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3 years ago

I NEED ANSWER ASAP

Physics

2 answers:

umka21 [38]3 years ago

7 0

Answer:

97% will sink below the water

Explanation:

Waters density is 1 g/cm3

If an object's density is greater than 1g/cm3, it will sink. If it's less then it would float. 1-0.97 = 0.03. Only 3% of the ice would show while 97 will be under.

IgorLugansk [536]3 years ago

5 0

Answer:

About 97 percent of the ice will sink below water.

Explanation:

Let the volume that is submerged into the water is given as V1

so at equilibrium condition we can say that buoyancy force on it is balanced by weight of the block

so we have

$\pho_{water} V_1 g = \rho_{ice} V g$

now we have

$\frac{V_1}{V} = \frac{\rho_{ice}}{\rho_{water}}$

$\frac{V_1}{V} = \frac{0.97}{1}$

so the percentage of block that is submerged into the water is given as

volume percent submerged = 97 %

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Which is usually greater for a pair of surfaces, the coefficient of static friction or the coefficient of sliding friction? why

lorasvet [3.4K]

Explanation :

Static friction is the frictional force between two objects that are at rest. While sliding friction is the frictional force between two objects in contact and are sliding w.r.t each other.

Static friction is usually greater than sliding friction because in static friction the contact forces is more and the interlocking between objects is tight as compared to sliding friction.

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3 years ago

At t=0 bullet A is fired vertically with an initial (muzzle) velocity of 450 m/s. When 3s. bullet B is fired upward with a muzzl

Debora [2.8K]

Answer:

At time 10.28 s after A is fired bullet B passes A.

Passing of B occurs at 4108.31 height.

Explanation:

Let h be the height at which this occurs and t be the time after second bullet fires.

Distance traveled by first bullet can be calculated using equation of motion

$s=ut+0.5at^2 \\$

Here s = h,u = 450m/s a = -g and t = t+3

Substituting

$h=450(t+3)-0.5\times 9.81\times (t+3)^2=450t+1350-4.9t^2-29.4t-44.1\\\\h=420.6t-4.9t^2+1305.9$

Distance traveled by second bullet

Here s = h,u = 600m/s a = -g and t = t

Substituting

$h=600t-0.5\times 9.81\times t^2=600t-4.9t^2\\\\h=600t-4.9t^2 \\$

Solving both equations

$600t-4.9t^2=420.6t-4.9t^2+1305.9\\\\179.4t=1305.9\\\\t=7.28s \\$

So at time 10.28 s after A is fired bullet B passes A.

Height at t = 7.28 s

$h=600\times 7.28-4.9\times 7.28^2\\\\h=4108.31m \\$

Passing of B occurs at 4108.31 height.

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3 years ago

A student throws a rock horizontally from the edge of a cliff that is 20 m high. The rock has an initial speed on 10 m/s. If air

fiasKO [112]

The distance of the rock from the base of the cliff is C) 20 m

Explanation:

The motion of the rock in this problem is a projectile motion, which consists of two independent motions:

- A uniform motion (constant velocity) along the horizontal direction

- An accelerated motion with constant acceleration (acceleration of gravity) in the vertical direction

We start by analyzing the vertical motion to find the time of flight of the rock (the time it takes to reach the ground). We can do it by using the suvat equation:

$s=u_y t+\frac{1}{2}at^2$