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3 years ago

In which situation would a space probe most likely experience centripetal acceleration?

Physics

1 answer:

tester [92]3 years ago

8 0

Question: In which situation would a space probe most likely experience centripetal acceleration?

as it revolves around a planet

as it flies straight past a moon

as it is pulled in a line toward the Sun

as it lifts off from Earth

Answer:

When "space probe revolves around a planet" most likely to experience centripetal acceleration

Explanation:

Centripetal acceleration defined as the rate in change of tangential velocity. Also, as per Newton's second law, any kind of force will be directly proportional to the acceleration attained by the object. So, for centripetal acceleration, the force will be the centripetal force. The centripetal force will be acting on an object rotating in a circular motion with its direction of force towards the center. Thus, centripetal acceleration will be experienced by an object or a space probe when it is in a circular motion. It means the space probe is revolving around a planet.

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A flea jumps by exerting a force of 1.09 ✕ 10−5 N straight down on the ground. A breeze blowing on the flea parallel to the grou

patriot [66]

Answer:

a= 17.877 m/s² : Magnitude of the acceleration of the flea

β = 88.21° : Direction of the acceleration of the flea

Explanation:

Conceptual analysis

We apply Newton's second law:

∑F = m*a (Formula 1)

∑F : algebraic sum of the forces in Newton (N)

m : mass in kilograms (kg)

a : acceleration in meters over second square (m/s²)

Problem development

Look at the flea free body diagram in the attached graphic

The acceleration is presented in the direction of the resultant force (R) applied over the flea .

$R= \sqrt{(F_{x})^{2} + (F_{y})^{2} }$

$R= \sqrt{(10.9)^{2}+(0.340)^{2} } *10^{-6} N$

R= 10.905*10⁻⁶ N

We apply the formula (1) to calculate the magnitude of the acceleration of the flea

∑F = m*a m = 6.1 * 10⁻⁷ kg

R = m*a

a= R/m

a= (10.905*10⁻⁶) / (6.1 * 10⁻⁷ )

a= 17.877 m/s²

β: Direction and magnitude of the acceleration of the flea

$\beta = tan^{-1} (\frac{F_{y} }{F_{x} } )$

$\beta = tan^{-1} (\frac{10.9*10^{-6} }{0.340*10^{-6} } )$

β = 88.21°

5 0

4 years ago

Why do astronauts feel weightless in space?

JulsSmile [24]

Weight = mass × gravitational field strength
At space gravitational field strength is very low so its"weightless".

6 0

3 years ago

Two like-charged particles are placed close to each other. How would the force of repulsion be affected if the charge on one of

astra-53 [7]

Answer: It will remain the same

Explanation:

According to Coulomb's Law:

"The electrostatic force $F_{E}$ between two point charges $q_{1}$ and $q_{2}$ is proportional to the product of the charges and inversely proportional to the square of the distance $d$ that separates them, and has the direction of the line that joins them".