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Ronch [10]
3 years ago
8

A stone is thrown up from the top of a tower. It reaches the ground in 5.64 seconds. A second stone is thrown down from the same

tower. They have the same initial velocity. This stone reaches the ground in 3.6 seconds. The height of the tower and the initial velocities aren't given. You need to find them.
Physics
1 answer:
Naddik [55]3 years ago
8 0

Answer:

add 5.64 to 3.6 where you'll find your awnser

Explanation:

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The __________ association area is responsible for perceiving and attending to stimuli, and the __________ association area is r
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Parietal on the first blank and temporal on the second

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If the vertical component of velocity for a projectile is 7.3 meters/second, find its hang time.
weqwewe [10]
The height of the projectile is given by:
h = ut - 0.5at²

The height will be 0 twice, once at t = 0 and the second time at the time when the journey has been completed, or t = hang time.
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t = 1.49 seconds
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3 years ago
A uniform metal meter-stick is balanced with a 1.0 kg rock attached to the left end of the stick. If the support is located 0.25
Dafna1 [17]

Answer:

c) 1.0 kg

Explanation:

The mass of the stick will be located at the centre of the metre rule. Since the rock is located 0.25m from the pivot, the mass of the meter rule is also 0.25m to the Right of the support

According to law of moment

Sum of clockwise moment = sum of anti clockwise moments

Clockwise moment = M×0.25(mass of metre rule is M)

CW moment = 0.25M

Anti clockwise moment = 0.25×1

ACW moments = 0.25kgm

Equate;

0.25M = 0.25

M = 0.25/0.25

M = 1.0kg

Hence the mass of the metre rule is 1.0kg

8 0
3 years ago
A planet orbits a star, in a year of length 2.35 x 107 s, in a nearly circular orbit of radius 3.49 x 1011 m. With respect to th
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Answer:

a)   w = 9.599 10⁴ rad / s , b)   v = 3.35 10¹⁶ m / s , c)    a = 3.22  10²¹ m / s²

Explanation:

For this exercise we must use the relation of angular kinematics

a) angular velocity, the distance remembered in orbit between time (period)

         w = 2π r / T

         w = 2 π 3.59 10¹¹ / 2.35 10⁷

         w = 9.599 10⁴ rad / s

b) linear and angular velocity are related by the equation

          v = w r

          v = 9,599 10⁴ 3.49 10¹¹

          v = 3.35 10¹⁶ m / s

c) the centripetal acceleration is

            a = v² / r = w² r

            a = (9,599 10⁴)²   3.49 10¹¹

            a = 3.22  10²¹ m / s²

7 0
3 years ago
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