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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

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A 100 cm3 block of lead weighs 11N is carefully submerged in water. One cm3 of water weighs 0.0098 N.

Pie

Volume of lead = 100 cm^3

density of lead = 11.34 g/cm^3

mass of the lead piece = density * volume

$m = 100 * 11.34 = 1134 g$

$m = 1.134 kg$

so its weight in air will be given as

$W = mg = 1.134* 9.8 = 11.11 N$

now the buoyant force on the lead is given by

$F_B = W - F_{net}$

$F_B = 11.11 - 11 = 0.11 N$

now as we know that

$F_B = \rho V g$

$0.11 = 1000* V * 9.8$

so by solving it we got

V = 11.22 cm^3

(ii) this volume of water will weigh same as the buoyant force so it is 0.11 N

(iii) Buoyant force = 0.11 N

(iv)since the density of lead block is more than density of water so it will sink inside the water

buoyant force on the lead block is balancing the weight of it

$F_B = W$

$\rho V g = W$

$13* 10^3 * V * 9.8 = 11.11$

$V = 87.2 cm^3$

(ii) So this volume of mercury will weigh same as buoyant force and since block is floating here inside mercury so it is same as its weight = 11.11 N

(iii) Buoyant force = 11.11 N

(iv) since the density of lead is less than the density of mercury so it will float inside mercury

Yes, if object density is less than the density of liquid then it will float otherwise it will sink inside the liquid

3 0

2 years ago

A spring is 14cm long. Three masses are hung from it and then it is measured again. Now it is 19.5cm long. What force did the th

lukranit [14]

Answer:

Gravity

Explanation:

The answer is gravity because when the 3 masses were hung from the spring, gravity pulled the spring towards the ground.

3 0

2 years ago

A force of 150 N accelerates a 25 kg wooden chair across a wood floor at 4.3 m/s2 . How big is the frictional force on the block

solniwko [45]

We can first calculate the net force using the given information.

By Newton's second law, F(net) = ma:

F(net) = 25 * 4.3 = 107.5

We can now calculate the frictional force, f, which is working against the applied force, F(app) (this is why the net force is a bit lower):

f = F(net) - F(app) = 150 - 107.5 = 42.5 N

Now we can calculate the coefficient of friction, u, using the normal force, F(N):

f = uF(n) --> u = f/F(N)
u = 42.5/[25(9.8)]
u = 0.17

4 0

2 years ago

The air in a pipe resonates at 150 Hz and 750 Hz, one of these resonances being the fundamental. If the pipe is open at both end

Xelga [282]

Answer:

Explanation:

Two frequencies with magnitude 150 Hz and 750 Hz are given

For Pipe open at both sides

fundamental frequency is 150 Hz as it is smaller

frequency of pipe is given by

$f=\frac{nv}{2L}$

where L=length of Pipe

v=velocity of sound

$f=150\ Hz$ for n=1

and f=750 is for n=5

thus there are three resonance frequencies for n=2,3 and 4

For Pipe closed at one end

frequency is given by

$f=\frac{(2n+1)}{4L}\cdot v$

for n=0

$f_1=\frac{v}{4L}$

$f_1=150\ Hz$

for n=2

$f_2=\frac{5v}{4L}$

Thus there is one additional resonance corresponding to n=1 , between $f_1$ and $f_2$

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2 years ago

Why is thesize saved prior to entering the for loop? 2. what is the running time of removefirsthalf if lst is an arraylist? 3. w

Mrac [35]

<span>After entering the loop, it should use the correct list size and the loop will be affected if the remove call changes the size of the list. If lst is an Arraylist the running time of removefirsthalf is O (n^2). So when the beginning is removed the next element will move forward. If lst is a LinkedList which is a dynamic structure the running is O (n) for removefirsthalf</span>

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2 years ago