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2 years ago

Define limitations in an experiment

Physics

1 answer:

laila [671]2 years ago

4 0

Answer:

The limitations of the experiment are those characteristics of design or methodology that impacted or influenced the application or interpretation of the results of your study.

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A tennis ball connected to a string is spun around in a vertical, circular path at a uniform speed. The ball has a mass m = 0.15

Oksanka [162]

1) 5.5 N

When the ball is at the bottom of the circle, the equation of the forces is the following:

$T-mg = m\frac{v^2}{R}$

where

T is the tension in the string, which points upward

mg is the weight of the string, which points downward, with

m = 0.158 kg being the mass of the ball

g = 9.8 m/s^2 being the acceleration due to gravity

$m \frac{v^2}{R}$ is the centripetal force, which points upward, with

v = 5.22 m/s being the speed of the ball

R = 1.1 m being the radius of the circular trajectory

Substituting numbers and re-arranging the formula, we find T:

$T=mg+m\frac{v^2}{R}=(0.158 kg)(9.8 m/s^2)+(0.158 kg)\frac{(5.22 m/s)^2}{1.1 m}=5.5 N$

2) 3.9 N

When the ball is at the side of the circle, the only force acting along the centripetal direction is the tension in the string, therefore the equation of the forces becomes:

$T=m\frac{v^2}{R}$

And by substituting the numerical values, we find

$T=(0.158 kg)\frac{(5.22 m/s)^2}{1.1 m}=3.9 N$

3) 2.3 N

When the ball is at the top of the circle, both the tension and the weight of the ball point downward, in the same direction of the centripetal force. Therefore, the equation of the force is

$T+mg=m\frac{v^2}{R}$

And substituting the numerical values and re-arranging it, we find

$T=m\frac{v^2}{R}-mg=(0.158 kg)\frac{5.22 m/s)^2}{1.1 m}-(0.158 kg)(9.8 m/s^2)=2.3 N$

4) 3.3 m/s

The minimum velocity for the ball to keep the circular motion occurs when the centripetal force is equal to the weight of the ball, and the tension in the string is zero; therefore:

$T=0\\mg = m\frac{v^2}{R}$

and re-arranging the equation, we find

$v=\sqrt{gR}=\sqrt{(9.8 m/s^2)(1.1 m)}=3.3 m/s$

7 0

2 years ago

Determine the angular velocity ω of the telescope as it orbits around the Sun.

lara31 [8.8K]

The JWST is postioned about 1.5 million kilometers from the earth on the side facing away from the sun

5 0

2 years ago

Which of the following is a derived unit? ampere newton second kilogram

Soloha48 [4]

Newton is your answer.

ampere, second, kilograms, are all base units.

hope this helps

5 0

2 years ago

The average power dissipated by a resistor connected to a sinusoidal emf is 5.0 W.

loris [4]

I this is tricky one sec.. I would go with b

4 0

3 years ago

How much work is done in holding a 15 N sack of potatoes while waiting in line at the

marta [7]

Answer: 2700 Joules

Explanation:

Given variables are:

Workdone (w) = ?

Weight of sack (w) = 15N

Time taken (t) = 3 minutes

The standard unit of time is seconds, so convert 3 minutes to seconds

(If 1 minute = 60 seconds

3 minutes = 3 x 60 = 180 seconds)

Recall that work is done is when force is applied on an object over a distance

i.e Workdone = force x distance.

(However, the force acting on the sack of potatoes is weight)

Hence, workdone = weight x time

w = 15N x 180 seconds

w = 2700 J

thus, 2700 joules of work was done.

4 0

3 years ago