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olga_2 [115]
3 years ago
8

A continental tropical air mass is _______ and _______.

Physics
1 answer:
Masteriza [31]3 years ago
6 0
A continental tropical air mass is dry and hot. It is formed over the land and in hot tropical areas. Summer months are when this kind of warm air mass is most noticeable and it can stay for a lengthy period of time.
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What is the density of a cube that has a mass of 3.75 g and a volume of 3 mL?
valina [46]

Answer:

\displaystyle \rho=1.25\ g/ml

Explanation:

<u>Density </u>

The density of a substance is the mass per unit volume. The density varies with temperature and pressure.

The formula to calculate the density of a substance of mass (m) and volume (V) is:

\displaystyle \rho=\frac{m}{V}

The cube has a mass of m=3.75 g and a volume of V=3 ml, thus the density is:

\displaystyle \rho=\frac{3.75\ g}{3\ ml}

\boxed{\displaystyle \rho=1.25\ g/ml}

Since 1 kg=1000 mg and 1 lt = 1000 ml, the density has the same value but with different units:

\displaystyle \rho=1.25\ kg/l

6 0
2 years ago
How are domains arranged in materials that are magnetic and in ones that are not
Valentin [98]

Answer:

In most materials, atoms are arranged in such a way that the magnetic orientation of one electron cancels out the orientation of another

8 0
3 years ago
An old grindstone, used for sharpening tools, is a solid cylindrical wheel that can rotate about its central axle with negligibl
krok68 [10]

(a) The moment of inertia of the wheel  is 78.2 kgm².

(b) The mass (in kg) of the wheel is 1,436.2 kg.

(c) The angular speed (in rad/s) of the wheel at the end of this time period is 3.376 rad/s.

<h3>Moment of inertia of the wheel</h3>

Apply principle of conservation of angular momentum;

Fr = Iα

where;

  • F is applied force
  • r is radius of the cylinder
  • α is angular acceleration
  • I is moment of inertia

I = Fr/α

I = (200 x 0.33) / (0.844)

I = 78.2 kgm²

<h3>Mass of the wheel</h3>

I = ¹/₂MR²

where;

  • M is mass of the solid cylinder
  • R is radius of the solid cylinder
  • I is moment of inertia of the solid cylinder

2I = MR²

M = 2I/R²

M = (2 x 78.2) / (0.33²)

M = 1,436.2 kg

<h3>Angular speed of the wheel after 4 seconds</h3>

ω = αt

ω = 0.844 x 4

ω = 3.376 rad/s

Thus, the moment of inertia of the wheel  is 78.2 kgm².

The mass (in kg) of the wheel is 1,436.2 kg.

The angular speed (in rad/s) of the wheel at the end of this time period is 3.376 rad/s.

Learn more about moment of inertia here: brainly.com/question/14839816

#SPJ1

7 0
1 year ago
A penny rides on top of a piston as it undergoes vertical simple harmonic motion with an amplitude of 4.0cm. If the frequency is
BlackZzzverrR [31]

Answer:

the penny loses contact at the piston's highest point.

f = 2.5 Hz

Explanation:

Concepts and Principles  

1- Newton's Second Law: The net force F on a body with mass m is related to the body's acceleration a by  

∑F = ma                                                          (1)  

2- The maximum transverse acceleration of a particle in simple harmonic motion is found in terms of the angular speed w and the amplitude A as follows:  

a_max = -w^2A                                                (2)  

3- The angular frequency w of a wave is related to the frequency f by:  

w = 2π f                                                              (3)  

Given Data  

- The amplitude of the piston is: A = (4.0 cm) ( 1/ 100 cm)=  0.04 m.  

- The frequency of oscillation of the piston is steadily increased.

Required Data

<em>In part (a), we are asked to determine the point at which the penny first loses contact with the piston.  </em>

<em>In part (b), we are asked to determine the maximum frequency for which the penny just barely remains in place for a full cycle.  </em>

Solution  

(a)  

The free-body diagram in Figure 1 shows the forces acting on the penny; mi is the gravitational force exerted by the Earth on the penny andrt is the normal contact force exerted by the piston on the penny.  

figure 1 is attached

Apply Newton's second law from Equation (1) in the vertical direction to the penny:  

∑F_y -mg= ma        

Solve for n=m(g+a) The penny loses contact with the surface of the oscillating piston when the normal force n exerted by the piston is zero. So  

0 = m(g + a)

a = —g  

Therefore, the penny loses contact with the piston when the piston starts accelerating downwards. The piston first acceleratesdownward at its highest point and hence the penny loses contact at the piston's highest point.

(b)  

The maximum acceleration of the penny at the highest point of the piston is found from Equation (2):  

a = —w^2A  

where a = —g at the highest point. So  

g = w^2A  

Solve for w:  

w =√g/A

Substitute for w from Equation (3):

2πf =  √g/A

Solve for f :  

f = 1/2π√g/A

Substitute numerical values:  

f = 1/2π√9.8 m/s^2/0.04

f = 2.5 Hz

6 0
3 years ago
The Kelly family and the Garcia family each used their sprinklers last summer. The water output rate for the Kelly family's spri
elena55 [62]

Answer:

30 hours

15 hours.

Explanation:

Let the Kelly used sprinkler for t hours and Gracia used the sprinkler for the rest ie (45 - t) hours .

Water output by Kelly = (35 x t) L

Water output by Gracia = 20 ( 45 - t ) L

Total output = 35t + 20(45 - t)

So,  35 t + 20 (45 - t ) = 1350

35 t + 900 - 20 t = 1350

15 t = 450

t = 30 hours

So Kelly used sprinkler for 30 hours and Gracia used it for 45 - 30 = 15 hours.

3 0
3 years ago
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