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SSSSS [86.1K]
2 years ago
9

Medium frequency waves are called what

Physics
1 answer:
USPshnik [31]2 years ago
5 0

Answer: The MF band is also known as the hectometer band as the wavelengths range from ten to one hectometer

Explanation:

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In an arcade game a 0.135 kg disk is shot across a frictionless horizontal surface by compressing it against a spring and releas
ioda

Answer:

v=2.42m/s

Explanation:

We use the energy conservation theorem in order to solve the problem. The energy when the spring is compressed is equal at the energy when the disk leaves the spring:

E_{initial}=E{final}\\E_{k-initial}+E_{p-initial}=E_{k-final}+E_{p-final}\\0+\frac{1}{2}kx^2=\frac{1}{2}mv^2+0

At the beginning the initial energy is totally potential, energy linked to the compressed spring. At the end the energy is totally kinetics

We solve the equation in order to find the speed.

k=162 N/m

x=7 cm=0.07m

m=0.135 kg

v=x*\sqrt{k/m}=0.07*\sqrt{162/0.135}=2.42m/s

6 0
3 years ago
Satellites, A and B, each of mass m, are to be launched into circular orbits about Earth's center. Satellite A is to orbit at an
Volgvan

Answer:

0.63

Explanation:

We are given that

Radius of earth,R_E=6370 km

Radius of orbit A,R_A=7590+6370=13960 km

Radius of orbit B,R_B=6370+15800=22170 km

We have to find the ratio of the potential energy of satellite B to that of satellite A in orbit.

Potential energy of orbit A=U_A=-\frac{GM}{R_A}=-\frac{GM}{13960}

Potential energy of orbit  B=U_B=-\frac{GM}{R_B}=-\frac{GM}{22170}

\frac{U_B}{U_A}=\frac{13960}{22170}=0.63

Hence,the ratio of the potential energy of satellite B to that of satellite A in orbit=0.63

8 0
4 years ago
Mass of mercury = 57 g Volume of mercury = 4.2 mL
kari74 [83]
24 beautiful women and women who have a good relationship and you can get a good laugh and
7 0
3 years ago
Si una bala de 122 g que viaja a 350 m/s golpea a un policía con chaleco antibalas el cual soporta 120 J de energía, ¿la bala at
Zanzabum

Answer:

Puesto que la energia cinética traslacional es mucho mayor que la capacidad del chaleco antibalas, la bala atravesaría el chaleco antibalas.

Explanation:

Un chaleco antibalas soporta el disparo de una bala disipando la energía de esta última a través de su propio material. Si sabemos que el chaleco antibalas soporta 120 joules de energía, cabe saber si la energía cinética traslacional es igual o inferior a ese límite, significando que la bala no atravesaría el chaleco.

La energía cinética traslacional de la bala (K), in joules, queda expresada con la siguiente fórmula:

K = \frac{1}{2}\cdot m\cdot v^{2} (1)

Donde:

m - Masa de la bala, en kilogramos.

v - Rapidez de la bala, en metros por segundo.

Si sabemos que m = 0.122\,kg y v = 350\,\frac{m}{s}, entonces la energía cinética traslacional de la bala es:

K = 7472.5\,J

Puesto que la energia cinética traslacional es mucho mayor que la capacidad del chaleco antibalas, la bala atravesaría el chaleco antibalas.

4 0
3 years ago
A horizontal force of magnitude 31.3 N pushes a block of mass 4.30 kg across a floor where the coefficient of kinetic friction i
Ostrovityanka [42]

Answer:

W_{F} =  +119.6 J : Work done by horizontal force

W_{Ff} =  -96.42 J : Work done by friction force

W_{n} =   23.18 J :  Net work

Explanation:

Work theory

Work is the product of a force applied to a body and the displacement of the body in the direction of this force.

Thus. The work is equal to the product of the force by the distance (d) and by the cosine of the angle ( α) that exists between the direction of the force (F) and the direction that travels the point or the object that moves:

W= F*d*cosα Formula (1)

The work is positive (W+) if the force goes in the same direction of movement.

The work is negative (W-)if the force goes in the opposite direction to the movement

Net work  (Wn)

Net work is defined as the sum of the work done from an initial place to an end point of the trajectory, for each of the individual forces to which the  system is subjected.

Problem development

We determine the forces that perform work:

in formula (1 )we observe that the forces that perform work are the forces parallel to displacement. that is α = 0.

For this case they are the horizontal force (F) and the frictional force (Ff)

F=31.3 N : Horizontal force

Ff=μk*N , μk:coefficient of kinetic friction  , N : Normal force

N=W  

W: block weight  W= m*g, m:  block mass , g: acceleration due to gravity

W= 4.30 kg* 9.8 m/s² = 42.14 N

Ff= 0.599*42.14 N= 25.24 N : Frictional force

Calculating of the Net work  (Wn)

W_{n} = W_{F} +W_{Ff}

W_{F} =   31.3N*3.82 m

W_{F} =  119.6 N*m

W_{F} =  +119.6 J

W_{Ff} =  -25.24N*3.82m

W_{Ff} =  -96.42 N* m

W_{Ff} =  -96.42 J

W_{n} = W_{F} +W_{Ff}

W_{n} = ( 119.6  -96.42) N*m

W_{n} =   23.18 N*m

W_{n} =   23.18 J

8 0
3 years ago
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