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3 years ago

12

En el MCU, la aceleración es: a) constante b) Varía en módulo, direccion, y sentido c) constante solo en el módulo d) constante

solo en dirección y sentido

1 answer:

JulijaS [17]3 years ago

7 0

If you put it into English in the comments I would be more then happy to help you! Thank you!

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Two 0.20-kg balls, moving at 4 m/s east, strike a wall. Ball A bounces backwards at the same speed. Ball B stops. Which statemen

muminat

Answer:

Option A

Explanation:

From the question we are told that:

Mass $m=0.20kg$

Velocity $v=4m/s$

Generally the equation for momentum for Ball A is mathematically given by

Initial Momentum

$M_{a1}=mV$

$M_{a1}=0.2*4$

$M_{a1}=0.8$

Final Momentum

$M_{a2}=-0.8kgm/s$

Therefore

$\triangle M_a=-1.6kgm/s$

Generally the equation for momentum for Ball B is mathematically given by

Initial Momentum

$M_{b1}=mV$

$M_{b1}=0.2*4$

$M_{b1}=0.8$

Final Momentum

$M_{b2}=-0 kgm/s$

Therefore

$|\triangle M_a|>|\triangle Mb|$

Option A

4 0

3 years ago

Which of the following phenomena are due to the electric interaction? (Select all that apply.) surface tension in water friction

Eddi Din [679]

Answer:

Surface tension in water

Friction between tires and pavement

Dissolution of salt in water

Explanation:

Surface tension in water: It is due to the electrostatic force of attraction (cohesive force) between water molecules.

Friction between tires and pavement: It is due to the attractive force between tires and pavement.

Dissolution of salt in water: The ions of $Na ^ +$ and $Cl ^ -$ separate due to the strong attraction of water molecules.

5 0

3 years ago

A 1.2 KG rubber ball is being thrown in the air if the ball is traveling at 2.0 M/S when it is 3.0 M off the ground what is the

Vitek1552 [10]

Answer:

37.7 J

Hope this helps! (see pictures)

6 0

3 years ago

5. What is the speed of a wave in a spring if it has a wavelength of 10 cm and a period of 0.2s

eimsori [14]

Explanation:

speed of wave

v = wavelength x frequency

since frequency is f = 1/Period then

v = wavelength : Period

v = 10 cm/ 0.2 s = 50 cm/s

v = 0.5 m/s

7 0

3 years ago

Estimate the electric field at a point 2.40 cm perpendicular to the midpoint of a uniformly charged 2.00-m-long thin wire carryi

nadya68 [22]

Answer:

$E = 1.85*10^{12}\frac{N}{C}$

Explanation:

Hi!

The perpendicular distance 2.4cm, is much less than the distance to both endpoints of the wire, which is aprox 1m. Then the edge effect is negligible at this field point, and we can aproximate the wire as infinitely long.

The electric filed of an infinitely long wire is easy to calculate. Let's call z the axis along the wire. Because of its simmetry (translational and rotational), the electric field E must point in the radial direction, and it cannot depende on coordinate z. To calculate the field Gauss law is used, as seen in the image, with a cylindrical gaussian surface. The result is:

$E = \frac{\lambda}{2\pi \epsilon_0 r}\\\lambda=\text{charge per unit length}=\frac{4.95 \mu C}{2 m} = 2.475 \frac{C}{m}\\r=\text{perpendicular distance to wire}\\\epsilon_0=8.85*10^{-12}\frac{C^2}{Nm^2}$

Then the electric field at the point of interest is estimated as:

$E = \frac{\22.475}{2\pi*( 8.85*10^{-12})*(2.4*10^{-2})}\frac{N}{C}=1.85*10^{12}\frac{N}{C}$

6 0

4 years ago