All categories

2 years ago

How do I solve this arithmetic sequence?

Physics

1 answer:

Serjik [45]2 years ago

4 0

Answer:

The 16ᵗʰ term of this sequence is 82

Step-by-step explanation:

Here,

First Term = a₁ = 9

Common Difference = (d) = 2

Now, For 16ᵗʰ term, n = 16

a₁₆ = 7 + (16 - 1) × 2

a₁₆ = 7 + 15 × 5

a₁₆ = 7 + 75

a₁₆ = 82

Thus, The 16ᵗʰ term of this sequence is 82

<u>-TheUnknownScientist</u>

You might be interested in

Two blocks are connected by a massless cable which goes through the center of a rotating turntable. The blocks have masses M1 =

Airida [17]

Answer:

If the final question is; at what velocity will the first block start to move outward in m/s?

$v = 3.5596 \frac{m}{s}$

Explanation:

The motion have the velocity that will make the block move using:

$F_{1}*F_{r}+ F_{2}= Ec \\Ec= \frac{M*v^{2} }{r}$

$m_{1} = 1.2 Kg$

$m_{2} = 2.8 Kg$

$r = 0,45 m$

μ $s= 0,54$

Resolving:

$\frac{m_{1}*v^{2} }{r} = us*m_{1}* g + m_{2} * g$

$v^{2} = \frac{((us *m_{1} + m_{2})*g )* r}{m_{1} }$

$v^{2} = \frac{((0.54 *1.2 + 2.8)*9.8 )* 0.45}{1.2}$

$v^{2} = 12.6714 \frac{m^{2} }{s^{2} }$

$v = 3.559 \frac{m}{s}$

4 0

3 years ago

when a electric current is passed through an insulated wire that is coiled around an iron core like a nail what is created?

Anika [276]

electromagnet

When a electric current is passed through an insulated wire that is coiled around an iron core, like a nail, an electromagnet is created.

8 0

2 years ago

A student drops two metallic objects into a 120-g steel container holding 150 g of water at 25°C. One object is a 206-g cube of

spayn [35]

Answer:

Mass of the aluminium chunk = 278.51 g

Explanation:

For an isolated system as given the energy lost and gains in the system will be zero therefore sum of all transfer of energy will be zero,as the temperature will also remain same

A specific heat formula is given as

Energy Change = Mass of liquid x Specific Heat Capacity x Change in temperature

Q = m×c×ΔT

Heat gain by aluminium + heat lost by copper = 0 (1)

For Aluminium:

Q = $m\times0.897\frac{J}{g.k}\times(25-5)$

Q = m x 17.94 joule

For Copper:

$Q= 206g\times0.385\frac{J}{g.k} \times(88-25)$

Q= 4996.53 Joule

from eq 1

m x 17.94 = 4996.53

$mass of aluminium = \frac{4996.53}{17.94} g$

Mass of the aluminium chunk = 278.51 g

3 0

3 years ago

What is the Morgan-Keenan Spectral Classification system?

Angelina_Jolie [31]

Answer: A method of classifying stars by their temperatures and compositions.

Explanation:

4 0

3 years ago

Steam is leaving a pressure cooker whose operating pressure is 20 psia. It is observed that the amount of liquid in the cooker h

Debora [2.8K]

Answer:

a) 34.05ft/s

b).1156.2BTU/lbm

c) 2.04BTU/s

Explanation:

Amount of liquid that has evaporated, m = ◇Vliq/ Vf

We replace the values to make conversion

m = (0.6gal/ 0.01683ft^3/lbm) × (0.13368ft^3/1gal)

m = 4.755lb

The mass flow rate of exit steam is given by:

m' = m/◇t

We replace values to make conversion

m' =( 4.766lb/45min) = 0.1059lb/min × 1min/60s

m' = 0.001765lb/s

The exit velocity V = m'/pA = m'Vg/A

We replace values to make conversion

V =[ (0.001765lbm/s)(20.093ft^3/lbm) /(0.15 in^2)]× (144in^2/1ft^2)

V = 34.05ft/s

b) The total and flow energies per unit mass is given by:

Eflow= Pv = h - u

We replace the values to make conversion

Eflow = 1156.2 - 1081.8

Eflow = 74.4BTU/lbm

Therefore theta= h + ke + pe

Theta approximately =h = 1156.2BTU/lbs

c) The rate at which energy is leaving the cooler by steam is given by:

Emass = m'theta

Emass = (0.001765)×(1156.2)

Emass = 2.04BTU/s

6 0

3 years ago