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4 years ago

3. Can there be forces acting on an object at rest? Explain why or why not.

1 answer:

3 0

Yes, because gravity always exerts a force whether an object is at rest or in motion. Gravity is the reason we aren’t all floating as cool as that would be we would probably be hit by a plane

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How is Newton’s third law of motion involved when you jump straight upward?

Bad White [126]

<span>Newtons 3rd law of motion states that action and reaction are equal and opposite. How it applies in this statement is that when you jump straight upward you hit the floor and you experience an equal magnitude of force against your feet, because force always acts as a result of two objects and whenever forces occur they do so in pairs of equal magnitude acting in opposite directions. I hope this helps!</span><span>
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3 0

3 years ago

Read 2 more answers

What can make an inclined plane more efficient?

Sophie [7]

Answer:

Efficiency can be increase by using rollers in conjunction with the inclined plane. Wedge. The wedge is an adaptation of the inclined plane. It can be used to raise a heavy load over a short distance or to split a log

3 0

3 years ago

a rocket burns propellant at a rate of dm/dt = 3.0 kg/s, ejecting gases with a speed of 8000 m/s relative to the rocket. Find th

elena-s [515]

Answer: 24 kN

Explanation:

Given

The rocket burns propellant at the rate of

$\dfrac{dm}{dt}=3\ kg/s$

Relative ejection of gases $v=8000\ m/s$

The magnitude of thrust force is given by

$F_t=v\dfrac{dm}{dt}\\\\F_t=8000\times 3=24,000\ N\ or\ 24\ kN$

5 0

3 years ago

The two speakers at S1 and S2 are adjusted so that the observer at O hears an intensity of 6 W/m² when either S1 or S2 is sounde

Zanzabum

Answer:

The minimum frequency is 702.22 Hz

Explanation:

The two speakers are adjusted as attached in the figure. From the given data we know that

$S_1 S_2$ =3m

$S_1 O$ =4m

By Pythagoras theorem

$S_2O=\sqrt{(S_1S_2)^2+(S_1O)^2}\\S_2O=\sqrt{(3)^2+(4)^2}\\S_2O=\sqrt{9+16}\\S_2O=\sqrt{25}\\S_2O=5m$

Now

The intensity at O when both speakers are on is given by

$I=4I_1 cos^2(\pi \frac{\delta}{\lambda})$

Here

I is the intensity at O when both speakers are on which is given as 6 $W/m^2$

I1 is the intensity of one speaker on which is 6 $W/m^2$

δ is the Path difference which is given as

$\delta=S_2O-S_1O\\\delta=5-4\\\delta=1 m$

λ is wavelength which is given as

$\lambda=\frac{v}{f}$

Here

v is the speed of sound which is 320 m/s.

f is the frequency of the sound which is to be calculated.

$16=4\times 6 \times cos^2(\pi \frac{1 \times f}{320})\\16/24= cos^2(\pi \frac{1f}{320})\\0.667= cos^2(\pi \frac{f}{320})\\cos(\pi \frac{f}{320})=\pm0.8165\\\pi \frac{f}{320}=\frac{7 \pi}{36}+2k\pi \\ \frac{f}{320}=\frac{7 }{36}+2k \\\\ {f}=320 \times (\frac{7 }{36}+2k )$

where k=0,1,2

for minimum frequency $f_1$ , k=1

${f}=320 \times (\frac{7 }{36}+2 \times 1 )\\\\{f}=320 \times (\frac{79 }{36} )\\\\ f=702.22 Hz$

So the minimum frequency is 702.22 Hz

5 0

3 years ago

julia throws a ball vertically upward from the ground with a speed of 5.89m/s. Andrew catches it when it is on its way down at a

aliya0001 [1]

Vo = 5.89 m/s Y = 1.27 m g = 9.81 m/s^2
Time to height
Tr = Vo / g Tr = (5.89 m/s) / (9.81 m/s^2) Tr = 0.60 s
Max height achieved is:
H = Vo^2 / [2g] H = (5.89 )^2 / [ 2 * (9.81) ] H = (34.69) / [19.62] H = 1.77 m
It falls that distance, minus Andrew's catch distance:
h = H - Y h = (1.77 m) - (1.27 m) h = 0.5 m
Time to descend is therefore:
Tf = √ { [2h] / g ] Tf = √ { [ 2 * (0.5 m) ] / (9.81 m/s^2) } Tf = √ { [ 1.0 m ] / (9.81 m/s^2) } Tf = √ { 0.102 s^2 } Tf = 0.32 s
Total time is rise plus fall therefore:
Tt = Tr + Tf Tt = (0.60 s) + (0.32 s) Tt = 0.92 s (ANSWER)

8 0

4 years ago