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fredd [130]
2 years ago
11

If air temperature increased how would it effect precipitation

Physics
1 answer:
Oduvanchick [21]2 years ago
6 0

Decrease Because Water Vapor Would Condense More Slowly

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6. A 145-g baseball moving 30.5 m/s strikes a stationary 5.75-kg brick resting on small rollers so it moves without significant
Sindrei [870]

Answer:

Explanation:

a )

momentum of baseball before collision

mass x velocity

= .145 x 30.5

= 4.4225 kg m /s

momentum of brick after collision

= 5.75 x 1.1

= 6.325 kg m/s

Applying conservation of momentum

4.4225 + 0 = .145 x v + 6.325 , v is velocity of baseball after collision.

v = - 13.12 m / s

b )

kinetic energy of baseball  before collision = 1/2 mv²

= .5 x .145 x 30.5²

= 67.44 J

Total kinetic energy before collision = 67.44 J

c )

kinetic energy of baseball after collision = 1/2 x .145 x 13.12²

= 12.48 J .

 kinetic energy of brick after collision

= .5 x 5.75 x 1.1²

= 3.48 J

Total kinetic energy after collision

= 15.96 J

3 0
3 years ago
What is the energy in joules of a photon with a frequency of 3.16e 12 s-1?
erica [24]
We have: Energy(E) = Planck's constant(h) × Frequency(∨)
Here, Planck's constant(h) = 6.626 × 10⁻³⁴ J/s
Frequency (∨) = 3.16 × 10¹² /s

Substitute the values into the expression:
E = (6.626 × 10⁻³⁴)(3.16 × 10¹²) J
E = 2.093 × 10⁻²¹ Joules

In short, Your Final answer would be 2.093 × 10⁻²¹ J

Hope this helps!
5 0
3 years ago
Read 2 more answers
Calculate the change in length of a 90.5 mm aluminum bar that has increased in temperature by from -14.4 oC to 154.6 oC
nignag [31]

Answer:

 ΔL = 3.82 10⁻⁴ m

Explanation:

This is a thermal expansion exercise

          ΔL = α L₀ ΔT

          ΔT = T_f - T₀

where ΔL is the change in length and ΔT is the change in temperature

Let's reduce the length to SI units

          L₀ = 90.5 mm (1m / 1000 mm) = 0.0905 m

let's calculate

          ΔL = 25.10⁻⁶ 0.0905 (154.6 - (14.4))

          ΔL = 3.8236 10⁻⁴ m

     

using the criterion of three significant figures

          ΔL = 3.82 10⁻⁴ m

5 0
2 years ago
If an object force of 50 N is used to move an object a distance of 20 m, what distance must the object be moved if the input for
steposvetlana [31]

Answer:

\ d_{out} = 100 \ m.

Explanation:

Given data:

F_{in} = 50 \ \rm N

F_{out} = 10 \ \rm N

d_{in} = 20 \ m

Let the distance traveled by the object in the second case be d_{out}.

In the given problem, work done by the forces are same in both the cases.

Thus,

W_{in} = W_{out}

F_{in}.d_{in} = F_{out}.d_{out}

\Rightarrow \ d_{out} = \frac{F_{in}.d_{in}}{F_{out}}

\ d_{out} = \frac{50 \times 20}{10}

\ d_{out} = 100 \ m.

5 0
3 years ago
In Lesson 20, a magnesium strip was used to ignite the thermite reaction. When magnesium is placed in a flame from a small blow
nadezda [96]

Answer:

2 electrons will be needed by unbound oxygen in order to fill its 2nd shell.

Explanation:

The chemical reaction between magnesium and oxygen gives magnesium oxide as a product.The reaction is chemically represented as:

2Mg(s)+O_2(g)\rightarrow 2MgO(s)

Magnesium is a metal of group-2 with 2 valence electrons.It has atomic number of 12.

[Mg]=1s^22s^22p^63s^2

In order to attain noble gas configuration it will loose two electrons.

[Mg]^{2+}=1s^22s^22p^6

Mg\rightarrow Mg^{2+}+2e^-...[1]

Oxygen is a non metal of group-16 with 6 valence electrons..It has atomic number of 8.

[O]=1s^22s^22p^4

In order to attain noble gas configuration it will gain two electrons.

[O]^{2-}=1s^22s^22p^6

O+2e^-\rightarrow O^{2-}..[2]

2 electrons will be needed by unbound oxygen in order to fill its 2nd shell.

6 0
3 years ago
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