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3 years ago

A group of elements are highly inactive and are used by chemists because of their chemical stability.

1 answer:

7 0

Those are the noble gases. I think they were called "noble" because
they behave like kings and other "noblemen" ... stuck up, and not
mixing with ordinary people.

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A girl pushes a 1.04 kg book across a table with a horizontal applied force 10 points

mr Goodwill [35]

Answer:

Approximately $11.0\; \rm m \cdot s^{-1}$ . (Assuming that $g = 9.81 \; \rm N \cdot kg^{-1}$ , and that the tabletop is level.)

Explanation:

Weight of the book:

$W = m \cdot g = 1.04 \; \rm kg \times 9.81\; \rm N \cdot kg^{-1} \approx 10.202\; \rm N$ .

If the tabletop is level, the normal force on the book will be equal (in magnitude) to weight of the book. Hence, $F(\text{normal force}) \approx 10.202\; \rm N$ .

As a side note, the $F_N$ and $W$ on this book are not equal- these two forces are equal in size but point in the opposite directions.

When the book is moving, the friction $F(\text{kinetic friction})$ on it will be equal to

$\mu_{\rm k}$ , the coefficient of kinetic friction, times
$F(\text{normal force})$ , the normal force that's acting on it.

That is:

$\begin{aligned}& F(\text{kinetic friction}) \\ &= \mu_{\rm k}\cdot F(\text{normal force})\\ &\approx 0.35 \times 10.202\; \rm N \approx 3.5708\; \rm N\end{aligned}$ .

Friction acts in the opposite direction of the object's motion. The friction here should act in the opposite direction of that $15.0\; \rm N$ applied force. The net force on the book shall be:

$\begin{aligned}& F(\text{net force}) \\ &= 15.0 \; \rm N - F(\text{kinetic friction}) \\& \approx 15.0 - 3.5708\; \rm N \approx 11.429\; \rm N\end{aligned}$ .

Apply Newton's Second Law to find the acceleration of this book:

$\displaystyle a = \frac{F(\text{net force})}{m} \approx \frac{11.429\; \rm N}{1.04\; \rm kg} \approx 11.0\; \rm m \cdot s^{-2}$ .

6 0

2 years ago

The energy an object acquires when it is exposed to a force is called _____ energy

r-ruslan [8.4K]

I'm pretty sure the energy an object acquires when exposed to a force is known was potential energy.

4 0

3 years ago

To practice Problem-Solving Strategy 15.1 Mechanical Waves. Waves on a string are described by the following general equation y(

DerKrebs [107]

Answer:

0.0549 m

Explanation:

Given that

equation y(x,t)=Acos(kx−ωt)

speed v = 8.5 m/s

amplitude A = 5.5*10^−2 m

wavelength λ = 0.5 m

transverse displacement = ?

v = angular frequency / wave number

and

wave number = 2π/ λ

wave number = 2 * 3.142 / 0.5

wave number = 12.568

angular frequency = v k

angular frequency = 8.5 * 12.568

angular frequency = 106.828 rad/sec ~= 107 rad/sec

equation y(x,t)=Acos(kx−ωt)

y(x,t)= 5.5*10^−2 cos(12.568 x−107t)

when x =0 and and t = 0

maximum y(x,t)= 5.5*10^−2 cos(12.568 (0) − 107 (0))

maximum y(x,t)= 5.5*10^−2 m

and when x = x = 1.52 m and t = 0.150 s

y(x,t)= 5.5*10^−2 cos(12.568 (1.52) −107(0.150) )

y(x,t)= 5.5*10^−2 × (0.9986)

y(x,t) = 0.0549 m

so the transverse displacement is 0.0549 m

5 0

3 years ago

44. A rescue helicopter is hovering over a person whose boat has sunk. One of the rescuers throws a life preserver straight down

Vladimir [108]

Answer:

18.4 m

Explanation:

(a)

The known variables in this problem are:

u = 1.40 m/s is the initial vertical velocity (we take downward direction as positive direction)

t = 1.8 s is the duration of the fall

a = g = 9.8 m/s^2 is the acceleration due to gravity

(b)

The vertical distance covered by the life preserver is given by

$d=ut + \frac{1}{2}at^2$

If we substitute all the values listed in part (a), we find

$d=(1.40 m/s)(1.8 s)+\frac{1}{2}(9.8 m/s^2)(1.8 s)^2=18.4m$

8 0

3 years ago

A sailfish swims 120 km/hr. How far will it travel in 8.0 minutes?

Rasek [7]

Answer:

16km

Explanation:

First change the minutes into hours then multiply by the distance.

(8÷60)×120=16km

5 0

3 years ago