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Maksim231197 [3]
2 years ago
9

Dimensionally, which of the following could be a velocity? *

Physics
1 answer:
astra-53 [7]2 years ago
5 0

Answer:

5 meters per second

Explanation:

5m is the distance

5m west is the vector

5m per second is the velocity

5m per second west is unknown

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True or False. The pressure inside a piston is the same in all directions.
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Answer:

True

Explanation:

Pascal's law says that pressure applied to an enclosed fluid will be transmitted without a change in magnitude to every point of the fluid and to the walls of the container. The pressure at any point in the fluid is equal in all directions.

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Explain the characteristics of the Gypsum Hills regionIf we have a sample of silicon (Si) atoms that has 14 protons, 14 electron
amid [387]

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second one

Explanation:

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3 years ago
On his fishing trip Justin takes the boat 25 km south. The fish aren’t biting so he goes 10 km west. He follows a school of fish
S_A_V [24]

Distance is 50 km

Displacement is 10 km

<u>Explanation:</u>

Given:

Distance toward south, x = 25 km

Distance towards west, y = 10 km

Distance towards north, z = 15 km

(a) Total distance, D = ?

Total distance, D = x + y + z

                       D = 25 + 10 + 15

                       D = 50km

(b) Displacement, d = ?

Displacement = final position - initial position

                      = 10 - 0 km

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8 0
2 years ago
There are no risks when taking medication to treat anxiety.
lidiya [134]

Answer:

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Explanation:

5 0
2 years ago
Read 2 more answers
A hot-air balloon consists of a basket hanging beneath a large envelope filled with hot air. A typical hot-air balloon has a tot
Deffense [45]

Answer:

1.05045 kg/m³

Explanation:

\rho_a = Density of air = 1.26 kg/m³

\rho_{ha} = Density of hot air

m_a = Mass of balloon = 539 kg

g = Acceleration due to gravity = 9.81 m/s²

v = Volume of air in balloon = 2.62\times 10^3\ m^3

The net force on the balloon will be

F_n=(\rho_a-\rho_{ha})vg

Also

F_n=m_ag

\\\Rightarrow 549g=(\rho_a-\rho_{ha})vg\\\Rightarrow 549=(1.26-\rho_{ha})2.62\times 10^3\\\Rightarrow -\rho_{ha}=\frac{549}{2.62\times 10^3}-1.26\\\Rightarrow \rho_{ha}=1.05045\ kg/m^3

The density of hot air inside the envelope is 1.05045 kg/m³

4 0
2 years ago
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