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lesya692 [45]
2 years ago
7

The measured value of mass M in an experiment is M = 0.743 ± 0.005kg. The error in 2M is

Physics
1 answer:
omeli [17]2 years ago
8 0

Given the measured value of mass M in the experiment the error in 2M is  ; dM

<u>Given data :</u>

Measured value of mass M = ( 0.743 ± 0.005 ) kg

<h3>The error in 2M in the measured value </h3>

The error in 2M in the measured value M = 0.743 ± 0.005 kg is dM while the error in M² will be 2dM

Hence we can conclude that the measured value of mass M in the experiment the error in 2M is  ; dM.

Learn more about error in 2M : brainly.com/question/6650225

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Eddi Din [679]

Answer:

Decrease the voltage,and do not change the resistance,the current will also decrease

Explanation:

Decrease the voltage,and do not change the resistance, the current will also decrease, because voltage is directly proportional to current

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3 years ago
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A cannon, positioned on a hill, shoots a cannonball horizontally at 23 m/s. The cannonball hits the stone wall 1.96 m below the
irina [24]

Answer: 14. 49 m

Explanation:

We can solve this problem with the following equations:

x=V_{o} cos \theta t (1)

y-y_{o}=V_{o} sin \theta t-\frac{1}{2}gt^{2} (2)

Where:

x is the horizontal distance between the cannon and the ball

V_{o}=23 m/s is the cannonball initial velocity

\theta=0\° since the cannonball was shoot horizontally

t is the time

y=0 is the final height of the cannonball

y_{o}=1.96 m is the initial height of the cannonball

g=9.8 m/s^{2} is the acceleration due gravity

Isolating t from (2):

t=\sqrt{-\frac{2(y-y_{o})}{g}} (3)

t=\sqrt{-\frac{2(0 m-1.96 m)}{9.8 m/s^{2}}} (4)

t=0.63 s (5)

Substituting (5) in (1):

x=(23 m/s) cos(0\°) 0.63 s (6)

Finally:

x=14.49 m

5 0
3 years ago
Which of the following is not an example of work?
barxatty [35]
<span>The following which is not an example of work is </span>C. holding a tray in the cafeteria line because <span>if force displaces an object it should work. I think it's clear and I am pretty sure this answer will help you.</span>
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Which county in Florida is most in need of safe rooms and hurricane ties?
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3 years ago
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Each plate of a parallel‑plate capacitor is a square of side 4.19 cm, 4.19 cm, and the plates are separated by 0.407 mm. 0.407 m
alexandr1967 [171]

Answer:

The electric field strength inside the capacitor is 49880.77 N/C.

Explanation:

Given:

Side length of the capacitor plate (a) = 4.19 cm = 0.0419 m

Separation between the plates (d) = 0.407 mm = 0.407\times 10^{-3}\ m

Energy stored in the capacitor (U) = 7.87\ nJ=7.87\times 10^{-9}\ J

Assuming the medium to be air.

So, permittivity of space (ε) = 8.854\times 10^{-12}\ F/m

Area of the square plates is given as:

A=a^2=(0.0419\ m)^2=1.75561\times 10^{-3}\ m^2

Capacitance of the capacitor is given as:

C=\dfrac{\epsilon A}{d}\\\\C=\frac{8.854\times 10^{-12}\ F/m\times 1.75561\times 10^{-3}\ m^2 }{0.407\times 10^{-3}\ m}\\\\C=3.819\times 10^{-11}\ F

Now, we know that, the energy stored in a parallel plate capacitor is given as:

U=\frac{CE^2d^2}{2}

Rewriting in terms of 'E', we get:

E=\sqrt{\frac{2U}{Cd^2}}

Now, plug in the given values and solve for 'E'. This gives,

E=\sqrt{\frac{2\times 7.87\times 10^{-9}\ J}{3.819\times 10^{-11}\ F\times (0.407\times 10^{-3})^2\ m^2}}\\\\E=49880.77\ N/C

Therefore, the electric field strength inside the capacitor is 49880.77 N/C

8 0
3 years ago
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