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sattari [20]
3 years ago
6

During which segments are two states of matter present? the answer is D

Physics
1 answer:
Tomtit [17]3 years ago
7 0

Answer:

2,4

Explanation:

LOL WHERE THE PICTURE GURL?

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Write the products for the following neutralization reactions
madam [21]

Answer:

answer 3

Explanation:

5 0
3 years ago
At higher speeds, how would you compensate for the decrease in field of vision
inysia [295]
When you ride a vehicle in a fast speed, then your peripheral vision will reduce that is why there is a need for you to follow the direction of the objects when you are travelling in order for you to compensate to the decrease in the field of vision.


5 0
4 years ago
Read 2 more answers
Find the kinetic energy of an electron whose de broglie wavelength is 34.0 nm.
dezoksy [38]
The De Broglie wavelength of the electron is
\lambda=34.0 nm=34 \cdot 10^{-9} m
And we can use De Broglie's relationship to find its momentum:
p= \frac{h}{\lambda}= \frac{6.6 \cdot 10^{-34} Js}{34 \cdot 10^{-9} m}=1.94 \cdot 10^{-26} kg m/s

Given p=mv, with m being the electron mass and v its velocity, we can find the electron's velocity:
v= \frac{p}{m}= \frac{1.94 \cdot 10^{-26} kgm/s}{9.1 \cdot 10^{-31} kg}=  2.13 \cdot 10^4 m/s

This velocity is quite small compared to the speed of light, so the electron is non-relativistic and we can find its kinetic energy by using the non-relativistic formula:
K= \frac{1}{2}mv^2= \frac{1}{2}(9.1 \cdot 10^{-31} kg)(2.13 \cdot 10^4 m/s)^2=2.06 \cdot 10^{-22} J
3 0
3 years ago
A 30.0-kg girl in a swing is pushed to one side and held at rest by a horizontal force \vec{F} ​F ​⃗ ​​ so that the swing ropes
Virty [35]

Answer:

169.74 N

Explanation:

Given,

Mass of the girl = 30 Kg

angle of the rope with vertical, θ = 30°

equating the vertical component of the tension

vertical component of the tension is equal to the weight of the girl.

 T cos θ = m g

 T cos 30° = 30 x 9.8

 T = 339.48 N

Tension on the two ropes is equal to 339.48 N

Tension in each of the rope = T/2

                                           = 339.48/2 = 169.74 N

Hence, the tension in each of the rope is equal to 169.74 N

7 0
3 years ago
A huge tank of glycerine with a density of 1.260 g/cm3 is vertically stationed on a platform which is 15 m above the ground. The
EleoNora [17]

Answer:

The tank is losing 4.976*10^{-4}  m^3/s

v_g = 19.81 \ m/s

Explanation:

According to the Bernoulli’s equation:

P_1 + 1 \frac{1}{2} \rho v_1^2 + \rho gh_1 = P_2 +  \frac{1}{2}  \rho v_2^2 + \rho gh_2

We are being informed that both the tank and the hole is being exposed to air :

∴ P₁ = P₂

Also as the tank is voluminous ; we take the initial volume  v_1 ≅ 0 ;

then v_2 can be determined as:\sqrt{[2g (h_1- h_2)]

h₁ = 5 + 15 = 20 m;

h₂ = 15 m

v_2 = \sqrt{[2*9.81*(20 - 15)]

v_2 = \sqrt{[2*9.81*(5)]

v_2= 9.9 \ m/s  as it leaves the hole at the base.

radius r = d/2  = 4/2 = 2.0 mm

(a) From the law of continuity; its equation can be expressed as:

J = A_1v_2

J = πr²v_2    

J =\pi *(2*10^{-3})^{2}*9.9

J =1.244*10^{-4}  m^3/s

b)

How fast is the water from the hole moving just as it reaches the ground?

In order to determine that; we use the relation of the velocity from the equation of motion which says:

v² = u² + 2gh ₂

v² = 9.9² + 2×9.81×15

v² = 392.31

The velocity of how fast the water from the hole is moving just as it reaches the ground is : v_g = \sqrt{392.31}

v_g = 19.81 \ m/s

4 0
3 years ago
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