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3 years ago

I need help ASAP. This is for 15 points

Physics

1 answer:

Sladkaya [172]3 years ago

8 0

Answer:

the Prime Meridian is the planets line of 0 degrees longitude sometimes called the Greenwich Meridian or the international Meridian

the great circle of Earth with the latitude of 0 degrees

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Water runs into a fountain, filling all the pipes, at a steady rate of 0.750 m3>s. (a) How fast will it shoot out of a hole 4

kati45 [8]

Answer:

velocity = 472 m/s

velocity = 52.4 m/s

Explanation:

given data

steady rate = 0.750 m³/s

diameter = 4.50 cm

solution

we use here flow rate formula that is

flow rate = Area × velocity .............1

0.750 = $\frac{\pi }{4}$ × (4.50× $10^{-2}$ )² × velocity

solve it we get

velocity = 472 m/s

and

when it 3 time diameter

put valuer in equation 1

0.750 = $\frac{\pi }{4}$ × 3 × (4.50× $10^{-2}$ )² × velocity

velocity = 52.4 m/s

5 0

3 years ago

What two characteristics do electromagnetic waves vary in?

rosijanka [135]

They size of the wave and the time of a certain wave.

4 0

3 years ago

Read 2 more answers

Wich of the following celestial bodies is most likely to have many craters

Ede4ka [16]

Where are the following awnsers?

6 0

2 years ago

A 2430 pound roller coaster starts from rest and is launched such that it crests a 105 ft high hill with a speed of 59 mph. The

Rudik [331]

Answer:

Explanation:

Given

Weight of roller coaster is $W=2430\ pound$

mass of roller coaster $m=\frac{W}{g}=\frac{2430}{32.2}=75.45$

Distance traveled by roller coaster $d=396\ ft$

drag force $f_d=85\ pounds$

velocity at top $v=59 mph\approx 86.53\ ft/s$

Suppose E is the initial energy

Conserving Energy at bottom and top

$E=\frac{1}{2}mv^2+mgh+f_d\cdot d$

$E=0.5\times 75.45\times 86.53^2+2430\times 105+85\times 396$

$E=2.9\times 10^5\ foot-pound$

5 0

3 years ago

A 3-kg wheel with a radius of 35 cm is spinning in the horizontal plane about a vertical axis through its center at 800 rev/s. A

Kruka [31]

Answer:

$\omega_f = 585.37 \ rev/s$

Explanation:

given,

mass of wheel(M) = 3 Kg

radius(r) = 35 cm

revolution (ω_i)= 800 rev/s

mass (m)= 1.1 Kg

I_{wheel} = Mr²

when mass attached at the edge

I' = Mr² + mr²

using conservation of angular momentum

$I \omega_i = I' \omega_f$

$(Mr^2) \times 800 = ( M r^2 + m r^2) \omega_f$

$M\times 800 = ( M + m )\omega_f$

$3\times 800 = (3+1.1)\times \omega_f$

$2400 = (4.1)\times \omega_f$

$\omega_f = 585.37 \ rev/s$

3 0

3 years ago