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slavikrds [6]
3 years ago
7

Three students derive the following equations in which x refers to distance traveled, v the speed, a the acceleration (m/s^2), a

nd t the time, and the subscript (0) means a quantity at time t=0: a) x=vt^2 + 2at, b) x=v0t+1/2at^2, and c) x=v0t+2at^2. Which of these could possibly be correct according to a dimensional check?
Physics
1 answer:
Lynna [10]3 years ago
3 0

Answer:

B) and C)

Explanation:

Dimensional check involves using the units of the fundamental quantities, like mass to verify the accuracy of a formula or model.

For this case, the quantities required are Length and Time. Their units are Metre (m) and Seconds (s) respectively.

Since x refers to Distance, v refers to velocity, a refers to acceleration and t refers to time,

A) x = vt² + 2at

=> m = (m/s * s²) + (m/s² * s)

m = ms + m/s

The units are unbalanced, hence, it is not dimensionally correct.

B) x = v₀t + 2at²

=> m = (m/s * s) + (m/s² * s²)

m = m + m

The units are balanced, hence, it is dimensionally correct.

C) x = v₀t + ¹/₂at²

=> m = (m/s * s) + (m/s² * s²)

m = m + m

The units are balanced, hence, it is dimensionally correct.

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Why would an asteroid have a difficult time of holding a probe or person?
babymother [125]

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5 0
3 years ago
A potential difference of 3.27 nV is set up across a 2.16 cm length of copper wire that has a radius of 2.33 mm. How much charge
miskamm [114]

Answer:

Charge = 4.9096 x 10⁻⁷ C

Explanation:

First, we find the resistance of the copper wire.

R = ρL/A

where,

R = resistance = ?

ρ = resistivity of copper = 1.69 x 10⁻⁸ Ω.m

L = Length of wire = 2.16 cm = 0.0216 m

A = Cross-sectional area of wire = πr² = π(0.00233 m)² = 1.7 x 10⁻⁵ m²

Therefore,

R = (1.69 x 10⁻⁸ Ω.m)(0.0216 m)/(1.7 x 10⁻⁵ m²)

R = 2.14 x 10⁻⁵ Ω

Now, we find the current from Ohm's Law:

V =IR

I = V/R

I = 3.27 x 10⁻⁹ V/2.14 x 10⁻⁵ Ω

I = 1.52 x 10⁻⁴ A

Now, for the charge:

I = Charge/Time

Charge = (I)(Time)

Charge = (1.52 x 10⁻⁴ A)(3.23 x 10⁻³ s)

<u>Charge = 4.9096 x 10⁻⁷ C</u>

8 0
4 years ago
A radioactive element undergoes decay via the loss of two alpha particles to form a stable daughter isotope. Following the decay
jeyben [28]

Answer:

Atomic number of the daughter isotope = (Atomic number of the parent isotope) - 4

Explanation:

An alpha particle has atomic number of 2 and a mass number of 4

Losing two alpha particles would lead to a reduction of 4 in atomic number and 8 in mass number for the radioactive isotope, to give the stable daughter isotope.

8 0
3 years ago
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