Which statement best describes the part of the atom that is shown by the arrow?

The least count of stopwatch is 0.2s.The time of 20 oscillations of a pendulum was measured to be 25s.Find the percentage error

Svet_ta [14]

Answer:

0.8%

Explanation:

We are given;

Number of oscillations; n = 20

Time taken; t = 25 s

Formula for period of oscillation;

T = t/n = 25/20 = 1.25 s

We are told that the least count is 0.2 s. Thus, error is; ΔT = 0.2 s

percentage error in the measurement of time is given by;

(0.2/(20 × 1.25)) × 100% = 0.8%

4 0

2 years ago

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

Which of the following is MOST useful to scientists in measuring the size of asteroids?

Alenkasestr [34]

Answer:c-The gravitational effect when spacecraft flies close to the asteriod

Explanation:

Gravitational effect on the spacecraft gives an estimate that how big is the asteroid by experiencing its gravitational pull.

The amount of extra thrust required to maintain the trajectory of the spacecraft during its motion hints at the scientist about the size of the asteroid.

Gravitational pull is directly proportional to the mass of object so greater the mass, greater will be the pull.

5 0

2 years ago

If you could help me out that would be fantastic.?!!!

Volgvan

It's either C. Or B.

3 0

3 years ago

A seagull flies at a velocity of 9.00 m/s straight into the wind. (a) if it takes the bird 20.0 min to travel 6.00 km relative t

enot [183]

Here we will the speed of seagull which is v = 9 m/s

this is the speed of seagull when there is no effect of wind on it