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3 years ago

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WHO EVER ANSWERS CORRECT I'LL GIVE BRAINLIEST!!!!!!! Why might it be helpful to us to measure gravity fluctuations on Earth? Wha

t is the most important type of information it gives us and why do you think it would or would not be useful?

1 answer:

Lera25 [3.4K]3 years ago

7 0

Answer:

Gravity varies when the Earth rotates and its mass and density differ based on where you are on the planet. Knowing how gravity affects sea level thus aids geodesists in making more precise calculations. The mean sea level would be higher in parts of the earth where gravitational forces are heavier.

Explanation:

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What is the ionosphere

ICE Princess25 [194]

A ionosphere is a layer of the earths atmosphere that is ionized by solar and cosmic radiation <span />

7 0

4 years ago

Lora (of mass 43.6 kg) is an expert skier. She starts at 3.6 m/s at the top of the lynx run, which is 67 m above the bottom. Wha

Likurg_2 [28]

Explanation:

As per the law of conservation of energy, the final mechanical energy of Lora is equal to its initial mechanical energy. So, when Lora is at the bottom of ski run then her potential energy will change into kinetic energy.

Hence, $E_{initial} = \frac{mv_{i}^{2}}{2} + mgh
$

$E_{final} = \frac{mv_{f}^{2}}{2}$

Now, final kinetic energy that will be at the bottom of the ski run is as follows.

Let, $E_{k} = E_{final}$

$E_{intial} = E_{final}$

$E_{k} = \frac{mv_{i}^{2}}{2} + mgh
$

= $\frac{(43.6 \times (3.6)^{2}}{2} + (43.6 \times 9.81 \times 67)$

= 282.53 + 28656.97

= 28939.502 J

Thus, we can conclude that her final kinetic energy at the bottom of the ski run is 28939.502 J.

3 0

3 years ago

Two particles with masses 2m and 9m are moving toward each other along the x axis with the same initial speeds vi. Particle 2m i

s2008m [1.1K]

Answer:

The final speed for the mass 2m is $v_{2y}=-1,51\ v_{i}$ and the final speed for the mass 9m is $v_{1f} =0,85\ v_{i}$ .

The angle at which the particle 9m is scattered is $\theta = -66,68^{o}$ with respect to the - y axis.

Explanation:

In an elastic collision the total linear momentum and the total kinetic energy is conserved.

<u>Conservation of linear momentum:</u>

Because the linear momentum is a vector quantity we consider the conservation of the components of momentum in the x and y axis.

The subindex 1 will refer to the particle 9m and the subindex 2 will refer to the particle 2m

$\vec{p}=m\vec{v}$

$p_{xi} =p_{xf}$

In the x axis before the collision we have

$p_{xi}=9m\ v_{i} - 2m\ v_{i}$

and after the collision we have that

$p_{xf} =9m\ v_{1x}$

In the y axis before the collision $p_{yi} =0$

after the collision we have that

$p_{yf} =9m\ v_{1y} - 2m\ v_{2y}$

so

$p_{xi} =p_{xf} \\7m\ v_{i} =9m\ v_{1x}\Rightarrow v_{1x} =\frac{7}{9}\ v_{i}$

then

$p_{yi} =p_{yf} \\0=9m\ v_{1y} -2m\ v_{2y} \\v_{1y}=\frac{2}{9} \ v_{2y}$

<u>Conservation of kinetic energy:</u>

$\frac{1}{2}\ 9m\ v_{i} ^{2} +\frac{1}{2}\ 2m\ v_{i} ^{2}=\frac{1}{2}\ 9m\ v_{1f} ^{2} +\frac{1}{2}\ 2m\ v_{2f} ^{2}$

so

$\frac{11}{2}\ m\ v_{i} ^{2} =\frac{1}{2} \ 9m\ [(\frac{7}{9}) ^{2}\ v_{i} ^{2}+ (\frac{2}{9}) ^{2}\ v_{2y} ^{2}]+ m\ v_{2y} ^{2}$

Putting in one side of the equation each speed we get

$\frac{25}{9}\ m\ v_{i} ^{2} =\frac{11}{9}\ m\ v_{2y} ^{2}\\v_{2y} =-1,51\ v_{i}$

We know that the particle 2m travels in the -y axis because it was stated in the question.

Now we can get the y component of the speed of the 9m particle:

$v_{1y} =\frac{2}{9}\ v_{2y} \\v_{1y} =-0,335\ v_{i}$

the magnitude of the final speed of the particle 9m is

$v_{1f} =\sqrt{v_{1x} ^{2}+v_{1y} ^{2} }$

$v_{1f} =\sqrt{(\frac{7}{9}) ^{2}\ v_{i} ^{2}+(-0,335)^{2}\ v_{i} ^{2} }\Rightarrow \ v_{1f} =0,85\ v_{i}$

The tangent that the speed of the particle 9m makes with the -y axis is

$tan(\theta)=\frac{v_{1x} }{v_{1y}} =-2,321 \Rightarrow\theta=-66,68^{o}$

As a vector the speed of the particle 9m is:

$\vec{v_{1f} }=\frac{7}{9} v_{i} \hat{x}-0,335\ v_{i}\ \hat{y}$

As a vector the speed of the particle 2m is:

$\vec{v_{2f} }=-1,51\ v_{i}\ \hat{y}$

8 0

3 years ago

What variable is represented on the y-axis?

gladu [14]

Distance/ Time which means Distance is on horizontal and time is on vertical

8 0

3 years ago

Matt plays badminton. He serves the birdie with a velocity of 4.25 m/s [right] covering a displacement of 5.2 m [forward]. How m

OLEGan [10]

Answer:

Time taken by birdie’s trip = 1.2235 s (Approx)

Explanation:

Given:

Velocity = 4.25 m/s

Displacement = 5.2 m

Find:

Time taken by birdie’s trip = ?

Computation:

⇒ Time taken = Displacement / Velocity

⇒ Time taken by birdie’s trip = Displacement / Velocity

⇒ Time taken by birdie’s trip = 5.2 m / 4.25 m/s

⇒ Time taken by birdie’s trip = 1.2235 s (Approx)

6 0

4 years ago