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3 years ago

I need help with number 1 and 2.

Physics

1 answer:

Vilka [71]3 years ago

4 0

For number one, use the equation of v= v initial+ at. Use v initial as zero for this situation and 9.8 m/s^2 as the acceleration due to gravity

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The forces acting on this object are _______ and the net force is ________ Balanced, 0 N Balanced, 10 N Unbalanced, 10 N Unbalan

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1st Blank Answer: Balanced
2nd Blank Answer: 0n

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3 years ago

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What is the pathway of sound through fluids starting at the oval window through to dissipation of the sound waves at the round w

ruslelena [56]

Perilymph of scala vestibule; endolymph of cochlear duct; perilymph of scala tympani

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3 years ago

A particle leaves the origin with a speed of 3 106 m/s at 38 degrees to the positive x axis. It moves in a uniform electric fiel

Salsk061 [2.6K]

Answer:

If the particle is an electron $E_y = 3.311 * 10^3 N/C$

If the particle is a proton, $E_y = 6.08 * 10^6 N/C$

Explanation:

Initial speed at the origin, $u = 3 * 10^6 m/s$

$\theta = 38^0$ to +ve x-axis

The particle crosses the x-axis at , x = 1.5 cm = 0.015 m

The particle can either be an electron or a proton:

Mass of an electron, $m_e = 9.1 * 10^{-31} kg$

Mass of a proton, $m_p = 1.67 * 10^{-27} kg$

The electric field intensity along the positive y axis $E_y$ , can be given by the formula:

$E_y = \frac{2 m u^2 sin \theta cos \theta}{qx} \\$

If the particle is an electron:

$E_y = \frac{2 m_e u^2 sin \theta cos \theta}{qx} \\$

$E_y = \frac{2 * 9.1 * 10^{-31} * (3*10^6)^2 *(sin38)( cos38)}{1.6*10^{-19} * 0.015} \\$

$E_y = 3311.13 N/C\\E_y = 3.311 * 10^3 N/C$

If the particle is a proton:

$E_y = \frac{2 m_p u^2 sin \theta cos \theta}{qx} \\$

$E_y = \frac{2 * 1.67 * 10^{-27} * (3*10^6)^2 *(sin38)( cos38)}{1.6*10^{-19} * 0.015} \\$

$E_y = 6.08 * 10^6 N/C$

8 0

3 years ago

A particular spiral galaxy can be approximated by a thin disk-like volume 62 Thousand Light Years in radius and 7 Hundred Light

SVETLANKA909090 [29]

Answer:

Approximate linear dimension is 2 light years.

Explanation:

Radius of the spiral galaxy r = 62000 LY

Thickness of the galaxy h = 700 LY

Volume of the galaxy = πr²h

= (3.14)(62000)²(700)

= (3.14)(62)²(7)(10)⁸

= 84568×10⁸

= $8.45\times 10^{12}$ (LY)³

Since galaxy contains number of stars = 1078 billion stars ≈ $1.078\times 10^{12}$

Now volume covered by each star of the galaxy = $\frac{\text{Total volume of the galaxy}}{\text{Number of stars}}$

= $\frac{8.45\times 10^{12} }{1.078\times 10^{12}}$

= 7.839 Light Years

Now the linear dimension across the volume

= $(\text{Average volume per star})^{\frac{1}{3}}$

= $(7.839)^{\frac{1}{3}}$

= 1.99 LY

≈ 2 Light Years

Therefore, approximate linear dimension is 2 light years.

8 0

3 years ago

Express the number in scientific notation: 4/100

ankoles [38]

Answer:

4*10^-2

Explanation:

for the scientific notation the first number must be between 1 and 10, so in this case it is 4. 4/100 is also equal to 0.04, and if we could the number of places before 4, there are two, therefore 4 times 10 to the power of -2

5 0

3 years ago