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3 years ago

Density what over what d=

Physics

2 answers:

lianna [129]3 years ago

7 0

The answer to your question is,

Density = mass divided by volume

-Mabel <3

mariarad [96]3 years ago

3 0

The answer to this question is that density is a degree of consistency measured by the quantity of mass per unit volume. Hope this helped!

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Why cant your rocket could never reach the speed of light?

Oksanka [162]

Answer:

The length of the object would shrink to zero which is not possible.

Explanation:

A rocket or any body cannot reach the speed of light because according to theory of relativity the and the Lorentz factor the length of the object would shrink to zero and the time dilation for that body would be infinite.

The Lorentz factor is given as:

$\gamma=\frac{1}{\sqrt{\frac{v^2}{c^2} } }$

where:

v = speed of the moving object

c = speed of light

4 0

3 years ago

If an object is moving up, is it considered to have a positive or negative velocity?

grin007 [14]

Answer:

A positive velocity

Explanation:

7 0

3 years ago

Imagine that you are working as a roller coaster designer. You want to build a record breaking coaster that goes 70.0 m/s at the

Rzqust [24]

Wow ! This is not simple. At first, it looks like there's not enough information, because we don't know the mass of the cars. But I"m pretty sure it turns out that we don't need to know it.

At the top of the first hill, the car's potential energy is

PE = (mass) x (gravity) x (height) .

At the bottom, the car's kinetic energy is

   KE = (1/2) (mass) (speed²) .

You said that the car's speed is 70 m/s at the bottom of the hill,
and you also said that 10% of the energy will be lost on the way
down. So now, here comes the big jump. Put a comment under
my answer if you don't see where I got this equation:

   KE = 0.9 PE

(1/2) (mass) (70 m/s)² = (0.9) (mass) (gravity) (height)

Divide each side by (mass):

   (0.5) (4900 m²/s²) = (0.9) (9.8 m/s²) (height)

(There goes the mass. As long as the whole thing is 90% efficient,
the solution will be the same for any number of cars, loaded with
any number of passengers.)

Divide each side by (0.9):

   (0.5/0.9) (4900 m²/s²) = (9.8 m/s²) (height)

Divide each side by (9.8 m/s²):

   Height = (5/9)(4900 m²/s²) / (9.8 m/s²)

      = (5 x 4900 m²/s²) / (9 x 9.8 m/s²)

      = (24,500 / 88.2) (m²/s²) / (m/s²)

      = 277-7/9 meters
      (about 911 feet)

3 0

2 years ago

A 1 530-kg automobile has a wheel base (the distance between the axles) of 2.70 m. The automobile's center of mass is on the cen

NeTakaya

Answer:

Force on front axle = 6392.85 N

Force on rear axle = 8616.45 N

Explanation:

As we know that the weight of the car is balanced by the normal force on the front wheel and rear wheels

Now we know that

$F_1 + F_2 = W$

$F_1 + F_2 = (1530\times 9.81)$

$F_1 + F_2 = 15009.3 N$

now we know that distance between the axis is 2.70 m and centre of mass is 1.15 m behind front axle

so we can write torque balance about its center of mass

$F_1(1.15) = F_2(2.70 - 1.15)$

$F_1 = 1.35 F_2$

now from above equation

$F_2 + 1.35F_2 = 15009.3$

now we have

$F_2 = 6392.85 N$

now the other force is given as

$F_1 = 8616.45 N$

4 0

3 years ago

A rough estimate of the radius of a nucleus is provided by the formula r 5 kA1/3, where k is approximately 1.3 × 10213 cm and A

Sphinxa [80]

Answer:

Density of 127 I = $\rm 1.79\times 10^{14}\ g/cm^3.$

Also, $\rm Density\ of\ ^{127}I=3.63\times 10^{13}\times Density\ of\ the\ solid\ iodine.$

Explanation:

Given, the radius of a nucleus is given as

$\rm r=kA^{1/3}$ .

where,

$\rm k = 1.3\times 10^{-13} cm.$

A is the mass number of the nucleus.

The density of the nucleus is defined as the mass of the nucleus M per unit volume V.

$\rm \rho = \dfrac{M}{V}=\dfrac{M}{\dfrac 43 \pi r^3}=\dfrac{M}{\dfrac 43 \pi (kA^{1/3})^3}=\dfrac{M}{\dfrac 43 \pi k^3A}.$

For the nucleus 127 I,

Mass, M = $\rm 2.1\times 10^{-22}\ g.$

Mass number, A = 127.

Therefore, the density of the 127 I nucleus is given by

$\rm \rho = \dfrac{2.1\times 10^{-22}\ g}{\dfrac 43 \times \pi \times (1.3\times 10^{-13})^3\times 127}=1.79\times 10^{14}\ g/cm^3.$

On comparing with the density of the solid iodine,

$\rm \dfrac{Density\ of\ ^{127}I}{Density\ of\ the\ solid\ iodine}=\dfrac{1.79\times 10^{14}\ g/cm^3}{4.93\ g/cm^3}=3.63\times 10^{13}.\\\\\Rightarrow Density\ of\ ^{127}I=3.63\times 10^{13}\times Density\ of\ the\ solid\ iodine.$

7 0

3 years ago