All categories

2 years ago

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

2 answers:

8 0

Yes a small child can play with fat child in the seesaw because if the the load distance is decreased the effort will increase. That's means if the distance between the fat boy and the fulcrum is decreased the small child needs less effort.so,he can play

Darya [45]2 years ago

6 0

The mass of the small child and that of the fat child is not the same. The small child would have a lesser mass compared to the fat child. This also implies that the weight force by the small child is smaller to the force of the fat child.

<h2>Further Explanation </h2>

Therefore, if both the small child and the fat child sit at equal distance from the pivot, the torques by them cannot be equal. The result is that the seesaw will rotate at the end of the child with a bigger weight (fat child).

Therefore, for both of them to play on the seesaw, they have to sit at an unequal distance from the pivot. This also implies that the child with a bigger weight (fat child) needs to sit near the pivot.

This question deals with torque. Torque is a vector quantity because it has both magnitude and direction. It can be described as the measure of a force that can make an object to rotate within an axis.

Torque can be classified into two and these include

Static
Dynamic

A static Torque is the type of force that does not produce an angular acceleration, that is, a static torque has nothing to with acceleration while dynamic torque produces angular acceleration.

LEARN MORE:

Can a small child play with a fat child on a see saw? explain how? brainly.com/question/4267445
Can a small child play with a fat child on a see saw? explain how? brainly.com/question/790091

KEYWORDS:

fat child
small child
seesaw
torque
pivot

You might be interested in

16) A wheel of moment of inertia of 5.00 kg-m2 starts from rest and accelerates under a constant torque of 3.00 N-m for 8.00 s.

KiRa [710]

Answer:

57.6Joules

Explanation:

Rotational kinetic energy of a body can be determined using the expression

Rotational kinetic energy = 1/2Iω²where;

I is the moment of inertia around axis of rotation. = 5kgm/s²

ω is the angular velocity = ?

Note that torque (T) = I¶ where;

¶ is the angular acceleration.

I is the moment of inertia

¶ = T/I

¶ = 3.0/5.0

¶ = 0.6rad/s²

Angular acceleration (¶) = ∆ω/∆t

∆ω = ¶∆t

ω = 0.6×8

ω = 4.8rad/s

Therefore, rotational kinetic energy = 1/2×5×4.8²

= 5×4.8×2.4

= 57.6Joules

6 0

2 years ago

Read 2 more answers

Onur drops a basketball from a height of 10\,\text{m}10m10, start text, m, end text on Mars, where the acceleration due to gravi

Doss [256]

Answer:

Explanation:

Given that,

Basket ball is drop from height

H=10m

It is dropped on planet mass

And the acceleration due to gravity on Mars is given as

g= 3.7m/s²

Time taken for the ball to reach the ground

Initial velocity of the body is zero

u=0m/s

Using equation of motion: free fall

H = ut + ½gt²

10 = 0•t + ½ × 3.7 ×t²

10 = 0 + 1.85t²

10 = 1.85t²

Then, t² =10/1.85

t² = 5.405

t = √ 5.405

t = 2.325seconds

So the time the ball spend on the air before reaching the ground is 2.325 seconds

5 0

2 years ago

Moving water, like that of a river, carries sediment as it moves along its bed. The faster the water flows, the more sediment th

katovenus [111]

Correct option: A

An object remains at rest until a force acts on it.

As the water moves faster, it applies greater force on the sediment, which over comes the frictional forces between the bed and the sediment. So, when the river flows faster, more and larger sediment particles are carried away. When the flow slows down, the river couldn't apply enough force on the larger sediments which can overcome the frictional force between the sediment and the river bed. So, the net force on the heavier particles become zero. Hence, the heavier particles of the load will settle out.

3 0

1 year ago

Read 2 more answers

A satellite completes one revolution of a planet in almost exactly one hour. At the end of one hour, the satellite has traveled

sesenic [268]

Velocity =

(distance between start point and end point, regardless of the route traveled) / (time spent traveling).

That distance (called the "displacement"), is 10 meters, and almost exactly 1 hour is almost exactly 3,600 seconds. So the numerical value of the velocity during that time is

(10) / (3,600) = almost exactly 0.00278 m/s

= 2.78 x 10^-3 m/s.

3 0

2 years ago

A positive point charge Q1 = 2.5 x 10-5 C is fixed at the origin of coordinates, and a negative point charge Q2 = -5.0 x 10-6 C

mario62 [17]

Answer:

3.62 m and - 1.4 m

Explanation:

Consider a location towards the positive side of x-axis beyond the location of charge Q₂

x = distance of the location from charge Q₂

d = distance between the two charges = 2 m

For the electric field to be zero at the location

E₁ = Electric field by charge Q₁ at the location = E₂ = Electric field by charge Q₂ at the location

$\frac{kQ_{1}}{(2 + x)^{2}}= \frac{kQ_{2}}{x^{2}}$

$\frac{2.5\times 10^{-5}}{(2 + x)^{2}}= \frac{5 \times 10^{-6}}{x^{2}}$

x = 1.62 m

So location is 2 + 1.62 = 3.62 m

Consider a location towards the negative side of x-axis beyond the location of charge Q₁

x = distance of the location from charge Q₁

d = distance between the two charges = 2 m

For the electric field to be zero at the location

E₁ = Electric field by charge Q₁ at the location = E₂ = Electric field by charge Q₂ at the location

$\frac{kQ_{1}}{(x)^{2}}= \frac{kQ_{2}}{ (2 + x)^{2}}$

$\frac{2.5\times 10^{-5}}{(x)^{2}}= \frac{5 \times 10^{-6}}{(2+x)^{2}}$

x = - 1.4 m

6 0

2 years ago