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katrin [286]
3 years ago
11

The specific heat of water is 4.2 J/g • °C. How much heat is required to raise the temperature of 100 g of water by 5°C? *

Physics
1 answer:
Svetach [21]3 years ago
7 0

Answer:

2100 J

Explanation:

The heat required to increase the temperature of a substance is given by

Q=mC\Delta T

where

m is the mass of the substance

C is its specific heat capacity

\Delta T is its change in temperature

For the water in this problem, we have:

m = 100 g is its mass

C = 4.2 J/g • °C is the specific heat capacity

\Delta T=5^{\circ}C is the increase in temperature

So, the amount of heat needed is:

Q=(100)(4.2)(5)=2100 J

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Answer:

7.1 m/s

Explanation:

First, find the time it takes for the fish to reach the water.

Given in the y direction:

Δy = 6.1 m

v₀ = 0 m/s

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Δy = v₀ t + ½ at²

6.1 m = (0 m/s) t + ½ (9.8 m/s²) t²

t = 1.12 s

Next, find the velocity needed to travel 7.9 m in that time.

Given in the x direction:

Δx = 7.9 m

a = 0 m/s²

t = 1.12 s

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Charge A and charge B are 3.00 m apart, and charge A is +1.44 C and charge B is +3.10 C. Charge C is located between them at a c
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Answer:

the distance from charge A to C is r₁₃= 1.216 m

Explanation:

following Coulomb's law , the force exerted by 2 point charges between themselves is:

F= k*q₁*q₂/r₁₂² , where q is charge , r is distance and 1 and 2 represents the charge A and charge B respectively , k=constant

since C ( denoted as 3) is at equilibrium

F₁₃=F₂₃

k*q₁*q₃/r₁₃²=k*q₂*q₃/r₂₃²

q₁/r₁₃²=q₂/r₂₃²

r₁₃²/q₁=r₂₃²/q₂

r₂₃=r₁₃*√(q₂/q₁)

since C is at rest and is co linear with A and B ( otherwise it would receive a net force in either vertical or horizontal direction) , we have

r₁₃+r₂₃=d=r₁₂

r₁₃+r₁₃*√(q₂/q₁)=d

r₁₃*(1+√(q₂/q₁))=d

r₁₃=d/(1+√(q₂/q₁))

replacing values

r₁₃=d/(1+√(q₂/q₁)) = 3.00 m/(1+√(3.10 C/1.44 C)) = 1.216 m

thus the distance from charge A to C is r₁₃= 1.216 m

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To store stacks of clean plates, a cafeteria uses a closed cart with a spring-loaded shelf inside. Customers can take plates off
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The answer for the following problem is mentioned below.

The option for the question is "A" approximately.

  • <u><em>Therefore the elastic potential energy of the string is 20 J.</em></u>

Explanation:

Given:

Spring constant (k) = 240 N/m

amount of the compression (x) = 0.40 m

To calculate:

Elastic potential energy (E)

We know;

<em>According to the formula;</em>

    E = \frac{1}{2} × k × x × x

   <u>E = </u>\frac{1}{2}<u> × k ×(x)²</u>

where;

E represents the elastic potential energy

K represents the spring constant

x represents amount of the compression in the string

So therefore,

Substituting the values in the above formula;

      E = \frac{1}{2} × 240 × (0.40)²

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KATRIN_1 [288]

Answer:

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See attached picture.

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