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3 years ago

13

the volume of ice block is 2400cm^3 and its density is 0.9 g/cm^3. how much part of it remains above the surface of water when i

t is kept in water?

1 answer:

stiks02 [169]3 years ago

4 0

Answer:

240 cm³

Explanation:

Weight = Buoyancy

mg = ρVg

m = ρV

(0.9 g/cm³ × 2400 cm³) = (1 g/cm³) V

V = 2160 cm³

The submerged volume is 2160 cm³, so the volume above the surface is 240 cm³.

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3 years ago

Consider this situation: Four ropes, each attached to the end

faust18 [17]

The forces acting on the elevator are:

Gravity force

Tension force

Air resistance

Explanation:

Let's go through each of the forces listed and see which ones are acting on the elevator.

Normal force: NO. The normal force is a force exerted by a surface whenever there is another object "pushing" on it. For instance, when a box is at rest on a table, the box is "pushing" on the table (due to its weight), and the table "pushes back" on the box, upward, in order to balance its weight: this is the normal force. In this case, the elevator is lifted, so it is not pushing on anything, therefore there is no normal force.
Gravity force: YES. The force of gravity acts on every object located in the gravitational field of the Earth; it pulls downward, and its magnitude is $mg$ , where m is the mass of the object and g is the acceleration of gravity.
Applied force: NO. Here there is no applied force, since there is nobody "pushing" or "pulling" the elevator.
Friction force: NO. As we are considering the forces on the elevator, and the elevator is not sliding against any surfaces, there is no force of friction. (The force of friction acts whenever there are two surfaces sliding against each other, which is not the case here)
Tension force: YES. The tension force is the force exerted by a rope or a string when pulling an object. In this case, there are four ropes pulling the elevator, therefore there are 4 forces of tension acting on the elevator, upward.
Air resistance: YES. As the elevator is moving through the air, the interaction between the molecules of air with the surface of the elevator produces a force (called air resistance) that "resists" the motion of the elevator, therefore pushing downward. However, the magnitude of this force is negligible in this case.

Learn more about forces:

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3 years ago

Why is fair play essential in sports?

Dmitry [639]

Answer: it teaches people tolerance and respect for others.

Explanation:

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2 years ago

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Juan is making ice tea. When he adds ice to the tea, why does the tea cool down?

Tresset [83]

Answer: because the ice was cold, and it melted in the tea. So the tea is now cooler.

Explanation:

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3 years ago

A parachutist falls 50.0 m without friction. When the parachute opens, he slows down at a rate of 67 m/s*2. If he reaches the gr

KIM [24]

Answer:

3.49 seconds

3.75 seconds

-43200 ft/s²

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

$s=ut+\frac{1}{2}at^2\\\Rightarrow 50=0t+\frac{1}{2}\times 9.81\times t^2\\\Rightarrow t=\sqrt{\frac{50\times 2}{9.81}}\\\Rightarrow t=3.19\ s$

Time the parachutist falls without friction is 3.19 seconds

$v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times 9.81\times 50+0^2}\\\Rightarrow v=31.32\ m/s$

Speed of the parachutist when he opens the parachute 31.32 m/s. Now, this will be considered as the initial velocity

$v=u+at\\\Rightarrow 11=31.32+9.81t\\\Rightarrow t=\frac{11-31.32}{-67}=0.3\ s$

So, time the parachutist stayed in the air was 3.19+0.3 = 3.49 seconds

$s=ut+\frac{1}{2}at^2\\\Rightarrow \frac{s}{2}=0t+\frac{1}{2}\times a\times t^2\\\Rightarrow \frac{s}{2}=\frac{1}{2}at^2$

$s=ut+\frac{1}{2}at^2\\\Rightarrow \frac{s}{2}=u1.1+\frac{1}{2}\times a\times 1.1^2$

Now the initial velocity of the last half height will be the final velocity of the first half height.

$v=u+at\\\Rightarrow v=at$

Since the height are equal

$\frac{1}{2}at^2=u1.1+\frac{1}{2}\times a\times 1.1^2\\\Rightarrow \frac{1}{2}at^2=at1.1+\frac{1}{2}\times a\times 1.1^2\\\Rightarrow 0.5t^2-1.1t-0.605=0\\\Rightarrow 500t^2-1100t-605=0$

$t=\frac{11\left(1+\sqrt{2}\right)}{10},\:t=\frac{11\left(1-\sqrt{2}\right)}{10}\\\Rightarrow t=2.65, -0.45$

Time taken to fall the first half is 2.65 seconds

Total time taken to fall is 2.65+1.1 = 3.75 seconds.

When an object is thrown with a velocity upwards then the velocity of the object at the point to where it was thrown becomes equal to the initial velocity.

$v^2-u^2=2as\\\Rightarrow a=\frac{v^2-u^2}{2s}\\\Rightarrow a=\frac{0^2-240^2}{2\times \frac{8}{12}}\\\Rightarrow a=-43200\ ft/s^2$

Magnitude of acceleration is -43200 ft/s²

5 0

3 years ago