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2 years ago

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This table shows the acceleration due to gravity on four planets. Planet Gravity (m/s2) Earth 9.8 Mercury 3.7 Neptune 11.2 Uranu

s 8.9 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. Mercury Neptune Earth Uranus < <

2 answers:

son4ous [18]2 years ago

6 0

Answer:

$\Large \boxed{\mathrm{Mercury, \ Uranus, \ Earth, \ Neptune}}$

Explanation:

Earth ⇒ 9.8 m/s²

Mercury ⇒ 3.7 m/s²

Neptune ⇒ 11.2 m/s²

Uranus ⇒ 8.9 m/s²

$W=m \times g$

$\sf {Weight=mass \times acceleration \ due \ to \ gravity}$

The weight of a person can differ on different planets, but the mass stays the same. The greater the acceleration due to gravity, the greater the weight.

Arranging the planets in increasing order based on a person’s weight on the planet:

Mercury, Uranus, Earth, Neptune

arsen [322]2 years ago

3 0

Answer:

Mercury, Uranus, Earth, Neptune

Explanation:

Weight is mass times gravity. The greater the gravity, the greater the weight.

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A basketball star covers 2.70 m horizontally in a jump to dunk the ball (see figure). His motion through space can be modeled pr

Likurg_2 [28]

Answer:

Part a)

$T = 0.81 s$

Part b)

$v_x = 3.33 m/s$

Part c)

$v_y = 3.91 m/s$

Part d)

$\theta = 49.55 degree$

Part e)

$T = 1.11 s$

Explanation:

Part a)

initial vertical position = 1.02 m

maximum height = 1.80 m

$\Delta y = 1.80 - 1.02$

$\Delta y = 0.78 m$

$v_f^2 - v_y^2 = 2a \Delta y$

$0 - v_y^2 = 2(-9.81)(0.78)$

$v_y = 3.91 m/s$

time taken by it to reach this height

$v_y = v_i + at$

$0 = 3.91 - 9.81 t_1$

$t_1 = 0.39 s$

Now when it again touch the ground then its speed is given as

$v_f^2 - v_y^2 = 2a \Delta y$

$v_f^2 - 0 = 2(9.81)(1.80 - 0.95)$

$v_y = 4.08 m/s$

time taken by it to reach this height

$4.08 = v_i + at$

$4.08 = 0 + 9.81 t_2$

$t_2 = 0.42 s$

$T = t_1 + t_2$

$T = 0.81 s$

Part b)

Horizontal velocity

$v_x = \frac{x}{t}$

$v_x = \frac{2.70}{0.81}$

$v_x = 3.33 m/s$

Part c)

vertical velocity is the intial y direction velocity

$v_y = 3.91 m/s$

Part d)

Take off angle is given as

$tan\theta = \frac{3.91}{3.33}$

$\theta = 49.55 degree$

Part e)

initial vertical position = 1.20 m

maximum height = 2.50 m

$\Delta y = 2.50 - 1.20$

$\Delta y = 1.30 m$

$v_f^2 - v_y^2 = 2a \Delta y$

$0 - v_y^2 = 2(-9.81)(1.30)$

$v_y = 5.05 m/s$

time taken by it to reach this height

$v_y = v_i + at$

$0 = 5.05 - 9.81 t_1$

$t_1 = 0.51 s$

Now when it again touch the ground then its speed is given as

$v_f^2 - v_y^2 = 2a \Delta y$

$v_f^2 - 0 = 2(9.81)(2.50 - 0.72)$

$v_y = 5.9 m/s$

time taken by it to reach this height

$5.9 = v_i + at$

$5.9 = 0 + 9.81 t_2$

$t_2 = 0.60 s$

$T = t_1 + t_2$

$T = 1.11 s$

5 0

2 years ago

This diagram shows the process that powers stars. This process is called?

viktelen [127]

The process that powers stars is C)fusion

4 0

2 years ago

Read 2 more answers

A sound source is located somewhere along the x-axis. Experiments show that the same wave front simultaneously reaches listeners

guapka [62]

Answer:

a) x₀ = - 2 m , b) y = 4.47 m

Explanation:

A wave travels in the middle with constant speed, let's use the equation of uniform motion

v = d / t

t = d / v

The distance to the first listeners, see attached

d₁ = x₀-x

t = (x₀ +7) / v

The distance to the second listener

d₂ = x - x₀

t = (+ 3- x₀) / v

As the wave arrives at the same time, we can equal the two equations

(x₀ +7) / v = (3 -x₀) / v

x₀ + 7 = 3 - x₀

2 x₀ = 3 - 7

x₀ = -4/2

x₀ = - 2 m

b) The time it takes for the wave to reach the listeners of the x-axis, where the speed of sound is 340 m / s

t = 5/340

t = 0.0147 s

Let's look for the distance the wave travels for the listener axis and

v = d₃ / t

d₃ = v.t

d₃ = 340 * 0.0147

d₃ = 5 m

For the distance component we use the Pythagorean triangle

d₃² = x₀² + y²

y² = d₃² - x₀²

y = √ (d₃² -4)

y = √ (5² -4)

y = 4.47 m

6 0

2 years ago

a powerboat accelerates along a straight path from 0 km/hr to 99.8 km/hr in 10.0 s.Find the average acceleration of the boat in

Alex777 [14]

(27.72-0)/10
= 2.772 m/s2

3 0

2 years ago

A ball is thrown upward. As it passes 5.0 m height it is traveling at 4.0 m/s up. What was its initial upward velocity? (a) 7.0

sveticcg [70]

Answer:

c) 10.7m/s

Explanation:

From the exercise we know that at 5m the ball is traveling at 4m/s

To calculate its initial velocity we need to solve the following equation:

$v_{y}^{2}=v_{oy}^{2}+2g(y-y_{o})$

Since the initial height is 0

Solving for $v_{o}$

$v_{oy}=\sqrt{v_{y}^{2}-2gy}=\sqrt{(4m/s)^2-2(-9.8m/s^2)(5m)}=10.7m/s$

5 0

2 years ago