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Flura [38]
3 years ago
5

(a) You wish to determine the height of the smokestack of a local coal burning power plant. You convince a member of the mainten

ance crew to mount the support for a simple pendulum at the top of the stack and you suspend a 1.00 kg mass that just misses the ground at its lowest point from the pendulum cord. If the period of the pendulum is 20.2 s, determine the height of the smokestack.
Physics
1 answer:
damaskus [11]3 years ago
4 0

SHİNE LİKE A BRİGHT STAR

❄NEVER GİVE UP❄

☻NEVER SAY NEVER☻

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What’s the unit for mass
KATRIN_1 [288]

the unit for mass is kilogram

8 0
3 years ago
How much work is needed to stop a 1,110 kg car that is moving straight down the
schepotkina [342]

Answer:

382.74 kJ.

Explanation:

The work that must be done to stop an 1100 kg car travelling at 59  km/h is  - 382.74 kJ.

3 0
3 years ago
Paco was driving his scooter west with an initial velocity of 4 m/s. He accelerates at 0.5 m/s2 for 30 seconds.
KonstantinChe [14]

Answer:

V = 19m/s

Explanation:

Given the following data;

Initial velocity, U = 4m/s

Acceleration, a = 0.5m/s²

Time, t = 30 seconds

To find the final velocity, we would use the first equation of motion;

V = U + at

Where;

V is the final velocity.

U is the initial velocity.

a is the acceleration.

t is the time measured in seconds.

V = 4 + 0.5*30

V = 4 + 15

V = 19m/s

Therefore, his final velocity is 19 meters per seconds.

6 0
3 years ago
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LiRa [457]

Answer:

different sample have different properties is not a characteristics of a compound ,

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because compound will always same properties no matter how quantity is

6 0
3 years ago
An object is traveling such that it has a momentum of magnitude 23.3 kg.m/s and a kinetic energy of 262 J. Determine the followi
Crazy boy [7]

Explanation:

It is given that,

Momentum of an object, p = 23.3 kg-m/s

Kinetic energy, E = 262 J

(a) Momentum is given by, p = mv

23.3 = mv...........(1)

Kinetic energy is given by, E=\dfrac{1}{2}mv^2

m = mass of the object

v = speed of the object

E=\dfrac{1}{2}\times (mv)\times v

262=\dfrac{1}{2}\times 23.3\times v

v = 22.48 m/s

(2) Momentum, p = mv

m=\dfrac{p}{v}

m=\dfrac{23.3\ kg-m/s}{22.48\ m/s}

m = 1.03 Kg

Hence, this is the required solution.

8 0
4 years ago
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