All categories

2 months ago

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

2 answers:

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 [3.2K]2 months 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

Static
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

You might be interested in

You and your friend Peter are putting new shingles on a roof pitched at 20degrees . You're sitting on the very top of the roof w

Keith_Richards [3271]

Answer:

v₀ = 3.8 m/s

Explanation:

According to Newton's second law relating to the box:

∑F = m*a Formula (1)

∑F: the net force in Newton (N)

m: mass expressed in kilograms (kg)

a: acceleration measured in meters per second squared (m/s²)

Information known:

m = 2.1 kg, the mass of the box

d = 5.4m, the length of the roof

θ = 20° is the angle between the roof and the horizontal

μk = 0.51, the coefficient of kinetic friction between the box and the roof

g = 9.8 m/s², gravitational acceleration

Forces influencing the box:

The x-axis is oriented parallel to the box's movement on the roof, and the y-axis is oriented perpendicularly.

W: Weight of the box: directed vertically

N: Normal force: perpendicular to the roof's angle

fk: Frictional force: parallel to the direction along the roof

Calculating the weight of the box:

W = m*g = (2.1 kg)*(9.8 m/s²)= 20.58 N

x-y components of weight:

Wx= Wsin θ=(20.58)*sin(20)°=7.039 N

Wy= Wcos θ=(20.58)*cos(20)°= 19.34 N

Finding the Normal force:

∑Fy = m*ay ay = 0

N-Wy = 0

N=Wy = 19.34 N

Calculating the Friction force:

fk=μk*N= 0.51* 19.34 N = 9.86 N

We substitute into Formula (1) to determine the box's acceleration:

∑Fx = m*ax ax=a: acceleration of the box

Wx-fk = (2.1)*a

7.039 - 9.86 = (2.1)*a

-2.821 = (2.1)*a

a=(-2.821)/(2.1)

a = -1.34 m/s²

Considering the box's Kinematics:

Since the box undergoes uniformly accelerated motion, we use the following to find the final speed of the box:

vf² = v₀² + 2*a*d Formula (2)

Where:

d refers to displacement = 5.4 m

v₀ is the initial speed

vf represents the final speed = 0

a is the box's acceleration = -1.34 m/s²

Plugging in the values into Formula (2):

0² = v₀² + 2*(-1.34)*(5.4)

2*(1.34)*(5.4) = v₀²

$v_{o} =\sqrt{14.472}$

v₀ = 3.8 m/s

7 0

2 months ago

During a snowball fight two balls with masses of 0.4 and 0.6 kg, respectively, are thrown in such a manner that they meet head-o

Yuliya22 [3333]

Answer:

The snowball's speed after the impact is 3 m/s

Explanation:

Given the following:

mass of each ball

m₁ = 0.4 Kg

m₂ = 0.6 Kg

initial speed of both balls = v₁ = 15 m/s

Speed of 1 Kg mass post-collision =?

Applying conservation of momentum

m₁ v₁ - m₂ v₁ = (m₁+m₂) V

A negative velocity indicates that the second ball moves in the opposite direction.

0.4 x 15 - 0.6 x 15 = (1) V

Therefore,

V = - 3 m/s

Consequently,

The snowball's speed following the collision is 3 m/s

8 0

2 months ago

A large box of mass m sits on a horizontal floor. You attach a lightweight rope to this box, hold the rope at an angle θ above t

inna [3103]

Answer:

The answer to the specified question will be " $\mu_{s}=\frac{T_{m}Cos\theta}{M_{g}-T_{m}Sin\theta}$ ".

Explanation:

Referring to the question,

$\sum F_{x}$

⇒ $TCos \theta-F_{s}=0$

⇒ $T_{m}Cos \theta =F_{s}$ ...(equation 1)

$\sum F_{y}$

⇒ $TSin \theta+F_{N}=m_{g}$

⇒ $M_{g}-TSin \theta=F_{N}$ ...(equation 2)

Now,

From equation 1 and equation 2, we conclude

⇒ $T_{m} Cos \theta = \mu_{s}F_{N}$

By substituting the value of $F_{N}$ , we derive

⇒ $T_{m} Cos\theta = \mu_{s}(M_{g}-T_{m}Sin \theta)$

⇒ $\mu_{s}=\frac{T_{m}Cos\theta}{M_{g}-T_{m}Sin\theta}$

4 0

2 months ago

Suppose an electrical wire is replaced with one having every linear dimension doubled (i.e., the length and radius have twice th

Softa [3030]

Response:

The new resistance is half of the original resistance.

Explanation:

Resistance in a wire is represented by:

$R=\dfrac{\rho L}{A}$

$\rho$ = resistivity of the material

L and A are the physical dimensions

If a wire is exchanged for one where all linear dimensions are doubled, i.e. l' = 2l and r' = 2r

The updated resistance of the wire can be calculated as follows:

$R'=\dfrac{\rho L'}{A'}$

$R'=\dfrac{\rho (2L)}{\pi (2r)^2}$

$R'=\dfrac{1}{2}\dfrac{\rho L}{A}$

$R'=\dfrac{1}{2}R$

The new resistance equals half of the original resistance. Thus, this provides the solution needed.

4 0

1 month ago

James Cameron piloted a submersible craft to the bottom of the Challenger Deep, the deepest point on the ocean's floor, 11,000 m

serg [3582]

Answer:

$4.1\cdot 10^8 N$

Explanation:

To begin with, we must determine the pressure acting on the sphere, which is calculated using:

$p=p_0 + \rho g h$

where

$p_0 =1.01\cdot 10^5 Pa$ denotes the atmospheric pressure

$\rho = 1000 kg/m^3$ represents the density of the water

$g=9.8 m/s^2$ signifies the acceleration due to gravity

$h=11,000 m$ indicates the depth

By substituting these values,

$p=1.01\cdot 10^5 Pa + (1000 kg/m^3)(9.8 m/s^2)(11,000 m)=1.08\cdot 10^8 Pa$

The sphere's radius is calculated as r = d/2 = 1.1 m/2 = 0.55 m

Thus, the sphere's total surface area can be expressed as

$A=4 \pi r^2 = 4 \pi (0.55 m)^2=3.8 m^2$

Consequently, the inward force acting on the sphere equals

$F=pA=(1.08\cdot 10^8 Pa)(3.8 m^2)=4.1\cdot 10^8 N$

8 0

1 month ago

Read 2 more answers