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svlad2 [7]
3 years ago
8

The diagram below shows a 1.0 x 105newton truck at rest on a hill that makes an angle of 8.0° with the

Physics
1 answer:
Digiron [165]3 years ago
7 0

Answer:

it \: is \: mg \sin(8.0 \degree)  \\  = 1.0 \times 10 {}^{5}  \times  \sin(8.0 \degree)  \\  = 9.9 \times 10 {}^{4}  newtons

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Asteroids are between 1000 km and less than 10 m in diameter. What is the diameter of most asteroids?
e-lub [12.9K]

This question needs research to be answered. From the given information alone it can't be answered without making wild assumptions.

Ideally, you need to take a look at a distribution (or a histogram) of asteroid diameters, identify the "mode" of such a distribution, and find the corresponding diameter. That value will be the answer.

I am attaching one such histogram on asteroid diameters from the IRAS asteroid catalog I could find online. (In order to get a single histogram, you need to add the individual curves in the figure first). Eyeballing this sample, I'd say the mode is somewhere around 10km, so the answer would be: the diameter of most asteroid from the IRAS asteroid catalog is about 10km.

7 0
3 years ago
A certain atom has atomic number Z = 25 and atomic mass number A = 52. a. What is the approximate radius of the nucleus of this
wlad13 [49]

Answer:

a)The approximate radius of the nucleus of this atom is 4.656 fermi.

b) The electrostatic force of repulsion between two protons on opposite sides of the diameter of the nucleus is 2.6527

Explanation:

r=r_o\times A^{\frac{1}{3}}

r_o=1.25 \times 10^{-15} m = Constant for all nuclei

r = Radius of the nucleus

A = Number of nucleons

a) Given atomic number of an element = 25

Atomic mass or nucleon number = 52

r=1.25 \times 10^{-15} m\times (52)^{\frac{1}{3}}

r=4.6656\times 10^{-15} m=4.6656 fm

The approximate radius of the nucleus of this atom is 4.656 fermi.

b) F=k\times \frac{q_1q_2}{a^2}

k=9\times 10^9 N m^2/C^2 = Coulombs constant

q_1,q_2 = charges kept at distance 'a' from each other

F = electrostatic force between charges

q_1=+1.602\times 10^{-19} C

q_2=+1.602\times 10^{-19} C

Force of repulsion between two protons on opposite sides of the diameter

a=2\times r=2\times 4.6656\times 10^{-15} m=9.3312\times 10^{-15} m

F=9\times 10^9 N m^2/C^2\times \frac{(+1.602\times 10^{-19} C)\times (+1.602\times 10^{-19} C)}{(9.3312\times 10^{-15} m)^2}

F=2.6527 N

The electrostatic force of repulsion between two protons on opposite sides of the diameter of the nucleus is 2.6527

6 0
2 years ago
Please help on this one someone
Nonamiya [84]

Valence electrons are the electrons in the outermost energy level of an atom — in the energy level that is farthest away from the nucleus.

I think it's A.

6 0
3 years ago
Draw the following patterns as they would appear when viewed through a compound microscope.:
goblinko [34]
2 j x b
H c e c hope this helppppsssssssss
3 0
3 years ago
Read 2 more answers
if somebody swim to the right for 30 meters then swims to left for 15 meters and then swims back to the right for 50 meters what
zubka84 [21]

50m

Explanation:

Displacement is the length of path traveled which is measured from start to the finishing of the path.

    Analysis of the journey;

 Starts from:

 0                                    30m                                              from right

                  15m                                                                       to left

                                                                       50m              to right

                 

The displacement is 50m from the starting point.

Distance is total path traveled and for this problem it is 30+ 15 + 50 = 95m

learn more:

displacement brainly.com/question/5461768

#learnwithBrainly

8 0
2 years ago
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