Why would a positive current flow from the positive end of a wire towards its negative end?

If you weighed 130 pounds on earth, you would weigh _____pounds on the moon

Aneli [31]

Answer:

152 pounds

Explanation:

4 0

3 years ago

An object moving forward at 10 m/s suddenly begins moving backward at 5 m/s. What is the most accurate explanation for this chan

Vinil7 [7]

My best guess would be C.

8 0

3 years ago

Read 2 more answers

Given the following situation of marble in motion on rolling 10 m/s horizontally from a height of 1.5-m with negligible friction

Norma-Jean [14]

Answer:

The ball would hit the floor approximately $0.55\; \rm s$ after leaving the table.

The ball would travel approximately $5.5\; \rm m$ horizontally after leaving the table.

(Assumption: $g = 9.81\; \rm m \cdot s^{-2}$ .)

Explanation:

Let $\Delta h$ denote the change to the height of the ball. Let $t$ denote the time (in seconds) it took for the ball to hit the floor after leaving the table. Let $v_0(\text{vertical})$ denote the initial vertical velocity of this ball.

If the air resistance on this ball is indeed negligible: $\displaystyle \Delta h = -\frac{1}{2}\, g\, t^{2} + v_0(\text{vertical}) \cdot t$ .

The ball was initially travelling horizontally. In other words, before leaving the table, the vertical velocity of the ball was $v_0(\text{vertical}) = 0 \; \rm m \cdot s^{-1}$ .

The height of the table was $1.5\; \rm m$ . Therefore, after hitting the floor, the ball would be $1.5\; \rm m \!$ below where it was before leaving the table. Hence, $\Delta h = -1.5\;\rm m$ .

The equation becomes:

$\displaystyle -1.5 = -\frac{9.81}{2} \, t^{2}$ .

Solve for $t$ :

$\displaystyle t = \sqrt{1.5 \times \frac{2}{9.81}} \approx 0.55$ .

In other words, it would take approximately $0.55\; \rm s$ for the ball to hit the floor after leaving the table.

Since the air resistance on the ball is negligible, the horizontal velocity of this ball would be constant (at $v(\text{horizontal}) =10\; \rm m \cdot s^{-1}$ ) until the ball hits the floor.

The ball was in the air for approximately $t = 0.55\; \rm s$ and would have travelled approximately $v(\text{horizontal})\cdot t \approx 5.5\;\rm m$ horizontally during the flight.

4 0

2 years ago

8. What type(s) of motion do the particles in a solid undergo? A. vibrational, rotational, and translational B. vibrational and

uranmaximum [27]

Explanation:

Particles in solids are always vibrating (moving back and forth) in place.

4 0

3 years ago

Ask Your Teacher A beam of light is incident from air on the surface of a liquid. If the angle of incidence is 33.5° and the ang

Oduvanchick [21]

Answer:

The critical angle for the liquid when surrounded by air is 30 degrees.

Explanation:

Given that,

The angle of incidence is 33.5° and the angle of refraction is 19.3°. Firstly, we can find the refractive index of the liquid. It can be calculated using Snell's law as :

$n=\dfrac{\sin i}{\sin r}\\\\n=\dfrac{\sin (33.5)}{\sin (19.3)}\\\\n=2$

The critical angle is given by :

$\theta_c=\sin^{-1}(\dfrac{n_1}{n_2})$

Here, $n_1$ is refractive index of air and $n_2$ is refractive index of liquid.

$\theta_c=\sin^{-1}(\dfrac{1}{2})\\\\\theta_c=30^{\circ}$

So, the critical angle for the liquid when surrounded by air is 30 degrees.

5 0

3 years ago