Answer:
Vf = 69.56 cm/s
Explanation:
In order to find the final speed of the ramp, we will use the equations of motion. First we use second equation of motion to find out the acceleration of marble:
s = Vi t + (1/2)at²
where,
s = distance traveled = 160 cm
Vi = Initial Speed = 0 cm/s (since, marble starts from rest)
t = time interval = 4.6 s
a = acceleration = ?
Therefore,
160 cm = (0 cm/s)(4.6 s) + (1/2)(a)(4.6 s)²
a = (320 cm)/(4.6 s)²
a = 15.12 cm/s²
Now, we use first equation of motion:
Vf = Vi + at
Vf = 0 cm/s + (15.12 cm/s²)(4.6 s)
<u>Vf = 69.56 cm/s</u>
The distance from the centre of the rule at which a 2N weight must be suspend from A is 29.3 cm.
<h3>Distance from the center of the meter rule</h3>
The distance from the centre of the rule at which a 2N weight must be suspend from A is calculated as follows;
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20 A (30 - x)↓ x ↓ 20 cm B 30 cm
2N 0.9N
Let the center of the meter rule = 50 cm
take moment about the center;
2(30 - x) + 0.9(x)(30 - x) = 0.9(20)
(30 - x)(2 + 0.9x) = 18
60 + 27x - 2x - 0.9x² = 18
60 + 25x - 0.9x² = 18
0.9x² - 25x - 42 = 0
x = 29.3 cm
Thus, the distance from the centre of the rule at which a 2N weight must be suspend from A is 29.3 cm.
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You would divide 6 into 1000 so your answer is B. 166
Primarily to survey and map the landscape of Mars for future missions. It also takes core samples and searches for water underground. This is important for many reasons, but most obviously for the upcoming missions to mars and the colonies that follow. Hope this helps?
