Don’t know how to do this can someone help

A 15 g ball at the end of 2.4 m string is swung in the horizontal circle. It revolves once every 2.09 s. What is the magnitude o

Mrrafil [7]

Answer:

0.325 N

Explanation:

From the question,

T = 4π²rm/t²............................ Equation 1

Where T = Tension, r = radius or length of the string, m = mass of the string, t = time.

Given: r = 2.4 m, m = 15 g = 0.015 kg, t = 2.09 s.

Constant: π = 3.14

Substitute these values into equation 1

T = 4(3.14²)(2.4)(0.015)/2.09²

T = 1.4198/4.3681

T = 0.325 N

3 0

3 years ago

Cohort studies allow the investigator to examine multiple outcomes and multiple exposures. Consider the following three exposure

mojhsa [17]

Answer:

See the explanation for the answer.

Explanation:

Cohort study for smoking

Outcomes that were examined

1. People at risk of lung cancer due to smoking

2. Influence of outdoor air pollution in lung cancer

3. Variation in tobacco smoking and heavy tobacco smoking was observed.

4. Occupational factors and housing factor were observed.

Cohort study for low vitamin D

Outcomes that were examined

1. People at risk of vitamin D deficiency.

2. Observing value in summer.

3. Correlating vitamin D value with weather and age

4. 25-hydroxy vitamin D was observed.

Cohort study for cold weather

Outcomes that were examined

1. Allergic and asthma risked people were observed.

2. effect on the people when temperature goes up.

3. Observation on the most harsh weather day.

4. Correlating it with age and analysis.

3 0

4 years ago

A 68.2 kg base runner begins his slide into second base while moving at a speed of 4.33 m/s. He slides so that his speed is zero

qwelly [4]

Answer:

$147.653J = W_f$

Explanation:

We know that:

$E_i - E_f = W_f$

where $E_i$ is the initial energy, $E_f$ is the final energy and $W_f$ is the energy lost due to friction.

so:

$\frac{1}{2}MV^2-0 = W_f$

Where M is the mass and V is the velocity of the runner, so:

$\frac{1}{2}(68.2kg)(4.33m/s)^2 = W_f$

$147.653J = W_f$

7 0

4 years ago

a stone is dropped from top of a tower of 50m high ,simltaneously another stone is thrown upward with a speed of 20m/s find the

aleksklad [387]

At the time that I'll call ' Q ', the height of the stone that was
dropped from the tower is

 H = 50 - (1/2 G Q²) ,

and the height of the stone that was tossed straight up
from the ground is

 H = 20Q - (1/2 G Q²) .

The stones meet when them's heights are equal,
so that's the time when

 50 - (1/2 G Q²) = 20Q - (1/2 G Q²) .

This is looking like it's going to be easy.

Add (1/2 G Q²) to each side.
Then it says
 50 = 20Q

Divide each side by 20: 2.5 = Q .

And there we are. The stones pass each other

 2.5 seconds

after they are simultaneously launched.

5 0

3 years ago

During operations outside controlled airspace at altitudes of more than 1,200 feet AGL, but less than 10,000 feet MSL, the minim

slavikrds [6]

Answer:

500 feet

Explanation:

It's stipulated that a minimum horizontal distance of 2000 feet from clouds should be allowed for VFR flights but at a night, the minimum distance below clouds requirement for VFR flight outside controlled airspace at altitudes of more than 1200 feet AGL, but less than 10000 feet MSL should be 500 feet.

3 0

3 years ago