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Oksi-84 [34.3K]
2 years ago
6

A small metallic bob is suspended from the ceiling by a thread of negligible mass. The ball is then set in motion in a horizonta

l circle so that the thread describes a cone. Calculate the magnitude of the angular momentum of the bob about a vertical axis through the supporting point. The acceleration of gravity is 9.8 m/s2. Answer in units of kg · m2/s
Physics
1 answer:
shusha [124]2 years ago
8 0

Answer:

19.99 kg m²/s

Explanation:

Angular Momentum (L) is defined as the product of the moment of Inertia (I) and angular velocity (w)

L = m r × v.

r and  v are perpendicular to each other,

where r = lsinθ.

l = 2.4 m

θ= 34°

g = 9.8 m/s²  and m = 5 kg

resolving using newtons second law in the vertical and horizontal components.

T cos θ − m g = 0

T sin θ − mw² lsin θ = 0

where T is the force with which the wire acts on the bob

w = √g / lcosθ

= √ 9.8 / 2.4 ×cos 34

= 2.2193 rad/s

the angular momentum  L = mr× v

= mw (lsin θ)²

= 5 × 2.2193 (2.4 ×sin 34°)²

=19.99 kg m²/s

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1 year ago
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a 1250 kg car accelerates from rest to 6.13m/s over a distance of 8.58m calculate the average force of traction
iogann1982 [59]
Use formula, v^2= u^2 + 2as.
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2 years ago
If a single constant force acts on an object that moves on a straight line, the object's velocity is a linear function of time.
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F=mkv

Explanation:

Given that

v = v_i - kx

We know that acceleration a given as

a=\dfrac{dv}{dt}

v = v_i - kx

\dfrac{dv}{dt}=\dfrac{dv_i}{dt}-k\dfrac{dx}{dt}

\dfrac{dv}{dt}=0-k\dfrac{dx}{dt}

We know that

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1 year ago
You want to move a heavy box with mass 30.0 kg across a carpeted floor. You pull hard on one of the edges of the box at an angle
charle [14.2K]

Answer:

a=5.54m/s^{2}

Explanation:

The net force, F_{net} of the box is expressed as a product of acceleration and mass hence

F_{net}=ma where m is mass and a is acceleration

Making a the subject, a= \frac {F_{net}}{m}

From the attached sketch,  

∑ F_{net}=Fcos\theta-F_{f} where F_{f} is frictional force and \theta is horizontal angle

Substituting ∑ F_{net} as F_{net} in the equation where we made a the subject

a= \frac {Fcos\theta-F_{f}}{m}

Since we’re given the value of F as 240N, F_{f} as 41.5N, \theta as 30^{o} and mass m as 30kg

a= \frac {240cos30-41.5}{30.0}=\frac {166.346}{30.0}=5.54m/s^{2}

6 0
2 years ago
A box with a mass of 100.0 kg slides down a ramp with a 50 degree angle. What is the weight of the box? N What is the value of t
nadya68 [22]

1) weight of the box: 980 N

The weight of the box is given by:

W=mg

where m=100.0 kg is the mass of the box, and g=9.8 m/s^2 is the acceleration due to gravity. Substituting in the formula, we find

W=(100.0 kg)(9.8 m/s^2)=980 N


2) Normal force: 630 N

The magnitude of the normal force is equal to the component of the weight which is perpendicular to the ramp, which is given by

N=W cos \theta

where W is the weight of the box, calculated in the previous step, and \theta=50^{\circ} is the angle of the ramp. Substituting, we find

N=(980 N)(cos 50^{\circ})=630 N


3) Acceleration: 7.5 m/s^2

The acceleration of the box along the ramp is equal to the component of the acceleration of gravity parallel to the ramp, which is given by

a_p = g sin \theta

Substituting, we find

W_p = (9.8 m/s^2)(sin 50^{\circ})=7.5 m/s^2

5 0
1 year ago
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