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Travka [436]
3 years ago
10

A person would have a different weight on each planet. Arrange the planets in increasing order based on a person’s weight on the

planet.
Planet Gravity (m/s2)
Earth 9.8
Mercury 3.7
Neptune 11.2
Uranus 8.9
Physics
2 answers:
JulsSmile [24]3 years ago
8 0

Answer: Mercury<Uranus<Earth<Neptune

Explanation: Weight is the downward force acting on an object due to gravity.

w=m\times g

w= weight of an object

m= mass of object

g= gravity

As the person has a constant mass, the weight differs due to the value of gravity (g). The planet having highest gravity would make a person weigh the most.

As the value of gravity varies: Mercury (3.7)<Uranus(8.9)<Earth(9.8)<neptune(11.2),  the weight of the person varies in the same order: Mercury<Uranus<Earth<neptune.

Mars2501 [29]3 years ago
5 0
Mercury, Uranus, Earth, Neptune.

Hope this helps!
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Answer:

1.4 * 10 ^-1 Ω

Explanation:

Hi,

For this question, we gotta use the formula

R = pL/A

p = The resistivity of your material at 20°C

L = length of the wire

A = cross-sectional area

The resistivity of tungsten is 5.60 * 10^-8 at 20°C

By plugging the values, we get:

R = (5.60 * 10^-8)(2.0)/(7.9*10^-7) = 1.4 * 10 ^-1 Ω

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3 years ago
Given. force of 88N and an acceleration of 4 m/s 2 what is the mass?
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2 years ago
In a fast-pitch softball game the pitcher is impressive to watch, as she delivers a pitch by rapidly whirling her arm around so
svetlana [45]

Answer:

(a). The magnitude of the total acceleration of the ball is 239.97 m/s².

(b). The angle of the total acceleration relative to the radial direction is 11.0°

Explanation:

Given that,

Radius of the circle = 0.681 m

Angular acceleration = 67.7 rad/s²

Angular speed =18.6 rad/s

We need to calculate the centripetal acceleration of the ball

Using formula of centripetal acceleration

a_{c}=\omega^2\times r

Put the value into the formula

a_{c}=(18.6)^2\times0.681

a_{c}=235.5\ m/s^2

We need to calculate the tangential acceleration of the ball

Using formula of tangential acceleration

a_{t}=r\alpha

Put the value into the formula

a_{t}=0.681\times67.7

a_{t}=46.104\ m/s^2

(a). We need to calculate the magnitude of the total acceleration of the ball

Using formula of total acceleration

a=\sqrt{a_{c}^2+a_{t}^2}

Put the value into the formula

a=\sqrt{(235.5)^2+(46.104)^2}

a=239.97\ m/s^2

(b). We need to calculate the angle of the total acceleration relative to the radial direction

Using formula of the direction

\theat=\tan^{-1}(\dfrac{a_{t}}{a_{c}})

Put the value into the formula

\theta=\tan^{-1}(\dfrac{46.104}{235.5})

\theta=11.0^{\circ}

Hence, (a). The magnitude of the total acceleration of the ball is 239.97 m/s².

(b). The angle of the total acceleration relative to the radial direction is 11.0°

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C. is the answer I believe
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