All categories

3 years ago

How does the electrical force relate to the charge of an object?

Physics

1 answer:

Akimi4 [234]3 years ago

6 0

Answer:

The electrical force is directly proportional to the charge

Explanation:

The electrical force between two object is directly proportional to the net charge on each object and inversely proportional to the square of the distance between them

You might be interested in

Is a cinder cone volcanoe constructive or deconstructive

Ksivusya [100]

Cinder cone volcanoes can be associated with either constructive or destructive margins.<span>
</span>

6 0

4 years ago

Why can rock avalanches move downslope at speeds exceeding 200 km/hr (125 mi/hr)?

Sati [7]

Rock avalanches move downslope at speeds exceeding 200 km/hr (125 mi/hr) because when the soil expands and compresses, it lifts certain particles and drops them downslope a short distance.

Landslides are the common name for soil or rock avalanches. The most frequent type of avalanche, snowslides, can travel downhill more quickly than a skier. When an unstable snow mass detaches from a slope, a snow avalanche starts.

Rock avalanches are caused by the rapid breakup of initially intact, highly fast-moving rock masses during transportation. In mountainous areas, rock avalanches, which are unexpected rock slope failures marked by high velocities, lengthy runouts, and massive volumes, are among the most deadly and costly geological hazards.

To learn more about Rock avalanches please visit -
brainly.com/question/12661230
#SPJ4

5 0

2 years ago

When isn't momentum conserved?

goblinko [34]

When there are external forces involved...

<span>F=dp/dt</span> so the change in momentum can only be 0 if F is 0

7 0

3 years ago

7. A ball is dropped from a height of 4.0 m. Just before it hits the ground it's momentum is

Gekata [30.6K]

Answer:

Approximately $1\; \rm kg$ . (Assumption: $g = 9.8\; \rm m \cdot s^{-2}$ .)

Explanation:

Let the mass of this ball be $x\; \rm kg$ .

Initial gravitational potential energy of this ball: $m \cdot g \cdot h \approx (39.2\, x) \; \rm J$ .

Just before the ball hits the ground, all $(39.2\, x)\; \rm J$ of gravitational potential energy would have been converted to kinetic energy. Calculate the velocity of the ball at that moment:

$\begin{aligned} \text{kinetic energy} = \frac{1}{2}\, m \cdot v^{2} \end{aligned}$ .

Therefore, right before hitting the ground, the velocity of the ball would be:

$\begin{aligned} v&= \sqrt{\frac{2\, (\text{kinetic energy})}{m}} \\ &= \sqrt{\frac{2 \times (39.2\, x)}{x}}\\ & \approx \sqrt{2 \times 39.2} \approx 8.85\; \rm m \cdot s^{-1}\end{aligned}$ .

The momentum $p$ of an object of mass $m$ and velocity $v$ would be $p = m \cdot v$ . Rewrite this equation to find an expression for mass $m\!$ given velocity $v\!$ and momentum $p\!$ :

$\displaystyle m = \frac{p}{v}$ .

Right before collision, the momentum of this ball is $p = 8.85\; \rm kg \cdot m \cdot s^{-1}$ while its velocity is $v \approx 8.85\; \rm m \cdot s^{-1}$ . Therefore, the mass of this ball would be:

$\begin{aligned}m &= \frac{p(\text{right before landing})}{v(\text{right before landing})} \\ &\approx \frac{8.85\; \rm kg \cdot m \cdot s^{-1}}{8.85\; \rm m \cdot s^{-1}} \approx 1\; \rm kg\end{aligned}$ .

7 0

3 years ago

4.) It was once recorded that a Jaguar left skid marks which were +300 m in length.

torisob [31]

Answer:

Initial velocity of the jaguar: 49 $\frac{m}{s^2}$ (answer d)

Explanation:

Considering that this uniformly accelerated problem (with negative acceleration since the jaguar was reducing its velocity to full stop), does not include the time the jaguar was skidding , we can use the kinematic equation that doesn't include time, but relates velocities (initial and final) with the acceleration ( $a$ ), and the distance "D" covered during the accelerated motion:

$v_f^2-v_i^2=2\,a\,D$

For our problem, the initial velocity ( $v_i$ is our unknown, the final velocity is zero ( $v_f = 0$ - since the jaguar stops in the process), the negative acceleration is given as $a=-4\,\frac{m}{s^2}$ , and the distance D of the skid marks is said to be 300 m in length. Therefore:

$v_f^2-v_i^2=2\,a\,D\\0-v_i^2=2\,(-4)\,(300)\\v_1^2=2400\\v_1=\sqrt{2400} \\v_1=48.99\,\frac{m}{s^2}$

Which we can round to 49 $\frac{m}{s^2}$

3 0

4 years ago