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kobusy [5.1K]
3 years ago
6

5. On the periodic table, which families are in the first period? a

Physics
2 answers:
kkurt [141]3 years ago
5 0

Answer:

The family in the first period is the alkali metal family.

fomenos3 years ago
4 0
Alkali metals

Alkali metals
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1 poi
tatuchka [14]

Answer:

true

Explanation:

for example assume you are setting in a moving bus and when someone see you from the ground you are in motion but for some who is with you in the bus you are not in motion.

6 0
2 years ago
Why can't 41 be represented in a binary table?
Airida [17]

Answer:

I think it is but I don't know for sure

Explanation:

41 101001

41 is 101001 on the binary table i think

8 0
3 years ago
Deprecion como la ejerzo
Mrac [35]

Solo dígales cómo se siente, dígales que no está contento porque está triste y no sabe qué hacer al respecto, que está deprimido, y que solo quiere que las cosas mejoren.

8 0
3 years ago
VELOCITY, SPEED, DISTANCE ETC..
vaieri [72.5K]

Answer:

Explanation:

Speed = distance / time

Velocity  = displacement / time

So ,

Speed = 50 km / 0.5 hr = 100 km/h

Velocity  = 40 km / 0.5hr = 80 km/h

4 0
2 years ago
If he leaves the ramp with a speed of 31.0 m/s and has a speed of 29.5 m/s at the top of his trajectory, determine his maximum h
raketka [301]

Answer:

The maximum height reached is 4.63 m.

Explanation:

Given:

Initial speed of the man (u) = 31.0 m/s

Speed at the top of trajectory (u_x) = 29.5 m/s

Acceleration due to gravity (g) = 9.8 m/s²

When the man reaches the top of the trajectory, the vertical component of velocity becomes zero and hence only horizontal component of velocity acts on him.

Also, since there is no net force acting in the horizontal direction, the acceleration is zero in the horizontal direction from Newton's second law. Thus, the horizontal component of velocity always remains the same.

So, speed at the top of trajectory is nothing but the horizontal component of initial velocity.

Now, initial velocity can be rewritten in terms of its components as:

u^2=u_x^2+u_y^2

Where, u_x\ and\ u_y are the initial horizontal and vertical velocities of the man.

Now, plug in the given values and simplify. This gives,

(31.0)^2=(29.5)^2+u_y^2\\\\961=870.25+u_y^2\\\\u_y^2=961-870.25\\\\u_y^2=90.75\ m^2/s^2--------1

Now, we know that, for a projectile motion, the maximum height is given as:

H=\frac{u_y^2}{2g}

Plug in the value from equation (1) and 9.8 for 'g' to solve for 'H'. This gives,

H=\frac{90.75}{2\times 9.8}\\\\H=4.63\ m

Therefore, the maximum height reached is 4.63 m.

3 0
3 years ago
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