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sp2606 [1]
2 years ago
9

How much magnification is needed to see a nanometer?.

Physics
1 answer:
ArbitrLikvidat [17]2 years ago
6 0

20,000,000



To see things at a nanometer, which is a trillionth of a meter, you would need to increase magnification nearly 20,000,000 times those are ionizing atoms
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A fast teaub abd a ski train driving from city a and city b face to face the fast train has a speed of 76km/h, the slow train ha
LekaFEV [45]

The distance between the two cities is 513.24 km.

<h3>Time of motion when the two trains meet</h3>

The time spent on the journey when the two trains meet is calculated as follows;

(Va - Vb)t = d

where;

  • d is the distance between the trains before meeting

(76 - 65)t = 40

11t = 40

t = 40/11

t = 3.64 hr

<h3>Distance traveled by the fast train</h3>

d1 = 76 km/h x 3.64 h

d1 = 276.64 km

<h3>Distance traveled by the slow train</h3>

d2 = 65 km/h x 3.64 h

d2 = 236.6 km

The distance between the two cities = 276.64 km + 236.6 km

                                                             = 513.24 km

Learn more about relative velocity here: brainly.com/question/17228388

7 0
2 years ago
Displacement vectors of 4 km north, 2km south, 5 km north, and 5km south combine to a total displacement of A 2 km south B 16 km
prohojiy [21]
I would prolly say C. 16 km south.
4 0
3 years ago
Read 2 more answers
An attack helicopter is equipped with a 20- mm cannon that fires 87 g shells in the forward direction with a muzzle speed of 853
uysha [10]

Answer

given,

mass of the shell = 87 g = 0.087 Kg

speed of the muzzle = 853 m/s

mass of the helicopter = 4410 kg

A burst of 176 shell fired in 2.93 s

resulting average force = ?

momentum of the shell = m v

                                       = 0.087 x 853

                                       = 74.21 kgm/s

momentum of 176 shell is = 176 p

                                          = 176 x 74.21

                                          = 13060.96

momentum of helicopter = - 13060.96 kgm/s

amount of speed reduce a = \dfrac{13060.96}{M}

                                          a= \dfrac{13060.96}{4410}

                                          a = 2.96 m/s²

velocity  = \dfrac{2.96}{2.93}

      v = 1.01 m/s

5 0
3 years ago
Read 2 more answers
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
9
laila [671]

Answer:

d

Explanation:

A dreams

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4 0
2 years ago
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