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Marina CMI [18]
4 years ago
7

Name a 3D shape with rectangular faces only?

Physics
2 answers:
Semenov [28]4 years ago
8 0
Triangular Prism Rectangular Prism Octagonal Prism
Lelechka [254]4 years ago
3 0
A cuboid has six. Squares are also rectangles.
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A ball of mass m is dropped from a height h above the ground. neglecting air resistance then determine the speed of the ball whe
MrMuchimi

Answer:

Explanation:

kinematic equation (g will have a negative value if we assume UP is positive)

v² = u² + 2as

a) v = √(0² + 2(g)(y - h))

b) v = √(vi² + 2(g)(y - h))

8 0
3 years ago
Gravity causes all falling objects to accelerate at a rate of 98 m/s2.<br> O True<br> O False
aleksandrvk [35]
It’s true the acceleration of falling objects on earth due to gravity is 98ms2
8 0
3 years ago
The graph below shows the variation with distance r from the nucleus of the square of the wave function, Ψ^2, of a hydrogen atom
Ludmilka [50]

The region a represents the distance of the electron from the nucleus.

According to the wave mechanical model of the atom, the probability of finding an electron within a given volume element (representing the atom) is the square of the wave function psi.

Since a is the region in space where there is the greatest probability of finding the electron in the atom, it follows that distance of the electron form the atom is always a.

Learn more about the wave mechanical model: brainly.com/question/1382157

3 0
3 years ago
6. An earthquake releases two types of traveling seismic waves, called transverse and longitudinal waves. The average speed of t
Ymorist [56]

Answer:

The distance away the center of the earthquake is 1083.24 km.

Explanation:

Given that,

Speed of transverse wave = 9.1\ km/s

Speed of longitudinal wave = 5.7 km/s

Time = 71 sec

We need to calculate the distance of transverse wave

Using formula of distance

d=v\times t

d=9.1\times t....(I)

The distance of longitudinal wave

d=5.7\times (t+71)....(II)

From the first equation

t=\dfrac{d}{9.1}

Put the value of t in equation (II)

d =5.7\times(\dfrac{d}{9.1}+71)

\dfrac{9.1d-5.7d}{9.1}=71\times5.7

d0.3736=404.7

d =1083.24\ km

Hence, The distance away the center of the earthquake is 1083.24 km.

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4 years ago
For a particle undergoing free-fall acceleration, the acceleration vs. time graph is
Sergeeva-Olga [200]
A straight line of y = 9.8
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