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3 years ago

2. An element is a mixture of two isotopes. One isotope has an atomic mass of 34.969 amu and an

Physics

2 answers:

yKpoI14uk [10]3 years ago

7 0

Answer:

0.3567

Explanation:

aliina [53]3 years ago

4 0

Answer: The average atomic mass of an element is 35.67 amu and the unknown element will be chlorine.

Explanation:

Mass of isotope 1 = 34.969 amu

% abundance of isotope 1 = 64.88% = $\frac{64.88}{100}=0.6488$

Mass of isotope 2 = 36.966 amu

% abundance of isotope 2 = (100-64.88)% = $\frac{100-64.88}{100}=0.3512$

Formula used for average atomic mass of an element :

$\text{ Average atomic mass of an element}=\sum(\text{atomic mass of an isotopes}\times {{\text { fractional abundance}})$

$A=\sum[34.969\times 0.6488)+(36.966 \times 0.3512]]$

$A=35.67amu$

Therefore, the average atomic mass of an element is 35.67 amu and the unknown element will be chlorine.

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In the diagram, a force of 20 newtons is applied to a block. The block is in dynamic equilibrium. What is the magnitude and dire

sladkih [1.3K]

Answer:B 20 newtons opposite to the direction of the applied force

Explanation:

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2 years ago

Read 2 more answers

the maximum range of a projectile is 2÷√3 times its actual range what is the angle of the projection for the actual range

Murrr4er [49]

Answer:

The actual angle is 30°

Explanation:

<h2>Equation of projectile:</h2><h2>y axis:</h2>

$v_y(t)=vo*sin(A)-g*t$

the velocity is Zero when the projectile reach in the maximum altitude:

$0=vo-gt\\t=\frac{vo}{g}$

When the time is vo/g the projectile are in the middle of the range.

$d_x(t)=vo*cos(A)*t\\$

R=Range

$R=d_x(t=2*\frac{vo}{g})$

$R=vo*cos(A)*2\frac{vo}{g} \\\\R=\frac{(vo)^{2}*2* sin(A)cos(A)}{g} \\\\R=\frac{(vo)^{2} sin(2A)}{g}$

**sin(2A)=2sin(A)cos(A)

<h2>The maximum range occurs when A=45°(because sin(90°)=1)</h2><h2>The actual range R'=(2/√3)R:</h2>

Let B the actual angle of projectile

$\frac{vo^{2} }{g} =(\frac{2}{\sqrt{3} }) \frac{vo^{2} *sin(2B)}{g}\\\\1= \frac{2 }{\sqrt{3}} *sin(2B)\\\\sin(2B)=\frac{\sqrt{3}}{2}\\\\$

2B=60°

B=30°

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3 years ago

A rocket on Earth experiences an upward applied force from its thrusters. As a result of this force, the rocket accelerates upwa

gayaneshka [121]

Answer:

F=m(11.8m/s²)

For example, if m=10,000kg, F=118,000N.

Explanation:

There are only two vertical forces acting on the rocket: the force applied from its thrusters F, and its weight mg. So, we can write the equation of motion of the rocket as:

$F-mg=ma$

Solving for the force F, we obtain that:

$F=ma+mg=m(a+g)$

Since we know the values for a (2m/s²) and g (9.8m/s²), we have that:

$F= m(2m/s^{2}+9.8m/s^{2})\\\\F=m(11.8m/s^{2})$

From this relationship, we can calculate some possible values for F and m. For example, if m=10,000kg, we can obtain F:

$F=(10,000kg)(11.8m/s^{2})\\\\F=118,000N$

In this case, the force from the rocket's thrusters is equal to 118,000N.

5 0

3 years ago

a spring is stretched 20 cm by a 30.0 n force. determine the work done in stretching the spring from 0 to 40 cm?

Alekssandra [29.7K]

The work done when a spring is stretched from 0 to 40cm is 4J.

What is work done?

Work done is the magnitude of force multiplied by displacement of an object. It is also the amount of energy transferred to an object when work is done on that.

The work done on the spring to stretch to 40cm is,

F = kx

where F is force, k is force constant.

k = F / x = 10 N / 20 * 10^-2 m = 50 N/m

W = 0.5 * k * (x)^2

where W = work done, k = force constant.

W = 0.5 x 50 x (40 x 10^-2)^2 = 4 J.

Therefore, the work done on the spring when it is stretched to 40cm is 4J.

To learn more about work done click on the given link brainly.com/question/25573309

#SPJ4

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1 year ago

A small bag of sand is released from an ascending hot-air balloon whose upward constant velocity is v0 = 2.45 m/s. knowing that

lakkis [162]

When you draw an illustration for this problem, you would come up with the same drawing as shown in the picture. As the hot-air balloon travels upwards, there is a slight time when the bag of sand rises up until it reaches the maximum height. Then, it goes back down to the ground. The total time would be t₁ + t₂. The solution is as follows:

H = v₀²/2g = (2.45)²/2(9.81) = 0.306 m
t₁ = H/v₀ = 0.306 m/2.45 m/s = 0.125 s

t₂ = √2(H + 98.8)/g = √2(0.306+ 98.8)/9.81
t₂ = 4.495 s

Total time = 0.125 s + 4.495 s = 4.62 seconds

8 0

3 years ago