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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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Answer:14.14 cm

Explanation:

Given

Spring Compression x=12 cm

Potential energy Stored in spring=72 J

Suppose k is the spring constant of spring

Potential Energy of spring is given by =\frac{kx^2}{2}

\frac{k(0.12)^2}{2}=72

k(0.12)^2=144

k=10,000 N/m

k=10 kN/m

for 100 J energy

\frac{k(x_0)^2}{2}=100

10\times 10^3\cdot (x_0)^2=200

(x_0)^2=2\times 10^{-2}

x_0=0.1414

x_0=14.14 cm

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What piece of equipment would a hospital most likely have as back up in case of a power failure?
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B. a generator

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4 years ago
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Why do waves in a tank are 10 cm long if they pass a point at a rate of 3.75 waves per second what is the speed in m/s?
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3 years ago
Two airplanes leave an airport at the same time. The velocity of the first airplane is 730 m/h at a heading of 65.3 ◦ . The velo
yanalaym [24]

Answer:

Plane will 741.6959 m apart after 1.7 hour                    

Explanation:

We have given time = 1.7 hr

So if we draw the vectors of a 2d graph we see that the difference in angles is   = 102^{\circ}-65.3^{\circ}=36.7^{\circ}

Speed of first plane  = 730 m/h

So distance traveled by first plane = 730×1.7 = 1241 m

Speed of second plane = 590 m/hr

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We represent these distances as two sides of the triangle, and the distance between the planes as the side opposing the angle 58.6.

Using the law of cosine, r^2 representing the distance between the planes, we see that:

r^2=1241^2+1003^2-2\times 1003\times 1241cos(36.7)=550112.8295

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3 years ago
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