What is the gravitational force between two identical 5000 kg asteroids whose centers of mass are separated by 100 m?

Convert 285,000 milliliters to decaliters.

Rudiy27

The answer is 28.5 decaliters. Hope this helps

5 0

3 years ago

A 2.0-kg projectile is fired with initial velocity components v0x = 30 m/s and v0y = 40 m/s from a point on the Earth's surface.

Alexxandr [17]

Answer:

Kinetic energy of the projectile at the vertex of the trajectory: $900\; {\rm J}$ .

Work done when firing this projectile: $2500\; {\rm J}$ .

Explanation:

Since the drag on this projectile is negligible, the horizontal velocity $v_{x}$ of this projectile would stay the same (at $30\; {\rm m\cdot s^{-1}}$ ) throughout the flight.

The vertical velocity $v_{y}$ of this projectile would be $0\; {\rm m\cdot s^{-1}}$ at the vertex (highest point) of its trajectory. (Otherwise, if $v_{y} > 0$ , this projectile would continue moving up and reach an even higher point. If $v_{y} < 0$ , the projectile would be moving downwards, meaning that its previous location was higher than the current one.)

Overall, the velocity of this projectile would be $v = 30\; {\rm m\cdot s^{-1}}\!$ when it is at the top of the trajectory. The kinetic energy $\text{KE}$ of this projectile (mass $m = 2.0\; {\rm kg}$ ) at the vertex of its trajectory would be:

$\begin{aligned} \text{KE} &= \frac{1}{2}\, m\, v^{2} \\ &= \frac{1}{2} \times 2.0\; {\rm kg} \times (30\; {\rm m\cdot s^{-1}})^{2} \\ &= 900\; {\rm J} \end{aligned}$ .

Apply the Pythagorean Theorem to find the initial speed of this projectile:

$\begin{aligned}v &= \sqrt{(v_{x})^{2} + (v_{y})^{2}} \\ &= \left(\sqrt{900 + 1600}\right)\; {\rm m\cdot s^{-1}} \\ &= 50\; {\rm m\cdot s^{-1}}\end{aligned}$ .

Hence, the initial kinetic energy $\text{KE}$ of this projectile would be:

$\begin{aligned} \text{KE} &= \frac{1}{2}\, m\, v^{2} \\ &= \frac{1}{2} \times 2.0\; {\rm kg} \times (50\; {\rm m\cdot s^{-1}})^{2} \\ &=2500\; {\rm J} \end{aligned}$ .

All that energy was from the work done in launching this projectile. Hence, the (useful) work done in launching this projectile would be $2500\; {\rm J}$ .

7 0

2 years ago

If you can answer both, please do. But if you can't, just answer one.

Setler [38]

Answer:

1.The force required to stop the shopping cart is, F = 12.25 N

Explanation:

Given data,

The mass of the shopping cart, m = 7 kg

The initial velocity of the shopping cart, u = 3.5 m/s

The final velocity of the shopping cart, v = 0 m/s

The time period of acceleration, t = 2 s

The change in momentum of the cart,

p = m(u - v)

= 7 (3.5 - 0)

= 24.5 kg m/s

The force is defined as the rate of change of momentum. To stop the shopping cart, the force required is given by the formula

F = p / t

= 24.5 / 2

= 12.25 N

Hence, the force required to stop the shopping cart is, F = 12.25 N

2.

We have: F = m × v/t

Here, m = 8500 Kg

v = 20 m/s

t = 10 s

Substitute their values into the expression,

F = 8500 × 20/10

F = 8500 × 2

F = 17000 N

In short, final answer would be 17000 N

Hope this helps!!

7 0

3 years ago

Light with a wavelength of 494 nm in vacuo travels from vacuum to water. Find the wavelength of the light inside the water in nm

Mamont248 [21]

Answer:

370.6 nm

Explanation:

wavelength in vacuum = 494 nm

refractive index of water with respect to air = 1.333

Let the wavelength of light in water is λ.

The frequency of the light remains same but the speed and the wavelength is changed as the light passes from one medium to another.

By using the definition of refractive index

$n = \frac{wavelength in air}{wavelength in water}$

where, n be the refractive index of water with respect to air

By substituting the values, we get

$1.333 = \frac{494}{\lambda }$

λ = 370.6 nm

Thus, the wavelength of light in water is 370.6 nm.

8 0

3 years ago

Pls help me with this problem!!

Olenka [21]

Answer:

v = 19.6 m/s.

Explanation:

Given that,

The radius of the circle, r = 5 m

The time period of the ball, T = 1.6s

We need to find the ball's tangential velocity.

The formula for the tangential velocity is given by :

$v=\dfrac{2\pi r}{T}$

Putting all the values in the above formula

$v=\dfrac{2\pi \times 5}{1.6}\\\\v=19.6\ m/s$

So, the tangential velocity of the ball is 19.6 m/s. Hence, the correct option is (c).

7 0

3 years ago