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goblinko [34]
4 years ago
15

The speed of a particle moving in a circle 2.0 m in radius increases at the constant rate of 4.4 m/s2. At an instant when the ma

gnitude of the total acceleration is 6.0 m/s2, what is the speed of the particle? Group of answer choices
Physics
1 answer:
Law Incorporation [45]4 years ago
6 0

Answer:

The speed of the particle is 2.86 m/s

Explanation:

Given;

radius of the circular path, r = 2.0 m

tangential acceleration,  a_t = 4.4 m/s²

total magnitude of the acceleration, a = 6.0 m/s²

Total acceleration is the vector sum of  tangential acceleration and radial acceleration

a = \sqrt{a_c^2 + a_t^2}\\\\

where;

a_c is the radial acceleration

a = \sqrt{a_c^2 + a_t^2}\\\\a^2 = a_c^2 + a_t^2\\\\a_c^2 = a^2 -a_t^2\\\\a_c = \sqrt{a^2 -a_t^2}\\\\a_c = \sqrt{6.0^2 -4.4^2}\\\\a_c = \sqrt{16.64}\\\\a_c = 4.08 \ m/s^2

The radial acceleration relates to speed of particle in the following equations;

a_c = \frac{v^2}{r}

where;

v is the speed of the particle

v^2 = a_c r\\\\v= \sqrt{a_c r} \\\\v = \sqrt{4.08 *2}\\\\v = 2.86 \ m/s

Therefore, the speed of the particle is 2.86 m/s

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8 0
3 years ago
A 900 kg car moves around a 500 m radius curve at 25.0 m/s. What is the centripetal force on
Gnesinka [82]

Answer:

9375 N

Explanation:

From the question,

Centripetal force (F) = mv²/r.................. Equation 1

Where m = mass of the car, v = velocity of the car, r = radius of the curve.

Given: m = 900 kg, r = 600 m, v = 25 m/s

Substitute these values into equation 1

F = (900×25²)/600

F = 9375 N.

Hence the centripetal force on the car is 9375 N

3 0
3 years ago
3) If an airplane is in flight cruising at constant velocity, what can be said about the net work
Crazy boy [7]

Explanation:

<h2>Yes!</h2>

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6 0
3 years ago
On a horizontal surface is located
Ierofanga [76]

By Newton's second law, the net vertical force acting on the object is 0, so that

<em>n</em> - <em>w</em> = 0

where <em>n</em> = magnitude of the normal force of the surface pushing up on the object, and <em>w</em> = weight of the object. Hence <em>n</em> = <em>w</em> = <em>mg</em> = 196 N, where <em>m</em> = 20 kg and <em>g</em> = 9.80 m/s².

The force of static friction exerts up to 80 N on the object, since that's the minimum required force needed to get it moving, which means the coefficient of <u>static</u> friction <em>µ</em> is such that

80 N = <em>µ</em> (196 N)   →   <em>µ</em> = (80 N)/(196 N) ≈ 0.408

Moving at constant speed, there is a kinetic friction force of 40 N opposing the object's motion, so that the coefficient of <u>kinetic</u> friction <em>ν</em> is

40 N = <em>ν</em> (196 N)   →   <em>ν</em> = (40 N)/(196 N) ≈ 0.204

And so the closest answer is C.

(Note: <em>µ</em> and <em>ν</em> are the Greek letters mu and nu)

3 0
3 years ago
The mass of a golf ball is 45.9
Tema [17]
The equation for the de Broglie wavelength is: 
<span>λ = (h/mv) √[1-(v²/c²)], </span>
<span>where h is Plank's Constant, m is the rest mass, v is velocity, and c is the velocity of light in vacuum. However, if c>>v (and it is, in this case) then the expression under the radical sign approaches 1, and the equation simplifies to: </span>
<span>λ = h/mv. </span>
<span>Substituting, (remember to convert the mass to kg, since 1 J = 1 kg·m²/s²): </span>
<span>λ = (6.63x10^-34 J·s) / (0.0459 kg) (72.0 m/s) = 2.00x10^-34 m.</span>
8 0
3 years ago
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