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2 years ago

Please Help me to find the answer...

Physics

1 answer:

KonstantinChe [14]2 years ago

3 0

Answer:

1. S-S Repulsion N-S Attraction

2. S-N Attraction N-N Repulsion

Explanation:

1. S-S Repulsion N-S Attraction

2. S-N Attraction N-N Repulsion

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What medical technique uses sound wave pulses and their reflections to map structures and organs inside the body?

garri49 [273]

The correct answer is “C” ultrasound. Hope this helps!

7 0

3 years ago

Read 2 more answers

h(t) = - 16t2 + 64t + 112 where t is the time in seconds. After how many seconds does the arrow reach it maximum height? Round t

laila [671]

Answer:

2 seconds

Explanation:

The function of height is given in form of time. For maximum height, we need to use the concept of maxima and minima of differentiation.

$h(t)=-16t^{2}+64t+112$

Differentiate with respect to t on both the sides, we get

$\frac{dh}{dt}=-32t+64$

For maxima and minima, put the value of dh / dt is equal to zero. we get

- 32 t + 64 = 0

t = 2 second

Thus, the arrow reaches at maximum height after 2 seconds.

8 0

3 years ago

An example of acceleration

TEA [102]

Answer:

Explanation:

Newton’s Second Law of Motion says that acceleration (gaining speed) happens when a force acts on a mass (object). Riding your bicycle is a good example of this law of motion at work. Your bicycle is the mass.

5 0

3 years ago

It is known that a shark can travel at a speed of 17 m/s. How far can a shark go in 8 seconds?

o-na [289]

Answer:136 m

Explanation:17*8

6 0

3 years ago

an empty density bottle weighed 23.5 and 48.4g when filled with water .The empty bottle was partially filled with sand until and

gavmur [86]

Answer:

Density of Sand is $2.653g/cm^{3}$ .

Explanation:

Given Empty Density bottle weighs 23.5gm(W=23.5gm)

Weight of bottle when completely filled water=48.4gm

So amount of water required to fill the bottle=Weight of bottle filled with water-W

Amount of water required to fill the bottle( $w_{max}$ )=48.4gm-23.5gm

$w_{max}=24.9$ g

Since we know density of water $d_{w} =1g/cm^{3}$ and $w_{max}$

We can calculate volume of empty space in the bottle(V).

$w_{max}=d_{w}$ $\times$ V

V= $\frac{w_{max} }{d_{w} }$

V= $\frac{24.9}{1}$

V=24.9 $g/cm^{3}$

Now bottle is partially filled with sand,and weight of bottle is ( $w_{s}$ )36.5gm

So,

Amount of sand added ( $m_{s}$ )=36.5-Weight of the bottle

$m_{s}$ =13g

After filling the bottle with water again,the weight of the bottle becomes ( $W_{2}$ =56.5g)

Therefore,

amount of water added to the bottle of sand in grams = $W_{2}$ -36.5gm

amount of water added =56.5g-36.5g

amount of water added =20g

As the density of water = 1g/ $cm^{3}$

Amount of water (in grams )=Volume of water occupied

20=volume of water added

Therfore volume of water added to the sand filled bottle( $V_{w}$ )=20 $cm^{3}$

As we know the total volume of the water bottle(V),

Volume of the sand occupied in the water bottle=V- $V_{w}$

$V_{s}$ =24.9g-20g

$V_{s}$ =4.9g

We know,

Density=Mass/Volume

Therefore,

density of sand = $\frac{m_{s} }{V_{s} }$

density of sand = $\frac{13}{4.9}$

density of sand = $2.653g/cm^{3}$

8 0

3 years ago

Please Help me to find the answer...​

Please Help me to find the answer...