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Ksju [112]
2 years ago
6

You probably noticed that the simulation returned varying values for the magnetic field. What other source(s) of error could occ

ur from the simulation
Physics
1 answer:
romanna [79]2 years ago
5 0

Physical and chemical laboratory experiments include three primary sources of error: systematic error, random error and human error. this simulation shows systematic error.

Systematic error applies to repeated measuring of the same object under repeated conditions of measurement. The amount of the error value is either positive or negative; which is called the fixed system error in laboratory experiments and laboratory tests.

Human error in laboratory experiments and lab tests primarily refers to the mistake in physical and chemical inspection phase caused by the factors of the inspector.

Random error is error caused by unknown factors. This error poses erratic changes at random, primarily due to a variety of small, independent, and accidental factors. The random error is atypical from the surface.

Learn more about systematic error here-

brainly.com/question/14149934

#SPJ4

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What does the big bang theory suggest
Alika [10]

Answer:

The Big Bang Theory is the leading explanation about how the universe began. At its simplest, it says the universe as we know it started with a small singularity, then inflated over the next 13.8 billion years to the cosmos that we know today.

Explanation: Hope this help! :)

7 0
3 years ago
A 2kg rock is moving at a speed of 6m/s. What constant force is needed to stop the rock in 7 x 10^-4?
dsp73

Explanation:

key to this problem is the impulse-momentum theorem which states that the change in the momentum of an object is equal to the impulse applied into it.

J

=

Δ

p

,

where

J

is the impulse and

Δ

p

is the change in momentum. Basically, the impulse is the product of force and time duration, that is,

J

=

F

Δ

t

In this problem, the impulse would be the product of the force stopping the rock and

0.7

s

.

On the other hand, momentum

p

is the product of the mass

m

and velocity

v

. Therefore, the change in momentum is given by

Δ

p

=

m

2

v

2

−

m

1

v

1

.

Starting with the impulse-momentum equation, we have

J

=

Δ

p

F

Δ

t

=

m

2

v

2

−

m

1

v

1

Divide both sides by

Δ

t

,

we get

F

Δ

t

Δ

t

=

m

2

v

2

−

m

1

v

1

Δ

t

F

=

m

2

v

2

−

m

1

v

1

Δ

t

Finally, substitute the values and we get

F

=

(

2

kg

)

(

0

)

−

(

2

kg

)

(

6

m

s

)

(

0.7

s

)

F

≈

−

20

kg

m

s

2

Since

1

N

=

1

kg

m

s

2

,

then

F

≈

−

20

N

Therefore, using the correct significant figures (in this case, we need one significant figure since 2 kg, 6 m/s and 0.7 s all have one) in the final answer, we would need to have approximately

20

N

force to stop the rock in

0.7

s

.

Note: The negative sign is referring to the direction of the force opposite of the direction of the velocity

v

1

.

6 0
3 years ago
A brave child decides to grab onto an already spinning merry‑go‑round. The child is initially at rest and has a mass of 25.5 kg.
Svetllana [295]

Answer:

72.75 kg m^2

Explanation:

initial angular velocity, ω = 35 rpm

final angular velocity, ω' =  19 rpm

mass of child, m = 15.5 kg

distance from the centre, d = 1.55 m

Let the moment of inertia of the merry go round is I.

Use the concept of conservation of angular momentum

I ω = I' ω'

where I' be the moment of inertia of merry go round and child

I x 35 = ( I + md^2) ω'

I x 35 = ( I + 25.5 x 1.55 x 1.55) x 19

35 I = 19 I + 1164

16 I = 1164

I = 72.75 kg m^2

Thus, the moment of inertia of the merry go round is 72.75 kg m^2.

5 0
3 years ago
Una muestra de 500g de agua se calienta desde 10°C hasta 80°C. Calcula la cantidad de calor absorbido por líquido si su calor es
weeeeeb [17]

Answer:

Q = 142.324kJ

Explanation:

Data:

M = 500g = 0.5kg

T1 = 10°C = (10 + 273.15)K = 285.15K

T2 = 80°C = (80 + 273.15)K = 353.15K

Q = ?

C = 4186J/kg.K

Q = mc(T2 - T1)

Q = 0.5 * 4186 * (353.15 - 285.15)

Q = 0.5 * 4186 * 68

Q = 142324J

Q = 142.324kJ.

7 0
3 years ago
What is the magnitude of the torque about his shoulder if he holds his arm straight out to his side, parallel to the floor
const2013 [10]

Complete Question

An athlete at the gym holds a 3.0 kg steel ball in his hand. His arm is 70 cm long and has a mass of 4.0 kg. Assume, a bit unrealistically, that the athlete's arm is uniform.

What is the magnitude of the torque about his shoulder if he holds his arm straight out to his side, parallel to the floor? Include the torque due to the steel ball, as well as the torque due to the arm's weight.

Answer:

The torque is  \tau = 34.3 \  N\cdot m

Explanation:

From the question we are told that

   The mass of the steel ball is  m  =  3.0 \  kg

    The length of arm is  l =  70 \ cm  = 0.7 \  m

    The mass of the arm is m_a  = 4.0 \  kg

Given that the arm of the athlete is uniform them the distance from the shoulder to the center of gravity of the arm is mathematically represented as

       r = \frac{l}{2}

=>    r = \frac{ 0.7}{2}  

=>    r = 0.35 \ m  

Generally the magnitude of torque about the athlete shoulder is mathematically represented as

      \tau =  m_a * g * r  + m * g *  L

=>    \tau =  4 * 9.8 * 0.35 + 3 * 9.8 *  0.70

=>    \tau = 34.3 \  N\cdot m

5 0
3 years ago
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