If Earth were twice as far from the sun, the force of gravity attracting the Earth to the Sun would be

If an object travels at a constant speed in a circular path, the acceleration of the object is:

Vilka [71]

Answer:

1- The acceleration of the object is larger in magnitude the smaller the radius of the circle.

Explanation:

The acceleration of an object in a circular path is

$a = \frac{v^2}{r}$

As can be seen from the equation, if the radius of the circle is decreases, the magnitude of the acceleration increases.

As for the direction of the acceleration, it is always towards the center, and it is always perpendicular to the direction of the velocity.

6 0

3 years ago

A proton is released from rest at the positive plate of a parallelplatecapacitor. It crosses the capacitor and reaches the negat

Triss [41]

Answer:

2.1406 × $10^6$ m/sec

Explanation:

we know that energy is always conserved

so from the law of energy conservation

$qV=\frac{1}{2}mv^2$

here V is the potential difference

we know that mass of proton = 1.67× $10^{-27}$ kg

we have given speed =50000m/sec

so potential difference $V=\frac{\frac{1}{2}\times 1.67\times 10^{-27}50000^2}{1.6\times 10^{-19}}=13.045$

now mass of electron =9.11× $10^{-31}$

so for electron

$\frac{1}{2}\times 9.11\times 10^{-31}v^2=1.6\times 10^{-19}\times 13.045=2.1406\times 10^6 m/sec$

so the velocity of electron will be 2.1406× $10^6$ m/sec

4 0

3 years ago

A mason is shot with a constant speed of 7.5 x 10²m/sinto a region, when an electric field produces acceleration on the mason of

nadya68 [22]

Answer:

Distance, d = 0.1 m

It is given that,

Initial velocity of meson,

Finally, the meson is coming to rest v = 0

Acceleration of the meson, (opposite to initial velocity)

Using third equation of motion as :

s is the distance the meson travelled before coming to rest.

So,

s = 0.1 m

The meson will cover the distance of 0.1 m before coming to rest. Hence, this is the required solution.

5 0

3 years ago

Through what vertical height is a 50 N object moved if 250 J of work is done lifting it against the gravitational field of Earth

zimovet [89]

5m

Explanation:

Given parameters:

Weight of object = 50N

Work done in lifting object = 250J

Unknown:

Vertical height = ?

Solution:

The work done on an object is the force applied to lift a body in a specific direction.

Work done = force x distance

Weight is a force in the presence of gravity;

Work done = weight x height of lifting

Height of lifting = $\frac{work done }{weight}$

Height of lifting = $\frac{250}{50}$ = 5m

The vertical height through which the object was lifted is 5m

learn more:

Work done brainly.com/question/9100769

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6 0

3 years ago

During which of the following chemical changes does a precipitate form?

babymother [125]

Answer is B.........

5 0

3 years ago

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