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Bumek [7]
3 years ago
6

A pointer is spun on a fair wheel of chance having its periphery labeled Trom 0 to 100. (a) Whhat is the sample space for this e

xperiment? (b)What is the probability that the pointer will stop between 20 and 35? (c) What is the probability that the wheel will stop on 58?
​
Engineering
2 answers:
liubo4ka [24]3 years ago
7 0

Answer:

A pointer is spun on a fair wheel of chance having its periphery labeled Trom 0 to 100. (a) Whhat is the sample space for this experiment? (b)What is the probability that the pointer will stop between 20 and 35? (c) What is the probability that the wheel will stop on 58?

​

Explanation:

thats all you said

Wewaii [24]3 years ago
5 0

Answer:

hii my name is RAGHAV what is your name

Explanation:

this question is which chapter

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Find the altitude of the right cylinder of maximum convex surface that can be inscribed in a given sphere.
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Answer:

The radius 4 is maximum in convex surface

5 0
2 years ago
About what thickness of aluminum is needed to stop a beam of (a) 2.5-MeV electrons, (b) 2.5-MeV protons, and (c) 10-MeV alpha pa
Nana76 [90]

The thickness of aluminium needed to stop the beam electrons, protons and alpha particles at the given dfferent kinetic energies is 1.5 x 10⁻¹⁴ m.

<h3>Thickness of the aluminum</h3>

The thickness of the aluminum can be determined using from distance of closest approach of the particle.

K.E = \frac{2KZe^2}{r}

where;

  • Z is the atomic number of aluminium  = 13
  • e is charge
  • r is distance of closest approach = thickness of aluminium
  • k is Coulomb's constant = 9 x 10⁹ Nm²/C²
<h3>For 2.5 MeV electrons</h3>

r = \frac{2KZe^2}{K.E} \\\\r = \frac{2 \times 9\times 10^9 \times 13\times (1.6\times 10^{-19})^2}{2.5 \times 10^6 \times 1.6 \times 10^{-19}} \\\\r = 1.5 \times 10^{-14} \ m

<h3>For 2.5 MeV protons</h3>

Since the magnitude of charge of electron and proton is the same, at equal kinetic energy, the thickness will be same. r = 1.5 x 10⁻¹⁴ m.

<h3>For 10 MeV alpha-particles</h3>

Charge of alpah particle = 2e

r = \frac{2KZe^2}{K.E} \\\\r = \frac{2 \times 9\times 10^9 \times 13\times (2 \times 1.6\times 10^{-19})^2}{10 \times 10^6 \times 1.6 \times 10^{-19}} \\\\r = 1.5 \times 10^{-14} \ m

Thus, the thickness of aluminium needed to stop the beam electrons, protons and alpha particles at the given dfferent kinetic energies is 1.5 x 10⁻¹⁴ m.

Learn more about closest distance of approach here: brainly.com/question/6426420

7 0
2 years ago
2. When manipulating your pedals, you should use your
astra-53 [7]

Answer:

D. left foot for the accelerator and your right foot for the brake.

* Hopefully this helps:) mark me the brainliest:)!!

7 0
3 years ago
Read 2 more answers
Engineers are designing a cylindrical air tank and are trying to determine the dimensions of the tank. The proposed material for
lana66690 [7]

Answer:

The length of tank is found to be 0.6 m or 600 mm

Explanation:

In order to determine the length, we need to find a volume for the tank.

For this purpose, we use ideal gas equation:

PV  = nRT

n = no. of moles = m/M

Therefore,

PV = (m/M)(RT)

V = (mRT)/(MP)

where,

V = volume of air = volume of container

m = mass of air = 4.64 kg

R = General Gas Constant = 8.314 J/mol.k

T = temperature of air = 10°C + 273 = 283 K

M = molecular mass of air = 0.02897 kg/mol

P = Pressure of Air = 20 MPa = 20 x 10^6 N/m²

V = (4.64 kg)(8.314 J/mol.k)(283 k)/(0.02897 kg/mol)(20 x 10^6 N/m²)

V = 0.01884 m³

Now, the volume of cylindrical tank is given as:

V = 0.01884 m³ = π(Diameter/2)²(Length)

Length = (0.01884 m³)(4)/π(0.2 m)²

<u>Length = 0.6 m = 600 mm</u>

4 0
3 years ago
The homogeneous, reversible, exothermic, liquid phase reaction: A근R Is being carried out in a reactor system consisting of an id
baherus [9]

Answer:

attached below

Explanation:

3 0
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