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2 years ago

Which factors could be potential sources of error in the experiment? check all that apply.

Physics

2 answers:

Vadim26 [7]2 years ago

6 0

(A)energy lost in the lever due to friction

(C) visual estimation of height of the beanbag

(E)position of the fulcrum for the lever affecting transfer of energy

WARRIOR [948]2 years ago

6 0

Answer:

The Correct Answer is A, C, and E.

Explanation:

There are many potential sources of errors when designing and carrying pout experiments.

Personal error: Inaccurate observation by observers, If the data collecting has two observers and they have collected two different data which is not matching with the experiment data then it would be considered as the Observer or personal error.
Instrumental errors: If Calibration of the instruments is not carried out correctly then it will produce incorrect data then it will be categorized under Instrumental error of experiment.

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A ski lift is used to transport people from the base of a hill to the top. If the lift leaves the base at a velocity of 15.5 m/s

Neko [114]

This can be calculated with the law of conservation of energy. The sky lift is starting with the speed v= 15.5 m/s and all of it's kinetic energy Ek is transformed to potential energy Ep so the energies have to be equal: Ep=Ek.
Since Ek=(1/2)*m*v² where m is mass and v is the speed, Ep=m*g*h, where m is mass, g= 9.81 m/s² and h is height. Now:

Ek=Ep

(1/2)*m*v²=m*g*h, masses cancel out,

(1/2)*v²=g*h, divide by g to get the height,

(1/2*g)*v²=h and now plug in the numbers:

h=12.245 m. Height of the hill rounded to the nearest tenth is h=12.25 m

5 0

2 years ago

Read 2 more answers

A system dissipates 12 J of heat into the surroundings; meanwhile, 28 J of work is done on the system. What is the change of the

timurjin [86]

Answer:

option C

Explanation:

given,

energy dissipated by the system to the surrounding = 12 J

Work done on the system = 28 J

change in internal energy of the system

Δ U = Q - W

system losses energy = - 12 J

work done = -28 J

Δ U = Q - W

Δ U = -12 -(-28)

Δ U = 16 J

hence, the correct answer is option C

6 0

2 years ago

The actions of an employee are not attributable to the employer if the employer has not directly or indirectly encouraged the em

zepelin [54]

Answer: the answer is d

Explanation: there are not more than 10 violations within a twelve month period hope this helps

4 0

2 years ago

A jogger accelerates from rest to 3.0 m/s in 2.0 s. A car accelerates from 38.0 to 41.0 m/s also in 2.0 s. (a) Find the accelera

Tom [10]

Answer:

a) The acceleration of the jogger is 1.5 m/s²

b) the acceleration of the car is also 1.5 m/s²

c) Yes, the car travels 76 m farther than the jogger.

Explanation:

a) The acceleration of an object is the variation of its velocity over time:

a = final velocity - initial velocity / time

for the jogger:

a = 3.0 m/s - 0 m/s / 2.0 s = 1.5 m/s ²

b) For the car:

a = 41.0 m/s - 38.0 m/s / 2.0 s = 1.5 m/s²

c) Let´s see the position of the car after 2 seconds.

The equation for the position of an accelerated object moving in a straight line is:

x = x0 + v0* t +1/2 * a * t²

Where:

x = position of the car at time "t"

x0 = initial position

v0 = initial velocity

t = time

a = acceleration

Let´s consider x0 = 0 because the origin of the reference system is located where the car starts accelerating. Then:

x = 38,0 m/s * 2 s + 1/2 * 1.5 m/s ² * (2.0 s)²

x = 79 m

In the same way, we can calculate the position of the jogger:

x = 0 m/s * t + 1/2 * 1.5 m/s ² * (2.0 s)²

x = 3 m

The car travels 79 m - 3 m = 76 m farther than the jogger

4 0

2 years ago

An green hoop with mass mh = 2.8 kg and radius rh = 0.13 m hangs from a string that goes over a blue solid disk pulley with mass

Otrada [13]

The only force on the system is the mass of the hoop F net = 2.8kg*9.81m/s^2 = 27.468 N The mass equal of the rolling sphere is found by: the sphere rotates around the contact point with the table.
So by applying the theorem of parallel axes, the moment of inertia of the sphere is computed by:I = 2/5*mR^2 for rotation about the center of mass + mR^2 for the distance of the axis of rotation from the center of mass of the sphere.
I = 7/5*mR^2 M = 7/5*m
Therefore, linear acceleration is computed by:F/m = 27.468 / (2.8 + 1/2*2 + 7/5*4) = 27.468/9.4 = 2.922 m/s^2

7 0

2 years ago