20. A mushroom is the

A person throws a 0.21-kg ball straight up into the air. It reaches a height of

Brrunno [24]

Answer:

A

Explanation:

As it begins to fall

F = ma a = 9.81

F = .21 * 9.81 = 2.06 N

7 0

2 years ago

A condition where warmer air lies above cooler air, limiting vertical circulation and trapping pollutants near the surface, is.

xxMikexx [17]

Answer: Temperature inversion

Explanation: Temperature inversion is a reverse of the normal temperature flow or distribution of the air,it usually causes temperature to rise as altitude increase towards the troposphere instead of reduction of temperature.

Temperature inversion is caused when warm,dense air flow above cold,less dense air. Temperature inversion is hazardous to man as it traps pollutants close to the Earth surface,this condition limits vertical circulation.

7 0

3 years ago

On an alien planet, a simple pendulum of length 0.300 m has oscillation frequency 0.709 Hz. Find the period of the pendulum. Fin

sweet [91]

Answer:

1.41s

5.95m/s

0.2746m

Explanation:

The time period

T = 1/f

= 1/0.709s

= 1.41 seconds

We have

T = 2π√l/g

T² = 4π²l/g

g = 4π²l/T²

g = 4x3.14²x0.300/1.41²

g = 5.95m/s² this is the acceleration due to gravity.

Then the time period of the glide

T2 = 2π√m/k

Length of pendulum = l

Time period T

T2 = 2π√l/g

Then T1 = T2

2π√m/k = 2π√l/g

M/k = l/g

L = g.m/k

L = 5.95x0.450/9.75

L = 0.2746

This must be the length of the simple pendulum

8 0

3 years ago

If zak's speed is 3.00 m/s when he starts to slide, what distance d will he slide before stopping? express your answer in meters

OverLord2011 [107]

That will depend on the coefficient of friction between the sliding surfaces, and also on Zak's weight. We don't have any of that information.

8 0

3 years ago

At t1 = 2.00 s, the acceleration of a particle in counterclockwise circular motion is 6.00 i + 4.00 j m/s2 . It moves at constan

Jet001 [13]

The particle moves with constant speed in a circular path, so its acceleration vector always points toward the circle's center.

At time $t_1$ , the acceleration vector has direction $\theta_1$ such that

$\tan\theta_1=\dfrac{4.00}{6.00}\implies\theta_1=33.7^\circ$

which indicates the particle is situated at a point on the lower left half of the circle, while at time $t_2$ the acceleration has direction $\theta_2$ such that

$\tan\theta_2=\dfrac{-6.00}{4.00}\implies\theta_2=-56.3^\circ$

which indicates the particle lies on the upper left half of the circle.

Notice that $\theta_1-\theta_2=90^\circ$ . That is, the measure of the major arc between the particle's positions at $t_1$ and $t_2$ is 270 degrees, which means that $t_2-t_1$ is the time it takes for the particle to traverse 3/4 of the circular path, or 3/4 its period.

Recall that

$\|\vec a_{\rm rad}\|=\dfrac{4\pi^2R}{T^2}$

where $R$ is the radius of the circle and $T$ is the period. We have

$t_2-t_1=(5.00-2.00)\,\mathrm s=3.00\,\mathrm s\implies T=\dfrac{3.00\,\rm s}{\frac34}=4.00\,\mathrm s$

and the magnitude of the particle's acceleration toward the center of the circle is

$\|\vec a_{\rm rad}\|=\sqrt{\left(6.00\dfrac{\rm m}{\mathrm s^2}\right)^2+\left(4.00\dfrac{\rm m}{\mathrm s^2}\right)^2}=7.21\dfrac{\rm m}{\mathrm s^2}$

So we find that the path has a radius $R$ of

$7.21\dfrac{\rm m}{\mathrm s^2}=\dfrac{4\pi^2R}{(4.00\,\mathrm s)^2}\implies\boxed{R=2.92\,\mathrm m}$

8 0

3 years ago