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Serhud [2]
3 years ago
8

Let's start with an example from history. Listed below are a series of claims regarding United States President John F. Kennedy

(1917-1963). Classify each statement according to whether or not it is falsifiable.
Physics
1 answer:
ruslelena [56]3 years ago
7 0

Answer:

Note that both of the falsifiable claims in this example happen to be true. The claim about Kennedy being the 35th President is falsifiable because it can be checked against historical records. The claim that Kennedy died from a bullet in his brain is falsifiable because it could have been shown false by the medical examiner. The remaining claims are not falsifiable: Statements that call on any type of supernatural being are by definition out of the realm of science. Similarly, a claim of something being "undetectable" could not be falsified, and a claim about what Kennedy would have done if he had lived is a conjecture that cannot be disproven.

Explanation:

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. Why are scientific models important? They help scientists prove scientific theories. They help scientists explain complex ques
sdas [7]
The best answer would be the 4th choice. "They help scientists explain concepts that are difficult to observe, this also covers the first answer which helps the scientist to answer complex questions. A scientific model is not used prove scientific laws as they may not always have all the data to prove so, instead it is used to allow them to explain better concepts revolving around science through research and may also allow them to predict results based on the accumulation of data and analyzing the trend of this resulting information.
3 0
3 years ago
565,900 seconds into days​
alukav5142 [94]

Answer:

6

Explanation:

4 0
3 years ago
A ball is dropped from the top of a building.After 2 seconds, it’s velocity is measured to be 19.6 m/s. Calculate the accelerati
zlopas [31]

Answer:

acceleration, a = 9.8 m/s²

Explanation:

'A ball is dropped from the top of a building' indicates that the initial velocity of the ball is zero.

u = 0 m/s

After 2 seconds, velocity of the ball is 19.6 m/s.

t = 2s, v = 19.6 m/s

Using

v = u + at

19.6 = 0 + 2a

a = 9.8 m/s²

6 0
3 years ago
Read 2 more answers
The Hall effect can be used to calculate the charge-carrier number density in a conductor. A conductor carrying current of 2.0 A
marysya [2.9K]

Answer:

option D

Explanation:

given,

A conductor is carrying current = 2.0 A is 0.5 mm thick

Hall voltage = 4.5 x 10-6 V

uniform magnetic field  =  1.2 T

density of the charge = n =?

hall voltage =V_h =\dfrac{i\ B}{n\ e\ L}

n = \dfrac{i\ B}{V\ e\ L}

n = \dfrac{2 \times 1.2 }{4.5 \times 10^{-6}\times 1.6 \times 10^{-19} \times 0.5 \times 10^{-3}}

n = 6.67 × 10²⁷ charges/m

hence the correct answer is option D

7 0
3 years ago
Three masses (3 kg, 5 kg, and 7 kg) are located in the xy-plane at the origin, (2.3 m, 0), and (0, 1.5 m), respectively.
Artist 52 [7]

Answer:

a) C.M =(\bar x, \bar y)=(0.767,0.7)m

b) (x_4,y_4)=(-1.917,-1.75)m

Explanation:

The center of mass "represent the unique point in an object or system which can be used to describe the system's response to external forces and torques"

The center of mass on a two dimensional plane is defined with the following formulas:

\bar x =\frac{\sum_{i=1}^N m_i x_i}{M}

\bar y =\frac{\sum_{i=1}^N m_i y_i}{M}

Where M represent the sum of all the masses on the system.

And the center of mass C.M =(\bar x, \bar y)

Part a

m_1= 3 kg, m_2=5kg,m_3=7kg represent the masses.

(x_1,y_1)=(0,0),(x_2,y_2)=(2.3,0),(x_3,y_3)=(0,1.5) represent the coordinates for the masses with the units on meters.

So we have everything in order to find the center of mass, if we begin with the x coordinate we have:

\bar x =\frac{(3kg*0m)+(5kg*2.3m)+(7kg*0m)}{3kg+5kg+7kg}=0.767m

\bar y =\frac{(3kg*0m)+(5kg*0m)+(7kg*1.5m)}{3kg+5kg+7kg}=0.7m

C.M =(\bar x, \bar y)=(0.767,0.7)m

Part b

For this case we have an additional mass m_4=6kg and we know that the resulting new center of mass it at the origin C.M =(\bar x, \bar y)=(0,0)m and we want to find the location for this new particle. Let the coordinates for this new particle given by (a,b)

\bar x =\frac{(3kg*0m)+(5kg*2.3m)+(7kg*0m)+(6kg*a)}{3kg+5kg+7kg+6kg}=0m

If we solve for a we got:

(3kg*0m)+(5kg*2.3m)+(7kg*0m)+(6kg*a)=0

a=-\frac{(5kg*2.3m)}{6kg}=-1.917m

\bar y =\frac{(3kg*0m)+(5kg*0m)+(7kg*1.5m)+(6kg*b)}{3kg+5kg+7kg+6kg}=0m

(3kg*0m)+(5kg*0m)+(7kg*1.5m)+(6kg*b)=0

And solving for b we got:

b=-\frac{(7kg*1.5m)}{6kg}=-1.75m

So the coordinates for this new particle are:

(x_4,y_4)=(-1.917,-1.75)m

5 0
3 years ago
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