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4 years ago

How does the study of social sciences affect our society?

Physics

1 answer:

Kruka [31]4 years ago

8 0

Explanation:

study of social sciencake us an effcient citizen of a democracy ,and it also help us to solve the practical problem in our daily life.

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A proton travels with a speed of 4.2×106 m/s at an angle of 30◦ west of north. A magnetic field of 2.5 T points to the north. Fi

Arada [10]

Answer:

$8.4\cdot 10^{-13} N$

Explanation:

The magnitude of the magnetic force on the proton is given by:

$F=qvB sin \theta$

where:

$q=1.6\cdot 10^{-19} C$ is the proton charge

$v=4.2\cdot 10^6 m/s$ is the proton velocity

$B=2.5 T$ is the magnetic field

$\theta=30^{\circ}$ is the angle between the direction of v and B

Substituting into the formula, we find

$F=(1.6\cdot 10^{-19}C)(4.2\cdot 10^6 m/s)(2.5 T) sin 30^{\circ}=8.4\cdot 10^{-13} N$

6 0

3 years ago

In a glider stunt at an air show, a towing airplane (motorized plane pulling the gliders) takes off from a level runway with two

irakobra [83]

Answer:

minimum length of runway is needed for take off 243.16 m

Explanation:

Given the data in the question;

mass of glider = 700 kg

Resisting force = 3700 N one one glider

Total resisting force on both glider = 2 × 3700 N = 7400 N

maximum allowed tension = 12000 N

from the image below, as we consider both gliders as a system

Equation force in x-direction

2ma = T -f

a = T-f / 2m

we substitute

a = (12000 - 7400 ) / (2 × 700 )

a = 4600/1400

a = 3.29 m/s²

Now, let Vf be the final speed and Ui = 0 ( as starts from rest )

Vf² = Ui² + 2as

solve for s

Vf² = 0 + 2as

2as = Vf²

s = Vf² / 2a

given that take of speed for the gliders and the plane is 40 m/s

we substitute

s = (40)² / 2×3.29

s = 1600 / 6.58

s = 243.16 m

Therefore, minimum length of runway is needed for take off 243.16 m

4 0

3 years ago

Earthquakes along the subduction zones near Sumatra and Chile are some of the strongest in

Lana71 [14]

Yes
The earthquakes in the san andrais fault line are one of the most dangerous.

6 0

3 years ago

Two large thin metal plates are parallel and close to each other. On their inner faces, the plates have excess surface charge of

wariber [46]

Answer:

For left = 0 N/C

For right = 0 N/C

At middle = $-7.6836 * 10^{-11} \vec{i}$ N/C

Explanation:

Given data :-

б = $6.8 * 10^{-22}$ C/ m²

Considering the two thin metal plates to be non conducting sheets of charges.

Electric field is given by

$E = \frac{\sigma }{2\varepsilon }$

1) To the left of the plate

$\vec{E}= (\frac{\sigma }{2\varepsilon })(-\vec{i})+ (\frac{\sigma }{2\varepsilon })(\vec{i})$ = 0 N/C.

2) To the right of them.

$\vec{E}= (\frac{\sigma }{2\varepsilon })(-\vec{i})+ (\frac{\sigma }{2\varepsilon })(\vec{i})$ = 0 N/C.

3) Between them.

$\vec{E}= (\frac{\sigma }{2\varepsilon })(-\vec{i})+ (\frac{\sigma }{2\varepsilon })(-\vec{i})$ = $(\frac{\sigma }{\varepsilon })(-\vec{i})$ = $-\frac{6.8 * 10^{-22} }{8.85 * 10 ^{-12} } \vec{i}$ = $-7.6836 * 10^{-11} \vec{i}$ N/C

5 0

3 years ago

What is the period of a wave that travels through a spring at 2.5 m/s and has a wavelength of 1.3 meters?

ruslelena [56]

Answer:

v=(1/f) ×λ

(1/f)= v/λ

(1/f)= 2.5 ÷ 1.3

(1/f)= 1.92 seconds

T = 1.92 seconds

Explanation:

period= 1÷ frequency

T=1/f

7 0

2 years ago