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3 years ago

The force exerted by the earth on a body is called ______of a body.

Physics

2 answers:

Leno4ka [110]3 years ago

7 0

Answer:weight

Explanation:

AVprozaik [17]3 years ago

4 0

Answer:

The force exerted by the earth on a body is called gravitational of a body.

i think this is the answer

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4. A light string is attatched to a heavy rope, and the whole thing is pulled tight. A wave is sent along the light string. When

ale4655 [162]

Answer:

The correct answer to the question is (A)

When it hits the heavy rope, compared to the wave on the string, the wave that propagates along the rope has the same (A) frequency

Explanation:

The speed of a wave in a string is dependent on the square root of the tension ad inversely proportional to the square root of the linear density of the string. Generally, the speed of a wave through a spring is dependent on the elastic and inertia properties of the string

$v = \sqrt{ \frac{T}{\mu } } = \sqrt{ \frac{T}{m/L } }$

Therefore if the linear density of the heavy rope is four times that of light rope the velocity is halved and since

v = f×λ therefore v/2 = f×λ/2

Therefore the wavelength is halved, however the frequency remains the same as continuity requires the frequency of the incident pulse vibration to be transmitted to the denser medium for the wave to continue as the wave is due to vibrating particles from a source for example

7 0

3 years ago

A nation undergoing the transition from undeveloped to industrialized is commonly dependent on this to supply its energy needs.

algol [13]

A) obviously :)))))))))

7 0

3 years ago

Read 2 more answers

Give a strong idea to protect , purify and the beautify of the Dal lake

Sholpan [36]

Answer:

Water in dal lake is test for any heavy metals and pollutant, sewage and drainage system are also monitored for the same.

Explanation:

Dal lake is located in Srinagar that is the state capital of Kashmir and is known for recreation and tourism purposes. The area covers about 18 km sq. and forms a part of natural wetlands.

The lake is prone to pollution and has recently undergone restoration measures. To address the problems of eutrophication algae and large-scale microplankton have been removed from the water.

The government of India has taken various measures to check the pollution by setting up a committee to monitor the proper use of allotted funds.

3 0

3 years ago

How much force is needed to accelerate a 68 kilogram-skier at a rate of 1.2 m/sec^2?

DiKsa [7]

<span>Answer: Force = 81.6 N

Explanation:
According to Newton's Second law:
F = ma --- (1)

Where F = Force = ?
m = Mass = 68 kg
a = Acceleration = 1.2 m/s^2

Plug in the values in (1):
(1) => F = 68 * 1.2
F = 81.6 N (The force needed to accelerate the skier at a rate of 1.2 m/s^2)</span>

4 0

3 years ago

Read 2 more answers

Two polarizing sheets have their transmission axes crossed so that no light is transmitted. A third sheet is inserted so that it

jek_recluse [69]

Answer:

a) I= I₀ (cos²θ - cos⁴θ) b) 75.5º

Explanation:

a) For this exercise we must use Malus's law

I = I₀ cos² θ

where tea is the angle between the two polarizers.

We apply this expression to our case

* Polarizer 1 suppose that it is vertical and polarizer 2 (intermediate) is at an angle θ with respect to the vertical

I₁ = I₀ cos² θ

* We analyze for the polarity 2 and the last polarizer 3 which indicate that it must be at 90º from the first one, therefore it must be horizontal.

The angle of polarizers 2 and 3 is θ' measured from the horizontal, if we measure with respect to the vertical

θ₂ = 90- θ’ = θ

fiate that in the exercise we must take a reference system and measure everything with respect to this system.

I = I₁ cos² θ'

we substitute

I = (I₀ cos² tea) cos² (θ - 90)

cos (θ -90) = cos θ cos 90 + sin θ sin 90 = sin θ

I = Io cos² θ sin² θ

1= cos²θ+ sin²θ

sin²θ = 1 - cos²θ

I= I₀ (cos²θ - cos⁴θ)

b) to find when the intensity is maximum,

we can use that we have an extreme point when the drift is zero

$\frac{dI}{d \theta}$ = 0

\frac{dI}{d \theta}= Io (2 cos θ - 4 cos³θ) = 0

whereby

cos θ - 2 cos³ θ = 0

cos θ ( 1 - 2 cos² θ) = 0

The zeros of this function are in

θ = 90º

1-2cos²θ =0 cos θ = 0.25 θ = 75.5º

Let's analyze this two results for the angle of 90º the intnesidd is zero with respect to the first polarizer, so it is not an acceptable solution.

Consequently, the angle that allows the maximum intensity to pass is 75.5º

5 0

3 years ago