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3 years ago

Do you think potassium is more or less reactive then sodium?

Physics

2 answers:

Anna35 [415]3 years ago

3 0

Answer:

Potassium is more reactive

Explanation:

Artist 52 [7]3 years ago

3 0

Answer:

more reactive

Explanation:

In potassium, the outermost electron is better shielded from the attractive force of the nucleus.

So potassium can be converted to ionic form more readily than sodium. Hence, potassium is more reactive than sodium.

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Suppose a log's mass is 5 kg. After burning, the mass of the ash is 1 kg. explain what May have happened to the other 4 kg.

ladessa [460]

The other 4 kg of mass may have departed the scene
of the fire, in the form of gases and smoke particles.

7 0

3 years ago

A whistling sound that is audible when a hearing aid is held in the hand with the power on and the volume high indicates that th

zzz [600]

The whistling sound from the hearing aids represents that your hearing aids is working perfectly ad is known as the "feedback". So, the given statement is true.

Answer: Option A

Explanation:

It's often sounds irritating when a hearing aids of your grandpa or Grandma whistles. especially, when they put them out of their ears. Actually, this feedback sound from hearing aids occur when the sounds from the outer side bounces back to the microphone of the hearing aids.

The sound bounces back when it doesn't gets inside of your ear canal so that one can hear the sound through the hearing aid. When the sounds bounces back in the hearing aid, it get re-amplified and thus we hear the whistle sound which is known as the feedback of the device.

It's not always the feedback sound though. Sometimes the device whistles when it has some mechanical defect or when one hugs the other one or water gets inside and damaged the whole system.

7 0

3 years ago

A baseball player hits a homerun, and the ball lands in the left field seats, which is 103m away from the point at which the bal

Sati [7]

(a) The ball has a final velocity vector

$\mathbf v_f=v_{x,f}\,\mathbf i+v_{y,f}\,\mathbf j$

with horizontal and vertical components, respectively,

$v_{x,f}=\left(20.5\dfrac{\rm m}{\rm s}\right)\cos(-38^\circ)\approx16.2\dfrac{\rm m}{\rm s}$

$v_{y,f}=\left(20.5\dfrac{\rm m}{\rm s}\right)\sin(-38^\circ)\approx-12.6\dfrac{\rm m}{\rm s}$

The horizontal component of the ball's velocity is constant throughout its trajectory, so $v_{x,i}=v_{x,f}$ , and the horizontal distance x that it covers after time t is

$x=v_{x,i}t=v_{x,f}t$

It lands 103 m away from where it's hit, so we can determine the time it it spends in the air:

$103\,\mathrm m=\left(16.2\dfrac{\rm m}{\rm s}\right)t\implies t\approx6.38\,\mathrm s$

The vertical component of the ball's velocity at time t is

$v_{y,f}=v_{y,i}-gt$

where g = 9.80 m/s² is the magnitude of the acceleration due to gravity. Solve for the vertical component of the initial velocity:

$-12.6\dfrac{\rm m}{\rm s}=v_{y,i}-\left(9.80\dfrac{\rm m}{\mathrm s^2}\right)(6.38\,\mathrm s)\implies v_{y,i}\approx49.9\dfrac{\rm m}{\rm s}$

So, the initial velocity vector is

$\mathbf v_i=v_{x,i}\,\mathbf i+v_{y,i}\,\mathbf j=\left(16.2\dfrac{\rm m}{\rm s}\right)\,\mathbf i+\left(49.9\dfrac{\rm m}{\rm s}\right)\,\mathbf j$

which carries an initial speed of

$\|\mathbf v_i\|=\sqrt{{v_{x,i}}^2+{v_{y,i}}^2}\approx\boxed{52.4\dfrac{\rm m}{\rm s}}$

and direction θ such that

$\tan\theta=\dfrac{v_{y,i}}{v_{x,i}}\implies\theta\approx\boxed{72.0^\circ}$

(b) I assume you're supposed to find the height of the ball when it lands in the seats. The ball's height y at time t is

$y=v_{y,i}t-\dfrac12gt^2$

so that when it lands in the seats at t ≈ 6.38 s, it has a height of

$y=\left(49.9\dfrac{\rm m}{\rm s}\right)(6.38\,\mathrm s)-\dfrac12\left(9.80\dfrac{\rm m}{\mathrm s^2}\right)(6.38\,\mathrm s)^2\approx\boxed{119\,\mathrm m}$

6 0

3 years ago

Someone please help me answer these

nekit [7.7K]

1. GPE

2. KE

3. KE

4. KE

5. Both

6. Both

7. Neither

8. Neither

Alright I think these should be right ;)

4 0

3 years ago

A boat, whose speed in still water is 8.0 m/s, crosses a river with a current of 6.0 m/s. If the boat heads perpendicular to the

kirill [66]

Answer: D

Rs = 10.0 m/s

The speed of the boat relative to an observer standing on the shore as it crosses the river is 10.0m/s

Explanation:

Since the boat is moving perpendicular to the current of the river, the speed of the boat has two components.

i. 8.0m/s in the direction perpendicular to the current

ii. 6.0m/s in the direction of the current.

So, the resultant speed can be derived by using the equation;

Rs = √(Rx^2 + Ry^2)

Taking

Ry = 8.0m/s

Rx = 6.0m/s

Substituting into the equation, we have;

Rs = √(6.0^2 + 8.0^2)

Rs = √(36+64) = √100

Rs = 10.0 m/s

The speed of the boat relative to an observer standing on the shore as it crosses the river is 10.0m/s

3 0

3 years ago

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