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sashaice [31]
2 years ago
8

(10p+15,15-10q)=(25,5)​

Physics
1 answer:
Harrizon [31]2 years ago
8 0
63783626736377474737377447
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Centripetal force is a centering force related to acceleration. The centripetal force when driving prevents which of the followi
aivan3 [116]

Answer:

C. The car from driving off the road on a curve

Explanation:

A centripetal force actually causes circular motion. This occurs when an object moves in a circular path or a circle,a force will definitely act on it.

For instance, a car travelling in a circular path must definitely experience this force acting on it, even when the car moves at a constant speed. If it does not exist the object will definitely spin off in a direction tangential to the circular path or curve.

4 0
3 years ago
What does Kepler's first law of planetary motion imply?
BlackZzzverrR [31]

<u>Answer:</u>

The correct answer option is D.  The distance between the planet and the Sun changes as the planet orbits the sun.

<u>Explanation:</u>

Kepler’s laws of planetary motion, derived by the German astronomer Johannes Kepler, are the laws of physics that describe the motions of the planets in the solar system.

According to the Kepler's first law of planetary motion: the path on which the planets orbit around the sun is elliptical in shape, with the center of the sun at one focus.

Therefore, the distance between the Sun and the planets vary as the planet orbit around the sun.

6 0
3 years ago
Calculate the orbital period of a dwarf planet found to have a semimajor axis of a = 4.0x 10^12 meters in seconds and years.
padilas [110]

Explanation:

We have,

Semimajor axis is 4\times 10^{12}\ m

It is required to find the orbital period of a dwarf planet. Let T is time period. The relation between the time period and the semi major axis is given by Kepler's third law. Its mathematical form is given by :

T^2=\dfrac{4\pi ^2}{GM}a^3

G is universal gravitational constant

M is solar mass

Plugging all the values,

T^2=\dfrac{4\pi ^2}{6.67\times 10^{-11}\times 1.98\times 10^{30}}\times (4\times 10^{12})^3\\\\T=\sqrt{\dfrac{4\pi^{2}}{6.67\times10^{-11}\times1.98\times10^{30}}\times(4\times10^{12})^{3}}\\\\T=4.37\times 10^9\ s

Since,

1\ s=3.17\times 10^{-8}\ \text{years}\\\\4.37\times 10^9\ s=4.37\cdot10^{9}\cdot3.17\cdot10^{-8}\\\\4.37\times 10^9\ s=138.52\ \text{years}

So, the orbital period of a dwarf planet is 138.52 years.

3 0
3 years ago
Which component of a galaxy is often found between the stars and looks like a cloud or smoke?
slava [35]

Answer:

Hey I would say Dust

hope this helped

5 0
2 years ago
the other end of the pipe. With the aide of the pipe, does the applied force produce a smaller torque, a greater torque, or the
pav-90 [236]

Answer:

With the addition of the pipe we have a greater torque.

Explanation:

We need to complete the description of the problem, searchin in internet we have:

"Sometimes, even with a wrench, one cannot loosen a nut that is frozen tightly to a bolt. It is often possible to loosen the nut by slipping one end of a long pipe over the wrench handle and pushing at the other end of the pipe. With the aid of the pipe, does the applied force produce a smaller torque, a greater torque, or the same torque on the nut?"

With the addition of the pipe we have a greater torque, as it increases the distance or radius of torque.

We know that torque is defined, as the product of force by distance, in this way we have:

T = F * d

where:

T = torque [N*m]

F = force [N]

d = distance [m]

We can see in the above equation, that increasing the distance increases torque proportionally.

3 0
3 years ago
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