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andrezito [222]
3 years ago
12

A nurse counts 66 heartbeats in one minute. What is the period of the hearts oscillation? In minutes

Physics
1 answer:
inn [45]3 years ago
7 0

The frequency is  66 / minute .

The period is 1/frequency = minute / 66 =

(60 seconds) / 66 = 0.91 second

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Two representative elements are in the same period of the periodic table. Which statement correctly describes the atoms of the t
oksian1 [2.3K]

Elements in the periodic table are arranged in the increasing order of their atomic number.

An atom will contain equal amount of protons and electrons. If two elements are in same period of periodic table then these elements will have same number of electrons.

Helium is one of the seven elements which are called as noble gases. This gas is odorless, tasteless and colorless which makes it unique from other elements.

It has two valence electrons while atoms of other elements in the group have 8 valence electrons.

Lithium has one valence electron, while calcium atom has 2 valence electrons.

The highest level of energy is due valence electrons of an atom.  The correct answer is Lithium.

Fluorine element has 9 electrons. There are 2 electrons in the first energy level and seven electrons in the second energy level.

Learn more at brainly.com/question/24373385

7 0
3 years ago
Un movil pasa por el punto A en direccion hacia B (350cm más adelante) y, luego, sigue hasta el punto C. Sabiendo que pasa por B
xeze [42]

Explanation:

PRIMERO HACES EL RECUENTO DEL TIEMPO Y LO CONVIERTES EN

SEGUNDOS Y ENTONCES

<em>t</em> = 227 s      t_{AB} = 227 S - 38 s = 189 s

t_{BC} = 38 s

LUEGO USANDO LA ECUACIÓN DE GALILEO GALILEI SSUPONIENDO

QUE EL MOVIL VIAJA A VELOCIDAD CONSTANTE

<em>v</em> = 3.50 m/189 s = 0.0185 m/s

PARA LA DISTANCIA NTRE B Y C

x_{BC} = 0.0185 m/S( 38 s) = 0.703 m

LA HORA EN QUE EL MOVIL PASA POR A ES

11:43:15 - 38 s - 189 s = 11:39:29

6 0
3 years ago
A satellite of mass m is in a circular orbit of radius R2 around a spherical planet of radius R1 made of a material with density
Anit [1.1K]

Answer: The gravitational force Fg exerted on the orbit by the planet is Fg = G 4/3πr3rhom/ (R1 + d+ R2)^2

Explanation:

Gravitational Force Fg = GMm/r2----1

Where G is gravitational constant

M Mass of the planet, m mass of the orbit and r is the distance between the masses.

Since the circular orbit move around the planet, it means they do not touch each other.

The distance between two points on the circumference of the two massesb is given by d, while the distance from the radius of each mass to the circumferences are R1 and R2 from the question.

Total distance r= (R1 + d + R2)^2---2

Recall, density rho =

Mass M/Volume V

Hence, mass of planet = rho × V

But volume of a sphere is 4/3πr3

Therefore,

Mass M of planet = rho × 4/3πr3

=4/3πr3rho in kg

From equation 1 and 2

Fg = G 4/3πr3rhom/ (R1 + d+ R2)^2

6 0
3 years ago
1. A 0.40 kg ball is launched at a speed of 16 m/s from a 22 m cliff.
Masja [62]

Answer:

51.2 J, 86.2 J, 137.4 J

Explanation:

The kinetic energy of the ball is given by:

K=\frac{1}{2}mv^2

where

m = 0.40 kg is its mass

v = 16 m/s is its speed

Substituting,

K=\frac{1}{2}(0.40)(16)^2=51.2  J

The potential energy of the ball is given by

U=mgh

where

m = 0.40 kg

g=9.8 m/s^2 is the acceleration of gravity

h = 22 m is the heigth of the cliff

Substituting,

U=(0.40)(9.8)(22)=86.2 J

Finally, the total mechanical energy is the sum of the kinetic energy and the potential energy:

E=K+U=51.2 + 86.2=137.4 J

4 0
3 years ago
Define friction State the law of frictionGive two factors that may affect friction
umka21 [38]

Friction:

Friction is the resistive force that acts on the body in a direction opposite to the direction of motion of the same object.

Laws of friction:

1. The frictional force acting on the object is proportional to the normal force acting on the object.

For example, if the object of mass m is moving along the horizontal surface, the normal force acting on the object is mg.

Thus, the frictional force acting on the object is,

\begin{gathered} F_r\propto F_N \\ F_r=\mu F_N \end{gathered}

where F_N is the normal force acting on the object,

\mu\text{ is the coefficient of friction}

The diagrammatic representation of the frictional force acting on the object is,

2. The frictional force acting on the object depends upon the nature of the surface in contact with the object.

3. The frictional force does not depend upon the area of contact of an object with the surface.

4. The kinetic friction acting on the object is independent of the velocity of the object.

5. The coefficient of static friction is more than the coefficient of kinetic friction.

Two factors on which the frictional force depends are:

1. The frictional force depends on the adhesion between the two bodies in contact.

2. The frictional force acting on the depends upon the roughness of the surface on which the object is moving.

3 0
1 year ago
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