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Scrat [10]
3 years ago
13

A student from Sidney, Australia compares the distance he obtained from the 1-st spark mark to the 25-th spark mark, with the si

milar result of a student who did the experiment in Gaithersburg, Maryland. Both students operated identical setups powered with 60 Hz AC, reported no missing spark marks, and achieved precision of their measurements in four significant figures. What difference in the results was found
Physics
1 answer:
Lady bird [3.3K]3 years ago
7 0

Answer:

The difference is obviously due tithe different gravitational constants as a result of variation in radius if the earth in Maryland and sydney

So

To calculate the difference

We use

distance s = 1/2 gt²

T = time difference between first spark and 25th spark will now be

24x (1/F)

= 24/60

= 0.4 sec

Therefore differnce in reading will be 0.5 x(9.8104 - 9.7967)x 0.42

= 0.002877 units

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Science requires theories that can be tested by research. true false
dezoksy [38]
True Requires the development of theories that can be tested by systematic research.
3 0
3 years ago
When Kevin pulls his cotton shirt off his body, the electrons get transferred from the (shirt or body) to the (shirt or body) .
Masja [62]
<span>When Kevin pulls his cotton shirt off his body, the electrons get transferred from the shirt (in form of static charges i.e. electrons to the body. So, the shirt becomes positively charged and Kevin’s body becomes negatively charged.

As a result of charge transfer from the shirt to the body, we can hear a crackling sound. or if observed in dark, a sparkle can be seen.</span>
6 0
3 years ago
A spring is 20cm long is stretched to 25cm by a load of 50N. What will be its length when stretched by 100N. assuming that the e
IgorLugansk [536]

Answer:

Final Length = 30 cm

Explanation:

The relationship between the force applied on a string and its stretching length, within the elastic limit, is given by Hooke's Law:

F = kΔx

where,

F = Force applied

k = spring constant

Δx = change in length of spring

First, we find the spring constant of the spring. For this purpose, we have the following data:

F = 50 N

Δx = change in length = 25 cm  - 20 cm = 5 cm = 0.05 m

Therefore,

50 N = k(0.05 m)

k = 50 N/0.05 m

k = 1000 N/m

Now, we find the change in its length for F = 100 N:

100 N = (1000 N/m)Δx

Δx = (100 N)/(1000 N/m)

Δx = 0.1 m = 10 cm

but,

Δx = Final Length - Initial Length

10 cm = Final Length - 20 cm

Final Length = 10 cm + 20 cm

<u>Final Length = 30 cm</u>

6 0
3 years ago
An object with an initial velocity of 10 m/s accelerates at a rate of 3.5 m/s2 for 8 seconds. How far will it have traveled duri
ICE Princess25 [194]

Answer:

Distance, d = 192 meters

Explanation:

We have,

Initial velocity of an object is 10 m/s

Acceleration of the object is 3.5 m/s²

Time, t = 8 s

We need to find the distance travelled by the object during that time. Second equation of motion gives the distance travelled by the object. It is given by :

d=ut+\dfrac{1}{2}at^2

d=10\times 8+\dfrac{1}{2}\times 3.5\times 8^2\\\\d=192\ m

So, the distance travelled by the object is 192 meters.

4 0
3 years ago
An astronaut on the moon places a package on a scale and finds its weight to be 10. N. () What would the weight of the package
jeka94

Answer:

(a) 61.25 N

(b) 6.25 kg

(c) 6.25 Kg

Explanation:

Weight on moon = 10 N

Acceleration due to gravity on moon = 1.6 m/s^2

Acceleration due to gravity on earth = 9.8 m/s^2

Let m be the mass of the package.

(a) Weight on earth = mass x acceleration due to gravity on earth

Weight on earth = 6.25 x 9.8 = 61.25 N

(b) Weight on moon = mass x acceleration due to gravity on moon

10 = m x 1.6

m = 6.25 kg

(c) Mass of the package remains same as mass does not change, so the mass of package on earth is 6.25 kg.

8 0
3 years ago
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