Ask question

Ask question

All categories

3 years ago

8

A block of wood 3 cm on each side has a mass of 27 g. what is the density of the block

1 answer:

butalik [34]3 years ago

7 0

Density: Mass/Volume

Volume: 3 x 3 x 3cm
= 27cm

27/27
= 1g/cm^3

You might be interested in

What can I do to increase range of motion in a joint

Phantasy [73]

Answer:

Damian here! (ﾉ◕ヮ◕)ﾉ*:･ﾟ✧

Stretching is used to improve range-of-motion (ROM) of a joint, but why? The most common reason is that the joint ROM is limited and is somehow affecting performance of a desired activity. Stretching is also used as a preventative measure.

Explanation:hope this helps? :))

8 0

3 years ago

A car at the top of a ramp starts from rest and rolls to the bottom of the ramp, achieving a certain final speed. If you instead

zimovet [89]

Answer:

It must be 4 times high.

Explanation:

Assuming that the car can be treated as a point mass, and that the ramp is frictionless, the total mechanical energy must be conserved.
This means, that at any time, the following must be true:
ΔK (change in kinetic energy) = ΔU (change in gravitational potential energy)

⇒ $m*g*h = \frac{1}{2} * m*v^{2}$

Let's call v₁, to the final speed of the car, and h₁ to the height of the ramp.

So, at the bottom of the ramp, all the gravitational potential energy

must be equal to the kinetic energy of the car (Defining the bottom of

the ramp as our zero reference for the gravitational potential energy):

$m*g*h_{1} = \frac{1}{2} * m*v_{1} ^{2}$ (1)

Now, let's do v₂ = 2* v₁
Replacing in (1) we get:

$m*g*h_{2} = \frac{1}{2} * m*(2*v_{1}) ^{2}$ (2)

Dividing (2) by (1), and rearranging terms, we get:
h₂ = 4* h₁

8 0

3 years ago

Calculate the Schwarzschild radius (in kilometers) for each of the following.1.) A 1 ×108MSun black hole in the center of a quas

Westkost [7]

Answer:

(I). The Schwarzschild radius is $2.94\times10^{8}\ km$

(II). The Schwarzschild radius is 17.7 km.

(III). The Schwarzschild radius is $1.1\times10^{-7}\ km$

(IV). The Schwarzschild radius is $7.4\times10^{-29}\ km$

Explanation:

Given that,

Mass of black hole $m= 1\times10^{8} M_{sun}$

(I). We need to calculate the Schwarzschild radius

Using formula of radius

$R_{g}=\dfrac{2MG}{c^2}$

Where, G = gravitational constant

M = mass

c = speed of light

Put the value into the formula

$R_{g}=\dfrac{2\times6.67\times10^{-11}\times1\times10^{8}\times1.989\times10^{30}}{(3\times10^{8})^2}$

$R_{g}=2.94\times10^{8}\ km$

(II). Mass of block hole $m= 6 M_{sun}$

We need to calculate the Schwarzschild radius

Using formula of radius

$R_{g}=\dfrac{2MG}{c^2}$

Put the value into the formula

$R_{g}=\dfrac{2\times6.67\times10^{-11}\times6\times1.989\times10^{30}}{(3\times10^{8})^2}$

$R_{g}=17.7\ km$

(III). Mass of block hole m= mass of moon

We need to calculate the Schwarzschild radius

Using formula of radius

$R_{g}=\dfrac{2MG}{c^2}$

Put the value into the formula

$R_{g}=\dfrac{2\times6.67\times10^{-11}\times7.35\times10^{22}}{(3\times10^{8})^2}$

$R_{g}=1.1\times10^{-7}\ km$

(IV). Mass = 50 kg

We need to calculate the Schwarzschild radius

Using formula of radius

$R_{g}=\dfrac{2MG}{c^2}$

Put the value into the formula

$R_{g}=\dfrac{2\times6.67\times10^{-11}\times50}{(3\times10^{8})^2}$

$R_{g}=7.4\times10^{-29}\ km$

Hence, (I). The Schwarzschild radius is $2.94\times10^{8}\ km$

(II). The Schwarzschild radius is 17.7 km.

(III). The Schwarzschild radius is $1.1\times10^{-7}\ km$

(IV). The Schwarzschild radius is $7.4\times10^{-29}\ km$

8 0

3 years ago

A 5.00 kg rock whose density is 4300 kg/m3 is suspended by a string such that half of the rock's volume is under water. You may

12345 [234]

Answer:

The tension on the string is $T = 43.302 \ N$

Explanation:

From the question we are told that

The mass of the rock is $m_r = 5.00 \ kg = 5000 \ g$

The density of the rock is $\rho = 4300 \ kg/m^3 = 4.3 g/dm^3$

Generally the volume of the rock is mathematically evaluated as

$V = \frac{m_r}{\rho}$

substituting values

$V = \frac{5000}{4.3}$

$V = 1162.7 \ dm^3$

The volume of the rock immersed in water is

$V_w = \frac{V}{2}$

substituting values

$V_w = \frac{1162.7 }{2}$

$V_w = 581.4 \ dm^3$

mass of water been displaced by the this volume is

$m_w = V_w$ According to Archimedes principle

=> $m_w = 581.4 \ g$

$m_w = 0.5814 \ kg$

The weight of the water displace is

$W _w = m_w * g$

$W _w = 0.5814 * 9.8$

$W _w = 5.698 \ N$

The actual weight of the rock is

$W_r = m_r * g$

$W_r = 5.0 * 9.8$

$W_r = 49.0 \ N$

The tension on the string is

$T = W_r - W_w$

substituting values

$T = 49.0 - 5.698$

$T = 43.302 \ N$

4 0

3 years ago

Two spherical conductors are separated by a distance much larger than either of their radii. Sphere A has a radius of 11.5 cm an

bonufazy [111]

Explanation:

As the given spheres are connected by a thin wire so, the potential on the spheres are the same.

$\frac{q_{1}}{r_{1}} = \frac{q_{2}}{r_{2}}$ ......... (1)

Hence, total charge will be as follows.

$q_{1} + q_{2}$ = Q = -95.5 nC .......... (2)

Using the above two equations, the final equation will be as follows.

$q_{2} = \frac{Qr_{2}}{r_{1} + r_{2}}$

and, $q_{1} = \frac{Qr_{1}}{r_{1} + r_{2}}$

Hence, we will calculate the charge on sphere B after the equilibrium is reached as follows.

$q_{2} = \frac{Qr_{2}}{r_{1} + r_{2}}$

= $\frac{-95.5 \times 74.4 cm}{(11.5 + 74.4) cm}$

= 82.714 nC

Thus, we can conclude that the charge on sphere B after equilibrium has been reached is 82.714 nC.

5 0

3 years ago