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2 years ago

Your teacher (175kg) and the lab (15kg) are 2m apart. What is the gravitational force between them? Draw a FBD, and label

1 answer:

7 0

The gravitational force between two objects is given by
$F=G \frac{m_1 m_2 }{r^2}$
where
G is the gravitational constant
m1 and m2 are the masses of the two objects
r is the separation between the two objects

In this problem, $m_1 = 175 kg$ , $m_2 =15 kg$ and $r=2 m$ , therefore the gravitational force between the two objects is
$F=(6.67 \cdot 10^{-11} m^3 kg^{-1} s^{-2}) \frac{(175 kg)(15 kg)}{(2m)^2}=4.38 \cdot10^{-8} N$

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Jerry runs 60 meters east and then 20 meters west in 10

liraira [26]

Answer: 8 meters per second

Explanation: If you add 60 to 20 you get 80 meters and since he ran those 80 meters in 10 seconds you divide 80 by ten and get 8 and then you get 8m/s

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3 years ago

-2 5/8 is bigger than -5/12

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3 years ago

Determine the velocity required for a moving object 2.00 x 10^4 m above the surface of Mars to escape from Mars's gravity. The m

Ket [755]

Answer:

Therefore the escape velocity from Mar's gravity is $15.88 \times 10^4$ m/s.

Explanation:

Escape velocity: Escape velocity is a the minimum velocity that a object needs to escape from the gravitational field of massive body.

$V_{escape}=\sqrt{\frac{2GM}{R}}$

$V_{escape}=$ Escape velocity

G=Universal gravitational constant = 6.673×10⁻¹¹N m²/Kg²

M= mass of Mars = 6.42×10²³ kg

R = Radius of the Mars = 3.40×10³m

The escape velocity does not depend on the velocity of a object.

$V_{escape}=\sqrt{\frac{2\times6.673\times 10^{-11}\times 6.42\times 10^{23}}{3.40\times10^3}}$

$=15.88 \times 10^4$ m/s

Therefore the escape velocity from Mar's gravity is $15.88 \times 10^4$ m/s.

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2 years ago

Identify the person who made the correct statement.

dmitriy555 [2]

The answer is: [C]: Neither Juan nor Christina are correct.
_________________________________________________
Explanation:
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Consider Juan's statement:
_______________________
"Juan said fossils deposits always contain only one type of organism."
__________________________
This statement is incorrect. There are many instances of fossil deposits containing more than one type of organism.
___________________________________
Consider Christina's statement.
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"Christina said fossil deposits never contain one type of organism."
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This statement is incorrect. Fossil deposits, by definition (of "fossil") ALWAYS contain <u>at least one type</u> of organism.
___________________________________________
As such, neither Juan nor Christina are correct.
And as such, Answer choice: [C] is the correct answer.
_____________________
Answer choices A, B, and D are ruled out.

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2 years ago

The launch velocity of a toy car launcher is determined to be 5 m/s. If the car is to be launched from a height of 0.5 m, where

lora16 [44]

Answer:

1.6 m

Explanation:

Given that the launch velocity of a toy car launcher is determined to be 5 m/s. If the car is to be launched from a height of 0.5 m.

The time for landing should be calculated by using the second equation of motion formula

h = Ut + 1/2gt^2

Let U = 0

0.5 = 1/2 × 9.8 × t^2

0.5 = 4.9t^2

t^2 = 0.5 / 4.9

t^2 = 0.102

t = 0.32 s

The target should be placed so that the toy car lands on it at:

Distance = 5 × 0.32

distance = 1.597 m

Distance = 1.6 m

Therefore, the target should be placed so that the toy car lands on it 1.6 metres away.

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3 years ago