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3 years ago

2.4.6.8.10th term (it's from qbasic)

Physics

1 answer:

IgorLugansk [536]3 years ago

8 0

Answer:

Explanation:

2,4,6,8..Are in A.P
Common Difference=d=4-2=3
First term=a=2

We know that

$\boxed{\sf a_n=a+(n-1)d}$

━☞a10=2+(10-1)×2

━☞2+9×2

━☞2+18

━☞18

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calculate the diameter of a silver wire of length 75cm , which is extended by 1.85mm when a 10kg mass is suspended from it's end

sdas [7]

Answer: $0.8\ mm$

Explanation:

Given

length of wire $l=75\ cm$

change in length $\Delta l=1.85\ mm$

mass of wire $m=10\ kg$

Young's modulus for silver $E=7.9\times 10^{10}\ N/m^2$

load on wire $F=mg$

$F=10\times 9.8=98\ kg$

change in length is given by

$\Delta l=\dfrac{Pl}{AE}$

Where A=area of cross-section

$A=\dfrac{Pl}{\Delta lE}$

$A=\dfrac{98\times 0.75}{1.85\times 10^{-3}\times 7.9\times 10^{10}}$

$A=\dfrac{73.5}{14.615\times 10^{7}}$

$A=5.029\times 10^{-7}\ m^2$

also wire is the shape of cylinder so cross-section is given by

$A=\dfrac{\pi d^2}{4}=5.029\times 10^{-7}\ m^2$

$\Rightarrow d^2=\dfrac{5.029\times 10^{-7}\times 4}{\pi }$

$\Rightarrow d^2=64.02\times 10^{-8}$

$\Rightarrow d=8\times 10^{-4}\ m$

$\Rightarrow d=0.8\ mm$

4 0

3 years ago

An elevator accelerates upward at 1.2 m/s 2 . the acceleration of gravity is 9.8 m/s 2 . what is the upward force exerted by the

Marysya12 [62]

We can find the force by using the following formula;
N = ma + mg
Fa = ma = 76 x 1.2 = 91.2
Fg = mg = 76 x 9.8 = 744.8
N = 91.2 + 744.8 = 836
So, the force is 836 N.

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4 years ago

1. What would be the force of attraction between any two bodies are made double and distance between them from their center is h

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3 years ago

a 2.4*10^2 N force is pulling an 85 kg refrigerator across a horizontal surface. the force acts at an angle of 20.0 above the su

shepuryov [24]

Answer:

a) The work done by pulling force, W = 1804.21 joules

b) The work done by the kinetic frictional force is, w = 1332.8 J

Explanation:

Given,

The pulling force, Fₐ = 2.4 x 10² newton

The mass of the refrigerator, m = 85 Kg

The angle of pulling force acting to the surface, ∅ = 20°

The coefficient of kinetic friction is, μₓ = 0.200

The distance covered by the refrigerator, s = 8 m

The force acting in the direction parallel to the surface is given by

F = Fₐ cos∅

∴ F = 2.4 x 10² x cos 20°

=225.53 N

The work done by the pulling force is given by

W = F · S

= 225.53 N x 8 m

= 1804.21 joules

The work done by the pulling force, W = 1804.21 joules

The kinetic friction force is given by the formula

Fₓ = μₓ · η

Where

μₓ - coeficient of kinetic frictoin

η - Normal force, (mg)

∴ Fₓ = 0.200 x 85 x 9.8

= 166.6 N

The work done by the kinetic frictional force is given by

w = Fₓ · S

= 166.6 N x 8 m

= 1332.8 J

Hence, the work done by the kinetic frictional force is, w = 1332.8 J

7 0

3 years ago

Read 2 more answers

) A skier starts down a frictionless 32° slope. After a vertical drop of 25 m, the slope temporarily levels out and then slopes

EastWind [94]

Answer:

The skier’s speed on the two level stretches are 22.13 m/s and 27.29 m/s.

Explanation:

Given that,

Slope down = 32°

Height = 25 m

Before leveling,

Slope down = 20°

Height = 38 m

We need to calculate the skier’s speed on the two level stretches

Using formula of energy

$P.E=K.E$

$mgh=\dfrac{1}{2}mv^2$

$v=\sqrt{2gh}$

For first stretch,

Height = 25 m

Put the value into the formula

$v=\sqrt{2\times9.8\times25}$

$v=22.13\ m/s$

For second stretch,

Height = 38 m

Put the value into the formula

$v=\sqrt{2\times9.8\times38}$

$v=27.29\ m/s$

Hence, The skier’s speed on the two level stretches are 22.13 m/s and 27.29 m/s.

5 0

4 years ago

2.4.6.8.10th term (it's from qbasic)​

2.4.6.8.10th term (it's from qbasic)