Physics College

Someone goes to lift a crate that is resting on the bottom of the pool filled with water (density of water is 1000 kg/m^3). While
still submerged, only 310 N is required to lift the crate. The crate is shaped like a cube with sides of 0.25 m. What is the density of
the cube? Numerical answer is assumed to be in units of kg/m^3

Answers

Answer 1

Answer:

the density of the cube is approximately 2016.07 kg/m^3.

Explanation:

The buoyant force acting on an object submerged in a fluid is equal to the weight of the fluid displaced by the object.

Let's first calculate the weight of the crate:

mass of crate = density * volume = density * (side length)^3 = density * 0.25^3 = 0.015625 * density

weight of crate = mass of crate * gravity = 0.015625 * density * 9.81 = 0.1530875 * density

where gravity is the acceleration due to gravity, which is approximately 9.81 m/s^2.

Since the crate is submerged in water, the buoyant force acting on it is:

buoyant force = weight of water displaced = density of water * volume of water displaced * gravity

The volume of water displaced is equal to the volume of the cube, which is 0.25^3 = 0.015625 m^3. Therefore, the buoyant force is:

buoyant force = 1000 kg/m^3 * 0.015625 m^3 * 9.81 m/s^2 = 1.534453125 N

According to the problem, it takes 310 N to lift the crate while it is still submerged. This means that the net force acting on the crate is:

net force = lifting force - buoyant force = 310 N - 1.534453125 N = 308.465546875 N

This net force is equal to the weight of the crate:

net force = weight of crate = 0.1530875 * density

Therefore, we can solve for the density of the crate:

density = net force / 0.1530875 = 308.465546875 / 0.1530875 = 2016.06666667 kg/m^3

Rounding to the nearest hundredth, we get:

density ≈ 2016.07 kg/m^3

Therefore, the density of the cube is approximately 2016.07 kg/m^3

Related Questions

In a water pistol, a piston drives water through a larger tube of radius 1.30 cm into a smaller tube of radius 1.10 mm as in the figure below. Answer parts a-f.

Answers

It takes 0.47 seconds for water to travel from the nozzle to the ground when the water pistol is fired horizontally.

What is the time it takes for water to travel from the nozzle to the ground?

We will denote the height of the water pistol above the ground as h and the initial velocity of water exiting the nozzle as v2. Assuming negligible air resistance, we will analyze the vertical motion of the water droplets.

The vertical displacement of the water droplets is calculated using equation: h = (1/2) * g * t^2.

Rearranging equation, we solve for time:

t = sqrt(2h / g).

Given data:

Height h = 1.10 m and the acceleration due to gravity g = 9.8 m/s^2, we get:

t = sqrt(2 * 1.10 / 9.8)

t = 0.47380354147

t = 0.47 seconds.

Answers

The magnitude of the angular acceleration of the roller is approximately 104.2 rad/s^2.

The magnitude of the angular acceleration of the roller can be determined using the torque equation and Newton's second law for rotational motion.
Step 1: Calculate the moment of inertia of the roller.
The moment of inertia (I) of a solid cylinder is given by the formula I = (1/2) * m * r^2, where m is the mass of the object and r is the radius.
In this case, the mass of the roller is 2.4 kg and the radius is 0.038 m.
So, I = (1/2) * 2.4 kg * (0.038 m)^2.
Step 2: Calculate the torque applied to the roller.
Torque (τ) is equal to the force (F) applied multiplied by the perpendicular distance (r) from the axis of rotation.
In this case, the force applied by Joe is 16 N and the distance is equal to the radius of the roller, 0.038 m.
So, τ = F * r.
Step 3: Use the torque equation.
The torque applied to the roller causes an angular acceleration (α) according to the equation τ = I * α.
Rearranging the equation, we get α = τ / I.
Step 4: Substitute the values into the equation.
Using the values we calculated earlier, we can substitute them into the equation α = τ / I.
α = (16 N * 0.038 m) / [(1/2) * 2.4 kg * (0.038 m)^2].
Step 5: Calculate the magnitude of the angular acceleration.
Evaluating the expression, we find that the magnitude of the angular acceleration of the roller is approximately 104.2 rad/s^2.
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. A 0.140 kg baseball is pitched toward home plate at 30.0 m/s.
The batter hits the ball back (opposite direction) to the pitcher at
44.0 m/s. Assume that towards home plate is positive. What is
the change in momentum for the ball?

Answers

The change in momentum for the baseball, which is hit back in the opposite direction by the batter, is -10.36 kg·m/s. This change in momentum is obtained by subtracting the initial momentum of 4.2 kg·m/s from the final momentum of -6.16 kg·m/s. The negative sign indicates the opposite direction of the momentum.

To find the change in momentum for the baseball, we can use the formula:

Change in momentum = Final momentum - Initial momentum

Momentum is defined as the product of mass and velocity.

Given data:

Mass of the baseball (m) = 0.140 kg

Initial velocity of the baseball ([tex]v_i_n_i_t_i_a_l)[/tex] = 30.0 m/s

Final velocity of the baseball ([tex]v_f_i_n_a_l_[/tex]) = -44.0 m/s (negative sign indicates opposite direction)

To calculate the initial momentum, we multiply the mass by the initial velocity:

Initial momentum = m * [tex]v_i_n_i_t_i_a_l[/tex] = 0.140 kg * 30.0 m/s = 4.2 kg·m/s

To calculate the final momentum, we multiply the mass by the final velocity:

Final momentum = m * [tex]v_f_i_n_a_l_[/tex] = 0.140 kg * (-44.0 m/s) = -6.16 kg·m/s

Now we can find the change in momentum:

Change in momentum = Final momentum - Initial momentum

Change in momentum = (-6.16 kg·m/s) - (4.2 kg·m/s)

Change in momentum = -10.36 kg·m/s

Therefore, the change in momentum for the baseball is -10.36 kg·m/s. The negative sign indicates a change in direction, as the ball is hit back in the opposite direction.

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