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3 years ago

You need to put a metal rod

1 answer:

8 0

-- Put the rod into the freezer for a while. As it cools,
it contracts (gets smaller) slightly.

-- Put the cylinder into hot hot water for a while. As it heats,
it expands (gets bigger) slightly.

-- Bring the rod and the cylinder togther quickly, before the
rod has a chance to warm up or the cylinder has a chance
to cool off.

-- I bet it'll fit now.

-- But be careful . . . get the rod exactly where you want it as fast
as you can. Once both pieces come back to the same temperature,
and the rod expands a little and the cylinder contracts a little, the fit
will be so tight that you'll probably never get them apart again, or even
move the rod.

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Someone please help me

Lapatulllka [165]

False... I hope that helps ;)

6 0

3 years ago

A hunter aims at a deer which is 40 yards away. Her cross- bow is at a height of 5ft, and she aims for a spot on the deer 4ft ab

shutvik [7]

Answer:

a) θ₁ = 0.487º , b) t = 0.400 s , x = 11.73 ft

Explanation:

For this exercise let's use the projectile launch relationships.

The initial height is I = 5 ft and the final height y = 4 ft

y = y₀ + $v_{oy}$ t - ½ g t²

The distance to the band is x = 40 yard (3 ft / 1 yard) = 120 ft

x = v₀ₓ t

t = x / v₀ₓ

We replace

y –y₀ = v_{oy} x / v₀ₓ - ½ g x² / v₀ₓ²

v_{oy} = v₀ sin θ

v₀ₓ = vo cos θ

y –y₀ = x tan θ - ½ g x² / v₀² cos² θ

5-4 = 120 tan θ - ½ 32 120 / (300 2 cos2 θ)

1 = 120 tan θ - 0.0213 sec² θ

Let's use the trigonometry relationship

Sec² θ = 1 - tan² θ

1 = 120 tan θ - 0.0213 (1 –tan²θ)

0.0213 tan²θ + 120 tanθ -1.0213 = 0

We change variables

u = tan θ

u² + 5633.8 u - 48.03 = 0

We solve the second degree equation

u = [-5633.8 ±√(5633.8 2 + 4 48.03)] / 2

u = [- 5633.8 ± 5633.82] / 2

u₁ = 0.0085

u₂= -5633.81

u = tan θ

θ = tan⁻¹ u

For u₁

θ₁ = tan⁻¹ 0.0085

θ₁ = 0.487º

For u₂

θ₂ = -89.99º

The launch angle must be 0.487º

b) let's look for the time it takes for the arrow to arrive

x = v₀ₓ t

t = x / v₀ cos θ

t = 120 / (300 cos 0.487)

t = 0.400 s

The deer must be at a distance of

v = 20 mph (5280 ft / 1 mi) (1 h / 3600s) = 29.33 ft / s

x = v t

x = 29.33 0.4

x = 11.73 ft

3 0

3 years ago

An object is dropped and is in free fall. Each second, the position of the object is marked. The distance between each mark is m

navik [9.2K]

C. hope this helps :)

8 0

3 years ago

Two resistors are to be combined in parallelto form an equivalent resistance of 400Ω. The resistors are takenfrom available stoc

vivado [14]

Answer:

Δ $R_{e}$ = 84 Ω, $R_{e}$ = (40 ± 8) 10¹ Ω

Explanation:

The formula for parallel equivalent resistance is

1 / $R_{e}$ = ∑ 1 / Ri

In our case we use a resistance of each

R₁ = 500 ± 50 Ω

R₂ = 2000 ± 5%

This percentage equals

0.05 = ΔR₂ / R₂

ΔR₂ = 0.05 R₂

ΔR₂ = 0.05 2000 = 100 Ω

We write the resistance

R₂ = 2000 ± 100 Ω

We apply the initial formula

1 / $R_{e}$ = 1 / R₁ + 1 / R₂

1 / $R_{e}$ = 1/500 + 1/2000 = 0.0025

$R_{e}$ = 400 Ω

Let's look for the error (uncertainly) of Re

$R_{e}$ = R₁R₂ / (R₁ + R₂)

R’= R₁ + R₂

$R_{e}$ = R₁R₂ / R’

Let's look for the uncertainty of this equation

Δ $R_{e}$ / $R_{e}$ = ΔR₁ / R₁ + ΔR₂ / R₂ + ΔR’/ R’

The uncertainty of a sum is

ΔR’= ΔR₁ + ΔR₂

We substitute the values

Δ $R_{e}$ / 400 = 50/500 + 100/2000 + (50 +100) / (500 + 2000)

Δ $R_{e}$ / 400 = 0.1 + 0.05 + 0.06

Δ $R_{e}$ = 0.21 400

Δ $R_{e}$ = 84 Ω

Let's write the resistance value with the correct significant figures

$R_{e}$ = (40 ± 8) 10¹ Ω

6 0

3 years ago

Please Help! Multiple Choice!

attashe74 [19]

Option A 3......
... ........

8 0

3 years ago