An adult dog can sit, rollover, and bark when needs to potty outside. these are a type of what behavior.

At each of the designated points, rotate the given vector to indicate the direction of the force exerted by the water on either

GrogVix [38]

Answer:

The direction of the force at A and B is perpendicular to the walls of the container.

The direction of the force at C is down.

The direction of the force in D is up

The direction of the force at E is to the left.

The attached figure shows the forces exerted by the water at points A, B, C, D and E.

Explanation:

The water is in contact with the bowl and with the fish. It exercises at points A, B, C, D and E, but the direction is different from the force.

The fish has a buoyant force on the water and that direction is up. The direction of at point D is up.

The column of water on the fish has a downward force, therefore the direction of the force at point C is down. The water column to the right of the fish has a force to the left, and the direction at point E is to the left.

The water will exert a force on the walls of the container and this force at points A and B is a on the walls of the container.

4 0

3 years ago

The Hubble Space Telescope orbits the Earth at approximately 612,000m altitude. Its mass is 11,100 kg and the mass of earth is 5

nexus9112 [7]

Answer:

7.55 km/s

Explanation:

The force of gravity between the Earth and the Hubble Telescope corresponds to the centripetal force that keeps the telescope in uniform circular motion around the Earth:

$G\frac{mM}{R^2}=m\frac{v^2}{R}$

where

$G=6.67\cdot 10^{-11} m^3 kg^{-1} s^{-2}$ is the gravitational constant

$m=11,100 kg$ is the mass of the telescope

$M=5.97\cdot 10^{24} kg$ is the mass of the Earth

$R=r+h=6.38\cdot 10^6 m+612,000 m=6.99\cdot 10^6 m$ is the distance between the telescope and the Earth's centre (given by the sum of the Earth's radius, r, and the telescope altitude, h)

v = ? is the orbital velocity of the Hubble telescope

Re-arranging the equation and substituting numbers, we find the orbital velocity:

$v=\sqrt{\frac{GM}{R}}=\sqrt{\frac{(6.67\cdot 10^{-11})(5.97\cdot 10^{24} kg)}{6.99\cdot 10^6 m}}=7548 m/s=7.55 km/s$

6 0

3 years ago

A cube has a drag coefficient of 0.8. what would be the terminal velocity of a sugar cube 1 cm on a side in air (ρ = 1.2 kg/m3)?

kramer

To answer this question, we should know the formula for the terminal velocity. The formula is written below:

v = √(2mg/ρAC)
where
m is the mass
g is 9.81 m/s²
ρ is density
A is area
C is the drag coefficient

Let's determine the mass, m, to be density*volume.
Volume = s³ = (1 cm*1 m/100 cm)³ = 10⁻⁶ m³
m = (1.6×10³ kg/m³)(10⁻⁶ m³) = 1.6×10⁻³ kg
A = (1 cm * 1 m/100 cm)² = 10⁻⁴ m²

v = √(2*1.6×10⁻³ kg*9.81 m/s²/1.6×10³ kg/m³*10⁻⁴ m²*0.8)
<em>v = 0.495 m/s</em>

7 0

3 years ago

A person jogs for 4.0 km in 32 mins,then 2.0km in 22mins,and finallu 1.0 km in 16 mins.whqt is the joggers average speed in km p

blondinia [14]

Average speed = (total distance covered) / (time to cover the distance)

total distance covered = (4km + 2km + 1km) = 7 km

time to cover the distance = (32min + 22min + 16min) = 70 min

Average speed = (7 km) / (70 min)

Average speed = 0.1 km/minute

5 0

3 years ago

Where are you likely to find a cooling plant located in the ductwork? A. A medium-sized office B. A large factory C. A single fa

Rufina [12.5K]

Answer:

i think it would be B, a large factory

Explanation:

8 0

3 years ago