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3 years ago

8

Electricity is the flow of electrons. The questions relate to how electricity is quantified. Electrons are charged particles. Th

e amount of charge that passes per unit time is called

1 answer:

xxTIMURxx [149]3 years ago

7 0

The amount of charge that passes per unit time is called <em>electric current</em> .

Current has dimensions of [Charge] / [Time] .

It's measured and described in units of ' Ampere ' .

1 Ampere means 1 Coulomb of charge passing a point every second.

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(3.16_Q2) Which weights would you use on a single thread to create a 6.86 N force? Question 2 options: Weight IDs A, B, C, D Wei

Tomtit [17]

1. E,F

2. D,E,F

3. B,C,E,G

4. A,B,C,D

I did the test! :)

5 0

3 years ago

Two kids are playing on a newly installed slide, which is 3 m long. John, whose mass is 30 kg, slides down into William (20 kg),

yuradex [85]

Answer:

$v=3.564\ m.s^{-1}$

$\Delta v =2.16\ m.s^{-1}$

Explanation:

Given:

mass of John, $m_J=30\ kg$

mass of William, $m_W=30\ kg$

length of slide, $l=3\ m$

(A)

height between John and William, $h=1.8\ m$

<u>Using the equation of motion:</u>

$v_J^2=u_J^2+2 (g.sin\theta).l$

where:

v_J = final velocity of John at the end of the slide

u_J = initial velocity of John at the top of the slide = 0

Now putting respective :

$v_J^2=0^2+2\times (9.8\times \frac{1.8}{3})\times 3$

$v_J=5.94\ m.s^{-1}$

<u>Now using the law of conservation of momentum at the bottom of the slide:</u>

<em>Sum of initial momentum of kids before & after collision must be equal.</em>

$m_J.v_J+m_w.v_w=(m_J+m_w).v$

where: v = velocity with which they move together after collision

$30\times 5.94+0=(30+20)v$

$v=3.564\ m.s^{-1}$ is the velocity with which they leave the slide.

(B)

frictional force due to mud, $f=105\ N$

<u>Now we find the force along the slide due to the body weight:</u>

$F=m_J.g.sin\theta$

$F=30\times 9.8\times \frac{1.8}{3}$

$F=176.4\ N$

<em><u>Hence the net force along the slide:</u></em>

$F_R=71.4\ N$

<em>Now the acceleration of John:</em>

$a_j=\frac{F_R}{m_J}$

$a_j=\frac{71.4}{30}$

$a_j=2.38\ m.s^{-2}$

<u>Now the new velocity:</u>

$v_J_n^2=u_J^2+2.(a_j).l$

$v_J_n^2=0^2+2\times 2.38\times 3$

$v_J_n=3.78\ m.s^{-1}$

Hence the new velocity is slower by

$\Delta v =(v_J-v_J_n)$

$\Delta v =5.94-3.78= 2.16\ m.s^{-1}$

8 0

3 years ago

How much power does it take to do 500 J of work in 10 seconds?

Dmitry_Shevchenko [17]

Power = work/time

= 500/10

= 50J/s or 50 watt

7 0

3 years ago

About how far does the S wave travel through Earth in 13 minutes?

Elina [12.6K]

Answer:

Explanation:

6000 km

6 0

3 years ago

Read 2 more answers

A 60 kg skier starts from rest at the top of a frictionless slope of height of 35 meters. The velocity at the bottom is v. If a

Solnce55 [7]

Answer:

Explanation:

Speed of skier without parachute

= √ 2gh

= √ 2 x 9.8 x 35

= 26.2 m / s

Speed of skier with parachute

net force downwards

mg - 200

= 60 x 9.8 -200

= 388 N

acceleration = 388 / 60

a = 6.47 m / s

v = √ 2ah

= √ 2 x 6.47 x 35

= 21.28 m / s

8 0

3 years ago

Read 2 more answers