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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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A stock person at the local grocery store has a job consisting of the following five segments:

vaieri [72.5K]

Answer:

Explanation:

Work done can be said to be positive if the applied force has a component to be in the direction of the displacement and when the angle between the applied force and displacement is positive.

In 1 and 2 work done is positive

6 0

2 years ago

The magnitude J(r) of the current density in a certain cylindrical wire is given as a function of radial distance from the cente

kipiarov [429]

Answer:

$I=68.31\times 10^{-6}\ A$

Explanation:

Given that

J(r) = Br

We know that area of small element

dA = 2 π dr

I = J A

dI = J dA

Now by putting the values

dI = B r . 2 π dr

dI= 2π Br² dr

Now by integrating above equation

$\int_{0}^{I}dI= \int_{r_1}^{r_2}2\pi Br^2 dr$

$I={2\pi B}\times \dfrac{r_2^3-r_1^3}{3}$

Given that

B= 2.35 x 10⁵ A/m³

r₁ = 2 mm

r₂ = 2+ 0.0115 mm

r₂ = 2.0115 mm

$I={2\pi B}\times \dfrac{r_2^3-r_1^3}{3}$

By putting the values

$I={2\pi \times 2.35 \times 10^5 }\times \dfrac{(2.0115\times 10^{-3})^3-(2\times 10^{-3})^3}{3}\ A$

$I=68.31\times 10^{-6}\ A$

7 0

2 years ago

Read 2 more answers

Tyler stands at rest on a skateboard. He has a mass of 120 kg. His friend (m = 60 kg) jumps into his arms at a speed of 2 m/s. I

Andrews [41]

Momentum question. This is an inelastic collision, so

m1v1+m2v2=Vf(m1+m2)
Vf=(m1v1+m2v2)/(m1+m2)=[(120kg)(0m/s)+(60kg)(2m/s)] / (120kg+60kg)
Vf=120kg m/s / 180kg
Vf=0.67m/s

0.67m/s

5 0

2 years ago

What is the final temperature when a 3.0 kg gold bar at 99 0C is dropped into 0.22 kg of water at 25oC?

slavikrds [6]

I will post my work, but is that 99 degrees Celsius and 25 degrees Celsius?

All you have to do is plug in the initial temperature for gold where it says Tg and the initial temperature for the water where it says Tw and then plug that in and you will have your answer.

8 0

2 years ago

Construction of a solar power plant is proposed for a desert area near a school. A student has hypothesized that the shade cast

hjlf

Answer:

4. The direct sunlight received by creosote bush in the desert area (in kWh/m2) during a 12 month period

Explanation:

The creosote bush depends on sunlight to produce the food they require through photosynthesis. The shade from the solar panels would reduce the amount of sunlight that the bush receives. This would increase the mortality of the bush.

In order to test the hypothesis the student must record the direct sunlight received by creosote bush in the desert area (in kWh/m2) during a 12 month period. If the plants receive sunlight less than the above amount the plants should start dying. If not then the hypothesis is false.

Hence, the answer is 4. The direct sunlight received by creosote bush in the desert area (in kWh/m2) during a 12 month period.

4 0

2 years ago