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NNADVOKAT [17]
3 years ago
8

Use the mass spectrum of europium to determine the atomic mass of europium. where the peak representing eu-151 has an exact mass

of 150.91986 amu and percent intensity is assumed be to exactly 91.610%, and the peak representing eu-153 has an exact mass of 152.92138 amu and percent intensity of 100.00%. express your answer to two decimal places with the appropriate units.
Physics
2 answers:
umka21 [38]3 years ago
4 0

Answer:

The atomic mass of a Europium atom is 151.96445 amu.

From the given information:

Percent intensity is 91.61% of Europium atom of molecular weight 150.91986 amu.

Percent intensity is 100.00% of Europium atom of molecular weight 152.92138 amu.

Abundance of Eu-151 atom:

Abundance of Eu-153 atom:

Atomic mass of Europium atom:

Therefore, the atomic mass of a Europium atom is 151.96445 amu.

Explanation:

slavikrds [6]3 years ago
3 0

Answer: The atomic mass of a Europium atom is 151.96445 amu.

From the given information:

Percent intensity is 91.61% of Europium atom of molecular weight 150.91986 amu.

Percent intensity is 100.00% of Europium atom of molecular weight 152.92138 amu.

Abundance of Eu-151 atom:

X_{Eu-151}=\frac{0.9161}{0.9161+1.000}=0.4781

Abundance of Eu-153 atom:

X_{Eu-153}=\frac{1.000}{0.9161+1.000}=0.5219

Atomic mass of Europium atom:

A=(X_{Eu-151}\times150.91986+X_{Eu-153}\times152.92138)amu\\A=(0.4781\times150.91986+0.5219\times152.92138)amu=151.96445 amu

Therefore, the atomic mass of a Europium atom is 151.96445 amu.

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Answer:

On the wavelength

Explanation:

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3 years ago
Bob walks 290 m south, then jogs 630 m southwest, then walks 290 m in a direction 26 ∘ east of north.
solong [7]

whos spying on bob O.o


3 0
3 years ago
How do I go about this?
Anna71 [15]

Hi there!

(a)

Recall that:
W = F \cdot d = Fdcos\theta

W = Work (J)
F = Force (N)
d = Displacement (m)

Since this is a dot product, we only use the component of force that is IN the direction of the displacement. We can use the horizontal component of the given force to solve for the work.

W =248(56)cos(30) = 12027.36 J

To the nearest multiple of ten:
W_A = \boxed{12030 J}

(b)
The object is not being displaced vertically. Since the displacement (horizontal) is perpendicular to the force of gravity (vertical), cos(90°) = 0, and there is NO work done by gravity.

Thus:
\boxed{W_g = 0 J}

(c)
Similarly, the normal force is perpendicular to the displacement, so:
\boxed{W_N = 0 J}

(d)

Recall that the force of kinetic friction is given by:
F_{f} =\mu_k mg

Since the force of friction resists the applied force (assigned the positive direction), the work due to friction is NEGATIVE because energy is being LOST. Thus:
W_f = -\mu_k mgd\\W_f = - (0.1)(56)(9.8)(56) = -3073.28 J

In multiples of ten:
\boxed{W_f = -3070 J}

(e)
Simply add up the above values of work to find the net work.

W_{net} = W_A + W_f \\\\W_{net} = 12027.36 + (-3073.28) = 8954.08 J

Nearest multiple of ten:
\boxed{W_{net} = 8950 J}}

(f)
Similarly, we can use a summation of forces in the HORIZONTAL direction. (cosine of the applied force)
F_{net} = F_{Ax} - F_f

W = F_{net} \cdot d = (F_{Ax} - F_f)

W = (F_Acos(30) - \mu_k mg)d\\W = (248cos(30) - 0.1(56)(9.8)) * 56 \\\\W = 8954.08 J

Nearest multiple of ten:
\boxed{W_{net} = 8950 J}

5 0
2 years ago
At the local playground, a 21-kg child sits on the right end of a horizontal teeter-totter, 1.8 m from the pivot point. On the l
Ksju [112]

Answer:

By convention a negative torque leads to clockwise rotation and a positive torque leads to counterclockwise rotation.

here weight of the child =21kgx9.8m/s2 = 205.8N

the torque exerted by the child Tc = - (1.8)(205.8) = -370.44N-m ,negative sign is inserted because this torque is clockwise and is therefore negative by convention.

torque exerted by adult Ta = 3(151) = 453N , counterclockwise torque.

net torque Tnet = -370.44+453 =82.56N , which is positive means counterclockwise rotation.

b) Ta = 2.5x151 = 377.5N-m

Tnet = -370.44+377.5 = 7.06N-m , positive ,counterclockwise rotation.

c)Ta = 2x151 = 302N-m

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3 years ago
Does the horizontal distance d travelled by the ball depend on the height of release? If it does depend on the height, what is t
elena-s [515]

Answer:

Explanation:

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depends upon the duration of time in which it remains in air . The longer it remains in air , the greater distance it can cover horizontally .

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If V be the velocity of throw and Vx be its horizontal component

Horizontal distance covered = t x Vx

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If h be the height from which the body is thrown , Vy be the vertical upward component of initial velocity

from the relation

s = ut + 1/2 at²

h = - Vy t  + 1/2 at²

As h increases , t will increase and therefore horizontal distance covered will increase. If the ball has only  horizontal velocity initially , Vy = 0

h = 1/2 gt²

t = \sqrt{\frac{2h}{g} }

Horizontal distance covered  = t x Vx

= \sqrt{\frac{2h}{g} } \times  V_x

From this expression also

Horizontal distance covered is proportional to \sqrt{h} .

7 0
3 years ago
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