The sun does not normally affect the tides to a noticeable degree. however, under certain circumstances the gravitational pull of the sun does affect the tides. under what two (2) conditions.the sun does not normally affect the tides to a noticeable degree. however, under certain circumstances the gravitational pull of the sun does affect the tides. under what two (2) conditions.

Answers

Answer 1

Answer: The answers are :
1) when the sun, moon, and earth are in a line only
2) when the gravitational forces of the Moon and the Sun are
perpendicular to one another with respect to the Earth.

Answer 2

Answer:

Answer:

High and low tides are result of combined effect of gravitational pull of the sun and the moon. When the two align in a straight line, the range of tides is maximum. This happens on new moon and full moon day.

On the other hand, when the sun and the moon align at right angles, the effect of gravity is minimum and the range of the tides is minimum.

Related Questions

an object down, but this is not true. If you place a box of mass 8 kg on a moving horizontal conveyor belt, the friction force of the belt acting on the bottom of the box speeds up the box. At first there is some slipping, until the speed of the box catches up to the speed of the belt, which is 5 m/s. The coefficient of kinetic friction between box and belt is 0.6. (a) How much time does it take for the box to reach this final speed
A 50.0-kg box is being pulled along a horizontal surface by means of a rope that exerts a force of 250 n at an angle of 32.0° above the horizontal. the coefficient of kinetic friction between the box and the surface is 0.350. what is the acceleration of the box?
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Feest. Fysics and motion11
Select the correct answer
You travel in a circle, whose circumference is 8 kilometers, at an average speed of 8 kilometers/hour. If you stop at the same point you started
from, what is your average velocity?
A
0 kilometers/hour
B.
2 kilometers/hour
4 kilometers/hour
D
8 kilometers/hour
E.
16 kilometers/hour
Rese

Answers

Velocity depends on the straight-line distance between your start-point and your end-point, regardless of what route you follow to get there.

If you stop at the same point where you started, then that distance is zero, no matter how far you drove before you returned to your start-point.

So the average velocity around any "CLOSED" path is zero. (A)

After a 0.320-kg rubber ball is dropped from a height of 19.0 m, it bounces off a concrete floor and rebounds to a height of 15.0 m. Determine the magnitude of the impulse delivered to the ball by the floor.

Answers

Given Information:

Mass of ball = m = 0.320 kg

Initial height = h₁ = 19 m

Final height = h₂ = 15 m

Required Information:

Impulse = I = ?

Answer:

Impulse = 11.77 kg.m/s

Explanation:l

We know that impulse is equal to change in momentum

I = Δp

I = p₁ - p₂

I = mv₁ - mv₂

I = m(v₁ - v₂)

Where m is the mass of ball, v₂ is the final velocity of the ball, and v₁ is the initial velocity of the ball.

So first we need to find the initial and final velocities of the ball

The relation between initial potential energy and final kinetic energy before the collision is given by

PE₁ = KE₂

mgh₁ = ½mv₂²

gh₁ = ½v₂²

v₂² = 2gh₁

v₂ = √2gh₁

v₂ = √2*9.8*19

v₂ = 19.3 m/s

The relation between initial kinetic energy and final potential energy after the collision is given by

KE₁ = PE₂

½mv₁² = mgh₂

½v₁² = gh₂

v₁² = 2gh₂

v₁ = √2gh₂

v₁ =√2*9.8*15

v₁ = 17.15 m/s

Finally, we can now find the magnitude of the impulse delivered to the ball by the floor.

I = 0.320(17.5 - (-19.3))

I = 11.77 kg.m/s

Answer:

$Imp = 11.666\,(kg\cdot m)/(s)$

Explanation:

Speed experimented by the ball before and after collision are determined by using Principle of Energy Conservation:

Before collision:

$(0.32\,kg)\cdot (9.807\,(m)/(s^(2)) )\cdot (19\,m) = (1)/(2)\cdot (0.320\,kg)\cdot v_(A)^(2)$

$v_(A) \approx 19.304\,(m)/(s)$

After collision:

$(1)/(2)\cdot (0.320\,kg)\cdot v_(B)^(2) = (0.32\,kg)\cdot (9.807\,(m)/(s^(2)) )\cdot (15\,m)$

$v_(B) \approx 17.153\,(m)/(s)$

The magnitude of the impulse delivered to the ball by the floor is calculated by the Impulse Theorem:

$Imp = (0.32\,kg)\cdot [(17.153\,(m)/(s) )-(-19.304\,(m)/(s) )]$

$Imp = 11.666\,(kg\cdot m)/(s)$

Which of these 23rd chromosomecombinations is likeliest to result in a
person with male and female traits?
ΧΟ
XXX
XXY
XY

Answers

Sorry if I’m wrong but I think it’s XO since o is not a sex chromosome

As a science project, you drop a watermelon off the top of the Empire State Building. 320 m above the sidewalk. It so happens that Superman flies by at the instant you release the watermelon. Superman is headed straight down with a constant speed of 30 m/s. A) How much time passes before the watermelon has the same velocity? B) How fast is the watermelon going when it passes Superman?C) How fast is the watermelon traveling when it hits the ground?

Answers

Answer:

3.06 seconds time passes before the watermelon has the same velocity

watermelon going at speed 59.9 m/s

watermelon traveling when it hits the ground at speed is 79.19 m/s

Explanation:

given data

height = 320 m

speed = 30 m/s

to find out

How much time passes before the watermelon has the same velocity and How fast is the watermelon going and How fast is the watermelon traveling

solution

we will use here equation of motion that is

v = u + at ....................1

here v is velocity 30 m/s and u is initial speed i.e zero and a is acceleration i.e 9.8 m/s²

put the value and find time t

30 = 0 + 9.8 (t)

t = 3.06 s

so 3.06 seconds time passes before the watermelon has the same velocity

and

we know superman cover distance is = velocity × time

so distance = 30 × t

and distance formula for watermelon is

distance = ut + 0.5×a×t² .............2

here u is initial speed i.e 0 and a is acceleration i.e 9.8 m/s² and h is 30 × t

30 × t = 0 + 0.5×9.8×t²

t = 6.12 s

so by equation 1

v = u + at

v = 0 + 9.8 ( 6.12)

v = 59.9 m/s

so watermelon going at speed 59.9 m/s

and

watermelon traveling speed formula is by equation of motion

v² - u² = 2as ......................3

here v is speed and u is initial speed i.e 0 and a is acceleration i.e 9.8 m/s² and s is distance i.e 320 m

v² - 0 = 2(9.8) 320

v = 79.19 m/s

so watermelon traveling when it hits the ground at speed is 79.19 m/s

A missile is moving 1350 m/s at a 25.0 angle

Answers

i will answer both versions assuming what you want to know is the distance it travels up from and over the ground. and how long until it reaches space. 540 meters per second up and over. to reach space which is 100km above sea level, it would take about 5400 minutes

You release a block from the top of a long, slippery inclined plane of length l that makes an angle θ with the horizontal. The magnitude of the block's acceleration is gsin(θ).1. For an x axis pointing down the incline and having its origin at the release position, derive an expression for the potential energy of the block-Earth system as a function of x. Suppose that the gravitational potential energy is measured relative to the ground at the bottom of the incline, UG(x=l)=0.Express your answer in terms of g and the variables m, l, x, and θ.(U^G=?)2. Use the expression you derived in the previous part to determine the speed of the block at the bottom of the incline. (Vx,f=?)Express your answer in terms of g and the variables m, l, and θ.

Answers

Answer:

UG (x) = m*g*x*sin(Q)

Vx,f (x)= sqrt (2*g*x*sin(Q))

Explanation:

Given:

- The length of the friction less surface L

- The angle Q is made with horizontal

- UG ( x = L ) = 0

- UK ( x = 0) = 0

Find:

derive an expression for the potential energy of the block-Earth system as a function of x.

determine the speed of the block at the bottom of the incline.

Solution:

- We know that the gravitational potential of an object relative to datum is given by:

UG = m*g*y

Where,

m is the mass of the object

g is the gravitational acceleration constant

y is the vertical distance from datum to the current position.

- We will consider a right angle triangle with hypotenuse x and angle Q with the base and y as the height. The relation between each variable can be given according to Pythagoras theorem as follows:

y = x*sin(Q)

- Substitute the above relationship in the expression for UG as follows:

UG = m*g*x*sin(Q)

- To formulate an expression of velocity at the bottom we can use an energy balance or law of conservation of energy on the block:

UG = UK

- Where UK is kinetic energy given by:

UK = 0.5*m*Vx,f^2

Where Vx,f is the final velocity of the object @ x:

m*g*x*sin(Q) = 0.5*m*Vx,f^2

-Simplify and solve for Vx,f:

Vx,f^2 = 2*g*x*sin(Q)

Hence, Velocity is given by: