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2 years ago

There are two main groups of recyclable material,

Physics

1 answer:

Illusion [34]2 years ago

3 0

Answer:Paper and plastic!

Explanation:

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A copper transmission cable 170 km long and 10.0 cm in diameter carries a current of 100 A .

Bond [772]

Answer:

a = = 37.2V

b = 13.39MJ

Explanation:

Given that

L = 170 × 10³

r = d/2

= 10cm / 2 = 5 cm

current I = 100A

we are to find the potential drop across the cable

so, we can use ohm’s law

V = IR = I (ρL/A)

ρ = resistivity of the copper

= 1.72 × 10⁻⁸ Ω.m

A = πr²

V = I(ρL/πr²)

= 100 (( 1.72 × 10⁻⁸ * 170 × 10³) / ( π * 0.05²))

= 37.2V

(b)

Energy (loss) = Pt

Enery (loss) = IVt

3600s per hour

= (100A)(37.2V)(3600s)

= 13.39MJ

8 0

3 years ago

What is the force needed to throw a ball 7 meters when 1778 J of work is done?

pogonyaev

Answer:

The answer is 254Newton

5 0

3 years ago

lakkis [162]

The right answer is “Atoms “

8 0

3 years ago

Read 2 more answers

A car is traveling at 16 m/s^2

Leya [2.2K]

Really? wow that's pretty cool

4 0

3 years ago

Spacetime interval: What is the interval between two events if in some given inertial reference frame the events are separated b

shepuryov [24]

Answer:

a. $\Delta s ^2 = 8.0888 \ 10^{17} m^2$

b. $\Delta s ^2 = 3.0234 \ 10^{16} m^2$

c. $\Delta s ^2 = 3.0234 \ 10^{20} m^2$

Explanation:

The spacetime interval $\Delta s^2$ is given by

$\Delta s ^2 = \Delta (c t) ^ 2 - \Delta \vec{x}^2$

please, be aware this is the definition for the signature ( + - - - ), for the signature (- + + + ) the spacetime interval is given by:

$\Delta s ^2 = - \Delta (c t) ^ 2 + \Delta \vec{x}^2$ .

Lets work with the signature ( + - - - ), and, if needed in the other signature, we can multiply our interval by -1.

$\Delta \vec{x}^2 = (7.5 \ 10 \ m)^2$

$\Delta \vec{x}^2 = 5,625 m^2$

$\Delta (c t) ^ 2 = (299,792,458 \frac{m}{s} \ 3 \ s)^2$

$\Delta (c t) ^ 2 = (899,377,374 \ m)^2$

$\Delta (c t) ^ 2 = 8.0888 \ 10^{17} m^2$

$\Delta s ^2 = 8.0888 \ 10^{17} m^2 - 5,625 m^2$

$\Delta s ^2 = 8.0888 \ 10^{17} m^2$

$\Delta \vec{x}^2 = (5 \ 10 \ m)^2$

$\Delta \vec{x}^2 = 2,500 m^2$

$\Delta (c t) ^ 2 = (299,792,458 \frac{m}{s} \ 0.58 \ s)^2$

$\Delta (c t) ^ 2 = (173,879,625.6 \ m)^2$

$\Delta (c t) ^ 2 = 3.0234 \ 10^{16} m^2$

$\Delta s ^2 = 3.0234 \ 10^{16} m^2 - 2,500 m^2$

$\Delta s ^2 = 3.0234 \ 10^{16} m^2$

$\Delta \vec{x}^2 = (5 \ 10 \ m)^2$

$\Delta \vec{x}^2 = 2,500 m^2$

$\Delta (c t) ^ 2 = (299,792,458 \frac{m}{s} \ 58 \ s)^2$

$\Delta (c t) ^ 2 = (1.73879 \ 10^{10} \ m)^2$

$\Delta (c t) ^ 2 = 3.0234 \ 10^{20} m^2$

$\Delta s ^2 = 3.0234 \ 10^{20} m^2 - 2,500 m^2$

$\Delta s ^2 = 3.0234 \ 10^{20} m^2$

5 0

2 years ago