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a_sh-v [17]
3 years ago
15

an object of mass 8 kg is whriled round in a vertical circle of radius 2m with a constant speed of 6m/s .Then the maximum and mi

nimum tensions in the string are.(g=10m/s^2)​
Physics
1 answer:
algol133 years ago
5 0

Answer:

Maximum Tension=224N

Minimum tension= 64N

Explanation:

Given

mass =8 kg

constant speed = 6m/s .

g=10m/s^2

Maximum Tension= [(mv^2/ r) + (mg)]

Minimum tension= [(mv^2/ r) - (mg)]

Then substitute the values,

Maximum Tension= [8 × 6^2)/2 +(8×9.8)] = 224N

Minimum tension= [8 × 6^2)/2 -(8×9.8)]

=64N

Hence, Minimum tension and maximum Tension are =64N and 2224N respectively

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On a trolley ride around an amusement park, a child travelled from one signpost to a second signpost at a constant speed of 125
Tomtit [17]

Answer:

Explanation:

Speed given = 125 m /min

125 /60 m /s

In 450 second it will travel

= 450 x 125 / 60

=937.5 m.  

As the distance  is covered in less than 450 seconds , The distance must be less than 937.5 m

In 400 seconds , it will travel

= 400 x 125 / 60

833.33 m

Since the distance is covered in more than 400 seconds , the distance must be more than ie 833.33 .

Hence the distance covered is more than .833 m but less than 937.5

In either case these distance are more than .8 km .

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3 years ago
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
Based on the free-body diagram, the net force acting<br> on this firework is
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0N. The net force acting on this firework is 0.

The key to solve this problem is using the net force formula based on the diagram shown in the image. Fnet = F1 + F2.....Fn.

Based on the free-body diagram, we have:

The force of gases is Fgases = 9,452N

The force of the rocket Frocket = -9452

Then, the net force acting is:

Fnet = Fgases + Frocket

Fnet = 9,452N - 9,452N = 0N

4 0
3 years ago
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ΑΣ Sign A. keeping everything grounded on Earth. C. planes not being able to fly into space. B. making tall skyscrapers safe, D.
stepladder [879]

Answer:

i think that it is c because it not only tells us about what the moon does to the earth it also tells us what the earth does to the moon.

Explanation:

3 0
3 years ago
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Sayid made a chart listing data of two colliding objects. A 5-column table titled Collision: Two Objects Stick Together with 2 r
Alborosie

Answer:

6 m/s is the missing final velocity

Explanation:

From the data table we extract that there were two objects (X and Y) that underwent an inelastic collision, moving together after the collision as a new object with mass equal the addition of the two original masses, and a new velocity which is the unknown in the problem).

Object X had a mass of 300 kg, while object Y had a mass of 100 kg.

Object's X initial velocity was positive (let's imagine it on a horizontal axis pointing to the right) of 10 m/s. Object Y had a negative velocity (imagine it as pointing to the left on the horizontal axis) of -6 m/s.

We can solve for the unknown, using conservation of momentum in the collision: Initial total momentum = Final total momentum (where momentum is defined as the product of the mass of the object times its velocity.

In numbers, and calling P_{xi} the initial momentum of object X and P_{yi} the initial momentum of object Y, we can derive the total initial momentum of the system: P_{total}_i=P_{xi}+P_{yi}= 300*10 \frac{kg*m}{s} -100*6\frac{kg*m}{s} =\\=(3000-600 )\frac{kg*m}{s} =2400 \frac{kg*m}{s}

Since in the collision there is conservation of the total momentum, this initial quantity should equal the quantity for the final mometum of the stack together system (that has a total mass of 400 kg):

Final momentum of the system: M * v_f=400kg * v_f

We then set the equality of the momenta (total initial equals final) and proceed to solve the equation for the unknown(final velocity of the system):

2400 \frac{kg*m}{s} =400kg*v_f\\\frac{2400}{400} \frac{m}{s} =v_f\\v_f=6 \frac{m}{s}

7 0
4 years ago
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