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igor_vitrenko [27]

3 years ago

11

If 2.0 j of work is requiered to move 4 c of chsrge between two points in an electric firld, what is the potential difference be

tween the two points

1 answer:

djverab [1.8K]3 years ago

5 0

2 J / 4 C = 1/2 joule per coulomb.

That's 1/2 volt.

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If vector C is added to vector D, the result is a third vector that is perpendicular to D and has a magnitude equal to 3D. What

Jet001 [13]

Answer:

(e) 3.2

Explanation:

We are given that vector C and D.

Let R be the magnitude of C+D.

According to question

R=3D

We have to find the ratio of the magnitude of C to that of D.

By using right triangle property

$C^2=R^2+D^2$

$C^2=(3D)^2+D^2$

$C^2=9D^2+D^2$

$C^2=10D^2$

$C=\sqrt{10D^2}=3.2D$

$\frac{C}{D}=3.2$

Hence, the ratio of the magnitude of C to that of D=3.2

(e) 3.2

5 0

3 years ago

Who might be experiencing psychosis?

aleksandrvk [35]

Answer:

D

Explanation:

I think the answer is D, because of hallucinations

4 0

3 years ago

A girl pushes a 1.04 kg book across a table with a horizontal applied force 10 points

mr Goodwill [35]

Answer:

Approximately $11.0\; \rm m \cdot s^{-1}$ . (Assuming that $g = 9.81 \; \rm N \cdot kg^{-1}$ , and that the tabletop is level.)

Explanation:

Weight of the book:

$W = m \cdot g = 1.04 \; \rm kg \times 9.81\; \rm N \cdot kg^{-1} \approx 10.202\; \rm N$ .

If the tabletop is level, the normal force on the book will be equal (in magnitude) to weight of the book. Hence, $F(\text{normal force}) \approx 10.202\; \rm N$ .

As a side note, the $F_N$ and $W$ on this book are not equal- these two forces are equal in size but point in the opposite directions.

When the book is moving, the friction $F(\text{kinetic friction})$ on it will be equal to

$\mu_{\rm k}$ , the coefficient of kinetic friction, times
$F(\text{normal force})$ , the normal force that's acting on it.

That is:

$\begin{aligned}& F(\text{kinetic friction}) \\ &= \mu_{\rm k}\cdot F(\text{normal force})\\ &\approx 0.35 \times 10.202\; \rm N \approx 3.5708\; \rm N\end{aligned}$ .

Friction acts in the opposite direction of the object's motion. The friction here should act in the opposite direction of that $15.0\; \rm N$ applied force. The net force on the book shall be:

$\begin{aligned}& F(\text{net force}) \\ &= 15.0 \; \rm N - F(\text{kinetic friction}) \\& \approx 15.0 - 3.5708\; \rm N \approx 11.429\; \rm N\end{aligned}$ .

Apply Newton's Second Law to find the acceleration of this book:

$\displaystyle a = \frac{F(\text{net force})}{m} \approx \frac{11.429\; \rm N}{1.04\; \rm kg} \approx 11.0\; \rm m \cdot s^{-2}$ .

6 0

3 years ago

Which is more bussin? mcdonald’s or taco bell

Maksim231197 [3]

Answer:

it depends on what you wanna get

if its chicken nuggies then mcdonalds

if its bomb a.ss tacos that taste pretty good but with meat that looks like literal sh.it then probably tacobell

7 0

3 years ago

Read 2 more answers

A 2430 pound roller coaster starts from rest and is launched such that it crests a 105 ft high hill with a speed of 59 mph. The

Rudik [331]

Answer:

Explanation:

Given

Weight of roller coaster is $W=2430\ pound$

mass of roller coaster $m=\frac{W}{g}=\frac{2430}{32.2}=75.45$

Distance traveled by roller coaster $d=396\ ft$

drag force $f_d=85\ pounds$

velocity at top $v=59 mph\approx 86.53\ ft/s$

Suppose E is the initial energy

Conserving Energy at bottom and top

$E=\frac{1}{2}mv^2+mgh+f_d\cdot d$

$E=0.5\times 75.45\times 86.53^2+2430\times 105+85\times 396$

$E=2.9\times 10^5\ foot-pound$

5 0

3 years ago