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2 years ago

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An airplane travels 2100 km at a speed of 1000 km/h. It then encounters a headwind which slows it to 800 km/h for the next 1300

km. What is the average speed for this plane trip?

1 answer:

Sidana [21]2 years ago

5 0

I don’t get it I need help

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The element in an incandescent light bulb that releases light energy is? A- a phosphor, B- mercury vapor, C- a thin tungsten fil

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C.) A Thin tungsten filament

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2 years ago

Identify the chemical change below

lesya [120]

Imma say go with C (:

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2 years ago

A piano string having a mass per unit length of 5.00 g/m is under a tension of 1350 N. Determine the speed of transverse waves i

padilas [110]

Answer:

The speed of transverse waves in this string is 519.61 m/s.

Explanation:

Given that,

Mass per unit length = 5.00 g/m

Tension = 1350 N

We need to calculate the speed of transverse waves in this string

Using formula of speed of the transverse waves

$v=\sqrt{\dfrac{T}{\mu}}$

Where, $\mu$ = mass per unit length

T = tension

Put the value into the formula

$v = \sqrt{\dfrac{1350}{5.00\times10^{-3}}}$

$v =519.61\ m/s$

Hence, The speed of transverse waves in this string is 519.61 m/s.

6 0

2 years ago

What is the acceleration of a 7 kg mass if the force of 70 N is used to move it toward the Earth?

Assoli18 [71]

Answer:

<h2>10 m/s²</h2>

Explanation:

The acceleration of an object given it's mass and the force acting on it can be found by using the formula

$a = \frac{f}{m} \\$

m is the mass

f is the force

From the question we have

$a = \frac{70}{7} = 10 \\$

We have the final answer as

<h3>10 m/s²</h3>

Hope this helps you

4 0

2 years ago

A ball with an initial velocity of 8.00 m/s rolls up a hill without slipping. (a) Treating the ball as a spherical shell, calcul

GrogVix [38]

Answer:

Part i)

h = 5.44 m

Part ii)

h = 3.16 m

Explanation:

Part i)

Since the ball is rolling so its total kinetic energy in this case will convert into gravitational potential energy

So we have

$\frac{1}{2}mv^2 + \frac{1}{2}I\omega^2 = mgh$

here we know that for spherical shell and pure rolling conditions

$v = R \omega$

$I = \frac{2}{3}mR^2$

$\frac{1}{2}mv^2 + \frac{1}{2}(\frac{2}{3}mR^2)(\frac{v^2}{R^2}) = mgh$

$\frac{5}{6}mv^2 = mgh$

$h = \frac{5v^2}{6g}$

$h = \frac{5(8^2)}{6(9.81)} = 5.44 m$

Part b)

If ball is not rolling and just sliding over the hill then in that case

$\frac{1}{2}mv^2 = mgh$

$h = \frac{v^2}{2g}$

$h = \frac{8^2}{2(9.81)} = 3.16 m$

3 0

2 years ago