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3 years ago

5

This is a substance that cannot be broken down further by chemical means.

2 answers:

viktelen [127]3 years ago

5 0

The answer is ELEMENT

Serggg [28]3 years ago

4 0

I think its an element

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Apilot of mass 70 kg rides a fighter jet The fighter jet moves in a vertical circle of radius 100 m at a constant

Cerrena [4.2K]

Answer:

the force exerted by the seat on the pilot is 10766.7 N

Explanation:

The computation of the force exerted by the seat on the pilot is as follows:

$F = Mg + \frac{MV^2}{R}\\\\= 70 \times 9.81 + \frac{70 \times 120^2}{100}\\\\= 10766.7 N$

Hence, the force exerted by the seat on the pilot is 10766.7 N

4 0

3 years ago

What must be the length of a simple pendulum if its oscillation frequency is to be equal to that of an air-track glider of mass

Anvisha [2.4K]

Answer:

the length of the simple pendulum is 0.25 m.

Explanation:

Given;

mass of the air-track glider, m = 0.25 kg

spring constant, k = 9.75 N/m

let the length of the simple pendulum = L

let the frequency of the air-track glider which is equal to frequency of simple pendulum = F

The oscillation frequency of air-track glider is calculated as;

$F = \frac{1}{2\pi } \sqrt{\frac{k}{m} } \\\\F = \frac{1}{2\pi } \sqrt{\frac{9.75}{0.25} } \\\\F = 0.994 \ Hz$

The frequency of the simple pendulum is given as;

$F = \frac{1}{2\pi} \sqrt{\frac{g}{l} } \\\\2\pi(F) = \sqrt{\frac{g}{l} } \\\\2\pi (0.994) = \sqrt{\frac{9.8}{l} } \\\\6.2455 = \sqrt{\frac{9.8}{l} } \\\\(6.2455)2 = \frac{9.8}{l} \\\\39.006 = \frac{9.8}{l} \\\\l = \frac{9.8}{39.006} \\\\l = 0.25 \ m$

Thus, the length of the simple pendulum is 0.25 m.

8 0

2 years ago

Find the position vector of a particle that has the given acceleration and the specified initial velocity and position. a(t) = 1

kondor19780726 [428]

Answer:

Explanation:

Given

Acceleration $a(t)=14t\hat{i]+\sin (t)\hat{j}+\cos (2t)\hat{k}$ [/tex]

and $v(0)=\hat{i}$

$r(0)=\hat{j}$

we know $a=\frac{\mathrm{d} v}{\mathrm{d} t}$

$\int dv=\int adt$

$v(t)=\int (14t\hat{i}+\sin (t)\hat{j}+\cos (2t)\hat{k})dt$

$v(t)=7t^2\hat{i}-\cos t\hat{j}+\frac{\sin (2t)\hat{k}}{2}+c$

at $t=0$

$v(0)=0-1\cdot \hat{j}+0+c$

$c=\hat{i}+\hat{j}$

$v(t)=(7t^2+1)\hat{i}+(1-\cos t)\hat{j}+\frac{\sin (2t)\hat{k}}{2}$

and $\frac{\mathrm{d} r}{\mathrm{d} t}=v(t)$

$\int dr=\int vdt$

$r(t)=\int ((7t^2+1)\hat{i}+(1-\cos t)\hat{j}+\frac{\sin (2t)\hat{k}}{2})dt$

$r(t)=(\frac{7}{2}t^3+t)\hat{i}+(t-\sin (t))\hat{j}+\frac{1}{2}\times (-\frac{1}{2}\cos 2t)\hat{k}+c_2$

at $t=0$

$r(0)=\hat{j}$

$r(t)=(\frac{7}{3}t^3+t)\hat{i}+(1+t-\sin t)\hat{j}+\frac{1}{4}(1-\cos 2t)\hat{k}$

4 0

3 years ago

Oooooooooooooooooooooooo

MaRussiya [10]

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8 0

3 years ago

Read 2 more answers

A soccer ball is released from rest at the top of a grassy incline. After 6.2 seconds, the ball travels 47 meters. One second la

alexandr1967 [171]

Answer:

(a) a = 2.44 m/s²

(b) s = 63.24 m

Explanation:

(a)

We will use the second equation of motion here:

$s = v_it+\frac{1}{2}at^2$

where,

s = distance covered = 47 m

vi = initial speed = 0 m/s

t = time taken = 6.2 s

a = acceleration = ?

Therefore,

$47\ m = (0\ m/s)(6.2\ s)+\frac{1}{2}a(6.2\ s)^2\\\\a = \frac{2(47\ m)}{(6.2\ s)^2}$

<u>a = 2.44 m/s²</u>

<u></u>

(b)

Now, we will again use the second equation of motion for the complete length of the inclined plane:

$s = v_it+\frac{1}{2}at^2$

where,

s = distance covered = ?

vi = initial speed = 0 m/s

t = time taken = 7.2 s

a = acceleration = 2.44 m/s²

Therefore,

$s = (0\ m/s)(6.2\ s)+\frac{1}{2}(2.44\ m/s^2)(7.2\ s)^2\\\\$

<u>s = 63.24 m</u>

6 0

3 years ago