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yanalaym [24]
3 years ago
15

Naturally occurring element X exists in three isotopic forms: X-28 (27.977 amu, 92.23% abundance), X-29 (28.976 amu, 4.67% abund

ance), and X-30 (29.974 amu, 3.10% abundance). Calculate the atomic weight of X. A. 86.93 amu B. 48.63 amu C. 28.09 amu D. 28.97 amu E. 27.16 amu
Physics
1 answer:
vova2212 [387]3 years ago
6 0

Answer:

C. 28.09 amu

Explanation:

The natural occurring element exist in 3 isotopic forms: namely X-28 (27.977 amu, 92.23% abundance),  X-29 (28.976 amu, 4.67% abundance) and  X-30 (29.974 amu, 3.10% abundance).

The atomic weight of elements depends on the isotopic abundance. If you know the fractional abundance and the mass of the isotopes the atomic weight can be computed.

The atomic weight is computed as follows:

atomic weight = mass of X-28 × fractional abundance + mass of X-29 × fractional abundance + mass of  X-30 × fractional abundance

atomic weight = 27.977 × 0.9223 + 28.976 × 0.0467 + 29.974 ×  0.0310

atomic weight = 25.8031871 + 1.3531792 + 0.929194

atomic weight = 28.0855603 amu

To 2 decimal place atomic weight = 28.09 amu

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Boats use echo sounders. Pulses of sound waves are reflected by the sea bed and
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Answer: depth = 3.75 m

Explanation:

Taken the speed of the sound in sea to be V = 1500 m/s

The time between an echo sounder sending and receiving a pulse is 0.005 s. That is,

Total time T = 0.005

Formula for echo is:

V = 2X/T

Where X = distance covered

Substitute V and T into the formula

1500 = 2X/0.005

Cross multiply

7.5 = 2X

Make X the subject of formula

X = 7.5/2

X = 3.75 m

Therefore, the depth of the sea is 3.75 m

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3 years ago
A circular loop of wire with a radius of 15.0 cm and oriented in the horizontal xy-plane is located in a region of uniform magne
bekas [8.4K]

Answer:

Explanation:

Given that,

Radius r = 15cm = 0.15m

Area of the circular loop can be determined using the formula for area of a circle

A = π r²

A = π × 0.15²

A = 0.0708 m²

Magnetic field B = 1.2T in positive z direction

B = 1.2 •k T.

If loop is remove from the field in the time interval

∆t = 2.3ms = 2.3×10^-3s

We want to find the average EMF and it is given as

ε = —∆Φ/∆t

The final flux is zero

Φf = 0

Where magnetic flux is given as

Φi = BA Cosθ

Where θ=0 since the area and the magnetic field point in the same direction.

Φi = BA Cos0

Φi = BA

Φi = 1.2 × 0.0708

Φi = 0.0848 Vs

Then, ε = —∆Φ/∆t

ε = —(Φf — Φi) / ∆t

ε = —(0-0.0848) / (2.3×10^-3)

ε = 0.0848 / (2.3×10^-3)

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3 years ago
Which statement is true? The speed of sound in air is inversely proportional to the temperature of the air. The speed of sound i
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6 0
2 years ago
Read 2 more answers
An object of mass 6 kg. is resting on a horizontal surface. A horizontal force
son4ous [18]

Answer:

a) The work done by the applied force is 1500 joules.

b) The kinetic energy of the block after 10 seconds is 1200 joules.

c) The magnitude of the force of friction is 3 newtons and its direction is against motion.

d) 300 joules of energy are lost during motion.

Explanation:

a) Since the object has a constant mass, on which a constant horizontal force is exerted. The work done by the force (W), measured in joules, is defined by the following expression:

W = F\cdot \Delta x (1)

Where:

F - Force, measured in newtons.

\Delta x - Distance, measured in meters.

If we know that F = 15\,N and \Delta x = 100\,m, then the work done by the force exerted on the object is:

W = (15\,N)\cdot (100\,m)

W = 1500\,J

The work done by the applied force is 1500 joules.

b) At first we need to calculate the net acceleration of the object (a), measured in meters per square second. By assuming a constant acceleration, we use the following kinematic formula:

\Delta x = v_{o}\cdot t +\frac{1}{2}\cdot a\cdot t^{2} (2)

Where v_{o} is the initial velocity of the object, measured in meters per second.

We clear the acceleration within the equation above:

\frac{1}{2}\cdot a \cdot t^{2} =  \Delta x-v_{o}\cdot t

a = \frac{2\cdot (\Delta x - v_{o}\cdot t)}{t^{2}}

If we know that \Delta x = 100\,m, v_{o} = 0\,\frac{m}{s} and t = 10\,s, then the net acceleration experimented by the object is:

a = \frac{2\cdot \left[100\,m-\left(0\,\frac{m}{s} \right)\cdot (10\,s)\right]}{(10\,s)^{2}}

a = 2\,\frac{m}{s^{2}}

By the 2nd Newton's Law, we construct the following equation of equilibrium under the consideration of a friction force acting against the motion of the object:

\Sigma F = F - f = m\cdot a (3)

Where:

F - External force exerted on the object, measured in newtons.

f - Kinetic friction force, measured in newtons.

If we know that F = 15\,N, m = 6\,kg and a = 2\,\frac{m}{s^{2}}, the kinetic friction force is:

f = F-m\cdot a

f = 15\,N-(6\,kg)\cdot \left(2\,\frac{m}{s^{2}} \right)

f = 3\,N

The work done by friction (W'), measured in joules, is:

W' = f\cdot \Delta x (4)

W' = (3\,N) \cdot (100\,m)

W' = 300\,J

And the net work experimented by the object is:

\Delta W = 1500\,J - 300\,J

\Delta W = 1200\,J

By the Work-Energy Theorem we understand that change in translational kinetic energy (\Delta K), measured in joules, is equal to the change in net work. That is:

\Delta K = \Delta W (5)

If we know that \Delta W = 1200\,J, then the change in translational kinetic energy is:

\Delta K = 1200\,J

The kinetic energy of the block after 10 seconds is 1200 joules.

c) The magnitude of the force of friction is 3 newtons and its direction is against motion.

d) The energy lost by the object is equal to the work done by the force of friction. Therefore, 300 joules of energy are lost during motion.

7 0
3 years ago
To lower the voltage from a "120 V" outlet, what is used (first, at least) in most situations
nasty-shy [4]

Answer:

B. Transformer

Explanation:

A transformer is a device that is used to either raise or lower voltages and currents in an electrical circuit. In modern electrical distribution systems, transformers are used to boost voltage levels so as to decrease line losses during transmission. It basically trades voltage for current in a circuit, while not affecting the total electrical power. This means it takes high-voltage electricity with a small current and changes it into low-voltage electricity with a large current, or vice versa.

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3 years ago
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