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Fynjy0 [20]
2 years ago
6

At the local playground, a 21-kg child sits on the right end of a horizontal teeter-totter, 1.8 m from the pivot point. On the l

eft side of the pivot an adult pushes straight down on the teeter-totter with a force of 151 N. Part A In which direction does the teeter-totter rotate if the adult applies the force at a distance of 3.0 m from the pivot?Part B
In which direction does the teeter-totter rotate if the adult applies the force at a distance of 2.5 m from the pivot?
(clockwise/counterclockwise)
Part C
In which direction does the teeter-totter rotate if the adult applies the force at a distance of 2.0 m from the pivot?
(clockwise/counterclockwise)
Physics
1 answer:
Ksju [112]2 years ago
5 0

Answer:

By convention a negative torque leads to clockwise rotation and a positive torque leads to counterclockwise rotation.

here weight of the child =21kgx9.8m/s2 = 205.8N

the torque exerted by the child Tc = - (1.8)(205.8) = -370.44N-m ,negative sign is inserted because this torque is clockwise and is therefore negative by convention.

torque exerted by adult Ta = 3(151) = 453N , counterclockwise torque.

net torque Tnet = -370.44+453 =82.56N , which is positive means counterclockwise rotation.

b) Ta = 2.5x151 = 377.5N-m

Tnet = -370.44+377.5 = 7.06N-m , positive ,counterclockwise rotation.

c)Ta = 2x151 = 302N-m

Tnet = -370.44+302 = -68.44N-m, negative,clockwise rotation.

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How fast would a 2-kilogram object need to move to have the same kinetic
siniylev [52]

Answer:

11.3 m/s

Explanation:

KE₁ = KE₂

½m₁v₁² = ½m₂v₂²

½ (2 kg) v² = ½ (4 kg) (8 m/s)²

v ≈ 11.3 m/s

8 0
2 years ago
Calculate the mass 9f the earth, assuring that uts is sphere with radius 6.67×10^6m.​
MA_775_DIABLO [31]

Answer:

6.86 × 10²⁴ kg

Explanation:

The mass of the earth m = density of earth, ρ × volume of earth, V

m = ρV

The density of the earth, ρ = 5515 kg/m³ and since the earth is a sphere, its volume is the volume of a sphere V = 4πr³/3 where r = radius of the earth = 6.67 × 10⁶ m

Since m = ρV

m = ρ4πr³/3

So, substituting the values of the variables into the equation for the mass of the earth, m, we have

m = 5515 kg/m³ × 4π(6.67 × 10⁶ m)³/3

m = 5515 kg/m³ × 4π × 296.741 × 10¹⁸ m³/3

m = 5515 kg/m³ × 1189.9639π × 10¹⁸ m³/3

m = 6546105.64378π × 10¹⁸ kg/3

m = 20565197.400122 × 10¹⁸ kg/3

m = 6855065.8 × 10¹⁸ kg

m = 6.8550658 × 10²⁴ kg

m ≅ 6.86 × 10²⁴ kg

8 0
2 years ago
An object with a 25 µC charge is 0.54 m away from a second charged object. They experience a force of 3250 N. What is the charge
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25 uc charge is 0.54 m away
6 0
2 years ago
Problem #2: An apple is thrown upward with an initial velocity of +24.0 m/s. a. Sketch the apple's trip and label what you know.
bogdanovich [222]

Answer:

The answer is below

Explanation:

a) The initial velocity (u) = 24 m/s

We can solve this problem using the formula:

v² = u² - 2gh

where v = final velocity, g= acceleration due to gravity = 9.8 m/s², h = height.

At maximum height, the final velocity = 0 m/s

v² = u² - 2gh

0² = 24² - 2(9.8)h

2(9.8)h = 24²

2(9.8)h = 576

19.6h = 576

h = 29.4 m

b) The time taken to reach the maximum height is given as:

v = u - gt

0 = 24 - 9.8t

9.8t = 24

t = 2.45 s

The total time needed for the apple to return to its original position = 2t = 2 * 2.45 = 4.9 s

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