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kherson [118]
3 years ago
9

How many calories are produced when 50 ml of water is raised from 22 degrees celsius to 52 degrees celsius?

Physics
1 answer:
vagabundo [1.1K]3 years ago
5 0

Answer:1500 calorie

Explanation:

Given

volume of water V=50\ ml

Initial temperature T_i=22^{\circ}C

Final temperature T_f=52^{\circ}C

We know,

specific heat of water is c=1\ calorie/gm-^{\circ}C

the density of water is \rho=1\ gm/ml

The heat required to raise the temperature

\Rightarrow Q=mc\Delta T\\\Rightarrow Q=\rho \cdot V\times c\times \Delta T\\\Rightarrow Q=1\times 50\times 1\times (52-22)\\\Rightarrow Q=1500\ cal.

1500 calories is required

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Answer:

the answer is D

Explanation:

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A long pipe of outer radius R 1 = 3.70 cm and inner radius R 2 = 3.15 cm carries a uniform charge density of 1.22 mC/m 3 . Assum
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Answer:

E = 4.72 * 10⁻⁶ Nm²

Explanation:

Parameters given:

Outer radius, R = 3.70cm = 0.037m

Inner radius, r = 3.15cm = 0.0315m

Permittivity of free space, ε₀ = 8.85 * 10⁻¹² C/Nm²

Charge density: 1.22 * 10⁻³ C/m³

The question requires that we solve using Gauss law which states that the net electric field through a closed surface is proportional to the enclosed electric charge.

Hence,

E = Q/Aε₀

Charge Q is given as

Q = ρπ(R² ⁻ r²)L

A = 2π(R - r)L

E = [ρπ(R² ⁻ r²)L]/[2π(R - r)ε₀L]

Using difference of two squares,

(R² ⁻ r²) = (R + r)(R - r)

E =[ρ(R + r)]/(2ε₀)

E = [1.22 * 10⁻³ *(0.0370 + 0.0315)]/(2 * 8.85 * 10⁻¹²)

E = 4.72 * 10⁻⁶ Nm²

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4 years ago
An object travels a given distance at a speed of 5 m/s. If the object travels the same distance at a higher speed, how will the
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a. it will take less time

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Mrs. Brown's class is studying magnets and electricity. At the end of the unit Claire states that magnets and electricity are bo
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Electromagnets use electricity to create a magnetic force.

Explanation:

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A bar-magnet with magnetic moment 2.5 Am^2 is placed in a homogeneous magnetic field (of 0.1 T that is directed along the z-axis
Angelina_Jolie [31]

Answer:

1)

Force on bar magnet  = 0

Torque on bar magnet = 0

2)

Force on bar magnet  = 0

Torque on bar magnet = 0.177 Nm

3)

Force on bar magnet  = 0

Torque on bar magnet = 0.25 Nm

Explanation:

Part 1)

net force on bar magnet in uniform magnetic field is always zero

Torque on bar magnet is given as

\tau = MBsin\theta

when bar magnet is inclined along z axis along magnetic field

then we will have

\tau = MBsin0 = 0

Part 2)

net force on bar magnet in uniform magnetic field is always zero

Torque on bar magnet is given as

\tau = MBsin\theta

when bar magnet is pointing 45 degree with z axis then we will have

\tau = MBsin45

\tau = (2.5)(0.1)sin45

\tau = 0.177 Nm

Part 3)

net force on bar magnet in uniform magnetic field is always zero

Torque on bar magnet is given as

\tau = MBsin\theta

when bar magnet is pointing 90 degree with z axis then we will have

\tau = MBsin90

\tau = (2.5)(0.1)sin90

\tau = 0.25 Nm

8 0
4 years ago
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