All categories

3 years ago

Nearly all scientists agree that fossil fuels will be used up in a few decades.

Physics

2 answers:

11Alexandr11 [23.1K]3 years ago

7 0

Answer:

it's true

mark me as brilliant please

STatiana [176]3 years ago

7 0

The answer is false because many believe that we will eventually stop using them,but the answer is false

You might be interested in

electricity flows back and forth in an lc circuit with a .33 f capacitor and a .1 h inductor. what is the angular frequency and

Shalnov [3]

The time it takes an object to complete one oscillation and return to its initial position is measured in terms of a period, or T. The formula for the angular frequency is = 2/T.

<h3>How is G determined in oscillation?</h3>

Use a stopwatch to calculate the oscillation's time period T. Calculate the pendulum's length L. Subtract the time period T's square from the length L.

<h3>How does oscillation's G work?</h3>

A mass attached to the end of a pendulum with a length of l causes it to oscillate with a period (T). T = 2(l/g), where g.

To know more about angular frequency visit:-

brainly.com/question/29107224

#SPJ4

4 0

1 year ago

A large blue marble of mass 3.5 g is moving to the right with a velocity of 15 cm/s. The large marble hits a small red marble of

neonofarm [45]

I hope this would help you

3 0

3 years ago

Please I need it now

Nana76 [90]

1. A vector $\vec x$ is a unit vector if its magnitude is 1. Given

$\vec A = \dfrac{\sqrt3}2\,\vec\imath - \dfrac12\,\vec\jmath \\\\ \vec B = -\dfrac{\sqrt2}2\,\vec\imath - \dfrac{\sqrt2}2\,\vec\jmath \\\\ \vec C = \dfrac12\,\vec\imath - \dfrac{\sqrt3}2\,\vec\jmath - \dfrac12\,\vec k$

$\vec A$ and $\vec B$ are unit vectors, while $\vec C$ is not, since

$\|\vec A\| = \sqrt{\left(\dfrac{\sqrt3}2\right)^2 + \left(-\dfrac12\right)^2} = 1 \\\\ \|\vec B\| = \sqrt{\left(-\dfrac{\sqrt2}2\right)^2 + \left(-\dfrac{\sqrt2}2\right)^2} = 1 \\\\ \|\vec C\| = \sqrt{\left(\dfrac12\right)^2+\left(-\dfrac{\sqrt3}2\right)^2+\left(-\dfrac12\right)^2} =\dfrac{\sqrt5}2 \neq 1$

2. Given some vector $\vec x$ , you can get the unit vector in the same direction as

$\vec A = -12\,\vec\imath + 9\,\vec\jmath \\\\ \vec B = 15\,\vec\imath-20\,\vec\jmath$

then the unit vectors in the direction of $\vec A$ and $\vec B$ , respectively, are

$\dfrac{\vec A}{\|\vec A\|} = \dfrac{-12\,\vec\imath+9\,\vec\jmath}{\sqrt{(-12)^2+9^2}} = \dfrac{-12\,\vec\imath+9\,\vec\jmath}{15} = -\dfrac45\,\vec\imath + \dfrac35\,\vec\jmath \\\\ \dfrac{\vec B}{\|\vec B\|} = \dfrac{15\,\vec\imath-20\,\vec\jmath}{\sqrt{15^2+(-20)^2}} = \dfrac{15\,\vec\imath-20\,\vec\jmath}{25} = \dfrac35\,\vec\imath - \dfrac45\,\vec\jmath$

$\vec A = \|\vec A\| \left(\cos(180^\circ-\theta)\,\vec\imath + \sin(180^\circ-\theta)\,\vec\jmath\right) \\\\ \vec A = 25 \left(-\sin(37^\circ)\,\vec\imath + \cos(37^\circ)\,\vec\jmath\right) \\\\ \vec A = -15\,\vec\imath + 20\,\vec\jmath$

$\vec B = \|\vec B\| \left(\cos(180^\circ + \alpha)\,\vec\imath + \sin(180^\circ+\alpha)\,\vec\jmath\right) \\\\ \vec B = 75 \left(-\sin(53^\circ)\,\vec\imath-\cos(53^\circ)\,\vec\jmath\right) \\\\ \vec B = -60\,\vec\imath - 45\,\vec\jmath$

6 0

3 years ago

A 6.00kg box is subjected to a force F=18.0N-(0.530N/m)x. Ignoring friction and using Work, find the speed of the box after it h

Nadusha1986 [10]

Answer:

Approximately $8.17\; \rm m \cdot s^{-1}$ assuming that the effect of gravity on the box can be ignored.

Explanation:

If the force $F$ is constant, then the work would be found with $W = F \cdot \Delta x$ . However, this equation won't work for this question since the

$\displaystyle W = \int\limits_{x_0}^{x_1} F\, d x$ ,

For this particular question, $x_0 = 0\; \rm m$ and $x_1 = 14.0\; \rm m$ . Apply this equation:

$\begin{aligned}W &= \int\limits_{x_0}^{x_1} F\, d x \\ &= \int\limits_{0\; \rm m}^{14.0\; \rm m} \left[{18.0\; \rm N} - {\left(0.530\; {\rm N \cdot m^{-1}}\right)}\cdot x \right]\, d x \\ &= \left[{(18.0\; \rm N)}\cdot x - \frac{1}{2}\;{\left(0.530\; {\rm N \cdot m^{-1}}\right)}\cdot x^2\right]_{x = 0\; \rm m}^{x = 14.0\; \rm m} \approx 200.06\; \rm N \cdot m\end{aligned}$ .

(Side note: keep in mind that $1\; \rm J = 1\; \rm N\cdot m$ .)

Since friction is ignored, all these work should have been converted to the mechanical energy of this object.

Assume that the effect of gravity on this box can also be ignored. That way, there won't be a change in the gravitational potential energy of this object. Hence, all these extra mechanical energy would be in the form of the kinetic energy of this box.

That is:

$\begin{aligned}& \text{Kinetic energy of this object} \\ =& \text{Initial Kinetic Energy} + \text{Change in Kinetic Energy} \\ =& \text{Initial Kinetic Energy} + \text{Change in Mechanical Energy} \\ =& \text{Initial Kinetic Energy} + \text{External Work} \\=& 0\; \rm N \cdot m + 200.06\; \rm N \cdot m \\ =& 200.06\; \rm N \cdot m \end{aligned}$ .

Keep in mind that the kinetic energy of an object of mass $m$ and speed $v$ is:

$\displaystyle \frac{1}{2}\, m \cdot v^{2}$ .

Therefore:

$\begin{aligned}v &= \sqrt{\frac{2\, (\text{Kinetic energy})}{m}} \\ &= \sqrt{\frac{2\times 200.06\; \rm N \cdot m}{6.00\; \rm kg}} \approx 8.17\; \rm m \cdot s^{-1}\end{aligned}$ .

7 0

3 years ago

When using a Microscope's high power objective the course adjustment knob:

kvasek [131]

Answer:

c. Can be used only with scanning

Explanation:

Coarse Adjustment Knob is a large knob on each side of the microscope, with a smaller knob in the middle. This large knob is used to bring an object into close focus as it moves the stage up or down. The coarse-adjustment knob is used ONLY with the scanning or low-power objective lenses.

5 0

3 years ago

Read 2 more answers