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joja [24]
3 years ago
13

Your friends tell you that all heavy objects sink in water. Do you agree or disagree? Explain your answer in terms of buoyant fo

rce.
Physics
2 answers:
Inessa [10]3 years ago
8 0

Answer:

Disagree

Explanation:

Concept of sink and float depends on the condition of water displaced by the objects

As we know that object will float when the weight of the object is less than the buoyancy force on it.

Also we can say that buoyancy force is the net upthrust applied by the water on the object.

This upthrust is equal to the weight of water that is displaced by the object.

So if heavy object displace more amount of water so that the weight of displaced water is more than the weight of the object then it must float on the water.

so this is not true that all heavy objects sink in water.

WITCHER [35]3 years ago
6 0
Disagree because its not about how heavy the object is and you don't know exactly how heavy, heavy is and the weight doesn't matter, its density that matters
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Answer: In your right wrist

Explanation:

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3 years ago
A diver leaves the end of a 4.0 m high diving board and strikes the water 1.3s later, 3.0m beyond the end of the board. Consider
shutvik [7]

Answer:

4.0 m/s

Explanation:

The motion of the diver is the motion of a projectile: so we need to find the horizontal and the vertical component of the initial velocity.

Let's consider the horizontal motion first. This motion occurs with constant speed, so the distance covered in a time t is

d=v_x t

where here we have

d = 3.0 m is the horizontal distance covered

vx is the horizontal velocity

t = 1.3 s is the duration of the fall

Solving for vx,

v_x = \frac{d}{t}=\frac{3.0 m}{1.3 s}=2.3 m/s

Now let's consider the vertical motion: this is an accelerated motion with constant acceleration g=9.8 m/s^2 towards the ground. The vertical position at time t is given by

y(t) = h + v_y t - \frac{1}{2}gt^2

where

h = 4.0 m is the initial height

vy is the initial vertical velocity

We know that at t = 1.3 s, the vertical position is zero: y = 0. Substituting these numbers, we can find vy

0=h+v_y t - \frac{1}{2}gt^2\\v_y = \frac{0.5gt^2-h}{t}=\frac{0.5(9.8 m/s^2)(1.3 s)^2-4.0 m}{1.3 s}=3.3 m/s

So now we can find the magnitude of the initial velocity:

v=\sqrt{v_x^2+v_y^2}=\sqrt{(2.3 m/s)^2+(3.3 m/s)^2}=4.0 m/s

4 0
3 years ago
5 milligrams into quintal​
Greeley [361]

Answer:

divide the mass value by 1e+8

4 0
3 years ago
3) A 30 kg child slides freely across a "Slip and Slide" on LEVEL GROUND. While the child slides, the force applied to keep them
zavuch27 [327]

Answer:

a = 0.1962 m/s^2

Explanation:

The magnitude of kinetic friction exerted is given by

F_k=\mu_kN

Where, μ_k= coefficient of kinetic friction= 0.02 and N = reaction force = mg

Where m= mass = 30 Kg and, g is acceleration due to gravity =9.81 m/s^2

F_k=0.02×30×9.81 =5.886 N

Now, since, there is no applied force this kinetic friction force will cause acceleration of the child

⇒ ma = F_k

here, a is the acceleration

⇒30a = 5.886

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4 0
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Keith_Richards [23]

Answer:

C.

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