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3 years ago

Which single force acts on an object in freefall?

Physics

2 answers:

Elena-2011 [213]3 years ago

6 0

gravity is the correct answer

jok3333 [9.3K]3 years ago

4 0

It has to be a)gravity

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The paper dielectric in a paper-and-foil capacitor is 0.0800 mm thick. Its dielectric constant is 2.50, and its dielectric stren

jeka94

Answer:

a) 0.723 m²

b) 2000V

Explanation:

Given that

Thickness of the capacitor, d = 0.08*10^-3 m

Dielectric constant of the capacitor, k = 2.5

Dielectric strength of the capacitor, E = 50*10^6

Capacitance of the capacitor, C = 0.2*10^-6

Permittivity of free space, E• = 8.85*10^-12

The area, A is given by the formula

A = (C * d) / (k * E•)

A = (0.2*10^-6 * 0.08*10^-3) / (2.5 * 8.85*10^-12)

A = 1.6*10^-11 / 2.213*10^-11

A = 0.723 m²

Potencial difference, V is given by the formula

V = E * d

V = 1/2 * 50*10^6 * 0.08*10^-3

V = 1/2 * 4000

V = 2000 V

8 0

4 years ago

A coin is placed 29 cm from the center of a horizontal turntable, initially at rest. The turntable then begins to rotate. When t

kobusy [5.1K]

Answer:

The coefficient of static friction between the coin and the turntable is 0.51

Explanation:

at the time of the slip:

centripetal force = frictional force

mv^2/r = x*m*980

v^2/r = 980x

x = v^2/980r

= [(120)^2]/[980*29]

= 0.51

Therefore, The coefficient of static friction between the coin and the turntable is 0.51

5 0

3 years ago

A velocity selector can be used to measure the speed of a charged particle. A beam of particles is directed along the axis of th

Contact [7]

Answer:

C. 450v

Explanation:

Using

Voltage= B*distance of separation*velocity

3mm x 0.3T x 5E5m/s

= 450v

5 0

3 years ago

The range of audible frequencies is from about 20.0 hz to 2.00×104 hz . what is range of the wavelengths of audible sound in air

TiliK225 [7]

The wavelength is related to the frequency by the relationship:
$\lambda= \frac{v}{f}$
where v is the wave speed and f is its frequency.

The speed of sound in air is v=344 m/s. The lowest frequency is f=20.0 Hz, so the corresponding wavelength is
$\lambda_1 = \frac{v}{f_1}= \frac{344 m/s}{20.0 Hz}=17.2 m$
The highest frequency is $f_2 = 2.00 \cdot 10^4 Hz$ , so the corresponding wavelength is
$\lambda_2 = \frac{v}{f_2}= \frac{344 m/s}{2.00 \cdot 10^4 Hz}=0.017 m$

Therefore, the range of wavelengths of audible sound in air is
[0.017 m - 17.2 m]

4 0

3 years ago

A mass of 10.0 kg is in a gravitational field of 3.50 N/kg. What force acts on the mass?

mash [69]

Answer:

Force=35 N

Explanation:

Given data

mass m=10.0 kg

Gravitational field E=3.50 N/kg

To find

Force

Solution

From definition of gravitational field intensity.

$E_{gravitational-field }=\frac{Force}{mass}\\ E=F/m\\F=mE\\F=(10.0 kg)*(3.50 N/kg)\\F=35N$

6 0

4 years ago

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