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3 years ago

What the answer for my amswer get a brainly list plessss

Physics

1 answer:

s344n2d4d5 [400]3 years ago

4 0

Answer:

Explanation:

to simplify let's make it apart the denominator into three part

$\frac{ {3xy}^{2} }{ {x}^{2} {y}^{2} } - \frac{ {8x}^{2} y}{ {x}^{2} {y}^{2} } + \frac{ {5x}^{2} {y}^{2} }{ {x}^{2} {y}^{2} }$

now let's simplify

$\frac{3}{x} - \frac{8}{y} + 5$

so the answer is A

hope it helps

for any question comment me ❤❤

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Two balloons are charged with an identical quantity and type of charge: -6.25 nanoCoulombs (10-9). They are held apart at a sepa

xenn [34]

Answer:

force between both the balloons is $9.23\times 10^{-7}N$

Explanation:

It is given charge on both balloons are same and equal to $q=-6.25\times 10^{-9}C$

Separation between both the balloons $r=61.7cm=0.617m$

According to Coulomb's law force between two charges is given by

$F=\frac{1}{4\pi \epsilon _0}\frac{q^2}{r^2}=\frac{Kq^2}{r^2}$

Therefore force is equal to

$F=\frac{9\times 10^9\times (-6.25\times 10^{-9})^2}{0.617^2}$

$F=9.23\times 10^{-7}N$

So force between both the balloons is $9.23\times 10^{-7}N$

3 0

3 years ago

A body moving with an initial velocity of 30m/s accelerates uniformly at the rate of 10m/s . what is the distance covered during

nikdorinn [45]

Answer:

The distance covered by the body is, S = 800 m

Explanation:

Given data,

The initial velocity of the body, u = 30 m/s

The acceleration of the body, a = 10 m/s²

Let the time period of travel be, t = 10 s

Using the II equations of motion,

S = ut + ½ at²

Substituting the given values,

S = 30 x 10 + ½ x 10 x 10²

S = 800 m

Hence, the distance covered by the body is, S = 800 m

5 0

3 years ago

How do clouds become stars ?

kenny6666 [7]

Stars form inside relatively dense concenstrations of interstellar gas and dust known as molecular clouds.

hope it helps

3 0

3 years ago

Read 2 more answers

A barrel 1 m tall and 60 cm in diameter is filled to the top with water. What is the pressure it exerts on the floor beneath it?

allsm [11]

Answer:

The pressure on the ground is about 9779.5 Pascal.

The pressure can be reduced by distributing the weight over a larger area using, for example, a thin plate with an area larger than the circular area of the barrel's bottom side. See more details further below.

Explanation:

Start with the formula for pressure

(pressure P) = (Force F) / (Area A)

In order to determine the pressure the barrel exerts on the floor area, we need the calculate the its weight first

$F_g = m \cdot g$

where m is the mass of the barrel and g the gravitational acceleration. We can estimate this mass using the volume of a cylinder with radius 30 cm and height 1m, the density of the water, and the assumption that the container mass is negligible:

$V = h\pi r^2=1m \cdot \pi\cdot 0.3^2 m^2\approx 0.283m^3$

The density of water is 997 kg/m^3, so the mass of the barrel is:

$m = V\cdot \rho = 0.283 m^3 \cdot 997 \frac{kg}{m^3}= 282.151kg$

and so the weight is

$F_g = 282.151kg\cdot 9.8\frac{m}{s^2}=2765.08N$

and so the pressure is

$P = \frac{F}{A} = \frac{F}{\pi r^2}= \frac{2765.08N}{\pi \cdot 0.3^2 m^2}\approx 9779.5 Pa$

This answers the first part of the question.

The second part of the question asks for ways to reduce the above pressure without changing the amount of water. Since the pressure is directly proportional to the weight (determined by the water) and indirectly proportional to the area, changing the area offers itself here. Specifically, we could insert a thin plate (of negligible additional weight) to spread the weight of the barrel over a larger area. Alternatively, the barrel could be reshaped (if this is allowed) into one with a larger diameter (and smaller height), which would achieve a reduction of the pressure.

7 0

3 years ago

An amateur player is about to throw a dart with an initial velocity of 15 meters/second onto a dartboard that is at a distance o

Minchanka [31]

Answer:

B. 0.16 m

Explanation:

The vertical distance by which the player will miss the target is equal to the vertical distance covered by the dart during its motion.

Since the dart is thrown horizontally, the initial vertical velocity is zero:

$v_y = 0$

While the horizontal velocity is

$v_x = 15 m/s$

The horizontal distance covered is

$d_x = 2.7 m$

Since the dart moves by uniform motion along the horizontal direction, the time it takes for covering this distance is

$t=\frac{d_x}{v_x}=\frac{2.7 m}{15 m/s}=0.18 s$

along the vertical direction, the motion is a uniformly accelerated motion with constant downward acceleration g=9.8 m/s^2, so the vertical distance covered is given by

$d_y = \frac{1}{2}gt^2=\frac{1}{2}(9.8 m/s^)(0.18 s)^2=0.16 m$

8 0

4 years ago