Ask question

Ask question

All categories

3 years ago

12

Would the tide be higher when the moon is on the same side of earth as New Brunswick or on the opposite side why

1 answer:

Sindrei [870]3 years ago

3 0

Tide is the alternating pattern of rising and falling sea level with respect to land.

There are three types of tide relative to the position of the moon and sun to that of the earth.

1. Spring tides. Occur when the moon sun and earth arrange themselves more or less in a straight line, like an arrow. These tides are the highest and lowest tides.

2. Perigean spring tides. Occur when the new moon or full moon closely aligns with perigee, the closest point to the earth in the moon's orbit. These tides are not as high or low as the spring tides.

3. Neap tides. Occur when the sun and the moon are at right angles as seen from earth.The tides are at minimum range.

You might be interested in

Brainlist nd 20 point

Alborosie

the same is the answer

ok

6 0

2 years ago

A comet is traveling through space with speed 3.01 ✕ 104 m/s when it encounters an asteroid that was at rest. The comet and the

Tcecarenko [31]

Answer: $8.493(10)^{-3} m/s$

Explanation:

According to the conservation of linear momentum principle, the initial momentum $p_{i}$ (before the collision) must be equal to the final momentum $p_{f}$ (after the collision):

$p_{i}=p_{f}$ (1)

In addition, the initial momentum is:

$p_{i}=m_{1}V_{1}+m_{2}V_{2}$ (2)

Where:

$m_{1}=1.71(10)^{14} kg$ is the mass of the comet

$m_{2}=6.06(10)^{20} kg$ is the mass of the asteroid

$V_{1}=3.01(10)^{4} m/s$ is the velocity of the comet, which is positive

$V_{2}=0 m/s$ is the velocity of the asteroid, since it is at rest

And the final momentum is:

$p_{f}=(m_{1}+m_{2})V_{f}$ (3)

Where:

$V_{f}$ is the final velocity

Then :

$m_{1}V_{1}+m_{2}V_{2}=(m_{1}+m_{2})V_{f}$ (4)

Isolating $V_{f}$ :

$V_{f}=\frac{m_{1}V_{1}}{m_{1}+m_{2}}$ (5)

$V_{f}=\frac{(1.71(10)^{14} kg)(3.01(10)^{4} m/s)}{1.71(10)^{14} kg+6.06(10)^{20} kg}$

Finally:

$V_{f}=8.493(10)^{-3} m/s$ This is the final velocity, which is also in the positive direction.

8 0

4 years ago

Calculate the potential energy of a 5.2 kg object positioned 5.8 m above the ground.

GrogVix [38]

Answer:

295.568J

Explanation:

use P=mgh

plug in givens

P= 5.2*9.8*5.8= 295.568J

4 0

3 years ago

Read 2 more answers

The diagram shows a box in the periodic table. 15 P Phosphorus 31.0 What is the atomic number of the element shown? A. 46.0 B. 1

natali 33 [55]

Answer:

b 15

Explanation:

the atomic number is the number of the element

4 0

3 years ago

In the Bohr model of the hydrogen atom, an electron({rm mass};m=9.1; times 10^{ - 31;}{rm kg}) orbits a proton at a distance of

max2010maxim [7]

Answer:

n=6.56×10¹⁵Hz

Explanation:

Given Data

Mass=9.1×10⁻³¹ kg

Radius distance=5.3×10⁻¹¹m

Electric Force=8.2×10⁻⁸N

To find

Revolutions per second

Solution

Let F be the force of attraction

let n be the number of revolutions per sec made by the electron around the nucleus then the centripetal force is given by

F=mω²r......................where ω=2π n

F=m4π²n²r...............eq(i)

as the values given where

Mass=9.1×10⁻³¹ kg

Radius distance=5.3×10⁻¹¹m

Electric Force=8.2×10⁻⁸N

we have to find n from eq(i)

n²=F/(m4π²r)

$n^{2} =\frac{8.2*10^{-8} }{9.11*10^{-31}* 4\pi^{2} *5.3*10^{-11} }\\ n^{2}=4.31*10^{31}\\ n=\sqrt{4.31*10^{31}}\\ n=6.56*10^{15}Hz$

8 0

3 years ago