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2 years ago

6

Natalia is studying a wave produced in her magnetics lab. This wave can move through the empty space in a vacuum and carries a l

ot of energy. What wave is Natalia most likely studying?

2 answers:

astra-53 [7]2 years ago

8 0

Answer:

A gamma ray

Explanation:

Someone else said this...

jarptica [38.1K]2 years ago

3 0

Answer:

a gammawave

Explanation

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Dane is holding an 8 kilogram box 2 metres above the ground. How much energy is in the box's gravitational potential energy stor

9966 [12]

156.8 Joules of energy is in the box's gravitational potential energy store

<u>Explanation</u>:

<em>Given:</em>

Mass of the box Dane is holding = 8 Kilograms

Height at which Dane is holding the box above the ground= 2 metres

<em>To Find:</em>

Gravitational potential energy in the box=?

<em>Solution:</em>

gravitational potential energy is the work done per mass on a object to move that object from one fixed location to to another location against gravity.Its unit is joules or J

Thus Gravitational potential energy is represented as,

$PE_g=mgh$

where

$PE_g$ is the gravitational potential energy

m is the mass

h is the height

g is the gravitational force( 9.8 $m/s^2$ )

Now substituting the given values,

$PE_g=8\times 9.8\times 2$

$PE_g=156.8 Joules$

4 0

2 years ago

Calculate the weight of a 4.5 kg rabbit.

solniwko [45]

The correct answer is: 13900589.

3 0

2 years ago

The gap between electrodes in a spark plug is 0.060 cm. Producing an electric spark in a gasoline-air mixture requires an electr

VladimirAG [237]

The minimum potential difference must be supplied by the ignition circuit to start a car is -1800 V

<u>Explanation:</u>

Given data,

E= 3 ×10 ⁶ Δx=0.06/100

We have to find the minimum potential difference

E= -ΔV/Δx

ΔV=- E × Δx

ΔV =-3 ×10 ⁶ . 0.06/100

ΔV=-1800 V

The minimum potential difference must be supplied by the ignition circuit to start a car is -1800 V

6 0

2 years 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 vessel of 0.25 m3 capacity is filled with saturated steam at 1500 kPa. If the vessel is cooled until 25% of the steam has cond

soldier1979 [14.2K]

Answer:

the final pressure is P=353.5 kPa and the heat lost by the system (heat transferred) is Q= -3614.7327 kJ ( negative sign means heat outflow)

Explanation:

since the steam is at saturated state at 1500 kPa , from tables of saturated steam

at P= 1500 kPa → specific volume v= 0.13225 m³/kg , ug=2593.3 kJ/kg

thus

m = V/v = 0.25 m³ / 0.13225 m³/kg = 1.89 kg

for condensation until 25% of steam, then mass of steam is

mg₂ = 0.25* 1.89 kg = 0.4725 kg

if we neglect the volume occupied by the liquid , then the steam occupies

v₂ = V/m = 0.25 m³ / 0.4725 kg =0.529 m³/kg

returning to the saturated steam table

at v₂=0.529 m³/kg → 353.5 kPa , uL₂= 584.1585 kJ/kg , ug₂= 2548.4965 kJ/kg

then from the first law of thermodynamics

ΔU= Q - W , but since V=constant , dV=0 and W=∫PdV =0

Q= ΔU

Q= (mg₂*ug₂+ml₂*uL₂) - (m*ug) = 1.89 kg* (0.25*2548.4965 kJ/kg + 0.75*584.1585 kJ/kg - 2593.3 kJ/kg) = -3614.7327 kJ

5 0

2 years ago