Convert 5.7 cm to mm:

Choose all facts that increase the orbital velocity of a vessel around planet B. Bigger mass of planet B smaller mass of planet

telo118 [61]

Answer:

- Bigger mass of planet B

- orbiting closer to planet B

Explanation:

The orbital velocity of the vessel around the planet can be found by equalizing the force of gravity between the vessel and the planet and the centripetal force:

$G\frac{mM}{r^2}=m\frac{v^2}{r}$

where

G is the gravitational constant

m is the mass of the vessel

M is the mass of the planet

r is the distance between the vessel and the centre of the planet

v is the orbital velocity of the vessel

Re-arranging the formula, we find an expression for v:

$v=\sqrt{\frac{GM}{r}}$

We see that:

- the bigger the mass of the planet, M, the bigger the velocity

- the bigger the distance between the vessel and the planet, r, the smaller the velocity

So, the correct choices that increase the orbital velocity are:

- Bigger mass of planet B

- orbiting closer to planet B

6 0

4 years ago

10. A hockey puck with mass 0.3 kg is sliding along ice that can be considered frictionless. The puck’s velocity is 20 m/s when

SVEN [57.7K]

Answer:

$d = \frac{v^2_i}{2a}= \frac{(20m/s)^2}{2* 3.43 m/s^2}=58.309m$

Explanation:

For this case we can use the second law of Newton given by:

$\sum F = ma$

The friction force on this case is defined as :

$F_f = \mu_k N = \mu_k mg$

Where N represent the normal force, $\mu_k$ the kinetic friction coeffient and a the acceleration.

For this case we can assume that the only force is the friction force and we have:

$F_f = ma$

Replacing the friction force we got:

$\mu_k mg = ma$

We can cancel the mass and we have:

$a = \mu_k g = 0.35 *9.8 \frac{m}{s^2}= 3.43 \frac{m}{s^2}$

And now we can use the following kinematic formula in order to find the distance travelled:

$v^2_f = v^2_i - 2ad$

Assuming the final velocity is 0 we can find the distance like this:

$d = \frac{v^2_i}{2a}= \frac{(20m/s)^2}{2* 3.43 m/s^2}=58.309m$

5 0

4 years ago

1.a car sitting at rest on the road <br>balanced or unbalanced<br>explain your answer

Paladinen [302]

Answer:

Balanced force

Explanation:

Balanced Forces, When forces are in balance, acceleration is zero. Velocity is constant and there is no net or unbalanced force. ... Although friction is acting on the person, there is no change in velocity and friction is not a net force in this case. Friction is only a net force if it changes the velocity of a mass.

7 0

4 years ago

Read 2 more answers

Light travels through a liquid at 2.25e8 m/s. What is the index of refraction of the liquid?

Pavlova-9 [17]

Answer:

The index of refraction of the liquid is n = 1.33 equivalent to that of water

Explanation:

Solution:-

- The index of refraction of light in a medium ( n ) determines the degree of "bending" of light in that medium.

- The index of refraction is material property and proportional to density of the material.

- The denser the material the slower the light will move through associated with considerable diffraction angles.

- The lighter the material the faster the light pass through the material without being diffracted as much.

- So, in the other words index of refraction can be expressed as how fast or slow light passes through a medium.

- The reference of comparison of how fast or slow the light is the value of c = 3.0*10^8 m/s i.e speed of light in vacuum or also assumed to be the case for air.

- so we can mathematically express the index of refraction as a ratio of light speed in the material specified and speed of light.

- The light passes through a liquid with speed v = 2.25*10^8 m/s :

$n = c / v\\\\n = \frac{ 3*10^8 }{2.25*10^8} \\\\n = 1.33$

- The index of refraction of the liquid is n = 1.33 equivalent to that of water.

8 0

3 years ago

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How much heat is required to raise the temperature of 0.25 kg of water from 20°C to 30°C

pentagon [3]

Answer:

10500 J/kg/*C

Explanation:

Quantity of heat required=mass of substance x specific heat capacity x change in temperature

Quantity of heat required=0.25 x 4200 x [30-20]

Quantity of heat required=0.25 x 4200 x 10

Quantity of heat required=10500 J/kg/*C

3 0

3 years ago