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2 years ago

8

A horizontal uniform meter stick supported at the 50-cm mark has a mass of 0.50 kg hanging from it at the 20-cm mark and a 0.30

kg mass hanging from it at the 60-cm mark. Determine the position on the meter stick at which one would hang a third mass of 0.60 kg to keep the meter stick balanced.

2 answers:

ElenaW [278]2 years ago

7 0

Answer:

70 cm

Explanation:

0.5 kg at 20 cm

0.3 kg at 60 cm

x = Distance of the third 0.6 kg mass

Meter stick hanging at 50 cm

Torque about the support point is given by (torque is conserved)

$0.5(50-20)=0.3(60-50)+0.6x\\\Rightarrow x=\dfrac{0.5(50-20)-0.3(60-50)}{0.6}\\\Rightarrow x=20\ cm$

The position of the third mass of 0.6 kg is at 20+50 = 70 cm

arsen [322]2 years ago

3 0

Answer:

70 cm mark

Explanation:

m1 = 0.5 kg

m2 = 0.3 kg

m3 = 0.6 kg

let the third mass is at d cm from 50 cm mark. take moments about the 50 cm mark.

Anticlockwise torque = clock wise torque

0.5 x ( 50 - 20) + 0.6 x d = 0.3 (60 - 50)

0.5 x 30 + 0.6 d = 0.3 x 10

15 + 0.6 d = 3

0.6 d = - 12

d = - 20 cm

So, it means third mass is at 20 cm right to the 50 cm mark. So, it is at 70 cm mark.

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Write the equivalent formulas for velocity, acceleration, and force using the relationships covered for UCM, Newton’s Laws, and

yKpoI14uk [10]

Answer:

The newton’s second law is F=ma

The Gravitational force is $F=\dfrac{Gm_{1}m_{2}}{r^2}$

Explanation:

Given that,

The equivalent formulas for velocity, acceleration, and force using the relationships covered for UCM, Newton’s Laws, and Gravitation.

We know that,

Velocity :

The velocity is equal to the rate of position of the object.

$v=\dfrac{dx}{dt}$ ....(I)

Acceleration :

The acceleration is equal to the rate of velocity of the object.

$a=\dfrac{dv}{dt}$ ....(II)

Newton’s second Laws

The force is equal to the change in momentum.

In mathematically,

$F=\dfrac{d(p)}{dt}$

Put the value of p

$F=\dfrac{d(mv)}{dt}$

$F=m\dfrac{dv}{dt}$

Put the value from equation (II)

$F=ma$

This is newton’s second laws.

Gravitational force :

The force is equal to the product of mass of objects and divided by square of distance.

In mathematically,

$F=\dfrac{Gm_{1}m_{2}}{r^2}$

Where, m₁₂ = mass of first object

m= mass of second object

r = distance between both objects

Hence, The newton’s second law is F=ma

The Gravitational force is $F=\dfrac{Gm_{1}m_{2}}{r^2}$

3 0

2 years ago

Official (Closed) - Non Sensitive

Pavlova-9 [17]

Answer:

The minimum running time is 319.47 s.

Explanation:

First we find the distance covered and time taken by the train to reach its maximum speed:

We have:

Initial Speed = Vi = 0 m/s (Since, train is initially at rest)

Final Speed = Vf = 29.17 m/s

Acceleration = a = 0.25 m/s²

Distance Covered to reach maximum speed = s₁

Time taken to reach maximum speed = t₁

Using 1st equation of motion:

Vf = Vi + at₁

t₁ = (Vf - Vi)/a

t₁ = (29.17 m/s - 0 m/s)/(0.25 m/s²)

t₁ = 116.68 s

Using 2nd equation of motion:

s₁ = (Vi)(t₁) + (0.5)(a)(t₁)²

s₁ = (0 m/s)(116.68 s) + (0.5)(0.25 m/s²)(116.68 s)²

s₁ = 1701.78 m = 1.7 km

Now, we shall calculate the end time and distance covered by train, when it comes to rest on next station.

We have:

Final Speed = Vf = 0 m/s (Since, train is finally stops)

Initial Speed = Vi = 29.17 m/s (The train must maintain max. speed for min time)

Deceleration = a = - 0.7 m/s²

Distance Covered to stop = s₂

Time taken to stop = t₂

Using 1st equation of motion:

Vf = Vi + at₂

t₂ = (Vf - Vi)/a

t₂ = (0 m/s - 29.17 m/s)/(- 0.7 m/s²)

t₂ = 41.67 s

Using 2nd equation of motion:

s₂ = (Vi)(t₂) + (0.5)(a)(t₂)²

s₂ = (29.17 m/s)(41.67 s) + (0.5)(- 0.7 m/s²)(41.67 s)²

s₂ = 607.78 m = 0.6 km

Since, we know that the rest of 7 km, the train must maintain the maximum speed to get to the next station in minimum time.

The remaining distance is:

s₃ = 7 km - s₂ - s₁

s₃ = 7 km - 0.6 km - 1.7 km

s₃ = 4.7 km

Now, for uniform speed we use the relation:

s₃ = vt₃

t₃ = s₃/v

t₃ = (4700 m)/(29.17 m/s)

t₃ = 161.12 s

So, the minimum running time will be:

t = t₁ + t₂ + t₃

t = 116.68 s + 41.67 s + 161.12 s

<u>t = 319.47 s</u>

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2 years ago

A spaceship flies from Earth to a distant star at a constant speed. Upon arrival, a clock on board the spaceship shows a total e

m_a_m_a [10]

Answer:

35 288 mile/sec

Explanation:

This is a problem of special relativity. The clocks start when the spaceship passes Earth with a velocity v, relative to the earth. So, out and back from the earth it will take:

$10 years = \frac{2d}{v}$

If we use the Lorentz factor, then, as observed by the crew of the ship, the arrival time will be:

$0.8 = \sqrt{1-\frac{v^{2} }{c^{2} } }$

Then the amount of time wil expressed as a reciprocal of the Lorentz factor. Thus:

$0.8 = \sqrt{1 - \frac{v^{2} }{c^{2} } }$

$0.64 = 1-\frac{v^{2} }{186282^{2} }$

solving for v, gives = 35 288 miles/s

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2 years ago

In a 5000 m race, the athletes run 12 1/2 laps; each lap is 400 m.Kara runs the race at a constant pace and finishes in 17.9 min

Ksju [112]

Answer:

No. of laps of Hannah are 7 (approx).

Solution:

According to the question:

The total distance to be covered, D = 5000 m

The distance for each lap, x = 400 m

Time taken by Kara, $t_{K} = 17.9 min = 17.9\times 60 = 1074 s$

Time taken by Hannah, $t_{H} = 15.3 min = 15.3\times 60 = 918 s$

Now, the speed of Kara and Hannah can be calculated respectively as:

$v_{K} = \frac{D}{t_{K}} = \frac{5000}{1074} = 4.65 m/s$

$v_{H} = \frac{D}{t_{H}} = \frac{5000}{918} = 5.45 m/s$

Time taken in each lap is given by:

$(v_{H} - v_{K})t = x$

$(5.45 - 4.65)\times t = 400$

$t = \frac{400}{0.8}$

t = 500 s

So, Distance covered by Hannah in 't' sec is given by:

$d_{H} = v_{H}\times t$

$d_{H} = 5.45\times 500 = 2725 m$

No. of laps taken by Hannah when she passes Kara:

$n_{H} = \frac{d_{H}}{x}$

$n_{H} = \frac{2725}{400} = 6.8$ ≈ 7 laps

3 0

1 year ago

You throw a ball of mass 1 kg straight up. You observe that it takes 2.2 s to go up and down, returning to your hand. Assuming w

Elina [12.6K]

Answer:

10.791 m/s

5.93505 m

Explanation:

m = Mass of ball

$v_f$ = Final velocity

$v_i$ = Initial velocity

$t_f$ = Final time

$t_i$ = Initial time

g = Acceleration due to gravity = 9.81 m/s²

From the momentum principle we have

$\Delta P=F\Delta t$

Force

$F=mg$

So,

$m(v_f-v_i)=mg(t_f-t_i)\\\Rightarrow v_i=v_f-g(t_f-t_i)\\\Rightarrow v_i=0-(-9.81)(1.1-0)\\\Rightarrow v_i=10.791\ m/s$

The speed that the ball had just after it left the hand is 10.791 m/s

As the energy of the system is conserved

$K_i=U\\\Rightarrow \dfrac{1}{2}mv_i^2=mgh\\\Rightarrow h=\dfrac{v_i^2}{2g}\\\Rightarrow h=\dfrac{10.791^2}{2\times 9.81}\\\Rightarrow h=5.93505\ m$

The maximum height above your hand reached by the ball is 5.93505 m

5 0

2 years ago