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3 years ago

An airplane traveling from Atlanta to New York travels 1,395 km in 2.5 h. What is the plane’s average speed?

1 answer:

3 0

D. 558 km/h because if you divide 1,395 by 2.5 you’ll get the answer and I know this is correct because the keyword for me to divide is average.

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Kepler's first law states that planets orbit the Sun in a path called a(n)

wariber [46]

The answer would be C.

3 0

3 years ago

Read 2 more answers

The intensity of a sound wave at a fixed distance from a speaker vibrating at 1.00 kHz is 0.750 W/m2. (a) Determine the intensit

sveticcg [70]

Answer:

a) I = 3.63 W / m² , b) I = 0.750 W / m²

Explanation:

The intensity of a sound wave is given by the relation

I = P / A = ½ ρ v (2π f $s_{max}$ )²

I = (½ ρ v 4π² s_{max}²) f²

a) with the initial condition let's call the intensity Io

cte = (½ ρ v 4π² s_{max}²)

I₀ = cte s² f₀²

I₀ = cte 10 6

If frequency is increase f = 2.20 10³ Hz

I = constant (2.20 10³) 2

I = cte 4.84 10⁶

let's find the relationship of the two quantities

I / Io = 4.84

I = 4.84 Io

I = 4.84 0.750

I = 3.63 W / m²

b) in this case the frequency is reduced to f = 0.250 10³ Hz and the displacement s = 4 s or

I = cte (f s)²

I = constant (0.250 10³ 4)²

I = cte 1 10⁶

the relationship

I / Io = 1

I = Io

I = 0.750 W / m²

6 0

2 years ago

A total charge of 4.70 is distributed on two metal spheres. When the spheres are 10 cm apart, they each feel a repulsive force o

Anna35 [415]

Answer:0.114 C

Explanation:

Given

Total 4.7 C is distributed in two spheres

Let $q_1$ and $q_2$ be the charges such that

$q_1+q_2=4.7$

and Force between charge particles is given by

$F=\frac{kq_1q_2}{r^2}$

$4.7\times 10^11=\frac{9\times 10^9\times q_1\cdot q_2}{0.1^2}$

$q_1\cdot q_2=0.522$

put the value of $q_1$

$q_2\left ( 4.7-q_2\right )=0.522$

$q_2^2-4.7q_2+0.522=0$

$q_2=\frac{4.7\pm \sqrt{4.7^2-4\times 1\times 0.522}}{2}$

$q_2=0.114 C$

thus $q_1=4.586 C$

3 0

3 years ago

Astronomers have observed a small, massive object at the center of our Milky Way Galaxy. A ring of material orbits this massive

Setler [38]

Answer:

The mass of the massive object at the center of the Milky Way galaxy is $3.44\times10^{37}\ Kg$

Explanation:

Given that,

Diameter = 10 light year

Orbital speed = 180 km/s

Suppose determine the mass of the massive object at the center of the Milky Way galaxy.

Take the distance of one light year to be 9.461×10¹⁵ m. I was able to get this it is 4.26×10³⁷ kg.

We need to calculate the radius of the orbit

Using formula of radius

$r=\dfrac{d}{2}$

$r=\dfrac{15\times9.461\times10^{15}}{2}$

$r=7.09\times10^{16}\ m$

We need to calculate the mass of the massive object at the center of the Milky Way galaxy

Using formula of mass

$M=\dfrac{v^2r}{G}$

Put the value into the formula

$M=\dfrac{(180\times10^3)^2\times7.09\times10^{16}}{6.67\times10^{-11}}$

$M=3.44\times10^{37}\ Kg$

Hence, The mass of the massive object at the center of the Milky Way galaxy is $3.44\times10^{37}\ Kg$

5 0

3 years ago

The MSDS for glacial acetic acid says that it is a flammable liquid that can severely burn any human tissue it comes in contact

Alex777 [14]

Answer: D) It should be handled in a fume hood, away from open flames.

3 0

3 years ago