Arrange the balls in order from greatest amount of gravitational potential energy to least.

A solid wooden cube, 30cm on each side can be totally submerged in water if it is pushed downward with a force of 54N. What is t

Stels [109]

Answer:

the density of the wooden cube is 204.1 kg/m³

Explanation:

Given;

applied force, F = 54 N

length of each side of the solid wooden cube, L = 30 cm = 0.3 m

mass of the wooden cube is calculated as;

F = mg

m = F/g

m = 54/9.8

m = 5.51 kg

The volume of the wooden cube is calculated as;

V = L³

V = (0.3)³

V = 0.027 m³

The density of the wooden cube is calculated as;

ρ = m/V

ρ = (5.51 kg) / (0.027 m³)

ρ = 204.1 kg/m³

Therefore, the density of the wooden cube is 204.1 kg/m³

4 0

2 years ago

A fixed mass of gas has a volume of gas of 25cm3. the pressure of the gas is 100kPA. the volume of the gas is slowly decreased b

Mekhanik [1.2K]

Answer:

the answer is alphabet A

8 0

3 years ago

Engineers and science fiction writers have proposed designing space stations in the shape of a rotating wheel or ring, which wou

Gennadij [26K]

Answer:

a. The station is rotating at $1.496 \frac{rev}{min}$

b. the rotation needed is $2.8502 \frac{rev}{min}$

Explanation:

We know that the centripetal acceleration is

$a_{c}= \omega ^2 r$

where $\omega$ is the rotational speed and r is the radius. As the centripetal acceleration is feel like an centrifugal acceleration in the rotating frame of reference (be careful, as the rotating frame of reference is <u>NOT INERTIAL,</u> the centrifugal force is a fictitious force, the real force is the centripetal).

The rotational speed is :

$2.7 \frac{m}{s^2} = \omega ^2 * 110 \ m$

$\omega ^2 = \frac{2.7 \frac{m}{s^2}} {110 \ m}$

$\omega = \sqrt{ 0.02454 \frac{rad^2}{s^2} }$

$\omega = 0.1567 \frac{rad}{s}$

Knowing that there are $2\pi \ rad$ in a revolution and 60 seconds in a minute.

$\omega = 0.1567 \frac{rad}{s} \frac{1 \ rev}{2\pi \ rad} \frac{60 \ s}{1 \ min}$

$\omega = 1.496 \frac{rev}{min}$

The rotational speed needed is :

$9.8 \frac{m}{s^2} = \omega ^2 * 110 \ m$

$\omega ^2 = \frac{9.8 \frac{m}{s^2}} {110 \ m}$

$\omega = \sqrt{ 0.08909 \frac{rad^2}{s^2} }$

$\omega = 0.2985 \frac{rad}{s}$

Knowing that there are $2\pi \ rad$ in a revolution and 60 seconds in a minute.

$\omega = 0.2985 \frac{rev}{min} \frac{1 \ rev}{2\pi \ rad} \frac{60 \ s}{1 \ min}$

$\omega = 2.8502 \frac{rev}{min}$

3 0

3 years ago

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To determine a waves frequency you must know the??

lutik1710 [3]

Answer: I think, the number of oscillations in a given period of time.

Explanation: Well I guess because in a period time is known as the rate of occurrence of the wave. Hope this helps!

7 0

3 years ago

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Which of the following are single-displacement reactions?

ololo11 [35]

Answer:

A & D

Explanation:

A single-displacement reaction is a chemical reaction whereby one element is substituted for another one in a compound and thereby generating a new element and also a new compound as products.

From the options, only options A & D fits this definition of single-displacement reactions.

For option D: Both left and hand and right hand sides each have one element and one compound. We can see that K is substituted from KBr to join Cl to form KCl and Br2 on the right hand side.

For option A: Both left and hand and right hand sides each have one element and one compound. We can see that OH is substituted from 2H2O to join Mg to form Mg(OH)2 and H2 on the right hand side.

The other options are not correct because they don't involve only and element and a compound on each side of the reaction.

6 0

2 years ago

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