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Julli
16 days ago
14

Can a small child play with fat child on the seesaw?Explain how?

Physics
2 answers:
inna [2.2K]16 days ago
8 0
Yes, a small child can indeed play with a heavier child on a seesaw. This is because reducing the distance from the fulcrum for the heavier child increases the effort required, allowing the smaller child to play without much strain.
Ostrovityanka [2.2K]16 days ago
6 0

The weights of the small child and the heavier child differ. The smaller child has less mass in comparison to the heavier one. This indicates that the force exerted by the smaller child is lesser than that of the heavier child.

Further Explanation

Consequently, when both children sit equidistant from the pivot, their torques will not balance. As a result, the seesaw will tilt towards the child with greater weight (the heavier child).

Thus, for the two of them to use the seesaw together, they must position themselves at different distances from the pivot. This means the heavier child should sit closer to the pivot.

This topic pertains to torque. Torque is defined as a vector quantity, since it comprises both magnitude and direction. It measures the effectiveness of a force in causing an object to rotate around an axis.

Torque can be divided into two categories, which are

  1. Static
  2. Dynamic

Static Torque refers to a force that does not result in angular acceleration, meaning static torque is not related to acceleration, while dynamic torque results in angular acceleration.

LEARN MORE:

  • Can a small child play with a fat child on a see saw? explain how?  
  • Can a small child play with a fat child on a see saw? explain how?  

KEYWORDS:

  • fat child
  • small child
  • seesaw
  • torque
  • pivot
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Keith_Richards [2256]

Answer:

Explanation:

The data indicates that point A is located midway between two charges.

To calculate the electric field at point A, we begin with the field produced by charge -Q ( 6e⁻ ) at A:

= 9 x 10⁹ x 6 x 1.6 x 10⁻¹⁹ / (2.5)² x 10⁻⁴

= 13.82 x 10⁻⁶ N/C

This field points towards Q⁻.

A similar field will arise from the charge Q⁺, but it will direct away from Q⁺ toward Q⁻.

To find the resultant field, we add these contributions:

= 2 x 13.82 x 10⁻⁶

= 27.64 x 10⁻⁶ N/C

For the force acting on an electron placed at A:

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8 0
17 days ago
A policeman in a stationary car measures the speed of approaching cars by means of an ultrasonic device that emits a sound with
Ostrovityanka [2204]

Answer:

The beats frequency measures approximately

4.4 kHz

Explanation:

The beat frequency arises from the original ultrasound frequency, f=41.2 kHz, and the frequency of the sound reflected off the car, f':

f_B = f'-f (1)

To calculate the frequency of the reflected sound, we apply the Doppler effect formula:

f'=\frac{v}{v-v_s}f

where

v = 340 m/s, the speed of sound

v_s =33.0 m/sis the velocity of the car

f=41.2 kHzis the frequency of the sound emitted

By substituting values,

f'=\frac{340 m/s}{340 m/s-33.0 m/s}(41.2 kHz)=45.6 kHz

Thus, the beat frequency (1) is

f_B = 45.6 kHz - 41.2 kHz=4.4 kHz

3 0
12 days ago
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Select the areas that would receive snowfall because of the lake effect.
Keith_Richards [2256]

Result:

- Grand Marais

- Two Harbors

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Explanation:

The locations likely to receive snowfall due to lake effect include Grand Marais, Two Harbors, and Duluth. These areas are situated directly along the shores of Lake Superior, one of the world’s largest lakes. Its substantial water volume significantly influences the local climate, generating a lot of humidity in the air and considerable evaporation, both of which lead to cloud formation and, when temperatures drop adequately, result in significant snowfall rather than rain.

8 0
6 days ago
1200 N-m of torque is used to drive a gear (A) of diameter 25 cm, which in turn drives another gear (B) of diameter 52 cm. What
Ostrovityanka [2204]

Response:

2.5kN.m

Details:

Torque relates directly to the pitch diameter

= Ta/Tb= Da/Db

For 120/Tb= 0.25/0.5

This gives Tb= 2.469kN.m, roughly 2.5kN.m


8 0
11 days ago
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A small 175-g ball on the end of a light string is revolving uniformly on a frictionless surface in a horizontal circle of diame
kicyunya [2264]
For motion in a circle.

Centripetal acceleration is calculated as mv²/r = mω²r

where v represents linear velocity, r equals radius which is diameter/2 equating to 1/2 or 0.5m

. Here, m is the mass of the object, which is 175g or 0.175kg.

The angular speed, ω, is derived from Angle covered / time

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                         = 4π  radians per second

Thus, Centripetal Acceleration = mω²r = 0.175*(4π)² * 0.5. Utilize a calculator

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. The acceleration's magnitude is approximately 13.817  m/s² and it is oriented towards the center of the circular path.

The tension in the string equates to m*a

                                   = 0.175*13.817

                                   = 2.418 N
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2 days ago
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