The earth's magnetic field deflects the flow of current from?

Please help me, I beg you

MA_775_DIABLO [31]

Answer:

40g

Explanation:

Solubility of Copper sulfate at 90°=60g

Solubility of potassium bromide at 90°=100g

100g-60g=40g

8 0

3 years ago

The velocity of a passenger relative to a boat is -vpb. The velocity of the boat relative to the river it is moving on is vbr. T

RideAnS [48]

Answer:

vps = vbr + vrs - vpb

Explanation:

If the passenger were at rest, his speed relative to the shore will be identical to the boat's, as follows:
vps = vbr + vrs
As he is moving in a direction opposite to the boat's, his velocity relative to the shore must be less than if he were at rest, in the same quantity that he was moving opposite to the boat, as follows:
vps = vbr+ vrs -vpb

5 0

2 years ago

A dockworker applies a constant horizontal force of 80.0 N to a block of ice on a smooth horizontal floor. The frictional force

Tamiku [17]

Answer:

(a) 91 kg (2 s.f.) (b) 22 m

Explanation:

Since it is stated that a constant horizontal force is applied to the block of ice, we know that the block of ice travels with a constant acceleration and but not with a constant velocity.

(a)

$s \ = \ ut \ + \ \displaystyle\frac{1}{2} at^{2} \\ \\ a \ = \ \displaystyle\frac{2(s \ - \ ut)}{t^{2}} \\ \\ a \ = \ \displaystyle\frac{2(11 \ - \ 0)}{5^{2}} \\ \\ a \ = \ \displaystyle\frac{22}{25} \\ \\ a \ = \ 0.88 \ \mathrm{m \ s^{-2}}$

Subsequently,

$F \ = \ ma \\ \\ m \ = \ \displaystyle\frac{F}{a} \\ \\ m \ = \ \displaystyle\frac{80 \ \mathrm{kg \ m \ s^{-2}}}{0.88 \ \mathrm{m \ s^{-2}}} \\ \\ m \ = \ 91 \mathrm{kg} \ \ \ \ \ \ (2 \ \mathrm{s.f.})$

*Note that the equations used above assume constant acceleration is being applied to the system. However, in the case of non-uniform motion, these equations will no longer be valid and in turn, calculus will be used to analyze such motions.

(b) To find the final velocity of the ice block at the end of the first 5 seconds,

$v \ = \ u \ + \ at \\ \\ v \ = \ 0 \ + \ (0.88 \mathrm{m \ s^{-2}})(5 \ \mathrm{s}) \\ \\ v \ = \ 4.4 \ \mathrm{m \ s^{-1}}$

According to Newton's First Law which states objects will remain at rest

or in uniform motion (moving at constant velocity) unless acted upon by

an external force. Hence, the block of ice by the end of the first 5

seconds, experiences no acceleration (a = 0) but travels with a constant

velocity of 4.4 $m \ s^{-1}$ .

$s \ = \ ut \ + \ \displaystyle\frac{1}{2}at^{2} \\ \\ s \ = \ (4.4 \ \mathrm{m \ s^{-2}})(5 \ \mathrm{s}) \ + \ \displaystyle\frac{1}{2}(0)(5^{2}) \\ \\ s \ = \ 22 \ \mathrm{m}$

Therefore, the ice block traveled 22 m in the next 5 seconds after the

worker stops pushing it.

4 0

3 years ago

I need someone to help me work through this

Sergio039 [100]

The average rate of change of distance over the time interval
3 ≤ t ≤ 6 represents the coin's average velocity over that interval.

3 0

3 years ago

In the Bohr model of hydrogen, the electron moves in a circular orbit around the nucleus. (a) Determine the orbital frequency of

Airida [17]

Answer:

(a) 6.567 * 10^15 rev/s or hertz

(b) 8.21 * 10^14 rev/s or hertz

Explanation:

Fn= 4π^2k^2e^4m * z^2/(h^3*n^3)

Where Fn is frequency at all levels of n.

Z = 1 (nucleus)

e = 1.6 * 10^-19c

m = 9.1 * 10^-31 kg

h = 6.62 * 10-34

K = 9 * 10^9 Nm2/c2

(a) for groundstate n = 1

Fn = 4 * π^2 * (9*10^9)^2*(1.6*10^-19)^4* (9.1 * 10^-31) * 1 / (6.62 * 10^-31)^3 = 6.567 * 10^15 rev/s

(b) first excited state

n = 1

We multiple the groundstate answer by 1/n^3

6.567 * 10^15 rev/s/ 2^3

F2 = 8.2 * 10^ 14 rev/s

3 0

3 years ago