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GenaCL600 [577]
3 years ago
15

Pls help i will give u brianliest!

Physics
1 answer:
Fofino [41]3 years ago
3 0

Answer:

Explanation:

neautrons and protons

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What information could you gather about a star if its light curve had multiple
yanalaym [24]

The information that could be gathered about a star whose light curve has multiple symmetrical depths is ; The shape and surface variegation of the star

The light curves of a KBO ( moons and stars ) are measured as a rate of the brightness of a star in relation to time. therefore the study of the light curve having multiple symmetrical depths ( depth of brightness ) will give an information about the shape/size and the surface variegation of the star

Hence we can conclude that The information that could be gathered about a star whose light curve has multiple symmetrical depths is ; The shape and surface variegation of the star

Learn more : brainly.com/question/19573734

7 0
2 years ago
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what is the kinetic energy of a 1 kilogram ball is thrown into the air with an initial velocity of 3 m/sec
s344n2d4d5 [400]
Data:
m (mass) = 1 Kg
s (speed) = 3 m/s
Kinetic energy = ? (Joule)


Formula (Kinetic energy)
E_{k} =  \frac{m*s^2}{2}

Solving:
E_{k} = \frac{m*s^2}{2}
E_{k} =  \frac{1*3^2}{2}
E_{k} =  \frac{1*9}{2}
E_{k} =  \frac{9}{2}
\boxed{\boxed{E_{k} = 4.5\:J}}\end{array}}\qquad\quad\checkmark
3 0
4 years ago
Một con lắc lò xo gồm 1 quả nặng có m= 0,2kg treo vào lò xo có độ cứng k= 100N/m, cho vật dao động điều hoà theo phương thẳng đứ
AleksandrR [38]
Honestly same man I think it’s b
4 0
3 years ago
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In a standard tensile test, a steel rod of 7 8-in. diameter is subjected to a tension force of 17 kips. Knowing that ν = 0.30 an
natali 33 [55]

Answer:

(a) Elongation of the rod==5.61×10⁻⁹m

(b) Change in diameter=1.640×10⁻⁸m

Explanation:

Given data

Diameter d=78 in=1.9812 m

Cross Area is:

A=(\pi /4)d^{2} \\A=(\pi /4)(1.9812m)^{2}\\A=3.08m^{2}

Applied Load P=17 KN=17×10³N

E=29 × 106 psi=1.99947961×10¹¹Pa

Stress and Strain in x direction

Stress in x direction

σ=P/A

=\frac{17*10^{3}N }{3.08m^{2} }\\ =5517.25Pa

σ=5517.25 Pa

Strain in x direction

ε=σ/E

=\frac{5517.25}{1.99947961*10^{11} } \\=2.76*10^{-8}

ε=2.76×10⁻⁸

Part (a)

Elongation of the rod=Lε

=(0.2032)(2.76×10⁻⁸)

Elongation of the rod==5.61×10⁻⁹m

Part(b) Change in diameter

Strain in y direction

ε₁= -vε

ε₁= -(0.30)(2.76×10⁻⁸)

ε₁=-8.28×10⁻⁹

Change in diameter=d×ε₁

Change in diameter=(1.9812m)×(-8.28×10⁻⁹)

Change in diameter=1.640×10⁻⁸m

4 0
3 years ago
Newly discovered planet has twice the mass and three times the radius of the earth. What is the free-fall acceleration at its su
skad [1K]

Answer:

g_n=\dfrac{2}{9}g

Explanation:

M = Mass of Earth

G = Gravitational constant

R = Radius of Earth

The acceleration due to gravity on Earth is

g=\dfrac{GM}{R^2}

On new planet

g_n=\dfrac{G2M}{(3R)^2}\\\Rightarrow g_n=\dfrac{2GM}{9R^2}

Dividing the two equations we get

\dfrac{g_n}{g}=\dfrac{\dfrac{2GM}{9R^2}}{\dfrac{GM}{R^2}}\\\Rightarrow \dfrac{g_n}{g}=\dfrac{2}{9}\\\Rightarrow g_n=\dfrac{2}{9}g

The acceleration due to gravity on the other planet is g_n=\dfrac{2}{9}g

4 0
3 years ago
Read 2 more answers
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