People who drink coffee are more likely to develop cancer

20 points, just need a basic understanding. You are on a mystery planet, what you know is that from a height of 10.0 meters, a d

ICE Princess25 [194]

(since you asked for basic understanding only, I am not including actual calculations. Please let me know in the comments section if you wish to verify your solution(s))

For (b): Use the formula for distance (s) made during an accelerated motion:

$s = \frac{1}{2}at^2+ v_0t+s_0= \frac{1}{2}at^2= \frac{1}{2}gt^2$

with v_0 and s_0 being the initial velocity and distance, both 0 in this case, and with "a" denoting the acceleration, in this case solely due to gravitational acceleration so: "g."

You are given the distance made, namely 10 m, and the duration t (0.88s) and so using the formula above you can solve for g.

For (c), to determine the final velocity at time 0.88s use the formula for the instantaneous velocity of an accelerated motion

(velocity at time t) = (acceleration) x (time)

again, with acceleration due to gravity, i.e., a = g and with g as determined under (b).

If my calculation is correct, this mystery planet could be the Jupiter.

7 0

4 years ago

A portable basketball set has a base and a post arrangement. The post arrangement consists of a post, backboard, hoop and net. T

Ray Of Light [21]

The rotational equilibrium condition allows finding the response to the minimum force of the wind and what happens when changing the water for sand, in the system

a) The minimum force of the wind that turns the system is Fw = 17.64 N

b) The system resists much greater forces because the base has more mass

Newton's Second Law can be applied to rotational motion in this case when the angular acceleration is zero we have the special case of rotational equilibrium

Σ τ = 0

Where τ is the torque

The reference system is a coordinate system with respect to which the torques are measured, in this case we will fix the system at the turning point, the junction of the base and the pole, we will assume that the counterclockwise rotations are positive.

For the torque the distance used is the perpendicular distance from the direction of the force to the axis of rotation, let's find this distance for each force

Wind force

cos 15 = $\frac{y_w}{2.35}$

$y_w$ = 2.35 cos 15

Post Weight

sin 15 = $\frac{x_p}{2.00}$

xp = 2.0 sin 15

Base weight

cos (90-15) = $\frac{x_b}{0.25}$

xB = 0.25 cos 75

Let's substitute in the rotational equilibrium equation

$F_w \ y_w + W_p \ x_p - W_b \ x_b = 0$

a) To calculate the minimum wind force we substitute the given values

They indicate the weight of the post is $W_p$ = 26.0 N and the weight of the base with water is $W_b$ = 810 N

$F_w = \frac{W_b \ x_b - W_p \ x_p }{y_w}$

$F_w = \frac{W_b \ 0.25 cos75 \ - W_p \ 2 sin 15}{2.35 cos 15}$

Let's calculate

$F_w = \frac{810 \ 0.25 \ cos75 \ - 26.0 \ 2 \ sin 15}{2.35 cos15}\\F_w = \frac{52.41 - 10.30}{2.3699}$

$F_w$ = 17.64 N

b) The water is exchanged for sand.

In this case, as the density of the sand is greater than that of the water, the base will have more weight, so it will resist stronger winds before turning over.

Using the rotational equilibrium condition we can find the response to the minimum force of the wind and what happens when changing the water for sand,

a) the minimum force of the wind that turns the system is Fw = 17.64 N

b) the system resists much greater forces because the base has more mass

Learn more here: brainly.com/question/7031958

3 0

3 years ago

Predict the date of the next full moon.

lapo4ka [179]

Answer:

The 19th

Explanation:

going based off of the calandar, you just start from the 4 at the bottom and then just count up as you go through it agains starting back at the top.

4 0

3 years ago

Read 2 more answers

The water in a tank is pressurized by air, and the pressure is measured by a multifluid manometer as shown in Figure below.. Det

Savatey [412]

Answer:

Spongebob: Bye Mr. Krabs! Bye Squidward! BYE SQUIDWARD!

Patrick: (clearly triggered) Why'd you say "bye squidward" twice?

Spongebob: I LiKe SqUiDwArD

5 0

3 years ago

Sources of error in a diverging lens experiment

Snezhnost [94]

Answer:

Lens and the virtual image was the image distance. Error was made when measuring the distances and determining values that were reasonable for focal length, so this error was accounted for when calculating the focal length by adding or subtracting uncertainties from the values.Plane mirrors, convex mirrors, and diverging lenses can never produce a real image. A concave mirror and a converging lens will only produce a real image if the object is located beyond the focal point (i.e., more than one focal length away). 5.

3 0

3 years ago