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nignag [31]
3 years ago
12

Explain why the elements of the same group exhibit the same chemical behavior​

Physics
1 answer:
ElenaW [278]3 years ago
4 0

Answer:

All the elements of a period have similar chemical properties because they have the same number of valence electrons in their outermost shell. Their atoms have the same number of electrons in the highest occupied energy level

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HELPPPP MEEE!!! 15 POINTS
GalinKa [24]

Answer:

I think it is eyeglasses

Explanation:

I think this because the glasses usually reflect the same thing sometimes and the light passes through really easy

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4 years ago
What is the mass and weight of a 10kg object on the moon where the force of gravity is 1/6-th that of the earth's?
sashaice [31]
<span>Mass on Earth = 10kg
Mass on Moon = 10kg
 
Weight on Earth=100N
Weight on Moon=100/6 = 16.66 N</span>
8 0
3 years ago
The way matter moves in a longitudinal wave
serg [7]

Answer:

A longitudinal wave is a type of mechanical wave, or wave that travels through matter, called the medium. In a longitudinal wave, particles of the medium vibrate in a direction that is parallel to the direction that the wave travels. Places where particles of the medium crowd closer together are called compression's.

Explanation:

6 0
4 years ago
A projectile of mass 9.6 kg is launched from the ground with an initial velocity of 12.4 m/s at angle of 54° above the horizonta
Temka [501]

Answer:

The location is at (3.535, 1.162) m

Solution:

As per the question:

Mass of the projectile, m = 9.6 kg

Initial velocity, v = 12.4 m/s

Angle, \theta = 54^{\circ}

Mass of one fragment, m = 6.5 kg

Time taken by the fragment, t = 1.42 s

Height of the fragment, y = 5.9 m

Horizontal distance, x = 13.6 m

Now,

To determine the location of the second fragment:

Horizontal Range, R = \frac{v^{2}sin2\theta}{g}

R = \frac{12.4^{2}sin2(54)}{9.8} = 14.92\ m

Time of flight, t' = \frac{2vsin\theta}{g} = \frac{2\times 12.4sin108}{9.8}= 2.406\ s

Now, for the fragments:

Mass of the other fragment, m' = M - m = 9.6 - 6.5 = 3.1 kg

Distance traveled horizontally:

s_{x} = vcos\theta = 12.4cos54^{\circ}\times 1.42 = 10.35\ m

Distance traveled vertically:

s_{y} = vcos\theta - \frac{1}{2}gt^{2}

s_{y} = 12.4sin54^{\circ}\times 1.42 -  \frac{1}{2}\times 9.8\times 1.42^{2} = 14.25 - 9.88 = 4.37\ m

Now,

s_{x} = \frac{mx + m'x'}{M}

10.35= \frac{6.5\times 13.6 + 3.1x'}{9.6}

x' = 3.535 m

Similarly,

s_{y} = \frac{my + m'y'}{M}

4.37= \frac{6.5\times 5.9 + 3.1y'}{9.6} = 1.162\ m

The location of the other fragment is at (3.535, 1.162)

5 0
4 years ago
A 10248 kg car is pulled by a low tow truck that has an acceleration of 2.0m/s what is the net force on the car
Ksenya-84 [330]

Answer: 20496N

Explanation:

The formula to calculate the net force will be given as:

Net force = Acceleration × Mass

Since 10248 kg car is pulled by a low tow truck that has an acceleration of 2.0m/s, the net force would be:

= 10248 × 2

= 20496N

8 0
3 years ago
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