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vampirchik [111]

3 years ago

10

Please Help!!!!! If you were in a completely weightless environment, would you need a force to make an object at rest start to m

ove? Explain you answer.

1 answer:

MrRissso [65]3 years ago

4 0

Answer:

Yes

Explanation:

Absolutely yes! Even in space objects have mass. And if they have mass, they have inertia.

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Calcula la resistencia de un conductor de cobre de 2 m de longitud y 0,1 m2 de sección. Resistividad del Cu = 1,7 · 10–8 Ω · m

Fudgin [204]

Answer:

an expression of more than two algebraic terms, especially the sum of several terms that contain different powers of the same variable(s).

5 0

2 years ago

A 2 kg block is pushed against a spring (k = 400 N/m), compressing it 0.3 m. When the block is released, it moves along a fricti

Kitty [74]

Answer:

$2.29 \mathrm{m} \text { the block slide if the } \mathrm{u}_{\mathrm{s}}=0.4$

$4.58 \mathrm{m} \text { the block slide if the } \mathrm{u}_{\mathrm{k}}=0.2$

Explanation:

Given values

Mass (m) = 2kg

K = 400 N/M

Compressing it 0.3 m

<u>The law of conservation of energy</u>:

$\frac{m v^{2}}{2}+\frac{k x^{2}}{2}=\text { constant }$

$\text { Where, } \frac{m v^{2}}{2} \text { is kinetic energy of the block. }$

$\frac{k \Delta l^{2}}{2}$ Energy of the spring deformation.

M mass of the block

x spring deformation

Therefore, if block left the spring (x = 0)

$\frac{m v^{2}}{2}+0=0+\frac{k \Delta l^{2}}{2}$

Where, Δl is initial spring deformation

$\frac{m v^{2}}{2}=\frac{k \Delta l^{2}}{2}$

$\mathrm{v}^{2}=\frac{k \Delta l^{2}}{m}$

$\mathrm{v}=\sqrt{\frac{k}{m} \times \Delta l^{2}}$

$v=\sqrt{\frac{400}{2} \times(0.3)^{2}}$

$\mathrm{v}=\sqrt{200 \times 0.09}$

<u>The law of conservation of energy</u>:

$\frac{m v^{2}}{2}+m g h=\text { constant }$

Where h is height

$\frac{m v^{2}}{2}+0=0+m g h$

$\frac{m v^{2}}{2}=m g h$

Cancel mass "m" each side

$\mathrm{h}=\frac{v^{2}}{2 g}$

Distance along incline equals

$\begin{array}{ll}{\text { For friction us }} & {\left(L=\frac{h}{u_{s}}\right)} \\ {\text { For friction } u_{k}} & {\left(L=\frac{h}{u_{k}}\right)}\end{array}$

$\begin{array}{l}{\mathrm{u}_{\mathrm{s}}=0.4} \\ {\mathrm{U}_{\mathrm{k}}=0.2} \\ {\text { For friction } \mathrm{u}_{\mathrm{s}}}\end{array}$

$\begin{array}{l}{\mathrm{h}=\frac{v^{2}}{2 g u_{s}}} \\ {\mathrm{L}=\frac{4.24^{2}}{2 \times 9.8 \times 0.4}} \\ {\mathrm{L}=\frac{17.9776}{784}}\end{array}$

$\begin{array}{l}{L=2.29 \mathrm{m}} \\ {2.29 \mathrm{m} \text { the block slide if the } \mathrm{u}_{5}=0.4} \\ {\text { For friction } \mathrm{u}_{\mathrm{k}}} \\ {\mathrm{L}=\frac{4.24^{2}}{2 \times 9.8 \times 0.2}}\end{array}$

$\begin{array}{l}{L=\frac{17.9776}{3.92}} \\ {L=4.58 \mathrm{m}} \\ {4.58 \mathrm{m} \text { the block slide if the } \mathrm{u}_{5}=0.4}\end{array}$

8 0

2 years ago

You want to determine whether the mass of an object attached to a parachute affects the time it takes to fall to the ground. In

hammer [34]

Answer:

The Answer is below!!

Explanation:

The larger the area of the parachute, the more air needs to be pushed out of the way, and so the slower it descends. the independent variable is the shape of the parachutes' canopies. The dependent variable is the drop speeds of the parachutes. How long does it take for the parachutes to reach the ground? Measure this using a stopwatch.

Hope I Helped!!

:)

5 0

2 years ago

What do you know about tides?

Neporo4naja [7]

Tides are the rise and fall of sea levels caused by the combined effects of the gravitational forces exerted by the Moon and the Sun, and the rotation of the Earth. Tide tables can be used to find the predicted times and amplitude (or "tidal range") of tides at any given locale.

For further expiation please contact me at 678-987-2411. Disclaimer(This is not a real number)

8 0

2 years ago

What is potentiometer

Mumz [18]

Answer:

i honestly don't know

Explanation:

but can you help me with a question

4 0

2 years ago

Read 2 more answers