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podryga [215]
3 years ago
11

The dimensions for force are the product of mass and length divided by time squared. Newton's second law states that force equal

s the product of mass and acceleration. What are the dimensions of acceleration?
T2
T2/L
L/T2
L/T
Physics
1 answer:
maw [93]3 years ago
5 0

Answer:

L/T^2.

Explanation:

Dimensional analysis is defined as the analysis of the relationships between different physical quantities by identifying their base quantities (such as length, mass, time, temperature etc.) and units of measure (such as miles vs. kilometers, pounds vs. kilograms etc.).

F = m * a

kg(m/s^2) = kg * a

In dimensions,

a = L/T^2.

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Calculate the lowest energy (in ev) for an electron in an infinite well having a width of 0.050 mm.
MA_775_DIABLO [31]

The lowest energy of electron in an infinite well is 1.2*10^-33J.

To find the answer, we have to know more about the infinite well.

<h3>What is the lowest energy of electron in an infinite well?</h3>
  • It is given that, the infinite well having a width of 0.050 mm.
  • We have the expression for energy of electron in an infinite well as,

                  E_n=\frac{n^2h^2}{8mL^2}

  • where;

                m=9.1*10^{-31}kg\\L=0.050*10^{-3}m\\h=6.63*10^{-34}Js\\n=1

  • Thus, the lowest energy of electron in an infinite well is,

                E_1=\frac{(6.63*10^{-34})^2}{8*9.1*10^{-31}*(0.050*10^{-3})}=1.2*10^{-33}J

Thus, we can conclude that, the lowest energy of electron in an infinite well is 1.2*10^-33J.

Learn more about the infinite well here:

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7 0
1 year ago
A 1300 kg steel beam is supported by two ropes. (Figure
Dmitriy789 [7]

Relative to the positive horizontal axis, rope 1 makes an angle of 90 + 20 = 110 degrees, while rope 2 makes an angle of 90 - 30 = 60 degrees.

By Newton's second law,

  • the net horizontal force acting on the beam is

R_1 \cos(110^\circ) + R_2 \cos(60^\circ) = 0

where R_1,R_2 are the magnitudes of the tensions in ropes 1 and 2, respectively;

  • the net vertical force acting on the beam is

R_1 \sin(110^\circ) + R_2 \sin(60^\circ) - mg = 0

where m=1300\,\rm kg and g=9.8\frac{\rm m}{\mathrm s^2}.

Eliminating R_2, we have

\sin(60^\circ) \bigg(R_1 \cos(110^\circ) + R_2 \cos(60^\circ)\bigg) - \cos(60^\circ) \bigg(R_1 \sin(110^\circ) + R_2 \sin(60^\circ)\bigg) = 0\sin(60^\circ) - mg\cos(60^\circ)

R_1 \bigg(\sin(60^\circ) \cos(110^\circ) - \cos(60^\circ) \sin(110^\circ)\bigg) = -\dfrac{mg}2

R_1 \sin(60^\circ - 110^\circ) = -\dfrac{mg}2

-R_1 \sin(50^\circ) = -\dfrac{mg}2

R_1 = \dfrac{mg}{2\sin(50^\circ)} \approx \boxed{8300\,\rm N}

Solve for R_2.

\dfrac{mg\cos(110^\circ)}{2\sin(50^\circ)} + R_2 \cos(60^\circ) = 0

\dfrac{R_2}2 = -mg\cot(110^\circ)

R_2 = -2mg\cot(110^\circ) \approx \boxed{9300\,\rm N}

8 0
1 year ago
Scientists measure the time between the arrival of an earthquake's _______ and _______ waves to help determine the distance betw
soldier1979 [14.2K]

Scientists measure the time between the arrival of an earthquake's __P____ and ___S____ waves to help determine the distance between the recording seismograph and the earthquake epicenter.

Explanation:

P- (compressional) and S- (shear) waves produced in earthquakes travel at different speeds. P waves are faster than S waves and hence will be detected first by a seismograph after an earthquake. The further away a seismograph is from the epicenter of an earthquake,  the longer the time difference between the two (2) waves will be.

Using several, at least 3, seismographs located at different geoghraphical locations and detecting earthquakes, geologists can extrapolate the epicenter of an earthquake using the time differences in arrivals of the two waves in each of the seismographs, using the mathematics of triangulation.

Learn More:

For more on P- and S-- waves check out;

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7 0
3 years ago
Two teams of nine members each engage in a tug of war. Each of the first team’s members has an average mass of 68 kg and exerts
olga2289 [7]

Answer:

a)  a = - 0.106 m/s^2  (←)

b) T = 12215.1064 N

Explanation:

If

F₁ = 9*1350 N = 12150 N   (→)

F₂ = 9*1365 N = 12285 N  (←)

∑Fx = M*a = (M₁  +M₂)*a           (→)

F₁ - F₂ = (M₁  +M₂)*a        

→       a = (F₁ - F₂) / (M₁  +M₂ ) = (12150-12285)N/(9*68+9*73)Kg

→       a = - 0.106 m/s^2            (←)

(b) What is the tension in the section of rope between the teams?

If we apply  ∑Fx = M*a   for the team 1

F₁ - T = - M₁*a  ⇒   T = F₁ + M₁*a  

⇒   T = 12150 N + (9 * 68 Kg)*(0.106 m/s^2)

⇒ T = 12215.1064 N

If we choose the team 2 we get

- F₂ + T = - M₂*a  ⇒   T = F₂ - M₂*a  

⇒   T = 12285 N - (9 * 73 Kg)*(0.106 m/s^2)

⇒ T = 12215.1064 N

4 0
3 years ago
A tortoise and hare start from rest and have a race. As the race begins, both accelerate forward. The hare accelerates uniformly
Mama L [17]

Answer:

Explanation:

To solve this, we start by using one of the equations of motion. The very first one, in fact

1

V = U + at.

V = 0 + 0.8 * 3.4 = 2.72 m/s.

2.

V = 0 + 0.8 * 4.3 = 3.44 m/s.

3.

d = ½ * 0.8 * 4.3² + 3.44 * 12.9

d = 7.396 + 44.376

d = 51.77 m.

4.

d = 62 - 51.77 = 10.23 m. = Distance

traveled during deceleration.

a = (V² - Vo²) / 2d.

a = (0² - 3.44²) / 20.46

a = -11.8336 / 20.46 = -0.58 m/s²

5.

t = (V - Vo)/a =(0 - 3.44) / -0.58

t = -3.44/-.58 = 5.93 s

= Stop time.

T = 4.3 + 12.9 + 5.93 = 23.13 s. = Total

time the hare was moving.

6.

d = Vo * t + ½ * a * t² = 62 m.

0 + 0.5 * (23.13)² * a = 61

267.5a = 61

a = 61/267.5

a = 0.23 m/s²

7 0
3 years ago
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