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3 years ago

If the moon disappeared what effect would this have on the earths tides

2 answers:

iogann1982 [59]3 years ago

4 0

The moon has a huge impact on the tides. With the Moon gone, the oceans would become much calmer. The Sun still has its effect on them known as the solar tides. Surfers wouldn't be completely devoid of waves.

Sliva [168]3 years ago

3 0

There would no longer be any tides.

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g You drop a 3.6-kg ball from a height of 3.5 m above one end of a uniform bar that pivots at its center. The bar has mass 9.9 k

Salsk061 [2.6K]

Answer:

h = 3.5 m

Explanation:

First, we will calculate the final speed of the ball when it collides with a seesaw. Using the third equation of motion:

$2gh = v_f^2 - v_i^2\\$

where,

g = acceleration due to gravity = 9.81 m/s²

h = height = 3.5 m

vf = final speed = ?

vi = initial speed = 0 m/s

Therefore,

$(2)(9.81\ m/s^2)(3.5\ m) = v_f^2 - (0\ m/s)^2\\v_f = \sqrt{68.67\ m^2/s^2}\\v_f = 8.3\ m/s$

Now, we will apply the law of conservation of momentum:

$m_1v_1 = m_2v_2$

where,

m₁ = mass of colliding ball = 3.6 kg

m₂ = mass of ball on the other end = 3.6 kg

v₁ = vf = final velocity of ball while collision = 8.3 m/s

v₂ = vi = initial velocity of other end ball = ?

Therefore,

$(3.6\ kg)(8.3\ m/s)=(3.6\ kg)(v_i)\\v_i = 8.3\ m/s$

Now, we again use the third equation of motion for the upward motion of the ball:

$2gh = v_f^2 - v_i^2\\$

where,

g = acceleration due to gravity = -9.81 m/s² (negative for upward motion)

h = height = ?

vf = final speed = 0 m/s

vi = initial speed = 8.3 m/s

Therefore,

$(2)(9.81\ m/s^2)h = (0\ m/s)^2-(8.3\ m/s)^2\\$

6 0

2 years ago

Betty is six years old and often asks her parents many questions. Her parents respond to her questions however trivial they may

Brums [2.3K]

This behavior helps Betty in <u>intellectual </u>development.

5 0

3 years ago

Read 2 more answers

What are four pieces of evidence for continental drift?

Tomtit [17]

continents, paleoclimate indicators, truncated geologic features, and fossils:D

5 0

3 years ago

Read 2 more answers

A car moves at a constant speed of 10 m/s. If the car doesn't accelerate during the next 40 s how far will it go?

tangare [24]

the answer is b. space = time * velocity

8 0

3 years ago

A 560-g squirrel with a surface area of 930cm2 falls from a 5.0-m tree to the ground. Estimate its terminal velocity. (Use a dra

bija089 [108]

Explanation:

Terminal velocity is given by:

$v_t=\sqrt{\frac{2mg}{\rho C_dA}}$

Here, m is the mass of the falling object, g is the gravitational acceleration, $C_d$ is the drag coefficient, $\rho$ is the fluid density through which the object is falling, and A is the projected area of the object. in this case the projected area is given by:

$A=\frac{A_s}{2}=\frac{930cm^2}{2}=465cm^2\\465cm^2*\frac{1m^2}{10^4cm^2}=0.0465m^2\\560g*\frac{1kg}{10^3g}=0.56kg$

Recall that drag coefficient for a horizontal skydiver is equal to 1 and air density is $1.28\frac{kg}{m^3}$ .

$v_t=\sqrt{\frac{2(0.56kg)(9.8\frac{m}{s^2})}{(1.28\frac{kg}{m^3}(1)(0.0465m^2)}}\\v_t=13.58\frac{m}{s}$

Without drag contribution the motion of the person is an uniformly accelerated motion, thus:

$v_f^2=v_o^2+2gh\\v_f=\sqrt{2gh}\\v_f=\sqrt{2(9.8\frac{m}{s^2})(5m)}\\v_f=9.9\frac{m}{s}$

7 0

3 years ago