For a neutral object, how many electrons must transfer to make a net charge of -2?

Tanzania [10]

X=r-p. Maybe I don't understand, but I am assuming that you need to isolate for X? you simply subtract p from both sides.<span />

4 0

2 years ago

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1.Suppose someone pulls a cart up a ramp a distance of 85cm along the ramp with a force of 15N.

Drupady [299]

1. 12.75 J

Assuming that the force applied is parallel to the ramp, so it is parallel to the displacement of the cart, the work done by the force is

$W=Fd$

where

F = 15 N is the magnitude of the force

d = 85 cm = 0.85 m is the displacement of the cart

Substituting in the formula, we get

$W=(15 N)(0.85 m)=12.75 J$

2. 10.6 N

In this part, the cart reaches the same vertical height as in part A. This means that the same work has been done (because the work done is equal to the gain in gravitational potential energy of the object: but if the vertical height reached is the same, then the gain in gravitational potential energy is the same, so the work done must be the same).

Therefore, the work done is

$W=Fd=12.75 J$

However, in this case the displacement is

d = 120 cm = 1.20 m

Therefore, the magnitude of the force in this case is

$F=\frac{W}{d}=\frac{12.75 J}{1.20 m}=10.6 N$

3 0

3 years ago

The potential energy of a body if its mass is 30 kg and height 30 m and gravity 10m/sec2<br><br>

Dafna1 [17]

Explanation:

potential energy= mgh

30 × 10 × 30 = 9000J or 9KJ

6 0

2 years ago

What must be done to the wires used to make an electromagnet before they are connected

dalvyx [7]

Answer:

Neatly wrap the wire around the nail

Explanation:

hope this helps ( not sure tho )

5 0

3 years ago

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A model rocket blasts off from the ground, rising straight upward with a constant acceleration that has a magnitude of 85.0 m/s2

sergey [27]

Answer:

Maximum height attained by the model rocket is 2172.87 m

Explanation:

Given,

Initial speed of the model rocket = u = 0
acceleration of the model rocket = $a\ =\ 85.0 m/s^2$
time during the acceleration = t = 2.30 s

We have to consider the whole motion into two parts

In first part the rocket is moving with an acceleration of a = 85.0 $m/s^2$ for the time t = 2.30 s before the fuel abruptly runs out.

Let $s_1$ be the height attained by the rocket during this time intervel,

$s_1\ =\ ut\ +\ \dfrac{1}{2}at^2\\\Rightarrow s_1\ =\ 0\ +\ 0.5\times 85\times 2.30^2\\\Rightarrow s_1\ =\ 224.825\ m$

And Final velocity at that point be v

$\therefore v\ =\ u\ +\ at\\\Rightarrow v\ =\ 0\ +\ 85.0\times 2.3\\\Rightarrow v\ =\ 195.5\ m/s.$

Now, in second part, after reaching the altitude of 224.825 m the fuel abruptly runs out. Therefore rocket is moving upward under the effect of gravitational acceleration,

Let ' $s_2$ ' be the altitude attained by the rocket to reach at the maximum point after the rocket's fuel runs out,

At that insitant,

initial velocity of the rocket = v = 195.5 m/s.

a = $-g\ =\ -9.81\ m/s^2$
Final velocity of the rocket at the maximum altitude = $v_f\ =\ 0$

From the kinematics,

$v^2\ =\ u^2\ +\ 2as\\\Rightarrow 0\ =\ u^2\ -\ 2gs_2\\\Rightarrow s_2\ =\ \dfrac{u^2}{2g}\\\Rightarrow s_2\ =\ \dfrac{195.5^2}{2\times 9.81}\\\Rightarrow s_2\ =\ 1948.02\ m$

Hence the maximum altitude attained by the rocket from the ground is

$s\ =\ s_1\ +\ s_2\ =\ 224.85\ +\ 1948.02\ =\ 2172.87\ m$

6 0

3 years ago