Marble and cotton ball friction assignment

A negatively charged rod is placed near two neutral metal spheres, as shown below. Which statements describe the charging method

grin007 [14]

<h3>Answer;</h3>

<em>The spheres develop opposite charges. </em>
<em>Electrons move from Sphere A to Sphere B. </em>
<em>The spheres are charged through induction.</em>

<h3><u>Explanation;</u></h3>

<u><em>When a negatively charged rod is placed near two neutral metal spheres, the spheres will develop opposite charges, because the neutral metal spheres have both negative and positive charges. </em></u>From the basic law of electrostatics unlike charges attracts and like charges repel.
Thus, <em><u>the sphere will develop opposite charges, electrons will move from Sphere A to sphere B,</u></em> hence we say that the spheres will be charged by induction such that sphere A will acquire a positive charge while sphere B will acquire negative charge.

3 0

4 years ago

Read 2 more answers

Which describes a feature of the physical changes of all substances?

JulijaS [17]

Answer:

A change in size, shape, or matter.

Explanation:

A physical change does not produce new substances. A physical change may only change the physical properties of a substance. a change in size, shape, or phase of matter.

8 0

2 years ago

How many chicken feet hatch from 3 dozen eggs? Explain.

Hunter-Best [27]

Answer: I think it would be 72 because 12x3=36 and then all the chick have 2 feet each so you would multiply t but 2 and that would be 72

Explanation:

5 0

3 years ago

A first order reaction, A -> products, has a rate reaction of .00250 Ms-1 when [A] = . 484 M. (a) What is the rate constant,

tamaranim1 [39]

Answer: a) The rate constant, k, for this reaction is $0.00516s^{-1}$

b) No $t_{\frac{1}{2}}$ does not depend on concentration.

Explanation:

Rate law says that rate of a reaction is directly proportional to the concentration of the reactants each raised to a stoichiometric coefficient determined experimentally called as order.

$A\rightarrow products$

Given: Order with respect to $A$ = 1

Thus rate law is:

a) $Rate=k[A]^1$

k= rate constant

$0.00250=k[0.484]^1$

$k=0.00516s^{-1}$

The rate constant, k, for this reaction is $0.00516s^{-1}$

b) Expression for rate law for first order kinetics is given by:

$t=\frac{2.303}{k}\log\frac{a}{a-x}$

where,

k = rate constant

t = age of sample

a = let initial amount of the reactant

a - x = amount left after decay process

Half life is the amount of time taken by a radioactive material to decay to half of its original value.

$t_{\frac{1}{2}}=\frac{2.303}{k}\log\frac{100}{50}$

$t_{\frac{1}{2}}=\frac{0.69}{k}$

Thus $t_{\frac{1}{2}}$ does not depend on concentration.

8 0

3 years ago

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

3 years ago