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Solution

a

Use the formula for mass:

\text{mass} = \rho \cdot V

where

\rho

is the density and

V

is the volume. Given

\rho = 950 \, \text{kg/m}^3

and

V = 82 \, \text{cm}^3 = 82 \times 10^{-6} \, \text{m}^3

, we get:

\text{mass} = 950 \, \text{kg/m}^3 \times 82 \times 10^{-6} \, \text{m}^3 = 0.0779 \, \text{kg}

b

Use the formula for pressure due to a liquid column:

P = \rho g h

where

\rho = 950 \, \text{kg/m}^3

,

g = 9.81 \, \text{m/s}^2

, and

h = 0.094 \, \text{m}

:

P = 950 \, \text{kg/m}^3 \times 9.81 \, \text{m/s}^2 \times 0.094 \, \text{m} = 875.37 \, \text{Pa}

c

The true pressure at point X is different from the calculated value because the calculated value only considers the pressure due to the liquid column. The true pressure also includes atmospheric pressure acting on the surface of the liquid

d

The object sinks because its density is greater than the density of the liquid. According to Archimedes' principle, an object will sink if its weight is greater than the buoyant force acting on it

e

To determine the volume of the object, submerge it completely in the liquid and measure the increase in the liquid level in the measuring cylinder. The increase in volume of the liquid is equal to the volume of the object

Answer

The mass of the liquid is 0.0779 kg. The pressure due to the liquid at point X is 875.37 Pa. The true pressure at point X is different because it includes atmospheric pressure. The object sinks because its density is greater than the liquid's density. The volume of the object can be determined by the increase in liquid level when the object is submerged.

Key Concept

Fluid Mechanics: Pressure and Archimedes' Principle

Explanation

The mass of the liquid is calculated using its density and volume. The pressure at a point in a liquid column is determined by the liquid's density, gravitational acceleration, and height of the liquid column. The true pressure includes atmospheric pressure. An object sinks if its density is greater than the liquid's density, and its volume can be measured by the

Solution

a

Deceleration is the rate at which an object slows down. It is the negative acceleration, meaning the velocity of the object decreases over time

b

To determine the distance traveled between

t=3.0 \, \text{s}

and

t=6.0 \, \text{s}

, we can use the area under the speed-time graph. The area under the graph represents the distance traveled. From

t=3.0 \, \text{s}

to

t=6.0 \, \text{s}

, the speed decreases from

11 \, \text{m/s}

to approximately

8 \, \text{m/s}

(assuming a linear decrease for simplicity). The area under the graph can be approximated as a trapezoid:

\text{Area} = \frac{1}{2} \times (\text{Base}_1 + \text{Base}_2) \times \text{Height}

\text{Base}_1 = 11 \, \text{m/s}

\text{Base}_2 = 8 \, \text{m/s}

\text{Height} = 3 \, \text{s}

\text{Distance} = \frac{1}{2} \times (11 + 8) \times 3 = \frac{1}{2} \times 19 \times 3 = 28.5 \, \text{m}

c

(i) The size of the deceleration after

t=6.0 \, \text{s}

increases as the speed continues to decrease more rapidly. (ii) The resultant force on the skater after

t=6.0 \, \text{s}

increases in magnitude because the deceleration (negative acceleration) increases, and according to Newton's Second Law,

F = ma

, the force is directly proportional to the acceleration

Answer

Deceleration is the rate at which an object slows down.

The distance traveled between

t=3.0 \, \text{s}

and

t=6.0 \, \text{s}

is

28.5 \, \text{m}

.

(i) The size of the deceleration increases after

t=6.0 \, \text{s}

.

(ii) The resultant force on the skater increases after

t=6.0 \, \text{s}

.

Key Concept

Newton's Second Law: The acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass. Equation:

F = ma

(Net force equals mass times acceleration)

Explanation

Given the force acting on the object, we can calculate the acceleration by using Newton's Second Law. The deceleration and resultant force are related

Solution

a

Use the definition of impulse:

\text{Impulse} = F \cdot \Delta t

. Given

F = 180 \, \text{N}

and

\Delta t = 0.050 \, \text{s}

, we calculate the impulse as follows:

\text{Impulse} = 180 \, \text{N} \times 0.050 \, \text{s} = 9 \, \text{Ns}

b

Use the relationship between impulse and change in momentum:

\text{Impulse} = \Delta p = m \cdot \Delta v

. Given

\Delta v = 20 \, \text{m/s}

and impulse

= 9 \, \text{Ns}

, we solve for mass

m

:

9 \, \text{Ns} = m \times 20 \, \text{m/s}

m = \frac{9 \, \text{Ns}}{20 \, \text{m/s}} = 0.45 \, \text{kg}

c

Use the kinematic equation for vertical motion to find the height:

v_f^2 = v_i^2 + 2a s

. Here,

v_f = 0 \, \text{m/s}

(at the highest point),

v_i = 20 \, \text{m/s}

, and

a = -9.8 \, \text{m/s}^2

(acceleration due to gravity). Solving for

s

(height):

0 = (20 \, \text{m/s})^2 + 2(-9.8 \, \text{m/s}^2) s

0 = 400 \, \text{m}^2/\text{s}^2 - 19.6 \, \text{m/s}^2 \cdot s

s = \frac{400 \, \text{m}^2/\text{s}^2}{19.6 \, \text{m/s}^2} \approx 20.41 \, \text{m}

d

The energy stored in the ball when it is no longer spherical is called elastic potential energy

Answer

Impulse: 9 Ns

Mass: 0.45 kg

Height: 20.41 m

Elastic potential energy

Key Concept

Newton's Second Law: The acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass. Equation:

F = ma

(Net force equals mass times acceleration)

Explanation

Given the force acting on the ball and the time of contact, we can calculate the impulse. Using the impulse and the change in velocity, we can determine the mass of the ball. Finally, using kinematic equations, we can find the height to which the ball rises

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