x= 6.6251 \,kg.m.s^{-1}

similarly

\Delta p_y= 7.65\times sin 30^{\circ}

\Delta p_y= 3.825 \,kg.m.s^{-1}

also, impulse

I=F\times t.........................(2)

where F is the force applied for t time.

Then from eq. (1) & (2)

F\times t=m.v

F\times 5.3\times 10^{-2}= 7.65

F=144.3396\,N

Now, the components

F_x=144.3396\times cos 30^{\circ}

F_x=125.0018\,N

&

F_y=144.3396\times sin 30^{\circ}

F_y=72.1698\,N

6 0
3 years ago
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