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Oliga [24]
3 years ago
7

How do Glaciers (located thousands of miles away) affect Florida's geography?

Physics
1 answer:
mr_godi [17]3 years ago
6 0

Answer:

They affected ocean levels, causing waves to affect different areas of the coastline at different times. Lake Eola is a sinkhole lake.

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The electron gun in an old TV picture tube accelerates electrons between two parallel plates 1.4cm apart with a 28kV potential d
yulyashka [42]

Answer:

a)   F = 3.2 10⁻¹⁰ N , b)       v = 9.9 10⁷ m / s

Explanation:

a) The electric force is

       F = q E

The electric field is related to the potential reference

     V = E d

     E = V / d

Let's replace

    F = e V / d

Let's calculate

    F = 1.6 10⁻¹⁹ 28 10³ / 1.4 10⁻²

    F = 3.2 10⁻¹⁰ N

b) For this part we can use kinematics

          v² = v₀ + 2 a d

          v = √ 2 ad

Acceleration can be found with Newton's second law

        e V / d = m a

        a = e / m V / d

        a = 1.6 10⁻¹⁹ / 9.1 10⁻³¹ 28 10³ / 1.4 10⁻²

        a = 3,516 10⁻¹⁷ m / s²

Let's calculate the speed

       v = √ (2 3,516 10¹⁷ 1.4 10⁻²)

       v = √ (98,448 10¹⁴)

       v = 9.9 10⁷ m / s

3 0
3 years ago
Communicating results allows scientists to learn from each other's results.
sveta [45]

True, scientists often talk to each other to figure out if their results were similar and what they could have done better.

Although, talking to other scientists does have risks, other scientists could copy your work and further better it.

So, your final answer is TRUE, sorry for the long answer, I needed to have a word count about 20 characters and then I got carried away! lol

3 0
3 years ago
Read 2 more answers
Answer fast !!!!!!! pleaseee
wariber [46]

Answer: Magnetic Fields

Explanation:

ig

5 0
2 years ago
Read 2 more answers
A 3.0-kg brick rests on a perfectly smooth ramp inclined at 34° above the horizontal. The brick is kept from sliding down the pl
Firdavs [7]

Answer:

d=0.137 m ⇒13.7 cm

Explanation:

Given data

m (Mass)=3.0 kg

α(incline) =34°

Spring Constant (force constant)=120 N/m

d (distance)=?

Solution

F=mg

F=(3.0)(9.8)

F=29.4 N

As we also know that

Force parallel to the incline=FSinα

F=29.4×Sin(34)

F=16.44 N

d(distance)=F/Spring Constant

d(distance)=16.44/120

d(distance)=0.137 m ⇒13.7 cm

4 0
3 years ago
On the Moon the acceleration due to gravity is about one sixth that on Earth. If a golfer on the Moon imparted the same initial
swat32

Explanation:

We know that that the range of the ball on the earth

R_{earth}=\frac{v_o^2sin2\theta}{g_{earth}}

therefore, range of the ball on moon

R_{moon}=\frac{v_o^2sin2\theta}{g_{moon}}

R_{moon}=\frac{v_o^2sin2\theta}{g_{earth}/12}

therefore,

R_{moon}=6R_{earth}

Therefore, the range of ball will be 6 times on the moon than that on earth

6 0
3 years ago
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