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2 years ago

Prima problema este urgent

Physics

1 answer:

Aleonysh [2.5K]2 years ago

5 0

Answerana alyom kint gahda amshi

Explanation:

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a certain car travels 20 km east then turns south for 13 km finally the car turns east again for 6 km

aliya0001 [1]

a^2+b^2=c^2

20km^2+13km^2=c^2

400km^2+169km^2=c^2

23.853720...km

7 0

3 years ago

Read 2 more answers

A sound with an intensity greater than_____ dB can cause damage to human ears.

sergiy2304 [10]

85 decibels or higher can cause damage to the human ear. Decibels is the value of the measurement of sound. Hearing loss can be caused by noise coming from a loud sound. Noise induced hearing loss maybe permanent loss or temporary loss. Examples of activities that can give you a noise induced hearing loss is target shooting, listening to music in your earphones that have a high volume, and using lawnmowers.

5 0

3 years ago

Assume the following values: d1 = 0.880 m , d2 = 1.11 m , d3 = 0.560 m , d4 = 2.08 m , F1 = 510 N , F2 = 306 N , F3 = 501 N , F4

dsp73

Answer:

= 2630.6 N.m

Explanation:

(FR)x = ΣFx = -F4 = -407 N

(FR)y = ΣFy =-F1-F2 -F3 = -510 - 306 - 501 = -1317 N

(MR)B =ΣM + Σ(±Fd)

= MA + F1(d1 +d2) + F2d2 - F4d3

= 1504 + 510(0.880+1.11) +306(1.11) - 407(0.560)

= 2630.64 N.m (counterclockwise)

6 0

3 years ago

What is the energy stored between 2 Carbon nuclei that are 1.00 nm apart from each other? HINT: Carbon nuclei have 6 protons and

Andrei [34K]

Answer:

$A. 8.29\times 10^{-18}\ J$

Explanation:

Given that:

p = magnitude of charge on a proton = $1.6\times 10^{-19}\ C$

k = Boltzmann constant = $9\times 10^{9}\ Nm^2/C^2$

r = distance between the two carbon nuclei = 1.00 nm = $1.00\times 10^{-9}\ m$

Since a carbon nucleus contains 6 protons.

So, charge on a carbon nucleus is $q = 6p=6\times 1.6\times 10^{-19}\ C=9.6\times 10^{-19}\ C$

We know that the electric potential energy between two charges q and Q separated by a distance r is given by:

$U = \dfrac{kQq}{r}$

So, the potential energy between the two nuclei of carbon is as below:

$U= \dfrac{kqq}{r}\\\Rightarrow U = \dfrac{kq^2}{r}\\\Rightarrow U = \dfrac{9\times10^9\times (9.6\times 10^{-19})^2}{1.0\times 10^{-9}}\\\Rightarrow U =8.29\times 10^{-18}\ J$