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3 years ago

If there is a gravitational force between all objects, why do we not feel or observe it?

Physics

2 answers:

Dmitry [639]3 years ago

4 0

We do not feel the gravitational forces from objects other than the Earth because they are weak.

Rudik [331]3 years ago

4 0

Answer:

We do. It's just way too small compared to the force between the objects we're observing and the earth. It's like looking inside a room with an elephant and a grain of rice. The rice is there, it's just too small compared to the room and the elephant inside for you to notice it. Or, if you ever traveled on a plane, you can easily see the towns, or the roads, but not the single people walking the street.

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At a depth of 10.9 km, the Challenger Deep in the Marianas Trench of the Pacific Ocean is the deepest site in any ocean. Yet, in

bearhunter [10]

Answer:

$P = 1.09 \times 10^8 Pa$

Explanation:

As we know that the pressure inside the liquid level is given as

$P = \rho g h + P_o$

here we have

$\rho = 1024 kg/m^3$

h = 10.9 km

also we know that

$P_o = 1.01 \times 10^5 Pa$

now we have

$P = (1.01 \times 10^5) + (1024)(9.81)(10.9 \times 10^3)$

$P = 1.09 \times 10^8 Pa$

8 0

3 years ago

The initial kinetic energy imparted to a 0.020 kg bullet is 1200 J. (a) Assuming it

Lilit [14]

Answer:

(a) Power= 207.97 kW

(b) Range= 5768.6 meter

Explanation:

Given,

Mass of bullet, $m=0.02 kg$

Kinetic energy imparted, $K=1200 J$

Length of rifle barrel, $d=1 m$

(a)

Let the speed of bullet when it leaves the barrel is $v$ .

Kinetic energy, $K=\frac{1}{2} mv^{2}$

$v=\sqrt{\frac{2K}{m} }$

$=\sqrt{\frac{2\times1200}{0.02} }$

$=346.4m/s$

Initial speed of bullet, $u=0$

The average speed in the barrel, $v_a_v_g=\frac{u+v}{2}$

$=\frac{0+346.4}{2} \\=173.2 m/s$

Time taken by bullet to cross the barrel, $t=\frac{d}{v}$

$=\frac{1}{173.2}\\ =0.00577 second$

Power, $P_a_v_g=\frac{W}{t}$

$=\frac{1200}{0.00577} \\=207.97kW$

(b)

In projectile motion,

Maximum height, $H_m=\frac{v^2\sin^2\theta}{2g} \\$

Range, $R=\frac{v^2\sin2\theta}{g}$

given that, $H_m=R$

then, $\frac{v^2\sin^2\theta}{2g}=\frac{v^2\sin2\theta}{g}\\\sin^2\theta=2\sin\theta\cos\theta\\\\\tan\theta=4\\\theta=\tan^-^14\\\theta=75.96^0\\R=\frac{v^2\sin2\theta}{g}\\=\frac{346.4^2\times\sin(2\times75.96)}{9.8}\\5768.6 meter$