Answer:
a) 2.5 m/s²
b) 6.12 m/s
Explanation:
Tension of rope = T = 356N
Weight of material = W = 478 N
Distance from the ground = s = 7.5 m
Acceleration due to gravity = g = 9.81 m/s²
Mass of material = m = 478/9.81 = 48.72
Final velocity before the bundle hits the ground = v
Initial velocity = u = 0
Acceleration experienced by the material when being lowered = a
a) W-T = ma
⇒478-356 = 48.72×a
⇒a = 2.5 m/s²
∴ Acceleration achieved by the material is 2.5 m/s²
b) v²-u² = 2as
⇒v²-0 = 2×2.5×7.5
⇒v² = 37.5
⇒v = 6.12 m/s
∴ Velocity of the material before hitting the ground is 6.12 m/s
<u>Answer:</u> The ball is travelling with a speed of 5.5 m/s after hitting the <u>bottle.</u>
<u>Explanation:</u>
To calculate the speed of ball after the collision, we use the equation of law of conservation of momentum, which is given by:
where,
are the mass, initial velocity and final velocity of ball.
are the mass, initial velocity and final velocity of bottle.
We are given:
Putting values in above equation, we get:
Hence, the ball is travelling with a speed of 5.5 m/s after hitting the bottle.
The number of grams of carbon that combine with 16 g of oxygen in the formation of CO₂ is 6g.
When two elements combine to make more than one compound, the masses of one element combined with a fixed amount of another element are in the ratio of whole numbers, according to the law of multiple proportions.
When combined with oxygen, carbon can produce two different compounds. They are referred to as carbon dioxide (CO₂) and carbon monoxide (CO).
Carbon monoxide is formed by combining 12 g of carbon with 16 g of oxygen whereas Carbon dioxide is formed when 12 g of carbon reacts with 32 g of oxygen. The amount of carbon is fixed at 12 g in each case. The mass ratio of carbon monoxide to carbon dioxide is 16: 32, or 1: 2.
But in the given case, 16g of oxygen is reacting instead of 32g. Therefore, the number of grams of carbon reacting will be:
Thus, 6g of carbon will react with 16g of oxygen to form carbon dioxide.
Read more about Law of Multiple Proportions:
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Weight of the child m = 50 kg
Radius of the merry -go-around r = 1.50 m
Angular speed w = 3.00 rad/s
(a)Child's centripetal acceleration will be a = w^2 x r = 3^2 x 1.50 => a = 9 x
1.5
Centripetal Acceleration a = 13.5m/sec^2
(b)The minimum force between her feet and the floor in circular path
Circular Path length C = 2 x 3.14 x 1.50 => c = 3 x 3.14 => C = 9.424
Time taken t = 2 x 3.14 / w => t = 6.28 / 3 => t = 2.09
Calculating velocity v = distance / time = 9.424 / 2.09 m/s => v = 4.5 m/s
Calculating force, from equation F x r = mv^2 => F = mv^2 / r => 50 x (4.5)^2
/ 1.5
F = 50 x 3 x 4.5 => F = 150 x 4.5 => F = 675 N
(c)Minimum coefficient of static friction u
F = u x m x g => u = F / m x g => u = 675/ 50 x 9.81 => 1.376
u = 1.376
Hence with the force and the friction coefficient she is likely to stay on merry-go-around.
