What do electricians call disconnects or disconnecting?

A typical male sprinter can maintain his maximum acceleration for 2.0 s, and his maximum speed is 10 m/s. After he reaches this

baherus [9]

20 characters longer later and woah

5 0

2 years ago

If you hold your hand over a pot of boiling water, it feels hot. Explain how the boiling water is transferring heat to your hand

disa [49]

by conduction, which is direct touch

5 0

2 years ago

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a ball rolls horizontally of the edge of the cliff at 4 m/s, if the ball lands at a distance of 30 m from the base of the vertic

algol13

Answer:

Approximately $281.25\; \rm m$ . (Assuming that the drag on this ball is negligible, and that $g = 10\; \rm m \cdot s^{-2}$ .)

Explanation:

Assume that the drag (air friction) on this ball is negligible. Motion of this ball during the descent:

Horizontal: no acceleration, velocity is constant (at $v(\text{horizontal})$ is constant throughout the descent.)
Vertical: constant downward acceleration at $g = 10\; \rm m \cdot s^{-2}$ , starting at $0\; \rm m \cdot s^{-1}$ .

The horizontal velocity of this ball is constant during the descent. The horizontal distance that the ball has travelled during the descent is also given: $x(\text{horizontal}) = 30\; \rm m$ . Combine these two quantities to find the duration of this descent:

$\begin{aligned}t &= \frac{x(\text{horizontal})}{v(\text{horizontal})} \\ &= \frac{30\; \rm m}{4\; \rm m \cdot s^{-1}} = 7.5\; \rm s\end{aligned}$ .

In other words, the ball in this question start at a vertical velocity of $u = 0\; \rm m \cdot s^{-1}$ , accelerated downwards at $g = 10\; \rm m \cdot s^{-2}$ , and reached the ground after $t = 7.5\; \rm s$ .

Apply the SUVAT equation $\displaystyle x(\text{vertical}) = -\frac{1}{2}\, g \cdot t^{2} + v_0\cdot t$ to find the vertical displacement of this ball.

$\begin{aligned}& x(\text{vertical}) \\[0.5em] &= -\frac{1}{2}\, g \cdot t^{2} + v_0\cdot t\\[0.5em] &= - \frac{1}{2} \times 10\; \rm m \cdot s^{-2} \times (7.5\; \rm s)^{2} \\ & \quad \quad + 0\; \rm m \cdot s^{-1} \times 7.5\; s \\[0.5em] &= -281.25\; \rm m\end{aligned}$ .

In other words, the ball is $281.25\; \rm m$ below where it was before the descent (hence the negative sign in front of the number.) The height of this cliff would be $281.25\; \rm m\!$ .

5 0

2 years ago

How does electroscope work ?

s2008m [1.1K]

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6 0

2 years ago

A horizontal uniform board of weight 125N and length 4 m is supported by vertical chains at each end. A person weighing 500N sit

Misha Larkins [42]

Let's apply an equation of equilibrium to the situation: The sum of the moments about the left end of the board must equal 0.

We have three moments to add. Positive force values indicate upward direction and negative values indicate downward direction. All distances given below are measured to the right side of the left end of the board:

The weight of the board, -125N, located at 2m (center of the board due to its uniform density)
The tension in the right chain, +250N, located at 4m
The weight of the person, -500N, located at a distance "x"

The sum of the moments must equal 0 and is given by:

ΣFx = 0

F is the magnitude of force, x = distance from the left end of the board

Plug in all of the force and distance values and solve for x:

ΣFx = 250(4) - 125(2) - 500x = 0

500x = 750

x = 1.5m

7 0

2 years ago