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Ad libitum [116K]
2 years ago
12

A uniform rod of length 50cm and mass 0.2kg is placed on a fulcrum at a distance of 40cm from the left end of the rod. At what d

istance from the left end of the rod should a 0.6kg mass be hung to balance the rod?
a.48 cm
b. 50 cm
c. 45 cm
d. the rod can not be balanced with this mass.
e.42 cm
show your work. NO LINKS. ​
Physics
1 answer:
oksano4ka [1.4K]2 years ago
5 0

Answer:

x = 45 cm

Explanation:

Given that,

The length of a rod, L = 50 cm

Mass, m₁ = 0.2 kg

It is at 40cm from the left end of the rod.

We need to find the distance from the left end of the rod should a 0.6kg mass be hung to balance the rod.

The centre of mass of the rod is at 25 cm.

Taking moments of both masses such that,

15\times 0.2=x\times 0.6\\\\x=\drac{3}{0.6}\\\\x=5\ cm

The distance from the left end is 40+5 = 45 cm.

Hence, at a distance of 45 cm from the left end it will balance the rod.

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Explanation:

We are told that its volume increases by 9.05%.

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To find the force per unit area which is also pressure, we will use bulk modulus formula;

B = Δp(V_o/ΔV)

Making Δp the subject gives;

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Now, B is bulk modulus of water with a value of 2.2 × 10^(9) N/m²

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A soft tennis ball is dropped onto a hard floor from a height of 1.50 m and rebounds to a height of 1.10 m. (a) Calculate its ve
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(a)   v = 5.42m/s

(b)   vo = 4.64m/s

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Explanation:

(a) The velocity of the ball before it hits the floor is given by:

v=\sqrt{2gh}        (1)

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v=\sqrt{2(9.8m/s^2)(1.50m)}=5.42\frac{m}{s}

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(b) To calculate the velocity of the ball, after it leaves the floor, you use the information of the maximum height reached by the ball after it leaves the floor.

You use the following formula:

h_{max}=\frac{v_o^2}{2g}       (2)

vo: velocity of the ball where it starts its motion upward

You solve for vo and replace the values of the parameters:

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(c) The acceleration is given by:

a=\frac{\Delta v}{\Delta t}=\frac{v_o-v}{3.50*10^{-3}s}=\frac{4.64m/s-(-5.42m)/s}{3.50*10^{-3}s}=2874.285\frac{m}{s^2}

a=\frac{\Delta v}{\Delta t}=\frac{v_o-v}{3.50*10^{-3}s}=\frac{4.64m/s-5.42m/s}{3.50*10^{-3}s}=-222.85\frac{m}{s^2}

The acceleration of the ball is 2874.28/s^2

(d) The compression of the ball is:

\Deta y=\frac{v^2}{2(a)}=\frac{(5.42m/s)^2}{2(2874.28m/s^2)}=5.11*10^{-3}m

THe compression of the ball when it strikes the floor is 5.11*10^-3m

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