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3 years ago

Match the letters to the numbers for 30 points!

2 answers:

7 0

-- 'Ca' (Calcium) is an element.
-- The proton has a positive charge.
-- Nuclear fusion results in the synthesis of atoms of new elements.
-- H₂O (water) is a chemical compound.
-- Nuclear fission is a decay of the nucleus.
-- The atomic number of an element is the number of protons
in each atom of it.
-- I suppose you're using the Greek letter <span>η ('eta', not 'nu')
to represent the neutron.
-- I suppose you're using ' e ' to represent the electron.
</span>

labwork [276]3 years ago

4 0

1-a
2-d
3-e
4-c
5-g
6-b
7-h
8-f

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You throw a rock up into the air as hard as you can. It stays in the air a total of 6.0 s. What

vaieri [72.5K]

Answer:

Explanation:

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3 years ago

Question:

exis [7]

Answer:

She can swing 1.0 m high.

Explanation:

Hi there!

The mechanical energy of Jane (ME) can be calculated by adding her gravitational potential (PE) plus her kinetic energy (KE).

The kinetic energy is calculated as follows:

KE = 1/2 · m · v²

And the potential energy:

PE = m · g · h

Where:

m = mass of Jane.

v = velocity.

g = acceleration due to gravity (9.8 m/s²).

h = height.

Then:

ME = KE + PE

Initially, Jane is running on the surface on which we assume that the gravitational potential energy of Jane is zero (the height is zero). Then:

ME = KE + PE (PE = 0)

ME = KE

ME = 1/2 · m · (4.5 m/s)²

ME = m · 10.125 m²/s²

When Jane reaches the maximum height, its velocity is zero (all the kinetic energy was converted into potential energy). Then, the mechanical energy will be:

ME = KE + PE (KE = 0)

ME = PE

ME = m · 9.8 m/s² · h

Then, equallizing both expressions of ME and solving for h:

m · 10.125 m²/s² = m · 9.8 m/s² · h

10.125 m²/s² / 9.8 m/s² = h

h = 1.0 m

She can swing 1.0 m high (if we neglect dissipative forces such as air resistance).

6 0

3 years ago

Suppose you are pushing a 3 kg box with a force of 25 N (directed parallel to the ground) over a distance of 15 m. Afterward, th

alex41 [277]

Answer: 321 J

Explanation:

Given

Mass of the box $m=3\ kg$

Force applied is $F=25\ N$

Displacement of the box is $s=15\ m$

Velocity acquired by the box is $v=6\ m/s$

acceleration associated with it is $a=\dfrac{F}{m}$

$\Rightarrow a=\dfrac{25}{3}\ m/s^2$

Work done by force is $W=F\cdot s$

$W=25\times 15\\W=375\ J$

change in kinetic energy is $\Delta K$

$\Rightarrow \Delta K=\dfrac{1}{2}m(v^2-0)\\\\\Rightarrow \Delta K=\dfrac{1}{2}\times 3\times 6^2\\\\\Rightarrow \Delta K=\dfrac{1}{2}\times 3\times 36\\\\\Rightarrow \Delta K=54\ J$

According to work-energy theorem, work done by all the forces is equal to the change in the kinetic energy

$\Rightarrow W+W_f=\Delta K\quad [W_f=\text{Work done by friction}]\\\\\Rightarrow 375+W_f=54\\\Rightarrow W_f=-321\ J$

Therefore, the magnitude of work done by friction is $321\ J$

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Answer:

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2 years ago