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masha68 [24]
3 years ago
11

A car with a mass of 2000 kg is moving around a circular curve at a uniform velocity of 25 m/s per second. The curve has a radiu

s of 80 m what is the centripedal force on the car
Physics
1 answer:
IrinaK [193]3 years ago
4 0
The centripetal force is:
 F = mv² / R
 Where:
 m: mass of the object
 v: object speed
 R: radius of the curve.
 We have to:
 m = 2000kg
 v = 25 m / s
 R = 80 meters.
 Then the centripetal force acting on the vehicle is:
 F = (2000kg * (25m / s) ²) / 80m
 F = 15625 N
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Answer:

4N

Explanation:

Force = mass x acceleration

Given

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4 years ago
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In a double slit experiment, if the separation between the two slits is 0.050 mm and the distance from the slits to a screen is
meriva

The spacing between the first-order and second-order bright fringes is 3 cm. Hence, this is the required solution.

<h3>What is double slit experiment?</h3>

The double-slit experiment serves as a proof in current physics that both light and matter may exhibit properties of classically defined waves and particles. It also illustrates the inherently probabilistic nature of quantum mechanical events. Thomas Young carried out the first experiment of this kind employing light in 1802, illustrating how light behaves like a wave. It was formerly believed that light was made up of either waves or particles. About a century later, with the advent of modern physics, it was discovered that light may in fact exhibit behaviour like that of both waves and particles. The identical behaviour of electrons was first shown by Davisson and Germer in 1927, and it was later extended to atoms and molecules.

The separation between the slits, d = 0.05mm = 5×10⁻⁵ m

The distance from the slits to a screen, D = 2.5 m

Let x is the spacing between the first-order and second-order bright fringes when coherent light of wavelength 600 nm illuminates the slits,

λ = 600nm = 6× 10⁻⁷ m  

We know that the bright fringe is given by :

y = nλD/d

So, the spacing between the first-order and second-order bright fringes is :

x = 2λD/d - λD/d

x =  λD/d

x = 6 × 10⁻⁷ × 2.5/5 × 10⁻⁵

x = 0.03 m

or

x = 3 cm

So, the spacing between the first-order and second-order bright fringes is 3 cm. Hence, this is the required solution.

to learn more about double slit experiment go to -

brainly.com/question/28108126

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3 years ago
A student decides to give his bicycle a tune up. He flips it upside down (so there's no friction with the ground) and applies a
aleksklad [387]

Answer:

V=9.2565m/s

Explanation:

From the question we are told that:

Force F = 34 N  

Time t = 0.6 s

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Radius of wheel r = 33 cm = 0.33 m

Moment of inertia, I = 1200 kgcm2 = 0.12 kg.m2

Generally the equation for Torque on pedal \mu is mathematically given by

\mu=F*L\\\mu=34*0.165

\mu=5.61N.m

Generally the equation for  angular acceleration \alpha is mathematically given by

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 \alpha=46.75

Therefore Angular speed is \omega

\omega=\alpha*t

\omega=(46.75)*(0.6)

\omega=28.05rad/s

Generally the equation for  Tangential velocity V is mathematically given by

V=r\omega

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5 0
3 years ago
The space shuttle is located exactly half way between the earth and the moon. Which statement is true regarding the gravitationa
sdas [7]

Answer:

The correct answer is option B)

Explanation:

Considering the given question as -

The space shuttle is located exactly half way between the earth and the moon. Which statement is true regarding the gravitational pull on the shuttle? A) The moon pulls more on the shuttle. B) The earth pulls more on the shuttle. C) Both are equal due to equal distances. D) Both are equal due to the mass of the shuttle.

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F = \dfrac{GM_{1}M_{2} }{r^{2} } where 'r' is the distance between the two bodies.

Let ,

M_{e} : Mass of the earth

M_{m} : Mass of the moon

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r_{e}    : Distance of satellite from earth

r_{m}   : Distance of satellite from moon

Given that r_{e}=r_{m}

Let r_{e}=r_{m}=r

Force on satellite by the earth is -

F_{e} = \dfrac{GM_{e}m }{r^{2} }

Force on satellite by the moon is -

F_{m} = \dfrac{GM_{m}m }{r^{2} }

∵ Mass of earth (M_{e}) > Mass of moon (M_{m})

∴ F_{e} > F_{m}

∴ The gravitational pull of earth on satellite is more than that of the moon.

4 0
4 years ago
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