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weeeeeb [17]
3 years ago
10

A spring tide occurs ________. twice a year at the new or full moon closest to the vernal and autumnal equinox once a month, or

12 times a year about 26 times per year, at every full and new moon once a year at the first full moon after the vernal equinox
Physics
2 answers:
lukranit [14]3 years ago
5 0

Answer: Option (1)

Explanation: A spring tide occurs when the sun, earth and the moon are collinear. It is the condition in which the gravitational pull is exerted on earth by both the sun as well as the moon, from opposite sides. Due to this, there occurs a high tide, in comparison to the regular tide, because of the combination of both the gravitational force. It occurs two time a year on each of the lunar month.

Thus, the correct answer is option (1).

zmey [24]3 years ago
4 0
I think the correct answer would be the third option. A spring tide occurs about 26 times per year, at every full and new moon. These tides happen during the new moon and the full moon of a year. It is a result of the position of the Moon, the Sun and the Earth and the gravitational effects of these systems. It is during the time when the moon's quarter would phase the sun and the moon work at right angles which will cause bulges cancelling the effect of each other. The tides of this type will have the greatest difference between the low and high tides of the water. <span />
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Answer:

How do mass and speed affect kinetic energy?

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8 0
3 years ago
At highway speeds, a particular automobile is capable of an acceleration of about 2.0 m/s2 . Part A At this rate, how long does
Natasha_Volkova [10]

Answer:

Time taken, t = 4.86 seconds

Explanation:

Given that,

Acceleration of a particular automobile, a=2\ m/s^2

Initial speed of the automobile, u = 75 km/h = 20.83 m/s

Final speed of the automobile, v = 110 km/h = 30.55 m/s

We need to find the time taken to accelerate from u to v. Let t is the time taken. It can be calculate as :

t=\dfrac{v-u}{a}

t=\dfrac{30.55-20.83}{2}

t = 4.86 seconds

So, the time taken by the automobile is 4.86 seconds. Hence, this is the required solution.

5 0
3 years ago
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8 0
3 years ago
The practical limit to an electric field in air is about 3.00 × 10^6 N/C . Above this strength, sparking takes place because air
AleksAgata [21]

Answer:

(a) x=157 m

(b) No

Explanation:

Given Data

Mass of proton m=1.67×10⁻²⁷kg

Charge of proton e=1.6×10⁻¹⁹C

Electric field E=3.00×10⁶ N/C

Speed of light c=3×10⁸ m/s

For part (a) distance would proton travel

Apply the third equation of motion

(v_{f})^{2} =(v_{i})^{2}+2ax

In this case vi=0 m/s and vf=c

so

c^{2}=(0)^{2}+2ax\\  c^{2}=2ax\\x=\frac{c^{2} }{2a}

x=\frac{c^{2}}{2a}--------Equation (i)

From the electric force on proton

F=qE\\where\\ F=ma\\so\\ma=qE\\a=\frac{qE}{m}\\

put this a(acceleration) in Equation (i)

So

x=\frac{c^{2} }{2(qE/m)}\\ x=\frac{mc^{2}}{2qE} \\x=\frac{(1.67*10^{-27})*(3*10^{8})^{2}  }{2*(1.6*10^{-19})*(3*10^{6})}\\ x=157m

For part (b)

No the proton would collide with air molecule

7 0
3 years ago
Calculate the maximum capillary rise/fall of mercury in a 0.5 mm radius glass capillary. Assume that the surface tension for mer
tekilochka [14]

Answer: 0.01 m

Explanation: The formulae for capillarity rise or fall is given below as

h = (2T×cosθ)/rpg

Where θ = angle mercury made with glass = 50°

T = surface tension = 0.51 N/m

g = acceleration due gravity = 9.8 m/s²

r = radius of tube = 0.5mm = 0.0005m

p = density of mercury.

h = height of rise or fall

From the question, specific gravity of density = 13.3

Where specific gravity = density of mercury/ density of water, where density of water = 1000 kg/m³

Hence density of mercury = 13.3×1000 = 13,300 kg/m³.

By substituting parameters, we have that

h = 2×0.51×cos 50/0.0005×9.8×13,300

h = 0.6556/65.17

h = 0.01 m

8 0
3 years ago
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