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3 years ago

What 2 properties are used to classify radiation

Physics

1 answer:

ahrayia [7]3 years ago

6 0

Large amount of energy is release

Temperature and pressure have no effect on the rate qt which radiation is emitted

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A car accelerates at a constant rate from 0 to 50 mph in three fourths min. How far does the car travel during that time?

Helen [10]

Answer:

the car have travelled 0.31 mile during that time

Explanation:

Applying the Equation of motion;

s = 0.5(u+v)t

Where;

s = distance travelled

u = initial speed = 0 mph

v = Final speed = 50 mph

t = time taken = 3/4 min = 3/4 ÷ 60 hours = 1/80 hour

Substituting the given values into the equation;

s = 0.5(0+50)×(1/80)

s = 0.3125 miles

s ~= 0.31 mile

the car have travelled 0.31 mile during that time

8 0

3 years ago

You hang a heavy ball with a mass of 10 kg from a gold wire 2.6 m long that is 1.6 mm in diameter. You measure the stretch of th

PolarNik [594]

<u>Answer:</u> The Young's modulus for the wire is $6.378\times 10^{10}N/m^2$

<u>Explanation:</u>

Young's Modulus is defined as the ratio of stress acting on a substance to the amount of strain produced.

The equation representing Young's Modulus is:

$Y=\frac{F/A}{\Delta l/l}=\frac{Fl}{A\Delta l}$

where,

Y = Young's Modulus

F = force exerted by the weight = $m\times g$

m = mass of the ball = 10 kg

g = acceleration due to gravity = $9.81m/s^2$

l = length of wire = 2.6 m

A = area of cross section = $\pi r^2$

r = radius of the wire = $\frac{d}{2}=\frac{1.6mm}{2}=0.8mm=8\times 10^{-4}m$ (Conversion factor: 1 m = 1000 mm)

$\Delta l$ = change in length = 1.99 mm = $1.99\times 10^{-3}m$

Putting values in above equation, we get:

$Y=\frac{10\times 9.81\times 2.6}{(3.14\times (8\times 10^{-4})^2)\times 1.99\times 10^{-3}}\\\\Y=6.378\times 10^{10}N/m^2$

Hence, the Young's modulus for the wire is $6.378\times 10^{10}N/m^2$

3 0

3 years ago

A large piece of jewelry has a mass of 130.8 g. A graduated cylinder initially contains 47.7 mL water. When the jewelry is subme

Shkiper50 [21]

Answer: The density of this piece of jewelry is $8.90g/cm^3$

Explanation:

To calculate the density, we use the equation:

$\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}$

Mass of piece of jewellery = 130.8 g

Density of piece of jewellery = ?

Volume of piece of jewellery =( 62.4-47.7 ) ml = 14.7 ml = $14.7cm^3$ $1cm^3=1ml$

Putting values in above equation, we get:

$\text{Density of piece of jewellery}=\frac{130.8g}{14.7cm^3}=8.90g/cm^3$

Thus density of this piece of jewelry is $8.90g/cm^3$

8 0

3 years ago

Using the scientific definition of work, does moving an object a greater amount of distance always require a greater amount of w

tester [92]

The answer is no. If you are dealing with a conservative force and the object begins and ends at the same potential then the work is zero, regardless of the distance travelled. This can be shown using the work-energy theorem which states that the work done by a force is equal to the change in kinetic energy of the object.
W=KEf−KEi
An example of this would be a mass moving on a frictionless curved track under the force of gravity.
The work done by the force of gravity in moving the objects in both case A and B is the same (=0, since the object begins and ends with zero velocity) but the object travels a much greater distance in case B, even though the force is constant in both cases.

3 0

3 years ago

a particle of mass2.5kg moves in a cnservative force field.its potential energy curveis dhwn below.from the curve,detrmine(a)the

Serhud [2]

Nothing can be determined from the curve without seeing the curve.

5 0

3 years ago