Ask question

Ask question

All categories

11Alexandr11 [23.1K]

3 years ago

10

Put the energy values in order from smallest at the top of your list to biggest at the bottom.

1 answer:

kodGreya [7K]3 years ago

3 0

2j
20k
100
2kj
20 000j
100kj

You might be interested in

The work done by the brakes during braking is equal to

kow [346]

Answer:

The amount of pressure delivered to them

Explanation:

6 0

2 years ago

A typical helium-neon laser found in supermarket checkout scanners emits 633-nm-wavelength light in a 1.0-mm-diameter beam with

Fofino [41]

Answer:

Eo = 9.796 x 10^2 N/C

Bo = 3.266 x 10^-6 T

Explanation:

Given

Wavelength λ = 633 nm

Diameter of the beam D = 1.0 mm

Power P = 1.0 mW

Solution

Radius of the beam r = D/2 = 0.5 mm = 0.0005 m

Area of cross section

$A = \pi r^{2} \\A = 3.15 \times 0.0005^{2}\\A = 7.58 \times 10^{-7} m^{2}\\$

Intensity

$I = \frac{P}{A} \\I = \frac{0.001}{7.85\times 10^{-7}} \\I = 1273.885 {W}/{m^{2} }$

Amplitude of Electric Field

$E_{o} = \sqrt{\frac{2I}{ \epsilon_{o}c } } \\E_{o} = \sqrt{\frac{2 \times 1273.88}{ 8.85 \times 10^{-12} \times 3 \times 10^{8} } }\\E_{o} = 9.796 \times 10^{2}N/C$

Amplitude of Magnetic Field

$B_{o} = \sqrt{\frac{2 \mu_{o}I}{c } } \\B_{o} = \sqrt{\frac{2 \times 4 \times \pi \times 10^{-7} \times 1273.88}{ 3 \times 10^{8} } }\\B_{o} = 3.266 \times 10^{-6} T$

5 0

4 years ago

Years ago, a block of ice with a mass of about 20.0 kg was used daily in a home icebox. The temperature of the ice was 0.0°C whe

m_a_m_a [10]

Answer: The ice absorb 6671.1 kJ of thermal energy.

Explanation:

The conversions involved in this process are :

$0.00^0C=273K$

$:H_2O(s)(273K)\rightarrow H_2O(l)(273K)$

Now we have to calculate the enthalpy change.

$\Delta H=n\times \Delta H_{fusion}$

where,

$\Delta H$ = enthalpy change = ?

m = mass of ice = 20.0 kg = $20.0\times 10^3g$ (1kg=1000g)

n = number of moles of ice= $\frac{\text{Mass of ice}}{\text{Molar mass of water}}=\frac{20.0\times 10^3g}{18g/mole}=1.11\times 10^3mole$

$\Delta H_{fusion}$ = enthalpy change for fusion = 6.01 KJ/mole = 6010 J/mole

Now put all the given values in the above expression, we get

$\Delta H=1.11\times 10^3mole\times 6010J/mole$

$\Delta H=6671100J=6671.1kJ$ (1 kJ = 1000 J)

Therefore, the enthalpy change is, 6671.1 kJ

6 0

3 years ago

Where is the electric field zero? a. region 2, 0.46 m from the +7 µC charge b. region 2, 0.46 m from the +5 µC charge c. the fie

Ratling [72]

Answer:

the field is not zero anywhere on the x axis (except at infinity)

Explanation:

From the Coulomb's law we have electric field intensity as:

$E=\frac{1}{4\pi.\epsilon_o} \frac{q}{r^2}$

where:

$\epsilon_o=$ permittivity of free space

$q=$ charge due to which the field is generated.

$r=$ distance from the charge

So, from the above relation:

Electric field due to a charge can only be zero at infinity.

3 0

4 years ago

A 15.0 kg load of bricks hangs from one end of a rope that passes over a small, frictionles pulley. A 28.0 kg counterweight is s

Talja [164]

Answer:

A) The free body diagrams for both the load of bricks and the counterweight are attached.

B) a = 2.96 m/s²

Explanation:

A)

The free body diagrams for both the load of bricks and the counterweight are attached.

B)

The acceleration of upward acceleration of the load of bricks is given by the following formula:

a = g(m₁ - m₂)/(m₁ + m₂)

where,

a = upward acceleration of load of bricks = ?

g = 9.8 m/s²

m₁ = heavier mass = mass of counterweight = 28 kg

m₂ = lighter mass = mass of load of bricks = 15 kg

Therefore, using these values in equation, we get:

a = (9.8 m/s²)(28 kg - 15 kg)/(28 kg + 15 kg)

<u>a = 2.96 m/s²</u>

3 0

4 years ago