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2 years ago

Compare tidal range to spring tide

1 answer:

7 0

Spring tides have higher high tides and lower low tides whereas neap tides have lower high tides and higher low tides. Hence, the range is much larger in a spring tide than in a low tide.

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Cheetahs the worlds fastest land animals can run up to about 125km/h. a cheetah chasing an impala runs 32m [N], then suddenly tu

AleksAgata [21]

Av Speed = total distance / time time = 32+ 46 / 2.7 = 28 m/sec
Av velocity = total displacement / time total = S / t
S = sqrt( 32^2 +46^2) = 56 m
Av Velocity = 56/ 2,7 = 20.75 m/sec
with angle tan^-1 = 0.7 north west ( about 35 degrees north west)

3 0

2 years ago

Explain three ways visual aids help you study

Katen [24]

Visual aids are tools that help to make an issue or lesson clearer or easier to understand and know (pictures, models, charts, maps, videos, slides, real objects etc.). ... Visual aids are those devices which are used in classrooms to encourage students learning process and make it easier and interesting.

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2 years ago

A spinning tossed baseball veers off course in the direction ofA) reduced air pressure on the ball.B) increased air pressure on

katovenus [111]

Answer:

A) reduced air pressure on the ball.

Explanation:

3 0

2 years ago

What is the length of the orbit of the earth around the sun?

Rzqust [24]

Answer:

S = 2 π R

R (mean) = 92.9E6 miles

S = 2 * 3.14 * 92.9E6 miles = 5.84E8 miles

5 0

2 years ago

An amateur player is about to throw a dart with an initial velocity of 15 meters/second onto a dartboard that is at a distance o

Minchanka [31]

Answer:

B. 0.16 m

Explanation:

The vertical distance by which the player will miss the target is equal to the vertical distance covered by the dart during its motion.

Since the dart is thrown horizontally, the initial vertical velocity is zero:

$v_y = 0$

While the horizontal velocity is

$v_x = 15 m/s$

The horizontal distance covered is

$d_x = 2.7 m$

Since the dart moves by uniform motion along the horizontal direction, the time it takes for covering this distance is

$t=\frac{d_x}{v_x}=\frac{2.7 m}{15 m/s}=0.18 s$

along the vertical direction, the motion is a uniformly accelerated motion with constant downward acceleration g=9.8 m/s^2, so the vertical distance covered is given by

$d_y = \frac{1}{2}gt^2=\frac{1}{2}(9.8 m/s^)(0.18 s)^2=0.16 m$

8 0

3 years ago