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7nadin3 [17]
2 years ago
14

a closed systems internal energy changes by 178 j as a result of being heated with 658 j of energy. the energy used to do work b

y the system is j
Physics
1 answer:
Akimi4 [234]2 years ago
7 0
<h3><u>Answer;</u></h3>

= 480 Joules

<h3><u>Explanation;</u></h3>

We use the formula, Q - W = ΔU

Where, Q = Heat transferred to the system

W = Work done by the system

ΔU = Change of internal energy.

As per the question, Q = 658 J

ΔU = 178 J

Thus, W = Q - ΔU = (658 - 178) J = 480 J.

The energy used to do work by the system is 480 J.

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8.4-1 Consider a magnetic field probe consisting of a flat circular loop of wire with radius 10 cm. The probe’s terminals corres
Vlad1618 [11]

Answer:

B_o = 1.013μT

Explanation:

To find B_o you take into account the formula for the emf:

\epsilon=-\frac{d\Phi_b}{dt}=-\frac{dBAcos\theta}{dt}=-Acos\theta\frac{dB}{dt}

where you used that A (area of the loop) is constant, an also the angle between the direction of B and the normal to A.

By applying the derivative you obtain:

\epsilon=-Acos\theta (2\pi f) B_ocos(2\pi f t+ \alpha)

when the emf is maximum the angle between B and the normal to A is zero, that is, cosθ = 1 or -1. Furthermore the cos function is 1 or -1. Hence:

\epsilon=2\pi fAB_o=2\pi (100*10^3Hz)(\pi (0.1m)^2)B_o=19739.20Hzm^2B_o\\\\B_o=\frac{20*10^{-3}V}{19739.20Hzm^2}=1.013*10^{-6}T=1.013\mu T

hence, B_o = 1.013μT

6 0
2 years ago
What is the weight of a 1-kilogram brick resting on a table?
MakcuM [25]

Answer:

The weight if the block is 10Newtons

Explanation:

The weight of any object is quantity of matter the object contains and it is always acting downwards on such body. This shows that the object is under the influence of gravity.

The weight of an object is calculated as mass of the object × its acceleration due to gravity

W = mg

Give the mass of the brick to be 1kg

g is the acceleration due to gravity = 10m/s²

Weight of the object = 1 × 10

= 10kgm/s² or 10Newtons

5 0
1 year ago
Calculate the current through a 10.0-m long 22-gauge nichrome wire with a radius of 0.321 mm if it is connected across a 12.0-V
Kipish [7]

Answer:

Therefore,

Current through Nichrome wire is 0.3879 Ampere.

Explanation:

Given:

Length = l = 10 meter

Radius = r = 0.321\ mm =0.321\times 10^{-3}\ meter

Resistivity=\rho=1.00\times 10^{-6}\ ohm\ meter

V = 12 Volt

To Find:

Current, I =?

Solution:

Resistance for 0.0-m long 22-gauge nichrome wire with a radius of 0.321 mm if it is connected across a 12.0-V battery given as

R=\dfrac{\rho\times l}{A}

Where,

R = Resistance

l = length

A = Area of cross section = πr²

\rho=Resistivity=1.00\times 10^{-6}\ ohm\ meter

Substituting the values we get

R=\dfrac{1\times 10^{-6}\times 10}{3.14\times (0.321\times 10^{-3})^{2}}

R=\dfrac{1\times 10^{-5}}{3.23\times 10^{-7}}

R=\dfrac{1\times 10^{2}}{3.23}

R=30.95\ ohm

Now by Ohm's Law,

V= I\times R

Substituting the values we get

I=\dfrac{V}{R}=\dfrac{12}{30.95}=0.3876\ Ampere

Therefore,

Current through Nichrome wire is 0.3879 Ampere.

4 0
1 year ago
A 1 kg coconut falls vertically from a height of 10 meters. Use the principle of conservation of mechanical
Kay [80]

Answer:

Explanation:

Question 1:

Mass=1kg

Acceleration due to gravity=9.8m/s^2

Height=10m

on the before falling it has potential energy

Potential energy=mass x acceleration due to gravity x height

Potential energy=1 x 9.8 x 10

Potential energy=98 joules

Question 2:

Potential energy=kinetic energy base base on energy transformation

Kinetic energy=(mass x (velocity)^2)➗2

98=(1 x(velocity))^2 ➗ 2

Cross multiplying

98 x 2=(velocity)^2

196=(velocity)^2

Velocity=√(196)

Velocity=14

Velocity=14m/s

5 0
2 years ago
What is the net force acting on the buggy?<br> N<br> The net force is pointing to the...
blagie [28]

Answer:

390, right

Explanation:

5 0
2 years ago
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