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3 years ago

What reasons forced

1 answer:

3 0

Gilbert had assumed that if it were an impact crater then the volume of the crater, as well as meteoritic material, should be present on the rim. Also, a large portion of the meteorite should be buried in the crater and that this would generate a large magnetic anomaly. Hope this helps. Have a nice day.

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If a chemical reaction has a AH value of -125 kJ, it would be considered what?

egoroff_w [7]

exothermic

Explanation:

negative means it's losing energy

4 0

2 years ago

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A simple pendulum is used to determine the acceleration due to gravity at the surface of a planet. The pendulum has a length of

Dafna1 [17]

Answer:

- pendulum use to determine acceleration

Explanation:

by calculating need to understand first the problem

the length is measure to be 2s

the lenght is also 2

2 obtainer on invistigation the most nearly is 8.

5 0

2 years ago

Convert 15 joule into erg.

GuDViN [60]

Answer:

<h2>Joule:</h2>

1 Joule of work is said to be done when a force of 1 Newton is applied to move/displace a body by 1 metre.

1 Joule= 1 Newton × 1 metre

1 Newton is the amount of force required to accelerate body of mass 1 kg by 1m/s²

So units of N is kgm/s²

So,

1 Joule

=1kgm/s² × m

=1kgm²/s²

1 erg is the amount of work done by a force of 1 dyne exerted for a distance of one centimetre.

1 Erg =1 Dyne × 1 cm

1 dyne is the force required to cause a mass of 1 gram to accelerate at a rate of 1cm/s².

1 Erg=1 gmcm/s² × cm

1 Erg=1 gmcm/s² × cm=1gmcm²/s²

this is what you need to convert 1gmcm²/s² to 1kgm²/s²

<h3>what you need to know for conversion</h3>

[1gm=0.001kg

1cm²

=1cm ×1cm

=0.01 m × 0.01 m

=0.0001m²

second remains constant

]

So,

1gmcm²/s²

=0.001kg×0.0001m²/s²

=0.001kg×0.0001m²/s² =0.0000001kgm²/s²

Hence,

<h3>1 Erg=0.0000001 Joule</h3><h3>1 Joule=10,000,000 Erg</h3>

3 0

2 years ago

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Someone please help me will give BRAILIEST!!!!!

Alex777 [14]

Uranium is the right answer. Scientists use 5 percent of the uranium after the bomb is refused to create stronger and better nuclear bombs.

6 0

3 years ago

In the figure, particle A moves along the line y = 31 m with a constant velocity v with arrow of magnitude 2.8 m/s and parallel

insens350 [35]

Answer:

59.26°

Explanation:

Since a is the acceleration of the particle B, the horizontal component of acceleration is a" = asinθ and the vertical component is a' = acosθ where θ angle between a with arrow and the positive direction of the y axis.

Now, for particle B to collide with particle A, it must move vertically the distance between A and B which is y = 31 m in time, t.

Using y = ut + 1/2a't² where u = initial velocity of particle B = 0 m/s, t = time taken for collision, a' = vertical component of particle B's acceleration = acosθ.

So, y = ut + 1/2a't²

y = 0 × t + 1/2(acosθ)t²

y = 0 + 1/2(acosθ)t²

y = 1/2(acosθ)t² (1)

Also, both particles must move the same horizontal distance to collide in time, t.

Let x be the horizontal distance,

x = vt (2)where v = velocity of particle A = 2.8 m/s and t = time for collision

Also, using x = ut + 1/2a"t² where u = initial velocity of particle B = 0 m/s, t = time taken for collision, a" = horizontal component of particle B's acceleration = asinθ.

So, x = ut + 1/2a"t²

x = 0 × t + 1/2(ainsθ)t²

x = 0 + 1/2(asinθ)t²

x = 1/2(asinθ)t² (3)

Equating (2) and (3), we have

vt = 1/2(asinθ)t² (4)

From (1) t = √[2y/(acosθ)]

Substituting t into (4), we have

v√[2y/(acosθ)] = 1/2(asinθ)(√[2y/(acosθ)])²

v√[2y/(acosθ)] = 1/2(asinθ)(2y/(acosθ)

v√[2y/(acosθ)] = ytanθ

√[2y/(acosθ)] = ytanθ/v

squaring both sides, we have

(√[2y/(acosθ)])² = (ytanθ/v)²

2y/acosθ = (ytanθ/v)²

2y/acosθ = y²tan²θ/v²

2/acosθ = ytan²θ/v²

1/cosθ = aytan²θ/2v²

Since 1/cosθ = secθ = √(1 + tan²θ) ⇒ sec²θ = 1 + tan²θ ⇒ tan²θ = sec²θ - 1

secθ = ay(sec²θ - 1)/2v²

2v²secθ = aysec²θ - ay

aysec²θ - 2v²secθ - ay = 0

Let secθ = p

ayp² - 2v²p - ay = 0

Substituting the values of a = 0.35 m/s, y = 31 m and v = 2.8 m/s into the equation, we have

ayp² - 2v²p - ay = 0

0.35 × 31p² - 2 × 2.8²p - 0.35 × 31 = 0

10.85p² - 15.68p - 10.85 = 0

dividing through by 10.85, we have

p² - 1.445p - 1 = 0

Using the quadratic formula to find p,

$p = \frac{-(-1.445) +/- \sqrt{(-1.445)^{2} - 4 X 1 X (-1)}}{2 X 1} \\p = \frac{1.445 +/- \sqrt{2.088 + 4}}{2} \\p = \frac{1.445 +/- \sqrt{6.088}}{2} \\p = \frac{1.445 +/- 2.4675}{2} \\p = \frac{1.445 + 2.4675}{2} or p = \frac{1.445 - 2.4675}{2} \\p = \frac{3.9125}{2} or p = \frac{-1.0225}{2} \\p = 1.95625 or -0.51125$

Since p = secθ

secθ = 1.95625 or secθ = -0.51125

cosθ = 1/1.95625 or cosθ = 1/-0.51125

cosθ = 0.5112 or cosθ = -1.9956

Since -1 ≤ cosθ ≤ 1 we ignore the second value since it is less than -1.

So, cosθ = 0.5112

θ = cos⁻¹(0.5112)

θ = 59.26°

So, the angle between a with arrow and the positive direction of the y axis would result in a collision is 59.26°.

5 0

3 years ago