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seropon [69]
3 years ago
8

When are tides highest? a. during the moon’s first quarter phase b. when the sun, Earth, and the moon are nearly in a line c. du

ring the moon’s third quarter phase d. when the moon is at a right angle to the sun
Physics
2 answers:
elena-14-01-66 [18.8K]3 years ago
7 0
<span>Ocean tides are highest when the sun, Earth, and the moon
are nearly in a line.  That means at the times of New Moon
and Full Moon.</span>
liubo4ka [24]3 years ago
3 0

Answer:

b. when the sun, Earth, and the moon are nearly in a line

Explanation:

Due to gravitational pull of sun and moon, the level of water in seas and oceans rises above its normal level. This is known as tides. Highest tides occur when the sun, the moon and the Earth align in a straight line i.e. during New moon and Full moon. This is because in this alignment, there is maximum effect of gravity. Tide occurrence during this phase are also referred as spring tides.

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Describe using examples how objects can be at rest and in motion simultaneously
elena-14-01-66 [18.8K]
An object can be at rest and still be in motion because the earth is always in motion.

5 0
3 years ago
Review
sleet_krkn [62]

Answer:

yes

Explanation:

8 0
3 years ago
Read 2 more answers
A spherical capacitor is formed from two concentric sphericalconducting shells separated by vacuum. The inner sphere has radius1
zubka84 [21]

Explanation:

(1).  Formula to calculate the potential difference is as follows.

       \Delta V = -\int E dr

                  = -\int \frac{kq}{r^{2}} dr

                 = \frac{kq}{r_{f}} - \frac{kq}{r_{i}}

                 = \frac{kq(r_{f} - r_{i})}{r_{f}r_{i}}

                 = \frac{9 \times 10^{9} \times 3.30 \times 10^{-9}(0.1 - 0.015)}{0.1 \times 0.015}

                = 38.7 volts

Therefore, magnitude of the potential difference between the two spheres is 38.7 volts.

(2).  Now, formula to calculate the energy stored in the capacitor is as follows.

           E = \frac{1}{2}QV

              = \frac{1}{2} \times 3.30 \times 10^{-9} \times 3.87 V

              = 6.39 \times 10^{-8} J

Thus, the electric-field energy stored in the capacitor is 6.39 \times 10^{-8} J.

7 0
3 years ago
Read 2 more answers
You throw a ball from the balcony onto the court in the basketball arena. You release the ball at a height of 7.00 m above the c
Mariana [72]

Answer:

Your friend has to wait 0.26 s after you throw the ball to start running.

Explanation:

The equation that gives the position vector of the ball is as follows:

r = (x0 + v0 · t · cos α, y0 + v0 · t ·sin α + 1/2 · g · t²)

Where:

x0 = initial horizontal positon

v0 = initial velocity

t = time

α = throwing angle

y0 = initial vertical position

g = acceleration due to gravity

The equation of displacement of your friend is as follows:

x = x0 + v0 · t + 1/2 · a · t²

Where:

x = position of your friend at time t

x0 = initial position

v0 = initial velocity

t = time

a = acceleration

Please, see the attached figure for a description of the situation. Notice that the frame of reference is located at the throwing point.

Let´s find the time of flight of the ball. We know that at the final time, the y-component of the vector r has to be -6.00 m (1 m above the ground). Then:

y = y0 + v0 · t ·sin α + 1/2 · g · t²

-6.00 m = 0 m + 9.00 m/s · t · sin 33.0° - 1/2 · 9.8 m/s² · t²

0 = -4.9 m/s² · t² + 9.00 m/s · sin 33.0° · t + 6.00 m

Solving the quadratic equation:

t = 1.71 s

Now that we have the time of flight, we can calculate the x-component of the vector r (the horizontal distance traveled by the ball):

x= x0 + v0 · t · cos α

x = 0m + 9.00 m/s · 1.71 s · cos 33°

x = 12.9 m

Then, your friend will have to run (12.9 m - 11.0 m) 1.9 m to catch the ball 1 m above the ground.

Let´s see, how much time it takes your friend to run that distance:

x = x0 + v0 · t + 1/2 · a · t²      (x0 = 0, v0 = 0)

x = 1/2 · a · t²

1.9 m = 1/2 · 1.80 m/s² · t²

Solving for t

t = 1.45 s

Then, since the time of flight of the ball is 1.71 s, your friend has to wait

1.71 s - 1.45 s = 0.26 s after you throw the ball to start running.

6 0
3 years ago
A galaxy spectrum has a redshift of 70,000 km per second. If the Hubble constant is 70 km per second per Mpc (megaparsec), how f
gogolik [260]

Using the Hubble law v = H₀d where v = recessional speed = 70,000 km per second H₀ = hubble constant = 70 km/s/Mpc and d = distance of galaxy.

Making d subject of the formula, we have

d = v/H₀

Substituting the values of the variables into the equation, we have

d = v/H₀

d = 70000 km/s/70 km/s/Mpc

d = 1000 Mpc

So, the galaxy is 1000 Mpc away from us.

Learn more about hubble law here:

brainly.com/question/18484687

8 0
3 years ago
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