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2 years ago

Noah drops a rock with a density of 1.73 g/cm3 into a pond. Will the rock float or sink?

Physics

2 answers:

pantera1 [17]2 years ago

6 0

<span>the rock will sink because it is more dense than water.</span>

bearhunter [10]2 years ago

5 0

Answer:

It will sink

Explanation:

An object in the water can float only if its density is lower than the density of the water.

In fact, for an object completely immersed in water, there are two forces acting on it:

- Its weight, $W=mg=\rho_o V g$ , downward, where $\rho_o$ is the density of the object, V its volume and g the gravitational acceleration

- The buoyant force, $B=\rho_w V g$ , upwards, there $\rho_w$ is the density of the water

We see that when the density of an object is larger than the density of the water, $\rho_o > \rho_w$ , the weight is greater than the buoyant force, $W>B$ , so the object sinks.

In this case, the rock has a density of 1.73 g/cm3, while water has a density of 1.0 g/cm^3, so the rock will sink.

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Mrs. Brown's class is studying magnets and electricity. At the end of the unit Claire states that magnets and electricity are bo

Gwar [14]

Answer:

Magnets can create electricity and electricity can create a magnetic force.

Both electric charges and magnets do not have to touch an object in order to exert a force on it.

Electromagnets use electricity to create a magnetic force.

Explanation:

7 0

2 years ago

Read 2 more answers

Ugonna stands at the top of an incline and pushes a 100−kg crate to get it started sliding down the incline. The crate slows to

Anna [14]

Answer:(a)891.64 N

(b)0.7

Explanation:

Mass of crate $m=100 kg$

Crate slows down in $s=1.5 m$

initial speed $u=1.77 m/s$

inclination $\theta =30^{\circ}$

From Work-Energy Principle

Work done by all the Forces is equal to change in Kinetic Energy

$W_{friction}+W_{gravity}=\frac{1}{2}mv_i^2-\frac{1}{2}mv_f^2$

$W_{gravity}=mg(0-h)=mgs\sin \theta$

$W_{gravity}=-mgs\sin \theta$

$W_{gravity}=-100\times 9.8\times 1.5\sin 30=-735 N$

change in kinetic energy $=\frac{1}{2}\times 100\times 1.77^2=156.64 J$

$W_{friction}=156.64+735=891.645$

(b)Coefficient of sliding friction

$f_r\cdot s=W_{friciton}$

$891.645=f_r\times 1.5$

$f_r=594.43 N$

and $f_r=\mu mg\cos \theta$

$\mu 100\times 9.8\times \cos 30=594.43$

$\mu =0.7$

5 0

2 years ago

A 5-kg concrete block is lowered with a downward acceleration of 2.8 m/s2 by means of a rope. The force of the block on the Eart

maksim [4K]

When the body touches the ground two types of Forces will be generated. The Force product of the weight and the Normal Force. This is basically explained in Newton's third law in which we have that for every action there must also be a reaction. If the Force of the weight is pointing towards the earth, the reaction Force of the block will be opposite, that is, upwards and will be equivalent to its weight:

F = mg

Where,

m = mass

g = Gravitational acceleration

F = 5*9.8

F = 49N

Therefore the correct answer is E.

5 0

2 years ago

A child is sliding a toy block (with mass = m) down a ramp. The coefficient of static friction between the block and the ramp is

tiny-mole [99]

Answer:

$F=mg(sin(\theta )-0.25 cos(\theta ))$

Explanation:

The free body diagram of the block on the slide is shown in the below figure

Since the block is in equilibrium we apply equations of statics to compute the necessary unknown forces

N is the reaction force between the block and the slide

For equilibrium along x-axis we have

$\sum F_{x}=0\\\\mgsin(\theta )-\mu N-F=0\\\therefore F=mgsin(\theta)-\mu N......(\alpha )\\Similarly\\\sum F_{y}=0\\\\N-mgcos(\theta )=0\\\therefore N=mgcos(\theta ).......(\beta )\\\\$

Using value of N from equation β in α we get value of force as

$F=mg(sin(\theta )-\mu cos(\theta ))$

Applying values we get

$F=mg(sin(\theta )-0.25 cos(\theta ))$

8 0

2 years ago

Read 2 more answers

A roller coaster car drops a maximum vertical distance of 35.4 m. Determine the maximum speed of the car at the bottom of that d

marissa [1.9K]

Answer:

The maximum speed of the car at the bottom of that drop is 26.34 m/s.

Explanation:

Given that,

The maximum vertical distance covered by the roller coaster, h = 35.4 m

We need to find the maximum speed of the car at the bottom of that drop. It is a case of conservation of energy. The energy at bottom is equal to the energy at top such that :

$mgh=\dfrac{1}{2}mv^2$

$v=\sqrt{2gh}$

$v=\sqrt{2\times 9.8\times 35.4}$

v = 26.34 m/s

So, the maximum speed of the car at the bottom of that drop is 26.34 m/s. Hence, this is the required solution.

8 0

2 years ago