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VARVARA [1.3K]
2 years ago
12

Forces of 11.9 N north, 19.1 N east, and 14.4 N south are simultaneously applied to a 3.77 kg mass as it rests on a frictionless

air table. What is the magnitude of its acceleration?
Physics
1 answer:
oksian1 [2.3K]2 years ago
4 0
Force Vector = <19.1i+11.9j-14.4j>N
Force Vector = <19.1i-2.5j>N
Acceleration Vector =(Force Vector)/mass
Acceleration Vector =<19.1i-2.5j>N/3.77
Acceleration Vector =<5.0663i-0.6631j>m/s^2
|a| = (5.0663^2+(-0.6631)^2)^0.5
|a| = 5.11 (m/s^2)
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True Or False? The tendency for an object in motion to remain in motion is called orbital speed.
Harlamova29_29 [7]

The correct answer to the question is False i.e the tendency of an object in motion to remain in motion is not called the orbital speed.

EXPLANATION:

Before going to answer this question, first we have to understand Newton's first laws of motion.

As per Newton's first laws of motion, every body continues to be in state of rest or of uniform motion in a straight line unless and until it is compelled by some external unbalanced forces.

Hence, as long as no unbalanced force is acting on a moving object, it will be in motion. This tendency of a moving object to be in motion is called inertia of motion of the body.

Inertia of motion is the property of the body by virtue of which a moving body always tries to be in motion.

Hence, the tendency of an object in motion to remain in motion is not called as the orbital speed.

7 0
3 years ago
Read 2 more answers
What is the ideal banking angle for a gentle turn of 1.20-km radius on a highway with a 105 km/h speed limit (about 65 mi/h), as
Mnenie [13.5K]

Answer:

4.14°

Explanation:

given:

r = 1.2 km

v = 105 km/h

1) <em>convert your given </em>

a) r = 1.2 km to m = 1200m

b) v = 105 km/h  to m/s = 29.2 m/s

2) <em>plug into your ideal banking angle equation</em>

tan^-1(\frac{v^2}{rg}) = \frac{29.2^2}{(1200)(9.8)} = 4.14°

8 0
2 years ago
Two satellites are orbiting earth at different altitudes. Which satellite orbits at a higher speed v around earth? Assume that t
alexandr1967 [171]

Answer:

a) the one with a lower orbit    b) the one with a higher  orbit

Explanation:

Let's consider orbital mechanics. To get an object in orbit, we need it to fall to earth parallel to the earth's surface. To understand it easily imagine a projectile thrown horizontally further and further away, at one point, the projectile hits the cannon from behind. Considering there is no wind resistance, that would be a projecile in orbit.

In other words, the circular orbits of some objects around a massive body are due to the equality between centrifugal acceleration and gravity acceleration.

\frac{v^2}{r} = \frac{GM}{r^2}.

so the velocity is

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

where "G" is the gravitational constant, "M" the mass of the massive body and "r" the distance between the object and the center of gravity of mass M. As you can note, if "r" increase, "v" decrease.

The orbital period of any object in orbit is

T = 2\pi \sqrt{\frac{a^3}{GM} }

where "a" is length of semi-major axis (a = r in circular orbits). So if "r" increase, "T" increase.

3 0
3 years ago
Intersystem crossing is:
Semenov [28]

An intersystem crossing (ISC) is a non-radiative process that involves the transition between two electronic states with different spin multiplicity. That is, when an electron is excited in a molecule in a basal singlet state (either by absorption or radiation) into a state of greater energy, an excited singlet or triplet state can be obtained.

Therefore, ISC is understood as an a non radio active transition between states with different spin multiplicity.

Correct answer is C:  a radiative transition between states with the same spin.

8 0
3 years ago
If you are in a car and your sibling is driving in the same car next to you at the same speed, does your brother appear to be mo
VLD [36.1K]

Answer:

yes, because you guy's are moving in a car. because if you think about it, your parents are driving, but the cars moving so basically you guys would be moving.

please mark as brainly

7 0
2 years ago
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