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3 years ago

What makes a grenade explode?

1 answer:

6 0

The striker throws the safety lever away from the grenade body as it rotates to detonate the primer. The primer explodes and ignites the fuse The fuse burns down to the detonator, which explodes the main charge

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A rectangular tank is filled to a depth of 10m with freshwater and open to air at atmospheric pressure.

charle [14.2K]

Answer:

1. 39068.07 N

2. 19534.036 N

Explanation:

depth of water h = 10 m

atmospheric pressure Patm = 101325 Pa

density of water p = 1000 kg/m^3

acceleration due to gravity g = 9.81 m/s^2

pressure due to depth of water = pgh

P = 1000 x 9.81 x 10 = 98100 Pa

total pressure on the bottom of the tank is Patm + p = 101325 + 98100 = 199425 Pa

Left plug has diameter = 50 cm = 0.5 m

radius = 0.5/2 = 0.25 m

height = 1 cm = 0.01 m

height below tank surface = 10 - 0.01 = 9.99

pressure at this depth = 1000 x 9.81 x 9.99 = 98001.9 Pa

total pressure = Patm + P = 101325 + 98001.9 = 199326.9 Pa

surface area of plug = π $r^{2}$ = 3.142 x $0.25^{2}$ = 0.196 $m^{2}$

force required to lift left plug = pressure x area

F = 199326.9 x 0.196 = 39068.07 N

The right side is a hemisphere with the same diameter, therefore surface area is half of the left plug

A = 0.196/2 = 0.098 $m^{2}$

force F required to lift right plug = 199326.9 x 0.098 = 19534.036 N

6 0

3 years ago

Approximately how many atoms are there along a 9.5 cm line

uranmaximum [27]

First convert 5.5 cm to meters.

(5.5 cm / 1) x (m / 100 cm) = 0.055 m

A typical atom is about 1.0E-10 m in diameter; thus:

0.055 m / 1.0E-10 m = 5.5E8 atoms or 550,000,000 end-to-end atoms in 5.5 cm

4 0

3 years ago

The Hubble Space Telescope has an aperture of 2.4 m and focuses visible light (400-700 nm). The Arecibo radio telescope in Puert

stealth61 [152]

Answer:

$y_{hubble} = 77\ \ m$

$y_{aceribo} = 1.1*10^6 \ \ m$

Explanation:

what is the smallest crater that each of these telescopes could resolve on our moon?

For moon ;

s = 3.8 × 10 ⁸ m

y = 1.22 λs/D

where;

λ = 400 nm = 400× 10 ⁻⁹

D = 2.4 m

The smallest crater for the hubble space is calculated as follows:

$y_{hubble} = 1.22*400*10^{-9}*3.8*10^8/2.4$

$y_{hubble} = 77\ \ m$

For Aceribo ;

y = 1.22 λs/D

where :

λ = 75 cm = 0.75 m

D = 305 m

$y_{acerbo} = 1.22*0.75 *3.8*10^8/305$

$y_{aceribo} = 1.1*10^6 \ \ m$

5 0

3 years ago

Just want a ride with

yaroslaw [1]

Answer:

My best friend lol cuz since quarantine i didn't see her

8 0

3 years ago

( timed for this!! please help!!) Which best defines scientific question?

Viefleur [7K]

My opinion, the answer is b

5 0

3 years ago