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ololo11 [35]
2 years ago
7

Which statement is true about the process shown?

Physics
2 answers:
Lady bird [3.3K]2 years ago
8 0

Answer:

The answer is A, I just took the quiz :)

ik you already got the answer, but just in case anyone else who gets this question looks here, the person who said A was right!!

erma4kov [3.2K]2 years ago
7 0
The correct answer should be a
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What is the force on a 15.5 kg ball that is falling freely due to the pull of gravity
velikii [3]

Answer:

151.9 N

Explanation:

Force = mass x acceleration

Acceleration due to gravity is 9.8 m/s^2 (you should memorize this number).

F = ma

F = (15.5)(9.8)

F = 151.9

3 0
2 years ago
3 track athletes run the 400 meter dash, Jackie 51 seconds and 1500 joules
Aleksandr [31]

Answer:

ish all usual isaiah or easier fall iF.Hg. diff potty off radial ish usually appraisal

7 0
2 years ago
The wave functions for states of the hydrogen atom with orbital quantum number l=0 are much simpler than for most other states,
Thepotemich [5.8K]
True. Fun fact. Hope this helps
4 0
3 years ago
A horizontal clothesline is tied between 2 poles, 16 meters apart. When a mass of 3 kilograms is tied to the middle of the cloth
Alla [95]

Answer:

The magnitude of the tension on the ends of the clothesline is 41.85 N.

Explanation:

Given that,

Poles = 2

Distance = 16 m

Mass = 3 kg

Sags distance = 3 m

We need to calculate the angle made with vertical by mass

Using formula of angle

\tan\theta=\dfrac{8}{3}

\thta=\tan^{-1}\dfrac{8}{3}

\theta=69.44^{\circ}

We need to calculate the magnitude of the tension on the ends of the clothesline

Using formula of tension

mg=2T\cos\theta

Put the value into the formula

3\times9.8=2T\times\cos69.44

T=\dfrac{3\times9.8}{2\times\cos69.44}

T=41.85\ N  

Hence, The magnitude of the tension on the ends of the clothesline is 41.85 N.

4 0
3 years ago
Read 2 more answers
In the sport of parasailing, a person is attached to a rope being pulled by a boat while hanging from a parachute-like sail. A r
IrinaK [193]

Answer:

570 N

Explanation:

Draw a free body diagram on the rider.  There are three forces: tension force 15° below the horizontal, drag force 30° above the horizontal, and weight downwards.

The rider is moving at constant speed, so acceleration is 0.

Sum of the forces in the x direction:

∑F = ma

F cos 30° - T cos 15° = 0

F = T cos 15° / cos 30°

Sum of the forces in the y direction:

∑F = ma

F sin 30° - W - T sin 15° = 0

W = F sin 30° - T sin 15°

Substituting:

W = (T cos 15° / cos 30°) sin 30° - T sin 15°

W = T cos 15° tan 30° - T sin 15°

W = T (cos 15° tan 30° - sin 15°)

Given T = 1900 N:

W = 1900 (cos 15° tan 30° - sin 15°)

W = 570 N

The rider weighs 570 N (which is about the same as 130 lb).

6 0
2 years ago
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