Which is an example of a mixture?

What are the factors that influence the force of gravity

blsea [12.9K]

The strength of the gravitational force between two objects depend on two factors, mass and distance.

5 0

4 years ago

In a playground, there is a small merry-go-round of radius 1.20 m and mass 160 kg. Its radius of gyration is 91.0 cm. (Radius of

aksik [14]

Answer:

a) 145.6kgm^2

b) 158.4kg-m^2/s

c) 0.76rads/s

Explanation:

Complete qestion: a) the rotational inertia of the merry-go-round about its axis of rotation

(b) the magnitude of the angular momentum of the child, while running, about the axis of rotation of the merry-go-round and

(c) the angular speed of the merry-go-round and child after the child has jumped on.

a) From I = MK^2

I = (160Kg)(0.91m)^2

I = 145.6kgm^2

b) The magnitude of the angular momentum is given by:

L= r × p The raduis and momentum are perpendicular.

L = r × mc

L = (1.20m)(44.0kg)(3.0m/s)

L = 158.4kg-m^2/s

c) The total moment of inertia comprises of the merry- go - round and the child. the angular speed is given by:

L = Iw

158.4kgm^2/s = [145kgm^2 + ( 44.0kg)(1.20)^2]

w = 158.6/208.96

w = 0.76rad/s

7 0

3 years ago

A 200 kg roller coaster is located at the top of a hill where the height is 30 m. What is the velocity when it reches the bottom

maw [93]

Assuming it starts at rest, the roller coaster only has potential energy at the top of the hill, which is

$E_P=(200\,\mathrm{kg})(30\,\mathrm m)g$

When it reaches the bottom, its potential energy will have converted to kinetic energy,

$E_K=\dfrac12(200\,\mathrm{kg})v^2$

where $v$ is its velocity at that point. By the law of conservation of energy, assuming no loss of energy due to other sources (e.g. sound, heat), we have

$E_P=E_K\iff(6000\,\mathrm{kg}\cdot\mathrm m)g=(100\,\mathrm kg})v^2$

$\implies v=\sqrt{60g}\,\dfrac{\rm m}{\rm s}\approx24.2\,\dfrac{\rm m}{\rm s}$

8 0

4 years ago

Read 2 more answers

Derive the dimension of Power

stellarik [79]

Answer:

The dimension of power is energy divided by the time or $[ML^2T^-3]$

Explanation:

Power = $\frac {Work Done}{Time}$

We can derive Dimensions of Power from both formula.

Power = Force * Velocity

As,

Force = mass * acceleration

Therefore, Dimensions of

Force = $[M]*[LT^-2] = [MLT^-2]$

Since,

Velocity = $\frac{Length}{Time}$

Now, Dimension of

Velocity = $[LT^-1]$

We have Both Dimensions,Now we can derive Dimensions Of Power,

Power = Force * Velocity

Power = $[MLT^-2] * [LT^-1]$

Power = $[ML^2T^-3]$

7 0

3 years ago

Two force A and B at a point at right angles. If their resultant is 50N and their sum is 70N,

kodGreya [7K]

Taking a wild guess here, but it sounds like you're asked to find |A| and |B| given the magnitude of the resultant A + B, i.e. |A + B| = 50 N, and the sum of the individual magnitudes, |A| + |B| = 70 N.

Recall that the dot product of a vector with itself is equal to the square of that vector's magnitude,

A • A = |A|²

Then

|A + B|² = (A + B) • (A + B)

|A + B|² = (A • A) + 2 (A • B) + (B • B)

|A + B|² = |A|² + 2 (A • B) + |B|²

Since A and B are perpendicular to one another, their dot product is

A • B = 0

So it follows that

|A + B|² = |A|² + |B|²

2500 N² = |A|² + |B|²

Substitute |B| = 70 N - |A| and solve for |A| :

2500 N² = |A|² + (70 N - |A|)²

2500 N² = |A|² + 4900 N² - (140 N) |A| + |A|²

2 |A|² - (140 N) |A| + 2400 N² = 0

|A|² - (70 N) |A| + 1200 N² = 0

|A|² - (70 N) |A| = - 1200 N²

|A|² - (70 N) |A| + 1225 N² = - 1200 N² + 1225 N²

(|A| - 35 N)² = 25 N²

|A| - 35 N = ± 5 N

|A| = 35 N ± 5 N

so that |A| = 30 N or |A| = 40 N. If we fix |A| to be one of these, then |B| will have the other value.

So, the magnitudes are |A| = 30 N and |B| = 40 N.

3 0

3 years ago