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son4ous [18]
3 years ago
5

How much heat is released to freeze 47.30 grams of copper at its freezing point of 1,085°C? The latent heat of fusion of copper

is 205.0 J/g.
Physics
2 answers:
krek1111 [17]3 years ago
4 0

the answer is -9,697

timofeeve [1]3 years ago
3 0

Answer:

9696.5 J of heat is released

Explanation:

Heat released= mass of substance× specific latent heat of fusion

                      = 47.3×205

                      = 9696.5 J

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Whatever that number is, the maximum hours per month is

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olya-2409 [2.1K]

Answer:

option 1

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4 0
3 years ago
Rony fills a bucket with water and whirls it in a vertical circle to demonstrate that the water will not spill from the bucket a
Romashka-Z-Leto [24]

Answer:

0 N, 3.49 m/s

Explanation:

Draw a free body diagram for the bucket at the top of the swing.  There are two forces acting on the bucket: weight and tension, both downwards.

If we take the sum of the forces in the radial direction, where towards the center is positive:

∑F = ma

W + T = m v² / r

The higher the velocity that Rony swings the bucket, the more tension there will be.  The slowest he can swing it is when the tension is 0.

W = m v² / r

mg = m v² / r

g = v² / r

v = √(gr)

Given that r = 1.24 m:

v = √(9.8 m/s² × 1.24 m)

v = 3.49 m/s

8 0
3 years ago
An insulating sphere is 8.00 cm in diameter and carries a 6.50 µC charge uniformly distributed throughout its interior volume.
Kobotan [32]

Explanation:

(a)   Formula to calculate the density is as follows.

            \rho = \frac{Q}{\frac{4}{3}\pi a^{3}}

                       = \frac{6.50 \times 10^{-6}}{\frac{4}{3} \times 3.14 \times (0.04)^{3}}

                     = 2.42 \times 10^{-2} C/m^{3}

Now, calculate the charge as follows.

            q_{in} = \rho(\frac{4}{3} \pi r^{3})

                      = 2.42 \times 10^{-2} C/m^{3} \times 4.1762 \times (0.01)^{3}

                      = 10.106 \times 10^{-8} C

or,                   = 101.06 nC

(b)  For r = 6.50 cm, the value of charge will be calculated as follows.

                q_{in} = \frac{Q}{\frac{4}{3}\pi a^{3}}

                          = \frac{6.50 \times 10^{-6}}{\frac{4}{3} \times 3.14 \times (0.065)^{3}}

                          = 7.454 \mu C

7 0
3 years ago
Jane, looking for tarzan, is running at top speed 5.3 m/s and grabs a vine hanging vertically from a tall tree in the jungle. ho
lutik1710 [3]

v₀ = initial speed as tarzan grabs the vine = 5.3 m/s

v = final speed as the tarzan reach the maximum height = 0 m/s

h = maximum height gained by the tarzan

m = mass of tarzan

using conservation of energy

initial kinetic energy = final kinetic energy + potential energy

(0.5) m v²₀ = (0.5) m v² + m g h

(0.5) v²₀ = (0.5) v² + g h

(0.5) (5.3)² = (0.5) (0)² + (9.8) h

h = 1.43 m


3 0
3 years ago
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