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3 years ago

In what frame of reference would you be at rest while riding in a car?

Physics

2 answers:

Alexeev081 [22]3 years ago

8 0

In the frame of reference of anybody in the car.

gulaghasi [49]3 years ago

8 0

<h2>Answer:</h2>

<u>We would be in the frame of reference to other people inside the car.</u>

<h2>Explanation:</h2>

In Physics the frame of reference is used for the observation and mathematical description of physical phenomena that occur around us and they formulate the physical laws with respect to others. In the given case all the people inside the car are the frame of reference to one another but if someone is outside of the car then his frame of reference will be changing as the car moves with respect to him.

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an astronaut on an eva has wandered dangerously far away from the shuttle. she has also exhausted all the fuel in her jet pack.

V125BC [204]

The conservation of the momentum allows to find the result of how the astronaut can return to the spacecraft is:

Throwing the thruster away from the ship.

The momentum is defined as the product of the mass and the velocity of the body, for isolated systems the momentum is conserved. If we define the system as consisting of the astronaut and the evo propellant, this system is isolated and the internal forces become zero. Let's find the moment in two moments.

Initial instant. Astronaut and thrust together.

p₀ = 0

Final moment. The astronaut now the thruster in the opposite direction of the ship.

$m_f$ = m v + M v '

where m is propellant mass and M the astronaut mass.

As the moment is preserved.

0 = m v + M v ’

v ’= $- \frac{m}{M} \ v$

We can see that the astronaut's speed is in the opposite direction to the propeller, that is, in the direction of the ship.

The magnitude of the velocity is given by the relationship between the masses.

In conclusion, using the conservation of the momentun we can find the result of how the astronaut can return to the ship is:

Throwing the thruster away from the ship.

Learn more here: brainly.com/question/14798485

5 0

2 years ago

Which of the following is an example of frontier research?

riadik2000 [5.3K]

Testing the electrical resistance of different materials. it is the answer

3 0

3 years ago

If a 2,000-kg car hits a tree with 500 n of force over a time of 0.5 seconds, what is the magnitude of its impulse?

Elden [556K]

Impulse = Force * time
Impulse = 500N *0.5 s =250 N*s

4 0

2 years ago

Read 2 more answers

Earth rotates on its axis once every 24 hours, so that objects on its surface execute uniform circular motion about the axis wit

cestrela7 [59]

Answer:

$v=1667.9km/h$

$a_{cp}=436.6km/h^2$

Explanation:

The speed is the distance traveled divided by the time taken. The distance traveled in 24hs while standing on the equator is the circumference of the Earth $C=2\pi R$ , where $R=6371km$ is the radius of the Earth.

We have then:

$v=\frac{C}{t}=\frac{2\pi R}{t}=\frac{2\pi (6371km)}{(24h)}=1667.9km/h$

And then we use the centripetal acceleration formula:

$a_{cp}=\frac{v^2}{R}=\frac{(1667.9km/h)^2}{(6371km)}=436.6km/h^2$

6 0

2 years ago

A 100g block lies on an inclined plane that makes an angle of 15 degrees with the horizontal. The coefficient of kinetic frictio

Fed [463]

Answer:

Mass that one should put in the container so that the 100 g block slides down the inclined plane at constant speed = 34.16 g

Explanation:

The vertical forces (with respect to the inclined plane) acting on the 100 g block include the component of the weight of the block in the direction vertical to the inclined plane and the normal reaction of the plane on the block.

And sum of upward forces = sum of downward forces.

N = mg cos θ

m = 100 g = 0.10 kg

g = acceleration due to gravity = 9.8 m/s²

θ = 15°

N = (0.1×9.8×cos 15°) = 0.946582 N

The horizontal forces (With respect to the inclined plane) include the frictional force (acting upwards for the inclined plane, opposite to the intended direction of motion), the Tension in the rope (acting downwards, away from the 100 g block) and the horizontal component (with respect to the inclined plane) of the weight of the block, F, (also acting downards).

For the body to slide down the inclined plane at constant speed, the downward sloping forces must balance the frictional force, that is, there will be no acceleration.

Frictional force = Tension + F

Frictional force = μN

where μ = coefficient of kinetic friction = 0.60

N = normal reaction = 0.9466 N

Frictional force = Fr = (0.60 × 0.9466) = 0.56796 N = 0.568 N

The horizontal component (with respect to the inclined plane) of the weight of the block (also acting downards) = mg sin θ

F = (0.10 × 9.8 × sin 15°) = 0.253624 N

Tension in the rope = T = ?

Fr = F + T

T = Fr - F = 0.568 - 0.253624 = 0.314376 N = 0.3144 N

But the balance on the rope now has the total weight on the container (weight of container + weight on the container) to be equal to 2T.

2T = mg

2 × 0.3144 = 9.8m

m = 0.06416 kg = 64.16 g.

Mass of the container = 30 g

So, mass that one should put in the container so that the 100 g block slides down the inclined plane at constant speed = 64.16 - 30 = 34.16 g

Hope this Helps!!!

8 0

3 years ago