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Anna35 [415]
3 years ago
11

Helen can write 15 postcards in one hour, while Kate can write

Physics
1 answer:
Akimi4 [234]3 years ago
4 0

Answer:5 hours

Explanation:

300 divided by 20 = 15

300 divided by 15 = 20

the difference is 5

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As waves get closer to a beach they:
hoa [83]

Answer:

As waves get closer to a beach they decrease in height.

Explanation:

As a wave crest approaches the shoreline, it is usual that one end of the line is closer to the shoreline than the other. The implication of this is that the energy in a wave is also spread over quite a larger area,this in turn reduces the height of the waves. In other words, refraction often makes waves smaller.

Waves are caused by wind. Wave height in the open ocean is determined by three factors. The greater the wind speed the larger the waves. The greater the duration of the wind (or storm) the larger the waves. The greater the fetch (area over which the wind is blowing - size of storm) the larger the waves.

3 0
3 years ago
Which is the most important factor in determining climate?
Leokris [45]

Answer:

B took the test (k12)

Explanation:

6 0
3 years ago
PLEASE I NEED HELP IS DUE IN A FEW MINUTES
motikmotik

Answer:

glutamate ,GABA

Explanation:

In the vertebrate central nervous system (CNS), glutamate serves as the major excitatory neurotransmitter, whereas GABA and glycine serve as the major inhibitory neurotransmitters.

7 0
3 years ago
A person walks in the following pattern: 3.1 km north, then 2.4 km west, and finally 5.2 km south. a) Sketch the vector diagram
zysi [14]

Answer:

d = 3.19 km

direction is given as

\theta = 41.2 degree South of West

Explanation:

Part b)

displacement is given as

d_1 = 3.1 \hat j

d_2 = 2.4 \hat i

d_3 = 5.2(-\hat j)

now we will have

d = d_1 + d_2 + d_3

d = 2.4 \hat i + (3.1 - 5.2)\hat j

d = 2.4 \hat i - 2.1 \hat j

total displacement is given as

d = \sqrt{2.4^2 + 2.1^2}

d = 3.19 km

direction is given as

tan\theta = \frac{-2.1}{2.4}

\theta = 41.2 degree South of West

3 0
4 years ago
One of the most important properties of materials in many applications is strength. Two of the qualitative measures of the stren
katrin2010 [14]

To solve this exercise it is necessary to take into account the concepts related to Tensile Strength and Shear Strenght.

In Materials Mechanics, generally the bodies under certain loads are subject to both Tensile and shear strenghts.

By definition we know that the tensile strength is defined as

\sigma = \frac{F}{A}

Where,

\sigma =Tensile strength

F = Tensile Force

A = Cross-sectional Area

In the other hand we have that the shear strength is defined as

\sigma_y = \frac{F_y}{A}

where,

\sigma_y =Shear strength

F_y = Shear Force

A_0 =Parallel Area

PART A) Replacing with our values in the equation of tensile strenght, then

311*10^6 = \frac{F}{(15*10^{-6})(30*10^{-2})}

Resolving for F,

F= 1399.5N

PART B) We need here to apply the shear strength equation, then

\sigma_y = \frac{F_y}{A}

210*10^6 = \frac{F_y}{15*10^{-6}30*10^{-2}}

F_y = 945N

In such a way that the material is more resistant to tensile strength than shear force.

6 0
3 years ago
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