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3 years ago

5

Which element in the earth’s atmosphere is essential for human life, is extremely rare or nonexistent in the atmospheres of othe

r planets in our solar system, and is primarily responsible for the development of life outside the oceans?

1 answer:

gayaneshka [121]3 years ago

3 0

That element is oxygen. It is also highly volatile since it reacts with other elements and is flammable.

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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

Violet light (410 nm) and red light (685 nm) pass through a diffraction grating with d = 3.33 x 10-6 m. What is the angular sepa

jeyben [28]

Answer:10.03

Explanation:

right on acellus

7 0

2 years ago

A cheerleader lifts his 79.4 kg partner straight up off the ground a distance of 0.945 m before releasing her. the acceleration

Oksi-84 [34.3K]

To find out how much work he has done, we must first calculate force using the force formula (F= Mass*Acceleration). In this case, mass is 79.4 and acceleration is the gravitational constant of 9.8m/s, plugging this into the formula we find that force is 778.12Newtons. Next, we need to multiply force by the distance to get the amount of energy used to lift his partner once. Which is 778.12 * .945 = 735.32. Finally, we need to multiply 735.32 by the number of times he lifts his partner, 33, to get 735.32 * 33 to find that the energy he has expended 24,265.56 Joules of energy.

5 0

2 years ago

A tree falls in a forest. How many years must pass before the 14C activity in 1.03 g of the tree's carbon drops to 1.02 decay pe

Illusion [34]

Answer:

t = 5.59x10⁴ y

Explanation:

To calculate the time for the ¹⁴C drops to 1.02 decays/h, we need to use the next equation:

$A_{t} = A_{0}\cdot e^{- \lambda t}$ (1)

<em>where $A_{t}$ : is the number of decays with time, A₀: is the initial activity, λ: is the decay constant and t: is the time.</em>

To find A₀ we can use the following equation:

$A_{0} = N_{0} \lambda$ (2)

<em>where N₀: is the initial number of particles of ¹⁴C in the 1.03g of the trees carbon </em>

From equation (2), the N₀ of the ¹⁴C in the trees carbon can be calculated as follows:

$N_{0} = \frac{m_{T} \cdot N_{A} \cdot abundance}{m_{^{12}C}}$

<em>where $m_{T}$ : is the tree's carbon mass, $N_{A}$ : is the Avogadro's number and $m_{^{12}C}$ : is the ¹²C mass. </em>

$N_{0} = \frac{1.03g \cdot 6.022\cdot 10^{23} \cdot 1.3\cdot 10^{-12}}{12} = 6.72 \cdot 10^{10} atoms ^{14}C$

Similarly, from equation (2) λ is:

$\lambda = \frac{Ln(2)}{t_{1/2}}$

<em>where t 1/2: is the half-life of ¹⁴C= 5700 years </em>

$\lambda = \frac{Ln(2)}{5700y} = 1.22 \cdot 10^{-4} y^{-1}$

So, the initial activity A₀ is:

$A_{0} = 6.72 \cdot 10^{10} \cdot 1.22 \cdot 10^{-4} = 8.20 \cdot 10^{6} decays/y$

Finally, we can calculate the time from equation (1):

$t = - \frac{Ln(A_{t}/A_{0})}{\lambda} = - \frac {Ln(\frac{1.02decays \cdot 24h \cdot 365d}{1h\cdot 1d \cdot 1y \cdot 8.20 \cdot 10^{6} decays/y})}{1.22 \cdot 10^{-4} y^{-1}} = 5.59 \cdot 10^{4} y$

I hope it helps you!

4 0

2 years ago

A vector → A has a magnitude of 56.0 m and points in a direction 30.0° below the negative x axis. A second vector, → B , has a m

MissTica

Answer:

The magnitude of the vector $\vec{C}$ is 107.76 m

Explanation:

To find the components of the vectors we can use:

$\vec{A} = | \vec{A} | \ ( \ cos(\theta) \ , \ sin (\theta) \ )$

where $| \vec{A} |$ is the magnitude of the vector, and θ is the angle over the positive x axis.

The negative x axis is displaced 180 ° over the positive x axis, so, we can take:

$\vec{A} = 56.0 \ m \ ( \ cos( 180 \° + 30 \°) \ , \ sin (180 \° + 30 \°) \ )$

$\vec{A} = 56.0 \ m \ ( \ cos( 210 \°) \ , \ sin (210 \°) \ )$

$\vec{A} = ( \ -48.497 \ m \ , \ - 28 \ m \ )$

$\vec{B} = 82.0 \ m \ ( \ cos( 180 \° - 49 \°) \ , \ sin (180 \° - 49 \°) \ )$

$\vec{B} = 82.0 \ m \ ( \ cos( 131 \°) \ , \ sin (131 \°) \ )$

$\vec{B} = ( \ -53.797 \ m \ , \ 61.886\ m \ )$

Now, we can perform vector addition. Taking two vectors, the vector addition is performed:

$(a_x,a_y) + (b_x,b_y) = (a_x+b_x,a_y+b_y)$

So, for our vectors:

$\vec{C} = ( \ -48.497 \ m \ , \ - 28 \ m \ ) + ( \ -53.797 \ m \ , ) = ( \ -48.497 \ m \ -53.797 \ m , \ - 28 \ m \ + \ 61.886\ m \ )$

$\vec{C} = ( \ - 102.294 \ m , \ 33.886 m \ )$

To find the magnitude of this vector, we can use the Pythagorean Theorem

$|\vec{C}| = \sqrt{C_x^2 + C_y^2}$

$|\vec{C}| = \sqrt{(- 102.294 \ m)^2 + (\ 33.886 m \)^2}$

$|\vec{C}| =107.76 m$

And this is the magnitude we are looking for.

5 0

3 years ago