All categories

3 years ago

Planet that is one astronomical unit from the sun

Physics

1 answer:

DIA [1.3K]3 years ago

3 0

That would be Earth, because astronomical unit is defined as distance between Earth and sun.

Hope this helps.

r3t40

You might be interested in

With its wheels locked, a van slides down a hill inclined at 40.0° to the horizonta l. Find the acceleration of this van A) if t

andreyandreev [35.5K]

Answer:

Explanation:

Check attachment for solution

7 0

3 years ago

Vivian went on a bicycle trip through Germany with her family. One afternoon, she rode her bicycle along a long flat road at a c

natima [27]

d = distance traveled by her on her bicycle on a long flat road = 24 kilometer

t = time taken by her to travel distance "d" on her bicycle on a long flat road = 1.2 hours

v = average speed of vivian = ?

we know that average speed is given as

v = d/t

inserting the values in the above formula

v = 24 kilometer / 1.2 hour

v = 20 kilometer/hour

hence the correct choice is

C) 20 km/h

8 0

2 years ago

At a certain elevation, the pilot of a balloon has a mass of 120 lb and a weight of 119 lbf. What is the local acceleration of g

Strike441 [17]

Answer:

31.905 ft/s²

Explanation:

Given that

Mass of the pilot, m = 120 lb

Weight of the pilot, w = 119 lbf

Acceleration due to gravity, g = 32.05 ft/s²

Local acceleration of gravity of found by using the relation

Weight in lbf = Mass in lb * (local acceleration/32.174 lbft/s²)

119 = 120 * a/32. 174

119 * 32.174 = 120a

a = 3828.706 / 120

a = 31.905 ft/s²

Therefore, the local acceleration due to gravity at that elevation is 31.905 ft/s²

3 0

2 years ago

When must scientific theories be changed

madreJ [45]

Answer:

when new information disproving the current theory becomes available.

Explanation:

hope this helps..

4 0

1 year ago

Water is pumped steadily out of a flooded basement at a speed of 5.4 m/s through a uniform hose of radius 0.83 cm. The hose pass

Gala2k [10]

To solve this problem it is necessary to apply the concepts related to the flow as a function of the volume in a certain time, as well as the potential and kinetic energy that act on the pump and the fluid.

The work done would be defined as

$\Delta W = \Delta PE + \Delta KE$

Where,

PE = Potential Energy

KE = Kinetic Energy

$\Delta W = (\Delta m)gh+\frac{1}{2}(\Delta m)v^2$

Where,

m = Mass

g = Gravitational energy

h = Height

v = Velocity

Considering power as the change of energy as a function of time we will then have to

$P = \frac{\Delta W}{\Delta t}$

$P = \frac{\Delta m}{\Delta t}(gh+\frac{1}{2}v^2)$

The rate of mass flow is,

$\frac{\Delta m}{\Delta t} = \rho_w Av$

Where,

$\rho_w$ = Density of water

A = Area of the hose $\rightarrow A=\pi r^2$

The given radius is 0.83cm or $0.83 * 10^{-2}$ m, so the Area would be

$A = \pi (0.83*10^{-2})^2$

$A = 0.0002164m^2$

We have then that,

$\frac{\Delta m}{\Delta t} = \rho_w Av$

$\frac{\Delta m}{\Delta t} = (1000)(0.0002164)(5.4)$

$\frac{\Delta m}{\Delta t} = 1.16856kg/s$

Final the power of the pump would be,

$P = \frac{\Delta m}{\Delta t}(gh+\frac{1}{2}v^2)$

$P = (1.16856)((9.8)(3.5)+\frac{1}{2}5.4^2)$

$P = 57.1192W$

Therefore the power of the pump is 57.11W

6 0

3 years ago