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velikii [3]
2 years ago
5

on 28 may _______ at Chaghi, Balochistan; Dr A.Q Khan commanded to perform six successful nuclear explsions.

Physics
1 answer:
KATRIN_1 [288]2 years ago
8 0
I think the answer would be 1998
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Twist-on connectors without the spring-steel coils (plastic threads only) are suitable for making branch-circuit connections.
Gnoma [55]

Answer:

if it is a plastic connector it wont work but if there is metal or steel it will work

Explanation:

6 0
3 years ago
30cm³ of brine of relative density 1.15 and 42cm³ of water are mixed. What is the density of the final solution​
Aleks04 [339]

Answer:

I think it's the most important part in this

7 0
3 years ago
A vertical spring gun is used to launch balls into the air. If the spring is compressed by 4.9 cm, the ball of mass 5.5 g is lau
AleksandrR [38]

We know, by conservation of energy :

\dfrac{kx^2}{2}=mgh

Therefore,

\dfrac{x_1^2}{x_2^2}=\dfrac{h_1}{h_2}

Putting given values, we get :

\dfrac{x_1^2}{x_2^2}=\dfrac{h_1}{h_2}\\\\\dfrac{4.9^2}{x_2^2}=\dfrac{50.2}{2\times 50.2}\\\\x_2^2=2\times 4.9^2\\\\x_2 = 4.9\times \sqrt{2}\\\\x_2=6.93\ cm

Therefore, the spring be compressed to 6.93 cm to send the ball twice as high.

Hence, this is the required solution.

6 0
2 years ago
A bag of sugar weighs 3.50 lb on Earth. What would it weigh in newtons on the Moon, where the free-fall acceleration isone-sixth
Alex73 [517]

Answer:

F_{Earth}= 15.57 N

F_{Moon}= 2.60 N

F_{Uranus}= 16.98 N

The mass of the bag is the same on the three planets. m=1.59 kg

Explanation:

The weight of the sugar bag on Earth is:

g=9.81 m/s²

m=3.50 lb=1.59 kg

F_{Earth}=m·g=1.59 kg×9.81 m/s²= 15.57 N

The weight of the sugar bag on the Moon is:

g=9.81 m/s²÷6= 1.635 m/s²

F_{Moon}=m·g=1.59 kg× 1.635 m/s²= 2.60 N

The weight of the sugar bag on the Uranus is:

g=9.81 m/s²×1.09=10.69 m/s²

F_{Uranus}=m·g=1.59 kg×10.69 m/s²= 16.98 N

The mass of the bag is the same on the three planets. m=1.59 kg

5 0
2 years ago
Two identical 7.10-gg metal spheres (small enough to be treated as particles) are hung from separate 700-mmmm strings attached t
nlexa [21]

Answer:

Explanation:

Let m be mass of each sphere and θ be angle, string makes with vertex in equilibrium.

Let T be tension in the hanging string

T cosθ = mg ( for balancing in vertical direction )

for balancing in horizontal direction

Tsinθ = F ( F is force of repulsion between two charges sphere)

Dividing the two equations

Tanθ = F / mg

tan17 = F / (7.1 x 10⁻³ x 9.8)

F = 21.27 x 10⁻³ N

if q be charge on each sphere , force of repulsion between the two

F = k q x q / r² ( r is distance between two sphere , r = 2 x .7 x sin17  = .41 m )

21.27 x 10⁻³  = (9 X 10⁹ x q²) / .41²

q² = .3973 x 10⁻¹²

q = .63 x 10⁻⁶ C

no of electrons required  = q / charge on a single electron

= .63 x 10⁻⁶ / 1.6 x 10⁻¹⁹

= .39375 x 10¹³

3.9375 x 10¹² .

4 0
3 years ago
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