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astraxan [27]
3 years ago
15

Pwease help me with thesse three

Physics
1 answer:
DiKsa [7]3 years ago
7 0

Answer:

1. Spring

2. 23.5

3. Summer

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A 2 kg, frictionless block is attached to a horizontal, ideal spring with spring constant 300 N/m. At t = 0 the spring is neithe
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Answer:

Explanation:

Given that,

Mass of block

M = 2kg

Spring constant k = 300N/m

Velocity v = 12m/s

At t = 0, the spring is neither stretched nor compressed. Then, it amplitude is zero at t=0

xo = 0

It velocity is 12m/s at t=0

Then, it initial velocity is

Vo = 12m/s

Then, amplitude is given as

A = √[xo + (Vo²/ω²)]

Where

xo is the initial amplitude =0

Vo is the initial velocity =12m/s

ω is the angular frequency and it can be determine using

ω = √(k/m)

Where

k is spring constant = 300N/m

m is the mass of object = 2kg

Then,

ω = √300/2 = √150

ω = 12.25 rad/s²

Then,

A = √[xo + (Vo²/ω²)]

A = √[0 + (12²/12.5²)]

A = √[0 + 0.96]

A = √0.96

A = 0.98m

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3 years ago
The wind increases to 14 mph from the north. Now what is your airspeed and what direction are you flying? If your destination is
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4 years ago
What systems of the body are involved in preparing and eating a sandwich
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<span>The systems of the body involved in preparing and eating a sandwich are : 
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3 years ago
A skier leaves the horizontal end of a ramp with a velocity of 31.0 m/s and lands 156.3 m from the base of a ramp how high is th
BartSMP [9]

<u>Answer:</u>

The height of ramp = 124.694 m

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Using second equation of motion,

s = ut + \frac{1}{2}at^2

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substituting values

156.3 = 31\times t + 0

t = \frac{156.3}{31 }

= 5.042 s

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t = 5.042 s; initial velocity, u =0

acceleration = acceleration due to gravity, g = 9.81 m/s^2

Substituting in h = ut + \frac{1}{2}gt^2

h = 0 + \frac{1}{2} \times 9.81 \times (5.042)^2

h = 124.694 m

So height of ramp = 124.694 m

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Does a car that is slowing down always have a negative acceleration explain
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