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valkas [14]
3 years ago
15

A combination of overplowing and drought caused the Dust Bowl. True or False.

Physics
2 answers:
slava [35]3 years ago
7 0
The answer would be true
OLga [1]3 years ago
6 0
TRUE>>>>>>>> brainliest plz
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How many miles does hubble circle above our planet?
Lyrx [107]
547 kilometres, or 340 miles
6 0
3 years ago
Can someone list 6 advantages of renewable energy.
igor_vitrenko [27]

Answer:

Solar energy,  Wind energy,  Hydro energy, Tidal energy, Geothermal energy,  Biomass Energy.

Explanation:

I hope that it helps you...

4 0
3 years ago
You need to focus a 10 mW, 632.8 nm Gaussian laser beam that is 5.0 mm in diameter into a sample. You have access to a lens with
Anna [14]

Answer:

ee that the lens with the shortest focal length has a smaller object

           

Explanation:

For this exercise we use the constructor equation or Gaussian equation

        \frac{1}{f}  = \frac{1}{p} + \frac{1}{q}

where f is the focal length, p and q are the distance to the object and the image respectively.

Magnification a lens system is

          m = \frac{h'}{h} = - \frac{q}{p}

             h ’= -\frac{h q}{p}

In the exercise give the value of the height of the object h = 0.50cm and the position of the object p =∞

Let's calculate the distance to the image for each lens

f = 6.0 cm

           \frac{1}{q} = \frac{1}{f }  - \frac{1}{p}

as they indicate that the light fills the entire lens, this indicates that the object is at infinity, remember that the light of the laser rays is almost parallel, therefore p = inf

          q = f = 6.0 cm

for the lens of f = 12.0 cm q = 12.0 cn

to find the size of the image we use

           h ’= h q / p

where p has a high value and is the same for all systems

           h ’= h / p q

Thus

f = 6 cm h ’= fo 6 cm

 

f = 12 cm h ’= fo 12  cm

therefore we see that the lens with the shortest focal length has a smaller object

8 0
3 years ago
These are example of foot and leg protection
Pavel [41]
Casts? When you get injured ?
6 0
4 years ago
On a one lane road, a person driving a car at v1 = 54 mi/h suddenly notices a truck 2.4 mi in front of him. That truck is moving
Tomtit [17]

Answer:

a) Δx₂ = 31*Δt

b) Δx₁ = 977.5 / a

c) a = 23 / Δt

e) Δx₁ = 42.5*Δt

g) Δt = 0.0565 h

i) a = 0.05 m/s²

Explanation:

Given

v₁ = 54 Mi/h

v₂ = 31 Mi/h

a)  We apply the formula

Δx₂ = v₂*Δt

⇒  Δx₂ = 31*Δt  (Assuming constant speed)

b) We use the formula

v₂² = v₁² - 2*a*Δx₁    ⇒   Δx₁ = (v₁² - v₂²) / (2*a)

⇒   Δx₁ = (54² - 31²) / (2*a)

⇒   Δx₁ = 977.5 / a

c) We use the equation

v₂ = v₁ - a*Δt   ⇒   a = (v₁ - v₂) / Δt

⇒   a = (54 - 31) / Δt

⇒   a = 23 / Δt

e)  We apply the formula

Δx₁ = v₁*Δt - 0.5*a*Δt²

Δx₁ = 54*Δt - 0.5*(23 / Δt)*Δt²

⇒   Δx₁ = 42.5*Δt

g) If   Δx₁ = 2.4 Mi    ⇒   2.4 = 42.5*Δt  ⇒   Δt = 0.0565 h

i) If  a = 23 / Δt  ⇒   a = 23 Mi / 0.0565 h = 407.29 Mi/h²

⇒   a = 0.05 m/s²

3 0
3 years ago
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