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3 years ago

An atom has 35 protons. what element is it?

Physics

2 answers:

GenaCL600 [577]3 years ago

5 0

bromine has and atomic number of 35 which means it has 35 protons

zvonat [6]3 years ago

4 0

The element that has 35 protons is Bromine

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In the 1920s what did Edmund hubble notice about the galaxies

zlopas [31]

Hubble noticed that the galaxies were moving away from us, which meant the universe was expanding.

This is why constellations change over time. In some years, the Big Dipper won't actually look like a dipper anymore.

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3 years ago

Please......... I need desperate help

PIT_PIT [208]

' A ' is one crest of the wave. After every wavelength, there's another one.

' B ' . . . the vertical arrow under B shows the amplitude of the wave

' C ' is one trough of the wave. After every wavelength, there's another one.

' D ' . . . the horizontal arrow over D shows the wavelength.

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3 years ago

A sphere has a radius of 3.9cm and a density of 7.58 g/cm cubed. What is the mass?

snow_lady [41]

Answer:

m=ρV

V=4/3 * pi * r3

V=1.3 * 3.14 * 3.9^3

V=242.14 cm^3

m=7.58 * 242.14

m=1.8 kG

Explanation:

1. We calculate volume for sphere.

2. Then we calculate mass of sphere.

3 0

3 years ago

find a magnitude of the force such that if the act at right angle there resultant is √10N but if the act of 50° the resultant is

Readme [11.4K]

Explanation:

Let magnitude of the two forces be x and y.

Resultant at right angle R1= √15N) and at

60 degrees be R2= √18N.

Now, R1 = √(x² + y²) = √15,

R2= √(x² + y² +2xycos50) = √18.

So x² + y² = 15,

and x² + y² + 1.29xy = 18,

therefore 1.29xy = 3,

y = 3/1.29x.

y = 2.33/x

Now, x2 + (2.33/x)2 = 15,

x² + 5.45/x² = 15

multiply through by x²

x⁴ + 5.45 = 15x²

x⁴ - 15x2 + 5.45 = 0

Now find the roots of the equation, and later y. The two values of x will correspond to the

magnitudes of the two vectors.

Good luck

7 0

3 years ago

Projectile Motion—A tennis ball is thrown out a window 28 m above the ground at an initial velocity of 15.0 m/s and 20.0° below

NNADVOKAT [17]

Answer:

The distance will be x = 41.7 [m]

Explanation:

We must first find the components in the x & y axes of the initial velocity.

$(v_{o})_{x} = 15*cos(20)= 14.09[m/s]\\(v_{o})_{y} = 15*sin(20)= 5.13[m/s]$

The acceleration is the gravity acceleration therefore.

g = 9.81 [m/s^2]

Now we can calculate how long it takes to fall.

$y=(v_{o})_{y}*t-0.5*g*t^2\\-28 = 5.13*t-0.5*9.81*t^2\\-28=-4.905*t^2+5.13*t\\4.905*t^2-5.13*t=28\\t = 2.96[s]$

With this time we can find the horizontal distance that runs the projectile.

$x=(v_{o})_{x}*t\\x=14.09*2.96\\x=41.7[m]$

5 0

3 years ago