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3 years ago

12

An astronaut in a spacesuit has a mass of 100 kilograms. What is the weight of this astronaut on the surface of the Moon where t

he strength of gravity is approximately 1/6 that of Earth?

1 answer:

arlik [135]3 years ago

3 0

Weight = (mass) x (gravity) = (100 kg) x (1.62 m/s^2) = 162 newtons.

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The slope of a linear velocity-time graph tells us the _____ of the object.

Andrew [12]

Speed of the object !!!!

4 0

3 years ago

An objects motion changes

NARA [144]

Answer:

uh finish the question please lol.

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2 years ago

Comets travel around the sun in elliptical orbits with large eccentricities. If a comet has speed 1.6×104 m/s when at a distance

xz_007 [3.2K]

Answer:

v₂ = 7.6 x 10⁴ m/s

Explanation:

given,

speed of comet(v₁) = 1.6 x 10⁴ m/s

distance (d₁)= 2.7 x 10¹¹ m

to find the speed when he is at distance of(d₂) 4.8 × 10¹⁰ m

v₂ = ?

speed of planet can be determine using conservation of energy

K.E₁ + P.E₁ = K.E₂ + P.E₂

$\dfrac{1}{2}mv_1^2-\dfrac{GMm}{r_1} = \dfrac{1}{2}mv_2^2-\dfrac{GMm}{r_2}$

$\dfrac{1}{2}v_1^2-\dfrac{GM}{r_1} = \dfrac{1}{2}v_2^2-\dfrac{GM}{r_2}$

$v_2^2= v_1^2 + \dfrac{2GM}{r_2}-\dfrac{2GM}{r_1}$

$v_2= \sqrt{v_1^2 +2GM(\dfrac{1}{r_2}-\dfrac{1}{r_1})}$

$v_2= \sqrt{(1.6\times 10^4)^2 +2\times 6.67 \times 10^{-11}\times 1.99 \times 10^{30}(\dfrac{1}{4.8\times 10^{10}}-\dfrac{1}{2.7\times 10^{11}})}$

v₂ = 7.6 x 10⁴ m/s

3 0

3 years ago

Which example has the least kinetic energy

Stells [14]

Answer:

D

Explanation:

Let’s calculate the kinetic energy for all of the choices.

a. (1/2)(100)(100)^2 = 50(10000)=500,000

b. (1/2)(100)(1)^2 = 50

c. (1/2)(10)(100)^2 = 5(10000) = 50,000

d. (1/2)(1)(1)^2 = 0.5

We can see that (d) has the least kinetic energy.

5 0

2 years ago

Dante is leading a parade across the main street in front of city hall. Starting at city hall, he marches the parade 4 blocks ea

Ipatiy [6.2K]

Answer:

The correct option is A)

Displacement: 6.71 m, Direction: 63.4 degrees north of east

Explanation:

Given that Dante is leading a parade across the main street in front of city hall.

Let, Initial location of parade is 0i+0j

One block of city is one units on the XY- graph

Statement 1: Parade marches the parade 4 blocks east, then 3 blocks south

New location of parade is 4i-3j

Statement 2: The parade marches 1 block west and 9 blocks north and finally stops.

Final location of parade is (4i-3j)+(-1i+9j)=3i+6j

Displacement is given by

Displacement = (Final destination)-(Initial destination)

Displacement = (3i+6j)-(0i+0j)=3i+6j

Thus,

Magnitude of displacement = $\sqrt{3^{2}+6^{2}}$

= 6.71 m

Direction of displacement = $tan^{-1}(\frac{Y}{X} )$

= $tan^{-1}(\frac{6}{3} )$

= 63.43 NE

Therefore, the correct option is A) Displacement: 6.71 m, Direction: 63.4 degrees north of east

5 0

3 years ago