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marshall27 [118]

2 years ago

14

A metal ion (X) with a charge of 4+ is attracted to a nonmetal ion (Z) with a

1 answer:

Keith_Richards [23]2 years ago

8 0

Answer:

A) X3Z4

Explanation:

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An object has the acceleration graph shown in (Figure 1). Its velocity at t=0s is vx=2.0m/s. Draw the object's velocity graph fo

timama [110]

Answer:

Explanation:

We may notice that change in velocity can be obtained by calculating areas between acceleration lines and horizontal axis ("Time"). Mathematically, we know that:

$v_{b}-v_{a} = \int\limits^{t_{b}}_{t_{a}} {a(t)} \, dt$

$v_{b} = v_{a}+ \int\limits^{t_{b}}_{t_{a}} {a(t)} \, dt$

Where:

$v_{a}$ , $v_{b}$ - Initial and final velocities, measured in meters per second.

$t_{a}$ , $t_{b}$ - Initial and final times, measured in seconds.

$a(t)$ - Acceleration, measured in meters per square second.

Acceleration is the slope of velocity, as we know that each line is an horizontal one, then, velocity curves are lines with slopes different of zero. There are three region where velocities should be found:

Region I (t = 0 s to t = 4 s)

$v_{4} = 2\,\frac{m}{s} +\int\limits^{4\,s}_{0\,s} {\left(-2\,\frac{m}{s^{2}} \right)} \, dt$

$v_{4} = 2\,\frac{m}{s}+\left(-2\,\frac{m}{s^{2}} \right) \cdot (4\,s-0\,s)$

$v_{4} = -6\,\frac{m}{s}$

Region II (t = 4 s to t = 6 s)

$v_{6} = -6\,\frac{m}{s} +\int\limits^{6\,s}_{4\,s} {\left(1\,\frac{m}{s^{2}} \right)} \, dt$

$v_{6} = -6\,\frac{m}{s}+\left(1\,\frac{m}{s^{2}} \right) \cdot (6\,s-4\,s)$

$v_{6} = -4\,\frac{m}{s}$

Region III (t = 6 s to t = 10 s)

$v_{10} = -4\,\frac{m}{s} +\int\limits^{10\,s}_{6\,s} {\left(2\,\frac{m}{s^{2}} \right)} \, dt$

$v_{10} = -4\,\frac{m}{s}+\left(2\,\frac{m}{s^{2}} \right) \cdot (10\,s-6\,s)$

$v_{10} = 4\,\frac{m}{s}$

Finally, we draw the object's velocity graph as follows. Graphic is attached below.

3 0

3 years ago

1)what is the momentum of a 40,000,000kg tanker traveling at 5 m/s?

Fittoniya [83]

Answer:

1)4*10^7 * 5= 2*10^8

2)50/10=5

5 0

3 years ago

The length of a football field is closest to<br> (1) 1000 cm (3) 1000 km<br> (2) 1000 dm (4) 1000 mm

Sergeu [11.5K]

For this case, the first thing you should know is that the length of a football field is around 100 meters.

We must then look for a measure close to this value.

We have the following unit conversion:

1 meter = 10 decimeters

Applying the conversion we have:

$(1000 dm)*(\frac{1}{10} \frac{m}{dm}) = 100 m$

Therefore, the measure closest to a soccer field is:

1000 dm

Answer:

The length of a football field is closest to:

(2) 1000 dm

5 0

3 years ago

Read 2 more answers

William Herschel thought he had found a comet when he spotted the green disk of: Group of answer choices Triton. Uranus. Oberon.

Reil [10]

Answer: Uranus

Explanation:

7 0

2 years ago

1. An object has a mass of 10g and a volume of 50mL. What is the density?

Alona [7]

1. 0.2 g/mL

The relationship between mass, density and volume of an object is

$d=\frac{m}{V}$

where

d is the density

m is the mass

V is the volume

For the object in this problem, we have

m = 10 g

V = 50 mL

Substituting into the equation,

$d=\frac{10}{50}=0.2 g/mL$

2. 10 mL

In this exercise we know:

- The density of the object: d = 2 g/mL

- The mass of the object: m = 20 g

Therefore, we can re-arrange the previous equation to find the volume:

$V=\frac{m}{d}$

And substituting values into the equation, we find

$V=\frac{20}{2}=10 mL$

7 0

3 years ago