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3 years ago

11

¿ Sobre que superficie se desplazara mas rápidamente un tejo, sobre cemento o sobre cerámica? ¿por que?

1 answer:

kvasek [131]3 years ago

4 0

Answer:

cemento potque es plano

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A 1.5 kilogram car is moving at 10 meters per second east. A braking force acts on the car for 5.0 seconds, reducing its velocit

Arada [10]

Answer:

I don't know

Explanation:

8 0

3 years ago

An asteroid orbits the Sun every 176 years. What is the asteroids average distance from the Sun? P ^ 2 = a ^ 3 where p = period

KatRina [158]

Answer:

The value is $x = 45.99 \ Au$

Explanation:

From the question we are told that

The period of the asteroid is $T = 176 \ years = 176 * 365 * 24 * 60* 60 = 5.55*10^{9}\ s$

Generally the average distance of the asteroid from the sun is mathematically represented as

$R = \sqrt[3]{ \frac{G M * T^2 }{4 \pi} }$

Here M is the mass of the sun with a value

$M = 1.99*10^{30} \ kg$

G is the gravitational constant with value $G = 6.67 *10^{-11} \ m^3 \cdot kg^{-1} \cdot s^{-2}$

$R = \sqrt[3]{ \frac{6.67 *10^{-11} * 1.99*10^{30} * [5.55 *10^{9}]^2 }{4 * 3.142 } }$

=> $R = 6.88 *10^{12} \ m$

Generally

$1.496* 10^{11} \ m \to 1 Au (Astronomical \ unit )$

So

$R = 6.88 *10^{12} \ m \ \ \ \ \to \ \ x \ Au$

=> $x = \frac{6.88 *10^{12}}{1.496 *10^{11}}$

=> $x = 45.99 \ Au$

7 0

2 years ago

Show all work please I am stuck

MissTica

Answer:

Explanation:

1 meter = 39.37 inches

meters x inches/meters = inches

6.23x10^-4 m x 39.37 in./m=0.245...=2.45x10^-2

5 0

2 years ago

Humans have three types of cone cells in their eyes, which are responsible for color vision. Each type absorbs a certain part of

xxMikexx [17]

Answer:

Frequency of the light will be equal to $6.97\times 10^{1}Hz$

Explanation:

We have given wavelength of the light $\lambda =430nm=430\times 10^{-9}m$

Velocity of light is equal to $v=3\times 10^8m/sec$

We have to find the frequency of light

We know that velocity is equal to $v=\lambda f$ , here $\lambda$ is wavelength and f is frequency of light

So frequency of light will be equal to $f=\frac{v}{\lambda }=\frac{3\times 10^8}{430\times 10^{-9}}=6.97\times 10^{1}Hz$

So frequency of the light will be equal to $6.97\times 10^{1}Hz$

5 0

3 years ago

On a frictionless horizontal air table, puck A (with mass 0.254 kg ) is moving toward puck B (with mass 0.367 kg ), which is ini

irinina [24]

Answer:

$v_a=0.8176 m/s$

$\Delta K=0.07969 J - 0.0849 J = -0.00521 J$

Explanation:

According to the law of conservation of linear momentum, the total momentum of both pucks won't be changed regardless of their interaction if no external forces are acting on the system.

Being $m_a$ and $m_b$ the masses of pucks a and b respectively, the initial momentum of the system is

$M_1=m_av_a+m_bv_b$

Since b is initially at rest

$M_1=m_av_a$

After the collision and being $v'_a$ and $v'_b$ the respective velocities, the total momentum is

$M_2=m_av'_a+m_bv'_b$

Both momentums are equal, thus

$m_av_a=m_av'_a+m_bv'_b$

Solving for $v_a$

$v_a=\frac{m_av'_a+m_bv'_b}{m_a}$

$v_a=\frac{0.254Kg\times (-0.123 m/s)+0.367Kg (0.651m/s)}{0.254Kg}$

$v_a=0.8176 m/s$

The initial kinetic energy can be found as (provided puck b is at rest)

$K_1=\frac{1}{2}m_av_a^2$

$K_1=\frac{1}{2}(0.254Kg) (0.8176m/s)^2=0.0849 J$

The final kinetic energy is

$K_2=\frac{1}{2}m_av_a'^2+\frac{1}{2}m_bv_b'^2$

$K_2=\frac{1}{2}0.254Kg (-0.123m/s)^2+\frac{1}{2}0.367Kg (0.651m/s)^2=0.07969 J$

The change of kinetic energy is

$\Delta K=0.07969 J - 0.0849 J = -0.00521 J$

3 0

3 years ago