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Natasha2012 [34]

3 years ago

6

A student claims that the Chernobyl accident proves that nuclear power is on safe for use in the United States which statement i

s the most valid counter argument for this claim

1 answer:

Setler79 [48]3 years ago

5 0

Hi! Check out my valid counter argument below!

"The accident only released harmless gamma rays."

Hope I helped!

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Each of 100 identical blocks sitting on a frictionless surface is connected to the next block by a massless string. The first bl

o-na [289]

Answer:

The tension in the string connecting block 50 to block 51 is 50 N.

Explanation:

Given that,

Number of block = 100

Force = 100 N

let m be the mass of each block.

We need to calculate the net force acting on the 100th block

Using second law of newton

$F=ma$

$100=100m\times a$

$ma=1\ N$

We need to calculate the tension in the string between blocks 99 and 100

Using formula of force

$F_{100-99}=ma$

$F_{100-99}=1$

We need to calculate the total number of masses attached to the string

Using formula for mass

$m'=(100-50)m$

$m'=50m$

We need to calculate the tension in the string connecting block 50 to block 51

Using formula of tension

$F_{50}=m'a$

Put the value into the formula

$F_{50}=50m\times a$

$F_{50}=50\times1$

$F_{50}=50\ N$

Hence, The tension in the string connecting block 50 to block 51 is 50 N.

8 0

3 years ago

Please help me......

Novosadov [1.4K]

The first one, Climate

6 0

3 years ago

A 4.4 kg mess kit sliding on a frictionless surface explodes into two 2.2 kg parts, one moving at 2.9 m/s, due north, and the ot

sp2606 [1]

Answer:

Original speed of the mess kit = 4.43 m/s at 50.67° north of east.

Explanation:

Let north represent positive y axis and east represent positive x axis.

Here momentum is conserved.

Let the initial velocity be v.

Initial momentum = 4.4 x v = 4.4v

Velocity of 2.2 kg moving at 2.9 m/s, due north = 2.9 j m/s

Velocity of 2.2 kg moving at 6.8 m/s, 35° north of east = 6.9 ( cos 35i + sin35 j ) = 5.62 i + 3.96 j m/s

Final momentum = 2.2 x 2.9 j + 2.2 x (5.62 i + 3.96 j) = 12.364 i + 15.092 j kgm/s

We have

Initial momentum = Final momentum

4.4v = 12.364 i + 15.092 j

v =2.81 i + 3.43 j

Magnitude

$v=\sqrt{2.81^2+3.43^2}=4.43m/s$

Direction

$\theta =tan^{-1}\left ( \frac{3.43}{2.81}\right )=50.67^0$

50.67° north of east.

Original speed of the mess kit = 4.43 m/s at 50.67° north of east.

6 0

3 years ago

How many hydrogen and carbon atoms in a diamond

Wewaii [24]

Answer:

Explanation:

Thus, total 4+4=8 C atoms are present per unit cell of diamond. Carbon has an electronic arrangement of 2,4. In diamond, each carbon shares electrons with four other carbon atoms - forming four single bonds.

4 0

3 years ago

Three identical resistors are connected in parallel. The equivalent resistance increases by 630 when one resistor is removed and

strojnjashka [21]

Answer:

each resistor is 540 Ω

Explanation:

Let's assign the letter R to the resistance of the three resistors involved in this problem. So, to start with, the three resistors are placed in parallel, which results in an equivalent resistance $R_e$ defined by the formula:

$\frac{1}{R_e}=\frac{1}{R} } +\frac{1}{R} } +\frac{1}{R} \\\frac{1}{R_e}=\frac{3}{R} \\R_e=\frac{R}{3}$

Therefore, R/3 is the equivalent resistance of the initial circuit.

In the second circuit, two of the resistors are in parallel, so they are equivalent to:

$\frac{1}{R'_e}=\frac{1}{R} +\frac{1}{R}\\\frac{1}{R'_e}=\frac{2}{R} \\R'_e=\frac{R}{2} \\$

and when this is combined with the third resistor in series, the equivalent resistance ( $R''_e$ ) of this new circuit becomes the addition of the above calculated resistance plus the resistor R (because these are connected in series):

$R''_e=R'_e+R\\R''_e=\frac{R}{2} +R\\R''_e=\frac{3R}{2}$

The problem states that the difference between the equivalent resistances in both circuits is given by:

$R''_e=R_e+630 \,\Omega$

so, we can replace our found values for the equivalent resistors (which are both in terms of R) and solve for R in this last equation:

$\frac{3R}{2} =\frac{R}{3} +630\,\Omega\\\frac{3R}{2} -\frac{R}{3} = 630\,\Omega\\\frac{7R}{6} = 630\,\Omega\\\\R=\frac{6}{7} *630\,\Omega\\R=540\,\Omega$

8 0

3 years ago