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3 years ago

HELPPPPPP ! Pleaseeee anyone ?

Physics

2 answers:

zvonat [6]3 years ago

7 0

1- periodic table 2- elements 3- atomic number 4- properties 5- periods 6- groups 7- group number 8- valence electrons, give me brainliest pleaseee !!

kenny6666 [7]3 years ago

4 0

Answer:

1 Periodic table

2 Elements

3 atomic number

4 properties

5 Periods

6 groups

7 group number

8 Valence electrons

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An overnight rainstorm has caused a major roadblock. Three massive rocks of mass m1=514 kg,m1=514 kg, m2=792 kg,m2=792 kg, and m

hjlf

Answer:

278.5 N

Explanation:

First, let us find the total mass to be moved. We add up all the given masses together.

The total mass to be moved, M

M = 514 + 792 + 322 = 1628 kg.

Next, we find the force that is applied to the blocks. Remember that F = ma, so

Force applied to all the blocks, F =

F = 1628 * 0.25 = 407 N

Now we find the force that was applied to only the first rock

F = 514 * 0.25 = 128.5 N

From this, we can get the remaining force that was applied to the last two rocks, by subtracting them from each other

F' = 407 - 128.5 = 278.5 N

3 0

3 years ago

A 410-g cylinder of brass is heated to 95.0*C and placed in a calorimeter containing 335 g of water at 25.0*C. The water is stir

yarga [219]

I don't know I am sorry

3 0

3 years ago

What is the speed in miles per second of a beam of light traveling at 3.00Ã—108m/s?

vlada-n [284]

Answer: 186411.3 mi/s

Explanation:

We have the speed of light in units of meters per second ( $\frac{m}{s}$ ):

$3(10)^{8} \frac{m}{s}$

And we need to express it in miles per second ( $\frac{mi}{s}$ ), knowing ( $1 m=0.000621371 miles$ ), then:

$3(10)^{8} \frac{m}{s} \frac{0.000621371 miles}{1 m}=186411.3 \frac{mi}{s}$

3 0

4 years ago

A uniform rod of mass 3.30×10−2 kg and length 0.450 m rotates in a horizontal plane about a fixed axis through its center and pe

Alex73 [517]

(a) 2.75 rev/min

The moment of inertia of the rod rotating about its center is:

$I_R=\frac{1}{12}ML^2$

where

$M=3.30\cdot 10^{-2} kg$ is its mass

L = 0.450 m is its length

Substituting,

$I_R=\frac{1}{12}(3.30\cdot 10^{-2})(0.450)^2=5.57\cdot 10^{-4} kg m^2$

The moment of inertia of the two rings at the beginning is

$I_r = 2mr^2$

where

m = 0.200 kg is the mass of each ring

$r=5.20\cdot 10^{-2} m$ is their distance from the center of the rod

Substituting,

$I_r=2(0.200)(5.20\cdot 10^{-2})^2=1.08\cdot 10^{-3} kg m^2$

So the total moment of inertia at the beginning is

$I_1=I_R+I_r = 5.57\cdot 10^{-4}+1.08\cdot 10^{-3}=1.64\cdot 10^{-3}kg m^2$

The initial angular velocity of the system is

$\omega_1 = 35.0 rev/min$

The angular momentum must be conserved, so we can write:

$L=I_1 \omega_1 = I_2 \omega_2$ (1)

where $I_2$ is the moment of inertia when the rings reach the end of the rod; in this case, the distance of the ring from the center is

$r=\frac{0.450 m}{2}=0.225 m$

so the moment of inertia of the rings is

$I_r=2(0.200)(0.225)^2=0.0203 kg m^2$

and the total moment of inertia is

$I_2 = I_R + I_r =5.57\cdot 10^{-4} + 0.0203 = 0.0209 kg m^2$

Substituting into (1), we find the final angular speed:

$\omega_2 = \frac{I_1 \omega_1}{I_2}=\frac{(1.64\cdot 10^{-3})(35.0)}{0.0209}=2.75 rev/min$

(b) 103.0 rev/min

When the rings leave the rod, the total moment of inertia is just equal to the moment of inertia of the rod, so:

$I_2 = I_R = 5.57\cdot 10^{-4}kg m^2$

So using again equation of conservation of the angular momentum:

$L=I_1 \omega_1 = I_2 \omega_2$

We find the new final angular speed:

$\omega_2 = \frac{I_1 \omega_1}{I_2}=\frac{(1.64\cdot 10^{-3})(35.0)}{5.57\cdot 10^{-4}}=103.0 rev/min$

7 0

3 years ago

What are the main causes of the convention currents in the asthenosphere?

Tju [1.3M]

I think it is c density and temperature

8 0

4 years ago

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