Ask question

Ask question

All categories

3 years ago

7

A person standing at the top of mountain aconcagua would be approximately 4.3 mi high. the radius of earth is 3959 mi. what is t

he distance to the horizon from this point?

2 answers:

Fed [463]3 years ago

6 0

Explanation :

It is given that,

The radius of the earth is, r = 3959 mi

From the attached figure, we have to find the distance to the horizon from this point.

So, using Pythagoras theorem:

$AP^2=OA^2-OP^2$

$x^2=(R+h)^2-R^2$

$x^2=(3959+4.3)^2-(3959)^2$

$x^2=34065.89\ mi$

$x=184.57\ mi$

The distance to the horizon from this point is 184.57 mi.

Hence, this is the required solution.

PilotLPTM [1.2K]3 years ago

5 0

Besides ambitious, it would be 184.6 mi

You might be interested in

Water falls without splashing at a rate of 0.200 L/s from a height of 3.60 m into a 0.730 kg bucket on a scale. If the bucket is

dimaraw [331]

Answer:

15.106 N

Explanation:

From the given information,

The weight of the bucket can be calculated as:

$W_b = m_bg = \\ \\ W_b = (0.730 \ kg) ( 9.80 \ m/s^2) \\ \\ W_b = 7.154 \ N$

The mass of the water accumulated in the bucket after 3.20s is:

$m_w= (0.20 \ L/s) ( 3.20)s$

$m _w=0.64 \ kg$

To determine the weight of the water accumulated in the bucket, we have:

$W_w = m_w g$

$W_w = ( 0.64 \ kg )(9.80\ m \ /s^2)$

$W_w = 6.272 \ N$

For the speed of the water before hitting the bucket; we have:

$v = \sqrt{2gh}$

$v = \sqrt{2*9.80 \ m/s^2 * 3.60 \ m}$

v = 8.4 m/s

Now, the force required to stop the water later when it already hit the bucket is:

$F = v ( \dfrac {dm}{dt} )$

$F = (8.4 \ m/s)( 0.200 \ L/s)$

F = 1.68 N

Finally, the reading scale is:

$F_{scale$ = 7.154 N + 6.272 N + 1.68 N

= 15.106 N

6 0

2 years ago

Read 2 more answers

I just need x isolated

mamaluj [8]

Answer:

$x=\frac{-y+\sqrt{y^2+4rt} }{2r}$

$x=\frac{-y-\sqrt{y^2+4rt} }{2r}$

Explanation:

$rx+y=\frac{t}{x}\\\\x(rx+y)=(\frac{t}{x})x\\\\rx^2+yx=t\\\\rx^2+yx-t=t-t\\\\rx^2+yx-t=0$

Solve using the quadratic formula.

$x=\frac{-y+\sqrt{y^2+4rt} }{2r}$

$x=\frac{-y-\sqrt{y^2+4rt} }{2r}$

7 0

2 years ago

Read 2 more answers

The current through a 0.2-H inductor is i(t) = 10te–5t A. What is the energy stored in the inductor?

lakkis [162]

Answer:

E = 10t^2e^-10t Joules

Explanation:

Given that the current through a 0.2-H inductor is i(t) = 10te–5t A.

The energy E stored in the inductor can be expressed as

E = 1/2Ll^2

Substitutes the inductor L and the current I into the formula

E = 1/2 × 0.2 × ( 10te^-5t )^2

E = 0.1 × 100t^2e^-10t

E = 10t^2e^-10t Joules

Therefore, the energy stored in the inductor is 10t^2e^-10t Joules

6 0

2 years ago

At what stage of a reaction do atoms have the highest energy?

alekssr [168]

The correct answer for the question that is being presented above is this one: "c. transition state stage." During the transition state stage, the reaction of the atoms have the highest energy. It is also <span>during the formation of the activated complex in the middle of the experiment.</span>

3 0

2 years ago

If the ball shown in the figure lands in 0.5 s, about what height was it thrown from?

Ede4ka [16]

Answer: 1.22 m

Explanation:

The equation of motion in this situation is:

$y=y_{o}+V_{oy}t-\frac{g}{2}t^{2}$ (1)

Where:

$y=0$ is the final height of the ball

$y_{o}=h$ is the initial height of the ball

$V_{oy}=V_{o}sin(0\°)=0$ is the vertical component of the initial velocity (assuming the ball was thrown vertically and there is no horizontal velocity)

$t=0.5 s$ is the time at which the ball lands

$g=9.8 m/s^{2}$ is the acceleration due gravity

So, with these conditions the equation is rewritten as:

$h=\frac{g}{2}t^{2}$ (2)

$h=\frac{9.8 m/s^{2}}{2}(0.5 s)^{2}$ (3)

Finally:

$h=1.22 m$

4 0

2 years ago