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TiliK225 [7]
4 years ago
15

What are the units of atmospheric pressure?

Physics
1 answer:
erastova [34]4 years ago
6 0

Answer:

a unit of pressure defined as 101325 Pa

Explanation:

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A ball rolls of a cliff and when it lands has a vertical velocity 35 ms ^-1. how high was the cliff?
Licemer1 [7]
Using V²=u² +2gH
let's assume the ball falls from rest, then it's initial velocity = 0ms-¹ (U=0)
then from the above info V= 35ms-¹
g= 10ms-² (value of acceleration due to gravity to the nearest tens)

so input your values above into the given formula to find height
35² = 0² +2×10×H
1225 = 20H
H = 1225÷20 = 61.25m
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4 years ago
How can we make a zero correction in a length of 5.30cm while the zero error is +0.05cm?​
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2 years ago
Determine the amount of potential energy of a 5 newton book that is moved to three different shelves on a bookcase. The height o
BigorU [14]
Gravitational Potential Energy = weight x height

for 1 meter:

GPE = 5 x 1
 = <u>5N</u>

for 1.5 metres:

GPE = 5 x 1.5
 = <u>7.5N</u>

for 2 metres:

GPE = 5 x 2
 = <u>10N</u>
5 0
4 years ago
In what sense is the earth a closed system?
I am Lyosha [343]

Answer:D

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4 0
3 years ago
Read 2 more answers
(1 point) A frictionless spring with a 3-kg mass can be held stretched 1.6 meters beyond its natural length by a force of 90 new
Nonamiya [84]

Answer:

x(t)=0.337sin((5.929t)

Explanation:

A frictionless spring with a 3-kg mass can be held stretched 1.6 meters beyond its natural length by a force of 90 newtons. If the spring begins at its equilibrium position, but a push gives it an initial velocity of 2 m/sec, find the position of the mass after t seconds.

Solution. Let x(t) denote the position of the mass at time t. Then x satisfies the differential equation  

m \frac{d^{2}x}{dt^{2}} +kx=0

Definition of parameters  

m=mass 3kg

k=force constant

e=extension ,m

ω =angular frequency

k=90/1.6=56.25N/m

ω^2=k/m= 56.25/1.6

ω^2=35.15625

ω=5.929

General solution will be

x(t)=c1cos(ωt)+c2Sin(ωt)

x(t)=c1cos(5.929t)+c2Sin(5.929t)

differentiating x(t)

dx(t)=-5.929c1sin(5.929t)+5.929c2cos(5.929t)

when x(0)=0, gives c1=0

dx(t0)=2m/s gives c2=0.337

Therefore, the position of the mass after t seconds is  

x(t)=0.337sin((5.929t)

6 0
3 years ago
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