How does velocity describe an objects motion?

A 30.0-kg child sits on one end of a long uniform beam having a mass of 20.0 kg, and a 40.0-kg child sits on the other end. The

qaws [65]

let the length of the beam be "L"

from the diagram

AD = length of beam = L

AC = CD = AD/2 = L/2

BC = AC - AB = (L/2) - 1.10

BD = AD - AB = L - 1.10

m = mass of beam = 20 kg

m₁ = mass of child on left end = 30 kg

m₂ = mass of child on right end = 40 kg

using equilibrium of torque about B

(m₁ g) (AB) = (mg) (BC) + (m₂ g) (BD)

30 (1.10) = (20) ((L/2) - 1.10) + (40) (L - 1.10)

L = 1.98 m

4 0

3 years ago

A yo-yo of mass M has an axle of radius b and a spool of radius R. Its moment of inertia can be taken to be MR2/2 and the thickn

kow [346]

Answer:

The tension in the cord is $T=\frac{MR^{2}g }{2b^{2}+R^{2} }$

Explanation:

Given:

M = mass

b = radius

R = spool of radius

The equation is:

$bT=(\frac{MR^{2} }{2} )(\frac{a}{b} )\\T=\frac{MR^{2}a }{2b^{2} }$ (eq. 1)

The sum of forces in y:

∑Fy = Mg - T = Ma

$Mg=(M+\frac{MR^{2} }{2b^{2} } )a\\a=\frac{2b^{2}g }{2b^{2}+R^{2} }$

Replacing in eq. 1

$T=\frac{MR^{2} }{2b^{2} } (\frac{2b^{2}g }{2b^{2} +R^{2} } )\\T=\frac{MR^{2}g }{2b^{2}+R^{2} }$

3 0

3 years ago

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A bowling ball with a mass of 9kg is thrown down a lane with a constant speed of 3 m/s. The ball hits the 1.5kg pin, initially a

olasank [31]

Answer:

M1 V1 = M1 V2 + M2 V3 conservation of momentum

V2 = (M1 V1 - M2 V3) / M1 where V2 = speed of M1 after impact

V2 = (3 * 9 - 1.5 * 5) / 9 = (27 - 7.5) / 9 = 2.17 m/s

Note: All speeds are in the same direction and have the same sign

7 0

2 years ago

A hawk is flying horizontally at 18.0 m/s in a straight line, 230 m above the ground. A mouse it has been carrying struggles fre

Lisa [10]

Answer:

a) vd = 47.88 m/s

b) θ = 80.9°

c) t = 6.8 s

Explanation:

In the situation of the problem, you can assume that the trajectory of the hawk and the trajectory of the mouse form a rectangle triangle.

One side of the triangle is the horizontal trajectory of the hawk after 2.00s of flight, the other side of the triangle is the distance traveled by the mouse when it is falling down. And the hypotenuse is the trajectory of the hawk when it is trying to recover the mouse.

(a) In order to calculate the diving speed of the hawk, you first calculate the hypotenuse of the triangle.

One side of the triangle is c1 = (18.0m/s)(2.0s) = 36m

The other side of the triangle is c2 = 230m - 3m = 227 m

Then, the hypotenuse is:

$h=\sqrt{(36m)^2+(227m)^2}=229.83m$ (1)

Next, it is necessary to calculate the falling down time of the mouse, this can be done by using the following formula:

$y=y_o+v_ot+\frac{1}{2}gt^2$ (2)

yo: initial height = 230m

vo: initial vertical speed of the mouse = 0m/s

g: gravitational acceleration = -9.8m/s^2

y: final height of the mouse = 3 m

You replace the values of the parameters in (2) and solve for t:

$3=230-4.9t^2\\\\t=\sqrt{\frac{227}{4.9}}=6.8s$

The hawk traveled during 2.00 second in the horizontal trajectory, hence, the hawk needed 6.8s - 2.0s = 4.8 s to travel the distance equivalent to the hypotenuse to catch the mouse.

You use the value of h and 4.8s to find the diving speed of the hawk:

$v_d=\frac{229.83m}{4.8s}=47.88\frac{m}{s}$

The diving speed of the Hawk is 47.88m/s

(b) The angle is given by:

$\theta=cos^{-1}(\frac{c_1}{h})=cos^{-1}(\frac{36m}{229.83m})=80.9 \°$

Then angle between the horizontal and the trajectory of the Hawk when it is descending is 80.9°

(c) The mouse is falling down during 6.8 s

4 0

3 years ago

Explain how do you determine the intensity of earthquake

Ivanshal [37]

Answer:

The Richter scale measures the largest wiggle (amplitude) on the recording, but other magnitude scales measure different parts of the earthquake. The USGS currently reports earthquake magnitudes using the Moment Magnitude scale, though many other magnitudes are calculated for research and comparison purposes.

6 0

3 years ago

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