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3 years ago

1. Since sleep is so important, we might wonder why people so often fail to get a sufficient amount

Physics

1 answer:

ziro4ka [17]3 years ago

8 0

Answer: 9/10

Explanation:

because it's really important and makes you energetic

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When tuning a guitar, by comparing the frequency of a string that is struck against a standard sound source (of known frequency)

lapo4ka [179]

Bdjsbevevfbfcnbfbffbs s kdkfc

5 0

3 years ago

Consider a situation where the acceleration of an object is always directed perpendicular to its velocity. This means that

Bond [772]

Answer:

this situation would not be physically possible

7 0

3 years ago

A horizontal spring with spring constant 950 N/m is attached to a wall. An athlete presses against the free end of the spring, c

OLga [1]

Answer:

66.5N

Explanation:

F = kx

Where F = force

K = spring constant

x = compression

Given

K = 950N/m

x = 7.0cm

F = ?

First convert the compression to meters .

7.0cm = 7.0 x 0.01

= 0.07 meters

Therefore

F = 950 x 0.07

= 66.5N

4 0

3 years ago

How much force is needed to keep the 750000 Newton Space Shuttle moving at a constant speed of 28000 km/h, in a straight line?

Alika [10]

The force needed to keep the space shuttle moving at constant speed is 0.

The given parameters;

<em>weight of the space shuttle, F = 750,000 N</em>
<em>constant speed of the space shuttle, v = 28,000 km/h</em>

The mass of the space shuttle is calculated as follows;

$W = mg\\\\m = \frac{W}{g} \\\\m = \frac{750,000}{9.8} \\\\m = 76,530.61 \ kg$

The force needed to keep the space shuttle moving at constant speed is calculated as follows;

$F = ma$

$F = 76,530.61 \times a$

where;

a is the acceleration of the space shuttle

At a constant speed, acceleration is zero.

F = 76,530.61 x 0

F = 0

Thus, the force needed to keep the space shuttle moving at constant speed is 0.

Learn more here:brainly.com/question/16374764

6 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