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elena55 [62]
4 years ago
10

The elevators in the Landmark Tower in Yokohama, Japan are among the fastest in the world. After starting from rest, they reach

a maximum speed of 120 meters per minute. It takes the elevator 4.0 seconds to reach maximum speed. What is its acceleration in m/s/s (meters per second per second)?
Physics
1 answer:
MariettaO [177]4 years ago
3 0

Answer:

The acceleration is 30 m/s²

Explanation:

Given that,

Maximum speed = 120 m/s

Time = 4.0 sec

We need to calculate the acceleration

Using equation of motion

v=u+at

Where, v = speed

u = initial speed

a = acceleration

t = time

Put the value into the formula

120=0+a\times4.0

a=\dfrac{120}{4.0}

a=30\ m/s^2

Hence, The acceleration is 30 m/s²

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How many moles of gas must be forced into a 4.6 l tire to give it a gauge pressure of 31.2 psi at 26 âc? the gauge pressure is r
Bess [88]
Use the Ideal Gas Law to the air in the tire : 

( P ) ( V ) = ( n ) ( R ) ( T ) 
n = ( P ) ( V ) / ( R ) ( T ) 
P = P gauge + P baro = 31.2 psig + 14.8 psia = 46 psia 
P = ( 46 psia ) ( 1 atm / 14.696 psia ) = 3.13 atm 
n = ( P ) ( V ) / ( R ) ( T ) 
n = ( 3.13 atm ) ( 4.6 L ) / ( 0.08206 atm - L / mol - K ) ( 26.0 + 273.2 K ) 
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7 0
3 years ago
What is the correct answer?
dexar [7]

Answer:

length=2

Explanation:

You simply find the zeros of the quartic polynomial

The factored form is (x+1)(x-5)(x-2)=0

The zeros become -1,5 and 2

Dimension 5 is already given and the question says length>width and 2>-1

Therefore length=2

5 0
3 years ago
A ballast is dropped from a stationary hot-air balloon that is at an altitude of 576 ft. Find (a) an expression for the altitude
Nadya [2.5K]

Answer:

<h2>a) S = \frac{1}{2}gt^2\\</h2><h2>b) 6secs</h2><h2>c) 192ft</h2>

Explanation:

If a ball dropped from a stationary hot-air balloon that is at an altitude of 576 ft, an expression for the altitude of the ballast after t seconds can be expressed using the equation of motion;

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

S is the altitude of the ballest

u is the initial velocity

a is the acceleration of the body

t is the time taken to strike the ground

Since the body is dropped from a stationary air balloon, the initial velocity u will be zero i.e u = 0m/s

Also, since the ballast is dropped from a stationary hot-air balloon, the body is under the influence of gravity, the acceleration will become acceleration due to gravity i.e a = +g

Substituting this values into the equation of the motion;

S = 0 + \frac{1}{2}gt^2\\ S = \frac{1}{2}gt^2\\

a) An expression for the altitude of the ballast after t seconds is therefore

S = \frac{1}{2}gt^2\\

b) Given S = 576ft and g = 32ft/s², substituting this into the formula in (a);

576 = \frac{1}{2}(32)t^2\\\\\\576*2 = 32t^2\\1152 = 32t^2\\t^2 = \frac{1152}{32} \\t^2 = 36\\t = \sqrt{36}\\ t = 6.0secs

This means that the ballast strikes the ground after 6secs

c) To get the velocity when it strikes the ground, we will use the equation of motion v = u + gt.

v = 0 + 32(6)

v = 192ft

7 0
3 years ago
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