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Hunter-Best [27]
4 years ago
14

A 0.120-kg, 50.0-cm-long uniform bar has a small 0.055-kg mass glued to its left end and a small 0.110-kg mass glued to the othe

r end. The two small masses can each be treated as point masses. You want to balance this system horizontally on a fulcrum placed just under its center of gravity. How far from the left end should the fulcrum be placed?
Physics
1 answer:
alukav5142 [94]4 years ago
7 0

Answer:

29.8 cm

Explanation:

For the system to be at equilibrium, the counterclockwise moments must be equal to clockwise moments. Let the fulcrum be placed x cm from the left end. If the fulcrum is placed in the middle of the bar, the clockwise moments are the net moments so the fulcrum must be moved closer to the right hand side to decrease the effects of the moment due to the mass on the right.  Let the fulcrum be placed x cm from the left end.

counterclockwise moments = clockwise moments

(0.055 kg * x) + (0.120*(x-25)) = 0.110 kg * (50-x)

0.055x + 0.12x - 3 = 5.5 - 0.11x

                            x = 29.8 cm

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A 0.50-kg toy is attached to the end of a 1.0-m very light string. The toy is whirled in a horizontal circular path on a frictio
Deffense [45]

Answer:

26.5 m/s

Explanation:

The tension in the string provides the centripetal force that keeps the toy in circular motion. So we can write:

T=m\frac{v^2}{r}

where:

T is the tension in the spring

m = 0.50 kg is the mass of the toy

r = 1.0 m is the radius of the circle (the length of the string)

v is the speed of the toy

The maximum tension in the string is

T = 350 N

If we substitute this value into the equation, we find the maximum speed that the mass can have before the string breaks:

v=\sqrt{\frac{Tr}{m}}=\sqrt{\frac{(350)(1.0)}{0.50}}=26.5 m/s

3 0
3 years ago
at a temperature of 10 c,700 ml of hydrogen is collected. if this gas is put into a 1000 ml container what will its new temperat
erastovalidia [21]

Answer:

14.3°C

Explanation:

Find the ratio of 10°C : 700ml then use the same ratio to 1000ml.

Have a great day <3

4 0
3 years ago
Read 2 more answers
Chris threw a basketball a distance of 27.5 m to score and win his
salantis [7]

Answer:

v₀ = 16.55 m/s

Explanation:

This motion of the ball can be modeled as a projectile motion with following data:

R = Range of Projectile = 27.5 m

θ = Launch Angle = 50°

g = acceleration due to gravity = 9.81 m/s²

v₀ = Initial Speed of Ball = ?

Therefore, using formula for range of projectile, we have:

R = \frac{v_{0}^2\ Sin2\theta}{g}\\\\v_{0}^2 = \frac{Rg}{Sin2\theta}\\\\v_{0}^2 = \frac{(27.5\ m)(9.81\ m/s^2)}{Sin100^o}\\\\v_{0} = \sqrt{273.93\ m^2/s^2}

<u>v₀ = 16.55 m/s</u>

8 0
3 years ago
List two examples of situations were neutral object is charged by conduction,induction and friction
katovenus [111]
1. Boiling water
2. Hair straightener or curler
5 0
3 years ago
The mean distance of an asteroid from the Sun is 2.98 times that of Earth from the Sun. From Kepler's law of periods, calculate
lutik1710 [3]

Answer:

The asteroid requires 5.14 years to make one revolution around the Sun.

Explanation:

Kepler's third law establishes that the square of the period of a planet will be proportional to the cube of the semi-major axis of its orbit:

T^{2} = a^{3} (1)

Where T is the period of revolution and a is the semi-major axis.

In the other hand, the distance between the Earth and the Sun has a value of 1.50x10^{8} Km. That value can be known as well as an astronomical unit (1AU).

But 1 year is equivalent to 1 AU according with Kepler's third law, since 1 year is the orbital period of the Earth.

For the special case of the asteroid the distance will be:

a = 2.98(1.50x10^{8}Km)

a = 4.47x10^{8}Km

That distance will be expressed in terms of astronomical units:

4.47x10^{8}Km.\frac{1AU}{1.50x10^{8}Km} ⇒ 2.98AU

Finally, from equation 1 the period T can be isolated:

T = \sqrt{a^{3}}

T = \sqrt{(2.98)^{3}}  

T = \sqrt{26.463592}

T = 5.14AU

Then, the period can be expressed in years:

5.14AU.\frac{1yr}{1AU} ⇒ 5.14 yr

T = 5.14 yr

Hence, the asteroid requires 5.14 years to make one revolution around the Sun.

8 0
3 years ago
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