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4 years ago

Which of the following group 7A elements is the most reactive?

1 answer:

8 0

Among the elements in group 7A, the most reactive would be Fluorine. Fluorine is a major component of toothpaste, what we use to brush and clean our teeth with. It is so reactive that it cannot be contained in glass.

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stellarik [79]

Answer:

the first one answer is no

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3 years ago

Which of the following is false

musickatia [10]

what are the options??

please include all of your question

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3 years ago

Read 2 more answers

Erase all the trajectories, and fire the pumpkin vertically again with an initial speed of 14 m/s. As you found earlier, the max

yanalaym [24]

Answer:

$\theta=39.49^{\circ}$

Explanation:

Maximum height of the pumpkin, $H_{max}=9.99\ m$

Initial speed, v = 22 m/s

We need to find the angle with which the pumpkin is fired. the maximum height of the projectile is given by :

$H_{max}=\dfrac{v^2\ sin^2\theta}{2g}$

On rearranging the above equation, to find the angle as :

$\theta=sin^{-1}(\dfrac{\sqrt{2gH_{max}}}{v})$

$\theta=sin^{-1}(\dfrac{\sqrt{2\times 9.8\times 9.99}}{22})$

$\theta=39.49^{\circ}$

So, the angle with which the pumpkin is fired is 39.49 degrees. Hence, this is the required solution.

8 0

3 years ago

One mole of iron (6 x 10^23 atoms) has a mass of 56 grams, and its density is 7.87 grams per cubic centimeter, so the center-to-

choli [55]

Answer:

Explanation:

Given that:

length l = 2.3 m

a = 0.12 cm = $0.12 \times 10^{-2} \ m$

$x = 1.17 \ cm = 1.17 \times 10^{-2}\ m$

m = 149 kg

$\delta = 7.87 \ g/cm^3$

$da = 2.28 \times 10^{-10}\ m$

$F_{net} = F-mg\\ \\0 = F - mg \\ \\ F = mg \\ \\ k_sx = mg \\ \\$

∴

$k_s = \dfrac{149(9.8)}{1.17 \times 10^{-2}} \\ \\ k_s = 124803.42 \ N /m$

$N_{chain} = \dfrac{A_{wire}}{A_{atom}} = \dfrac{A_w}{da^2}$

$N_{chain} = \dfrac{(a)^2}{(da)^2} = (\dfrac{a}{da})^2$

$N_{chain} = (\dfrac{0.12 \times 10^{-2} }{2.28 \times 10^{-10}})^2$

$N_{chain} = 2.77 \times 10^{13}$

$N_{bond} = \dfrac{L}{da} \\ \\ = \dfrac{2.3}{2.28 \times 10^{-10}} \\ \\ N_{bond} = 1.009 \times 10^{10}$

$\text{Finally; the stiffness of a single interatomic spring is:}$

$k_{si} =\dfrac{N_{bond}}{N_{chain}}\times k_s$

$k_{si} =\dfrac{(1.009 \times 10^{10})}{2.77*10^{13}}}\times (124803.42)$

$\mathbf{k_{si} =45.46 \ N/m}$

4 0

3 years ago

If a car is moving to the left with constantvelocity, one can conclude thatthere mustbe no forces applied to the car.the netforc

Allisa [31]

Answer:

the net force applied to the car is zero.

Explanation:

According to Newton's second law, the acceleration of an object (a) is directly proportional to the net force applied (F):

$a=\frac{F}{m}$

where m is the object's mass.

In this problem, the car is moving with constant velocity: this means that the acceleration is zero, a = 0. Therefore, according to the previous equation, the net force must also be zero: F = 0. So, the correct answer is

the net force applied to the car is zero.

3 0

4 years ago