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dezoksy [38]
3 years ago
5

When measurements are close to their true values, they are said to be

Physics
2 answers:
Nikitich [7]3 years ago
8 0

Answer:

A) accurate

Explanation:

Accuracy refers to how close a measurement is to the true value of what is being measured.

katovenus [111]3 years ago
3 0

Answer:

When measurements are close to true

value they are called to be accurate

Explanation:

You might be interested in
5 points
Drupady [299]

Answer:

distance is not a vector. it is scalar

5 0
3 years ago
Rutherford's gold-foil experiment led him to conclude that
erica [24]

Answer:

C. a dense region of positive charge existed somewhere in the atom.

Explanation:

Physicist Ernest Rutherford created the gold foil experiment in which he shot a beam of alpha particles at a sheet of gold foil, which then sent a  few of the particles flying after they were deflected. Based on the information gathered after completing this experiment, Rutherford concluded that a dense region of positive charge existed somewhere in the atom.

I hope this answered your question. If you have any more questions feel free to ask away at Brainly.

3 0
3 years ago
An object travels along a straight, horizontal surface with an initial speed of 2 ms. The position of the object as a function o
pashok25 [27]

Answer:

The options are not provided, so i will answer in a general way.

We know that:

The movement is along a straight horizontal surface, then we have one-dimensional motion.

The speed is 2m/s

We want a graph of position vs time.

Now, remember the relation:

Distance = Speed*Time

Then we can write the position as a function of time as:

P(t) = 2m/s*t + P0

Where t is our variable, that represents time in seconds, and P0 is the position at time t = 0seconds, we can assume that this is zero.

Then the equation is:

P(t) = 2m/s*t

And the graph is something like:

3 0
3 years ago
An unstable particle is created in the upper atmosphere from a cosmic ray and travels straight down toward the surface of the ea
german

Answer:

Q1)  Time taking by particle to travel the 40 km wrt. earth = 1.34\times10^{-6} sec.

Q2) The distance traveled by particle where the particle created to surface of earth wrt. particle's frame = 3.84 km.

Q3) The time taking by particle to travel from where it is created to the surface of the earth = 1.285\times10^{-5} sec.

Explanation:

Given :

Speed of particle wrt. earth v=0.99537c

Distance between where particle is created and earth surface = 40 km

we know that,

⇒       v = \frac{x}{t}

Where x = 40\times10^{3} m, v = 0.99537c, we know speed of light c = 3 \times10^{8}

∴      t = \frac{x}{v}

         = \frac{40 \times10^{3} }{0.99537\times3\times10^{8} }

      t = 1.34\times10^{-6} sec

∴ Thus, time taking by particle to travel the 40 km wrt. earth t = 1.34\times10^{-6} sec

According to the lorentz transformation,

⇒    l = l_{o} \sqrt{1-\frac{v^2}{c^2} }

Where l = improper length, l_{o} =proper length (distance measured wrt. rest frame) = 40 km

     l = 40 \sqrt{1-\frac{v^2}{c^2 }

     l = 40 \times 0.096

     l = 3.84 km

∴ Thus, the distance traveled by particle where the particle created to surface of earth wrt. particle's frame = 3.84 km.

According to the time dilation,

   \Delta t = \frac{\Delta t_{o} }{\sqrt{1-\frac{v^2}{c^2} } }

Where \Delta t = improper time (wrt. earth frame time) =1.34\times10^{-6} sec ,  \Delta t _{o} = proper time (wrt. particle frame).

 1.34\times10^{-6} = \frac{ \Delta t_{o}}{0.096}

 \Delta t_{o} = 1.285 \times10^{-5} sec

Thus, the time taking by particle to travel from where it is created to the surface of the earth = 1.285 \times10^{-5} sec.

6 0
3 years ago
A string under a tension of 36 N is used to whirl a rock in a horizontal circle of radius 3.6 m at a speed of 16.12 m/s. The str
ExtremeBDS [4]

Answer:

6010.457N

Explanation:

Centripetal acceleration = a= V²/R

At a radius of 3.6m and velocity of 16.12m/s,

Acceleration is

a = 16.12²/ 3.6 = 72.182 m/s²

Force = Mass (m) * Acceleration (a)

36 = m * 72.182

m = 36/72.182

At breaking point

Radius = 0.468 m and Velocity = 75.1 m/s

a = V²/R = 75.1²/0.468

a = 12051.3 m/s

F = Mass(m) * Acceleration (a)

F = m * 12051.3

m = F/ 12051.3

Settings the ratio of mass equal

m = m

=> 36/72.182 = F/12051.3

F = 12051.3 * 36/72.182

F = 6010.457N

3 0
3 years ago
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