Feeling slightly uncomfortable during exercise

The use of force to move an object is <br> .

Ne4ueva [31]

The use of force to move an object is called work. This only applies if the object moves.

4 0

2 years ago

The charges on two metallic balls are 5.0 and 7.0 coulombs respectively. They are kept 1.2 meters apart. What is the force of in

Marysya12 [62]

We know, F = k * q₁ * q₂ / r²

Substitute the known values,
F = 9 * 10⁹ * 5 * 7 / (1.2)²

F = 315 * 10⁹ / 1.44

F = 218.75 * 10⁹ N

F = 2.1875 * 10¹¹ N [ Final Answer ]

Hope this helps!

5 0

3 years ago

Do you think a baseball curves better at the top of a high mountain or down on a flat plain

Serhud [2]

A baseball will curve better on the flat plain if it is higher than sea level but low elevation.

Hope this helped!

4 0

2 years ago

A car speeds up from 18.54 m/s to<br> 29.52 m/s in 13.84 s.<br> The acceleration of the car is:

valkas [14]

Answer:

.7934 $m/s^{2}$

Explanation:

Acceleration = change in velocity / change in time

A = 10.98 $m/s$ / 13.84 $s$

A = .7934 $m/s^{2}$

5 0

2 years ago

Read 2 more answers

A 2.93 kg particle has a velocity of (2.98 i hat - 3.98 j) m/s.

cupoosta [38]

Answer:

a) The x and y components of the momentum are $8.731\,\frac{kg\cdot m}{s}$ and $-11.661\,\frac{kg\cdot m}{s}$ , respectively.

b) The magnitude and direction of its momentum are approximately 14.567 kilogram-meters per second and 306.823º.

Explanation:

a) The vectorial equation of momentum is represented by the following expression:

$\vec p = m\cdot \vec v$ (1)

Where:

$\vec p$ - Vector momentum, measured in kilogram-meters per second.

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

$\vec v$ - Vector velocity, measured in meters per second.

If we know that $m = 2.93\,kg$ and $\vec v = 2.98\,\hat{i}-3.98\,\hat{j}\,\,\,\left[\frac{m}{s} \right]$ , then the momentum is:

$\vec p = (2.93)\cdot (2.98\,\hat{i}-3.98\,\hat{j})\,\,\,\left[\frac{kg\cdot m}{s} \right]$

$\vec p = 8.731\,\hat{i}-11.661\,\hat{j}\,\,\,\left[\frac{kg\cdot m}{s} \right]$

The x and y components of the momentum are $8.731\,\frac{kg\cdot m}{s}$ and $-11.661\,\frac{kg\cdot m}{s}$ , respectively.

b) The magnitude and direction of momentum are represented by the following expressions:

$\|\vec p \| = \sqrt{p_{x}^{2}+p_{y}^{2}}$ (2)

$\theta = \tan^{-1}\left(\frac{p_{y}}{p_{x}} \right)$ (3)

Where:

$\|\vec p\|$ - Magnitude of momentum, measured in kilogram-meters per second.

$\theta$ - Direction of momentum, measured in sexagesimal degrees.

If we know that $p_{x} = 8.731\,\frac{kg\cdot m}{s}$ and $p_{y} = -11.661\,\frac{kg\cdot m}{s}$ , then the magnitude and direction of momentum are, respectively:

$\|\vec p\| = \sqrt{\left(8.731\,\frac{kg\cdot m}{s} \right)^{2}+\left(-11.661\,\frac{kg\cdot m}{s} \right)^{2}}$

$\|\vec p\| \approx 14.567\,\frac{kg\cdot m}{s}$

$\theta =\tan^{-1}\left(\frac{-11.661\,\frac{kg\cdot m}{s} }{8.731\,\frac{kg\cdot m}{s} } \right)$

$\theta \approx 306.823^{\circ}$

The magnitude and direction of its momentum are approximately 14.567 kilogram-meters per second and 306.823º.

6 0

3 years ago