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Tema [17]
3 years ago
12

Objects accelerate because

Physics
1 answer:
erica [24]3 years ago
3 0
Friction I think soooooooooo
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A force of 250N is applied to the rope on a pulley system. The box you are
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The force would be 1000
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Water flows through a first pipe of diameter 3 inches. If it is desired to use another pipe for the same flow rate such that the
Alborosie

Answer:

the diameter of the second pipe is 2.52 in

Explanation:  

Given the data in the question;

We know that; the rate of flow is the same;

so

Av1 = Av2

v ∝ √h

\frac{A1}{A2} = \frac{V2}{V1}

\frac{A1}{A2}  = √(  \frac{h2}{h1} )

( π/4.D1² / π/4.D2² ) = √(  \frac{h2}{h1} )

( D1² / D2² ) =  √(  \frac{2h1}{h1} ) since second is double of first

so

( D1² / D2² ) =  √(  \frac{2}{1} )  

3² / D2² =  √2

D2²√2  = 9

D2² = 9/√2

D2² = 9 / 1.4142

D2² = 6.364

D2 = √ 6.364

D2 = 2.52 in

Therefore, the diameter of the second pipe is 2.52 in

3 0
3 years ago
When a physical change in a sample occurs, which of the following would NOT change?
GREYUIT [131]
It's Composition. (C)
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3 years ago
In order to get the most accurate mass measurement, you must __(blank)__.
pshichka [43]

Answer:

c

Explanation:

electronic balance must always be at zero to prevent errors in measured value

4 0
3 years ago
A rocket, initially at rest on the ground, accelerates straight upward from rest with constant acceleration 34.3 m/s^2 . The acc
Eddi Din [679]

Answer:

The maximum height reached by the rocket is 1.94 × 10³ m.

Explanation:

The height of the rocket can be calculated using the following equations:

y = y0 + v0 · t + 1/2 · a · t²    (when the rocket is accelerated upward).

y = y0 +  v0 · t + 1/2 · g · t² (after the rocket runs out of fuel).

Where:

y = height at time t.

y0 = initial height.

v0 = initial velocity.

t = time.

a = acceleration due to engines of the rocket.

g = acceleration due to gravity.

In the same way, the velocity of the rocket can be calculated as follows:

v = v0 + a · t  (when the rocket has fuel)

v = v0 + g · t   (when the rocket runs out of fuel)

Where "v" is the velocity at time "t"

First, let´s find the height reached until the rocket runs out of fuel.

y = y0 + v0 · t + 1/2 · a · t²

y = 0 m + 0 m/s · t + 1/2 · 34.3 m/s² · (5.00 s)²

y = 429 m

And now, let´s find the velocity reached in that time of upward acceleration:

v = v0 + a · t

v = 0 m/s + 34.3 m/s² · 5.00 s

v = 172 m/s

When the rocket runs out of fuel, it is accelerated downward due to gravity. But, since the rocket has initially an upward velocity (172 m/s), it will not fall immediately and will continue to go up until the velocity becomes 0. In that instant, the rocket is at its maximum height and thereafter it will start to fall with negative velocity.

Then, using the equation for velocity, we can calculate the time it takes the rocket to reach its maximum height:

v = v0 + g · t

0 = 172 m/s - 9.80 m/s² · t

-172 m/s / -9.80 m/s² = t

t = 17.6 s

With this time, we can now calcualte the maximum height. Notice that the initial velocity and height are the ones reached during the upward acceleration phase:

y = y0 +  v0 · t + 1/2 · g · t²

ymax = 429 m + 172 m/s · 17.6 s - 1/2 · 9.80 m/s² · (17.6 s)²

ymax = 1.94 × 10³ m

4 0
4 years ago
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