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riadik2000 [5.3K]
3 years ago
5

Why is a car rounding a curve accelerating, even if it is moving at a constant speed?

Physics
2 answers:
LiRa [457]3 years ago
8 0
Because "acceleration" does NOT mean "speeding up". It means ANY change in the speed OR direction of motion. Speeding up, slowing down, and changing direction are all 'accelerations'. If a car is not moving in a straight line, then its direction is changing and it is accelerating, even if its speed is constant.
leonid [27]3 years ago
5 0
Automobile speedometers give an instantaneous velocity. It is important to remember that because the velocity has a direction, a change in either the speed or the direction results in a change in velocity. This leads to an interesting fact, a car going around a curve can have a constant speed and yet still have a changing velocity. The car in the following diagram is traveling around a curve. Suppose the driver looks at his speedometer when the car is at points "A", "B", and "C". At each of these points he notices that the speedometer reads 30km/h. His speed is a constant 30km/h as he rounds the curve. Now look at what is happening to his direction of motion. As he goes around the curve, his direction changes from East to South. Because of this change in direction, we have to conclude that his velocity is changing even though his speed is constant.


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3 years ago
(a) Suppose that your measured weight at the equator is one-half your measured weight at the pole on a planet whose mass and dia
stealth61 [152]

Answer:

7160.2812 s or 1.988 hours

Explanation:

m = Mass of person

R = Radius of Earth = 6.37\times 10^{6}\ m

g = Acceleration due to gravity = 9.81 m/s²

\omega = Angular speed

Force at equator would be

F_e=m(g-\omega^2R)

Force at pole

F_p=mg

From the question

F_e=\dfrac{1}{2}F_p\\\Rightarrow m(g-\omega^2R)=\dfrac{1}{2}F_p\\\Rightarrow \omega=\sqrt{\dfrac{g}{2R}}

Time period is given by

T=\dfrac{2\pi}{\omega}\\\Rightarrow T=2\pi\sqrt{\dfrac{2R}{g}}\\\Rightarrow T=2\pi\sqrt{\dfrac{2\times 6.37\times 10^6}{9.81}}\\\Rightarrow T=7160.2812\ s=1.988\ hours

The rotational period of the planet is 7160.2812 s or 1.988 hours

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3 years ago
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La cantidad de materia en un objeto.
Scrat [10]

La respuesta correcta es Masa

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Reproduction ,if I’m not mistaking if I get u the wrong answer just report my comment
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