4. The fastest runner in a track and field event is the one with the<br>

A bomb at rest at the origin of an xy-coordinate system explodes into three pieces. Just after the explosion, one piece, of mass

ValentinkaMS [17]

Answer:

Explanation:

a ) It is given that bomb was at rest initially , so , its momentum before the explosion was zero.

b ) We shall apply law of conservation of momentum along x and y direction separately because no external force acts on the bomb.

If v be the velocity of the third part along a direction making angle θ

with x axis ,

x component of v = vcosθ

So momentum along x axis after explosion of third part = mv cosθ

= 10 v cosθ

Momentum along x of first part = - 5 x 42 m/s

momentum of second part along x direction =0

total momentum along x direction before explosion = total momentum along x direction after explosion

0 = - 5 x 42 + 10 v cosθ

v cosθ = 21

Similarly

total momentum along y direction before explosion = total momentum along y direction after explosion

0 = - 5 x 38 + 10 v sinθ

v sinθ= 21

squaring and and then adding the above equation

v² cos²θ +v² sin²θ = 21² +19²

v² = 441 + 361

v = 28.31 m/s

Tanθ = 21 / 19

θ = 48°

6 0

2 years ago

What is the quantity of heat energy required to convert 10g cube of ice at -30oC to steam at 120oC. also draw a graph of tempera

julia-pushkina [17]

30,869.2 J

Explanation:

Given parameters:

Mass of ice cube = 10g

Initial temperature = -30°C

Final temperature = 120°C

Specific heat capacity of water = 4.2J/g°C

Specific heat capacity of ice = 2.1J/g°C

Specific heat capacity of steam = 1.996J/g°C

Latent heat of fusion of water(l) = 334J/g

Latent heat of vaporization = 2230J/g

Unknown:

Quantity of heat required = ?

Temperature-energy graph = ?

Solution:

The temperature energy profile is attached to this solution.

q = mc∅ₓ + mlₓ + mc∅ₙ + mlₙ + mc∅ₐ

qₓ = mc∅ₓ in converting ice from -30°C to ice at 0°C

qₓ is the amount of heat supplied to the ice that changes its temperature from -30°C to that at freezing point:

qₓ = 10 x 4.2 x (0-(-30)) = 10 x 2.1 x 30 = 630J

qₓ = mlₓ in converting ice to water

qₓ here is the latent heat used to break the ice bonds without a change in temperature:

qₓ = ml = 10 x 334 = 3340J

qₙ = mc∅ₙ is the heat from water at 0°C to boiling point.

This is the heat required to take water from freezing temperature to its boiling point

qₙ = mc∅ₙ = 10 x 4.2 x (100 - 0) = 4200J

qₙ = mlₙ is the heat in vaporizing water

In vaporizing water, there is no temperature change when the hydrogen bonds are broken. The heat supplied is not used to raise temperature. We use the latent heat of vaporization:

qₙ = 10 x 2230 = 22300J

qₐ = mc∅ₐ from vapor at 100°C to 120°C

This is the heat used to raise the temperature of vapor:

qₐ = 10 x 1.996 x (120-100) = 399.2J

The total heat:

q = qₓ + qₙ + qₐ = 630J + 3340J + 4200J + 22300J + 399.2J =30,869.2 J

Learn more:

Specific heat brainly.com/question/7210400

#learnwithBrainly

7 0

2 years ago

Josh has a toy car of mass 3 kg tied to a string of length 2 m. He ties the string to a pole and has the toy car drive in a circ

gulaghasi [49]

Part a)

Centripetal acceleration is defined as

$a_c = \frac{v^2}{R}$

now here we know that

$v = 3m/s$

$R = 2 m$

m = 3 kg

now from above formula we have

$a_c = \frac{3^2}{2}$

$a_c = 4.5 m/s^2$

Part b)

Maximum possible tension in the string is given as

$T = 50 N$

now by force equation we have

$F = ma$

$50 = 3 a$

$a = \frac{50}{3} m/s^2$

now again by above formula

$\frac{v^2}{R} = \frac{50}{3}$

$v = \sqrt{\frac{50 \times 2}{3}}$

$v = 5.77 m/s$

5 0

3 years ago

Read 2 more answers

What units must the constant G have in order for the u it’s F to be newtons (N)?

EleoNora [17]

The units for G must be $[N][m^2][kg^{-2}]$

Explanation:

The magnitude of the gravitational force between two objects is given by:

$F=G\frac{m_1 m_2}{r^2}$

where

F is the force

G is the gravitational constant

$m_1, m_2$ are the masses of the two objects

$r$ is the separation between the objects

We know that:

The units of F are Newtons (N)
The units of $m_1,m_2$ are kilograms (kg)
The units of $r$ are metres (m)

So, we can rewrite the equation in terms of G, to find its units:

$G=\frac{Fr^2}{m_1 m_2}=\frac{[N][m]^2}{[kg][kg]}=[N][m^2][kg^{-2}]$

Learn more about gravitational force:

brainly.com/question/1724648

brainly.com/question/12785992

#LearnwithBrainly

5 0

2 years ago

Help meh in this question plzzz <br>

iragen [17]

The Moment of Inertia of the Disc is represented by $I = \frac{15}{32}\cdot M\cdot R^{2}$ . (Correct answer: A)

Let suppose that the Disk is a Rigid Body whose mass is uniformly distributed. The Moment of Inertia of the element is equal to the Moment of Inertia of the entire Disk minus the Moment of Inertia of the Hole, that is to say:

$I = I_{D} - I_{H}$ (1)

Where:

$I_{D}$ - Moment of inertia of the Disk.

$I_{H}$ - Moment of inertia of the Hole.

Then, this formula is expanded as follows:

$I = \frac{1}{2}\cdot M\cdot R^{2} - \frac{1}{2}\cdot m\cdot \left(\frac{1}{2}\cdot R^{2} \right)$ (1b)

Dimensionally speaking, Mass is directly proportional to the square of the Radius, then we derive the following expression for the Mass removed by the Hole ( $m$ ):