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bulgar [2K]
3 years ago
6

A 10kg crate sits on a distant planet. The planet has the same mass as Earth, but 80% of the radius of Earth (Earth radius = 637

0km.) What is the approximate weight of the object on the distant planet?
A) 120N
B) 160N
C) 80N
D) 100N
Physics
1 answer:
Nataliya [291]3 years ago
8 0
The answer is d! Hope I helped you
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brings together the unmarried girls and young women of the country to cut reeds for the annual repairs

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Which property of a substance is not altered by a physical change?
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Composition Is the property
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A child aims his air rifle at an angle of 10 degrees above the horizontal. he fires a bb while simultaneously dropping another b
Gwar [14]

Answer:

The b-b at rest will hit the ground first

Explanation:

When the first b-b is shot at angle it will first experience some horizontal force of air resistance before it experience another vertical force of resistance when going done while the second b-b only experiences the vertical force of resistance making reach the ground faster.

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4 years ago
Two children of mass 18 kg and 29 kg sit balanced on a seesaw with the pivot point located at the center of the seesaw. If the c
DedPeter [7]

Answer:

3.085 [m].

Explanation:

1) The rule:

m₁*g*l₁=m₂*g*l₂, where m₁ and l₁ - the mass and distance for the small child, m₂ and l₂ - for the big child;

2) according to the condigion l₁+l₂=5, then

3) it is possible to make up the system:

\left \{ {{l_1+l_2=5} \atop {m_1*l_1=m_2*l_2}} \right. \ = > \ \left \{ {{l_1=5-l_2} \atop {18*(5-l_2)=29*l_2}} \right. \ = > \ \left \{ {{l_1=\frac{145}{47}} \atop {l_2=\frac{90}{47}}} \right.

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7 0
2 years ago
A rock is thrown vertically upward with a speed of 18.0 m/s from the roof of a building that is 50.0 m above the ground. Assume
Andreyy89

Answer:

(a) 5.7 s

(b) 39 m/s

Explanation:

(a) u = 18 m/s

At the maximum height, the final velocity of ball is zero. lte teh time taken by the ball to go from 50 m height to maximum height is t.

use first equation of motion.

v = u + g t

0 = 18 - 10 x t

t = 1.8 s

Let the maximum height attained by the ball when it thrown from 50 m height is h'.

Use third equation of motion

v^2 = u^2 + 2 g h'

0 = 18^2 - 2 x 10 x h'

h' = 16.2 m

Total height from the ground H = h + h' = 50 + 16.2 = 76.2 m

Let t' be the time taken by the ball to hit the ground as it falls from maximum height.

use third equation of motion

H = ut + 1/2 x g t'^2

76.2 = 0 + 1/2 x 10 x t'^2

t' = 3.9 s

Total time taken by the ball to hit the ground = T = t + t' = 1.8 + 3.9 = 5.7 s

(b) Let v be the velocity with which the ball strikes the ground.

v^2 = u^2 + 2 g H

v^2 = 0 + 2 x 10 x 76.2

v = 39 m/s

4 0
4 years ago
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