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3 years ago

A Los Angeles business man has booked a flight from Madrid to New York, in which city he will get off from the train?

Physics

1 answer:

maxonik [38]3 years ago

4 0

Answer: Los Angeles.

Explanation: We know that the businessman is from Los Angeles, and we also know that the flight from Madrid to New York, so it is safe to say that the man is currently in the city of Madrid, in Spain.

Then, after the flight, he will be in New York, and from there hey may take a train to return to his home, so the city in which he gets off from the train maybe Los Angeles, where he lives.

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ndicate whether the statement is true or false. Whenever we move, we alter the rate at which we move into the future. A. True B.

Ksju [112]

Whenever we move, we alter the rate at which we move into the future. This statement is true.

4 0

3 years ago

I WILL MARK YOU THE BRAINLIEST

mr Goodwill [35]

Explanation:

700N right

to get the net force

you gotta let one direction be the negative ( the smaller force)

so the total force towards the left is 100N ( 60 + 40= 100)

which is smaller than the right force which is 800 N so you let 100 N be negative

so without even calculating , you can know that it will be moving towards the right because right force > left force

your add both forces ( remember 100 N is negative)

so 800N + ( - 100N)

= 700N

towards the right

hope this helps

this is just one method that helped me understand

please mark it brainliest

3 0

3 years ago

Read 2 more answers

A car travels at a steady 40.0 m/s around a horizontal curve of radius 200 m. What is the minimumcoefficient of static friction

zhenek [66]

Answer:

c. 0.816

Explanation:

Let the mass of car be 'm' and coefficient of static friction be 'μ'.

Given:

Speed of the car (v) = 40.0 m/s

Radius of the curve (R) = 200 m

As the car is making a circular turn, the force acting on it is centripetal force which is given as:

Centripetal force is, $F_c=\frac{mv^2}{R}$

The frictional force is given as:

Friction = Normal force × Coefficient of static friction

$f=\mu N$

As there is no vertical motion, therefore, $N=mg$ . So,

$f=\mu mg$

Now, the centripetal force is provided by the frictional force. Therefore,

Frictional force = Centripetal force

$f=F_c\\\\\mu mg=\frac{mv^2}{R}\\\\\mu=\frac{v^2}{Rg}$

Plug in the given values and solve for 'μ'. This gives,

$\mu = \frac{(40\ m/s)^2}{200\ m\times 9.8\ m/s^2}\\\\\mu=\frac{1600\ m^2/s^2}{1960\ m^2/s^2}\\\\\mu=0.816$

Therefore, option (c) is correct.

7 0

3 years ago

On a frictionless horizontal air table, puck A (with mass 0.254 kg ) is moving toward puck B (with mass 0.367 kg ), which is ini

irinina [24]

Answer:

$v_a=0.8176 m/s$

$\Delta K=0.07969 J - 0.0849 J = -0.00521 J$

Explanation:

According to the law of conservation of linear momentum, the total momentum of both pucks won't be changed regardless of their interaction if no external forces are acting on the system.

Being $m_a$ and $m_b$ the masses of pucks a and b respectively, the initial momentum of the system is

$M_1=m_av_a+m_bv_b$

Since b is initially at rest

$M_1=m_av_a$

After the collision and being $v'_a$ and $v'_b$ the respective velocities, the total momentum is

$M_2=m_av'_a+m_bv'_b$

Both momentums are equal, thus

$m_av_a=m_av'_a+m_bv'_b$

Solving for $v_a$

$v_a=\frac{m_av'_a+m_bv'_b}{m_a}$

$v_a=\frac{0.254Kg\times (-0.123 m/s)+0.367Kg (0.651m/s)}{0.254Kg}$

$v_a=0.8176 m/s$

The initial kinetic energy can be found as (provided puck b is at rest)

$K_1=\frac{1}{2}m_av_a^2$

$K_1=\frac{1}{2}(0.254Kg) (0.8176m/s)^2=0.0849 J$

The final kinetic energy is

$K_2=\frac{1}{2}m_av_a'^2+\frac{1}{2}m_bv_b'^2$

$K_2=\frac{1}{2}0.254Kg (-0.123m/s)^2+\frac{1}{2}0.367Kg (0.651m/s)^2=0.07969 J$

The change of kinetic energy is

$\Delta K=0.07969 J - 0.0849 J = -0.00521 J$

3 0

3 years ago

I need help with 2 questions . Please help c:

dezoksy [38]

A) According to the nebular theory, the Solar System formed from a huge gaseous nebula which at a certain point was perturbated. Atoms and molecules started colliding, forming planetesimals (a sort of big rocks). The planetesimals were attracted to each other by gravity, forming bigger warm almost spherical objects called protoplanets, which at the end cooled down forming planets.
Therefore the correct answer is "all of the above".

b) The planets closer to the Sun were (and still are) subject to higher temperatures, due to their close distance to the Sun. In these conditions, rocky materials undergo condensation, while iced gaseous materials undergo vaporization. In the outer parts of the Solar System temperatures are too low to allow these transformations.
The correct answer is again "all of the above".

8 0

3 years ago