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Bess [88]
1 year ago
15

If something is a good conductor, what type of insulator is it?

Physics
2 answers:
nlexa [21]1 year ago
8 0

4. a poor insulator

If rest other things are kept constant or unchanged then a good conductor can be termed as a poor insulator.

Katarina [22]1 year ago
8 0

Answer:

4 conductors and insulators are opposites of each other.

Explanation:

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A pyrotechnical releases a 3 kg firecracker from rest. at t=0.4 s, the firecracker is moving downward with a speed 4 m/s. At the
olga2289 [7]

Answer:

a) F = 30 N, b)   I = 12 N s , c)  I = -12 N s , d) ΔI = 0 N s

Explanation:

This exercise is a case at the moment, let's define the system formed by the firecracker and its two parts, in this case the forces during the explosion are internal and the moment is conserved

Initial, before the explosion

     p₀ = m v

The speed can be found by kinematics

     v = v₀ - g t

     v = 0 - 10 0.4

     v = -4.0 m / s

Final after division

     pf = m₁ v₁f + m₂ v₂f

    p₀ = pf

    M v = m₁ v₁f + m₂ v₂f

Where M is the initial mass (M = 3 kg), m₁ is the mass mtop (m₁ = 1 kg) and m₂ in the mass m botton (m₂ = 2kg) and the piece that moves up (v₁f = 6m/s )

a) before the explosion the only force acting on the body is gravity

     F = mg

     F = 3 10 = 30 N

b) The expression for momentum is

     I = Ft

Before the explosion the only force that acts is the weight

    I = mg t

    I = 3 10 0.4

    I = 12 N s

c) To calculate this part we use the conservation of the moment and calculate the speed of the body that descends body 2

    M v = m₁ v₁f + m₂ v₂f

    v₂f = (M v - m₁ v₁f) / m₂

    v₂f = (3 (-4) - 1 6) / 2

   v₂f = - 9 m / 2

The negative sign indicates that body 2 (botton) is descending

Now we can use the momentum and momentum relationship for the body during the explosion

    I = F t = Dp

   F t = pf –po)

   F t= [m₁ v₁f + m₂ v₂f]

   

   I = [1 6 + 2 (-9) -0]

   I = -12 N s

This is the impulse during the explosion the negative sign indicates that it is headed down

d) impulse change

I₀ = Mv

I₀ = 3 *4

I₀ =-12 N s

 ΔI =If – I₀  

ΔI = - 12 – (-12)

ΔI = -0 N s

3 0
3 years ago
Which of these BEST describes the term ecliptic? A. the circle where "Earth meets the sky" B. the change in azimuth and altitude
jeyben [28]
I'll say B ....hope that helps...
4 0
3 years ago
Read 2 more answers
Experiment with the battery voltage set to 15 volts, measure the current in a parallel circuit with 1,2,3, and 4 light bulbs. (I
koban [17]

Answer:

1) current = I

2) Resistance = V/I

3) current = 2I

4) resistance = V/2I

5) current = 3I

6) Resistance = V/3I

7) Current = 4I

8) Resistance = V/4I

Explanation:

When one bulb is connected across the battery then let say the current is given as I

Then resistance is given as

R = \frac{V}{I}

When two bulbs are in parallel with the battery then

total current becomes twice of initial current

so we have

current = 2I

Resistance of the circuit is now

R = \frac{V}{2I}

When three bulbs are in parallel with the battery then

total current becomes three times of initial current

so we have

current = 3I

Resistance of the circuit is now

R = \frac{V}{3I}

When four bulbs are in parallel with the battery then

total current becomes four times of initial current

so we have

current = 4I

Resistance of the circuit is now

R = \frac{V}{4I}

3 0
3 years ago
A line _____ on a typical Speed vs. Time graph means an object is experiencing a constant acceleration.
Savatey [412]

Answer:

The answer should be C. slanted upward to the right.

Hope this helps. :-)

7 0
3 years ago
Read 2 more answers
Calculate the velocity of a 1650kg satellite that is in a circular orbit of 4.2 x 10^6m above the surface of a planet which has
Anastasy [175]
-- We're going to be talking about the satellite's speed. 
"Velocity" would include its direction at any instant, and
in a circular orbit, that's constantly changing.

-- The mass of the satellite makes no difference.

Since the planet's radius is  3.95 x 10⁵m  and the satellite is
orbiting  4.2 x 10⁶m  above the surface, the radius of the
orbital path itself is

                               (3.95 x 10⁵m) + (4.2 x 10⁶m)

                     =        (3.95 x 10⁵m) + (42 x 10⁵m)

                     =           45.95 x 10⁵ m

The circumference of the orbit is  (2 π R) =  91.9 π x 10⁵ m.

The bird completes a revolution every 2.0 hours,
so its speed in orbit is

                                     (91.9 π x 10⁵ m) / 2 hr

                        =        45.95 π x 10⁵  m/hr  x  (1 hr / 3,600 sec)

                        =           0.04 x 10⁵      m/sec

                        =              4 x 10³      m/sec  

                                     (4 kilometers per second)
6 0
3 years ago
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