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3 years ago

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The cup was filled at a constant rate. It started empty. After 2 seconds, it held 70 cm cubed of water.Find the percent of the c

up that is filled with water after 3 seconds.

1 answer:

Helen [10]3 years ago

6 0

Answer:

In 3 seconds the cup is filled with $105\ cm^3$ <u>.</u>

Step-by-step explanation:

Given:

2 seconds = $70\ cm^3$ water

We need to find the percent of the cup that is filled with water after 3 seconds.

Solution:

Now Given:

The cup was filled at a constant rate.

First we will find the volume of water filled in the cup after 1 seconds.

2 seconds = $70\ cm^3$ water

1 seconds = Volume of water filled in 1 second.

By Using Unitary method we get;

Volume of water filled in 1 second = $\frac{70}{2}= 35\ cm^3$

Now we will find the volume of water filled in the cup after 3 seconds.

1 second = $35\ cm^3$ of water is filled

3 seconds = volume of water filled in the cup after 3 seconds

Again by using Unitary method we get;

volume of water filled in the cup after 3 seconds = $35 \times 3 = 105\ cm^3$

Hence In 3 seconds the cup is filled with $105\ cm^3$ .

To find the percent of the cup filled with water after 3 seconds we would required the dimensions of the cup which is not available in the question.

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What is the value of x in the equation 2x – 3 = 9 – 4x?

strojnjashka [21]

Move the 4x over to the left side and get:
6x - 3 = 9

Then move the 3 over to the right....
6x = 12

Divide by 6

x = 2

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3 years ago

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What is the value of the arithmetic series below?

xxTIMURxx [149]

Answer:

(A)-494

Step-by-step explanation:

Given the arithmetic series

$S_{19}=\sum_{k=1}^{19}4-3k$

The terms in the sequence are:

When k=1, 4-3k=4-3(1)=1
When k=2, 4-3k=4-3(2)=-2
When k=3, 4-3k=4-3(3)=-5

Therefore, the terms in the sequence are: 1, -2, -5, ...

First term, a =1

Common difference, d=-2-1=-3

The sum of an arithmetic series, $S_n=\dfrac{n}{2}[2a+(n-1)d]$

Therefore:

$S_{19}=\dfrac{19}{2}[2(1)+(19-1)(-3)]\\=9.5[2+18*-3]\\=9.5[2-54]\\=9.5*-52\\=-494$

The correct option is A.

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3 years ago

I need help please I think the answer is 9 but I'm not sure

Marysya12 [62]

Answer:

let ratio be 3x,11xand13x

perimeter=298cm

3x+11x+13x=298

27x=298

x=298/27=11approax

shortest side=3×11=33

Step-by-step explanation:

you are right .

3 0

3 years ago

I need help In this please

Shalnov [3]

The answer is x < 12

1) Minus 6 from both sides of the inequality

2) Multiply both sizes by 4

3) Divide both sides by -3

4) Swap the direction of the inequality (You always do this when you multiply/divide both sizes of an inequality)

5) You should have x < 12

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4 years ago

A ball is launched upward at 14 m/s from a platform 30 m high.Find the maximum height the ball will reach and how long it will t

BartSMP [9]

Answer:

The ball will reach a maximum height of 39.993 meters after 1.428 seconds.

Step-by-step explanation:

Let suppose that no non-conservative forces acts on the ball during its motion, then we can determine the maximum height reached by the Principle of Energy Conservation, which states that:

$K_{1}+U_{g,1} = K_{2}+U_{g,2}$ (1)

Where:

$K_{1}$ , $K_{2}$ - Initial and final translational kinetic energies, measured in joules.

$U_{g,1}$ , $U_{g,2}$ - Initial and final gravitational potential energies, measured in joules.

By definition of translational kinetic energy and gravitational potential energy we expand and simplify the expression above:

$\frac{1}{2}\cdot m\cdot v_{2}^{2}+m\cdot g\cdot y_{2}= \frac{1}{2}\cdot m\cdot v_{1}^{2}+m\cdot g\cdot y_{1}$ (2)

Where:

$m$ - Mass of the ball, measured in kilograms.

$g$ - Gravitational acceleration, measured in meters per square second.

$v_{1}$ , $v_{2}$ - Initial and final speed of the ball, measured in meters per second.

$y_{1}$ , $y_{2}$ - Initial and final heights of the ball, measured in meters.

The final height of the ball is determined by the following formula:

$v_{2}^{2}+2\cdot g\cdot y_{2} = v_{1}^{2}+2\cdot g\cdot y_{1}$

$v_{1}^{2}-v_{2}^{2}+2\cdot g \cdot y_{1}=2\cdot g\cdot y_{2}$

$y_{2} = y_{1}+\frac{v_{1}^{2}-v_{2}^{2}}{2\cdot g}$ (3)

If we know that $y_{1} = 30\,m$ , $v_{1} = 14\,\frac{m}{s}$ , $v_{2} = 0\,\frac{m}{s}$ and $g = 9.807\,\frac{m}{s^{2}}$ , the maximum height that the ball will reach is:

$y_{2} = 30\,m + \frac{\left(14\,\frac{m}{s} \right)^{2}-\left(0\,\frac{m}{s} \right)^{2}}{2\cdot \left(9.807\,\frac{m}{s^{2}} \right)}$

$y_{2} = 39.993\,m$

The ball will reach a maximum height of 39.993 meters.

Given the absence of non-conservative forces, the ball exhibits a free fall. The time needed for the ball to reach its maximum height is computed from the following kinematic formula:

$t = \frac{v_{2}-v_{1}}{-g}$ (4)

If we know that $v_{1} = 14\,\frac{m}{s}$ , $v_{2} = 0\,\frac{m}{s}$ and $g = 9.807\,\frac{m}{s^{2}}$ , then:

$t = \frac{0\,\frac{m}{s}-14\,\frac{m}{s} }{-9.807\,\frac{m}{s^{2}} }$

$t = 1.428\,s$

The ball will take 1.428 seconds to reach its maximum height.

6 0

3 years ago