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Ede4ka [16]
1 year ago
11

The coal with the highest energy available per unit burned is:________

Physics
1 answer:
Elan Coil [88]1 year ago
7 0
A Ignite. Because. When you put coal in a gas pit, it ignites and burns. However, there is a chance that all of us fed
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Two samples of dirt are collected from a suspect's tread in his shoe and a crime scene. She notes very similar characteristics.
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8 0
3 years ago
In springboard diving, the diver strides out to the end of the board, takes a jump onto its end, and uses the resultant spring-l
Ksenya-84 [330]

Answer:

10.4 m/s

Explanation:

The problem can be solved by using the following SUVAT equation:

v=u+at

where

v is the final velocity

u is the initial velocity

a is the acceleration

t is the time

For the diver in the problem, we have:

u=+6.3 m/s is the initial velocity (positive because it is upward)

a=g=-9.8 m/s^2 is the acceleration of gravity (negative because it is downward)

By substituting t = 1.7 s, we find the velocity when the diver reaches the water:

v=+6.3 + (-9.8)(1.7)=-10.4 m/s

And the negative sign means that the direction is downward: so, the speed is 10.4 m/s.

3 0
3 years ago
Read 2 more answers
How long does it take the earth to complete one revolution
serious [3.7K]
It takes 365 1/4 days for the Earth to complete 1 full revolution.

Please mark me as brainliest. I really need it.
7 0
2 years ago
A paper clip that has a mass of 1.5 grams is thrown into the air and initially has a kinetic energy
Vsevolod [243]

Answer:

v = 4.2 \ m/s

Explanation:

Given data:

Mass of the paper clip, m = 1.5 \ g = 0.0015 \ kg

Kinetic energy, K = 0.013 \ \rm J

Let the velocity of the paper clip when it is thrown be <em>v</em>.

Thus,

K = \frac{1}{2}mv^{2}

0.013 = 0.5 \times 0.0015 \times v^{2}

\Rightarrow \ v = 4.16 \ m/s

v = 4.2 \ m/s.  (rounding to nearest tenth)

3 0
3 years ago
Two long, parallel wires separated by 3.50 cm carry currents in opposite directions. The current in one wire is 1.55 A, and the
vaieri [72.5K]

Answer:

Therefore,

The magnitude of the force per unit length that one wire exerts on the other is

\dfrac{F}{l}=2.79\times 10^{-5}\ N/m

Explanation:

Given:

Two long, parallel wires separated by a distance,

d = 3.50 cm = 0.035 meter

Currents,

I_{1}=1.55\ A\\I_{2}=3.15\ A

To Find:

Magnitude of the force per unit length that one wire exerts on the other,

\dfrac{F}{l}=?

Solution:

Magnitude of the force per unit length on each of @ parallel wires seperated by the distance d and carrying currents I₁ and I₂ is given by,

\dfrac{F}{l}=\dfrac{\mu_{0}\times I_{1}\times I_{2}}{2\pi\times d}

where,

\mu_{0}=permeability\ of\ free\ space =4\pi\times 10^{-7}

Substituting the values we get

\dfrac{F}{l}=\dfrac{4\pi\times 10^{-7}\times 1.55\times 3.15}{2\pi\times 0.035}

\dfrac{F}{l}=2.79\times 10^{-5}\ N/m

Therefore,

The magnitude of the force per unit length that one wire exerts on the other is

\dfrac{F}{l}=2.79\times 10^{-5}\ N/m

7 0
3 years ago
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