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sasho [114]
2 years ago
15

If a special type of ice has a specific heat capacity of 2060 J/kg K. How

Physics
1 answer:
KengaRu [80]2 years ago
4 0

Answer:

Q = 144612 Joules.

Explanation:

Given the following data;

Mass = 2.6 kg

Initial temperature = -27°C to Kelvin = 273 + (-27) = 246K

Final temperature = 0°C to Kelvin = 273K

Specific heat capacity = 2060 J/kgK.

To find the quantity of heat absorbed;

Heat capacity is given by the formula;

Q = mcdt

Where;

Q represents the heat capacity or quantity of heat.

m represents the mass of an object.

c represents the specific heat capacity of water.

dt represents the change in temperature.

dt = T2 - T1

dt = 273 - 246

dt = 27 K

Substituting the values into the equation, we have;

Q = 2.6*2060*27

Q = 144612 Joules.

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The battery for a certain cell phone is rated at 3.70 V. According to the manufacturer it can produce math]3.15 \times 10^{4} J[
AysviL [449]

Answer:

The average current that this cell phone draws when turned on is 0.451 A.

Explanation:

Given;

voltage of the phone, V = 3.7 V

electrical energy of the phone battery, E = 3.15 x 10⁴ J

duration of battery energy, t = 5.25 h

The power the cell phone draws when turned on, is the rate of energy consumption, and this is calculated as follows;

P = \frac{E}{t}

where;

P is power in watts

E is energy in Joules

t is time in seconds

P = \frac{3.15*10^4}{5.25*3600s} = 1.667 \ W

The average current that this cell phone draws when turned on:

P = IV

I = \frac{P}{V} =\frac{1.667}{3.7} = 0.451 \ A

Therefore, the average current that this cell phone draws when turned on is 0.451 A.

5 0
3 years ago
Arm ab has a constant angular velocity of 16 rad/s counterclockwise. At the instant when theta = 60
geniusboy [140]

The <em>linear</em> acceleration of collar D when <em>θ = 60°</em> is - 693.867 inches per square second.

<h3>How to determine the angular velocity of a collar</h3>

In this question we have a system formed by three elements, the element AB experiments a <em>pure</em> rotation at <em>constant</em> velocity, the element BD has a <em>general plane</em> motion, which is a combination of rotation and traslation, and the ruff experiments a <em>pure</em> translation.

To determine the <em>linear</em> acceleration of the collar (a_{D}), in inches per square second, we need to determine first all <em>linear</em> and <em>angular</em> velocities (v_{D}, \omega_{BD}), in inches per second and radians per second, respectively, and later all <em>linear</em> and <em>angular</em> accelerations (a_{D}, \alpha_{BD}), the latter in radians per square second.

By definitions of <em>relative</em> velocity and <em>relative</em> acceleration we build the following two systems of <em>linear</em> equations:

<h3>Velocities</h3>

v_{D} + \omega_{BD}\cdot r_{BD}\cdot \sin \gamma = -\omega_{AB}\cdot r_{AB}\cdot \sin \theta   (1)

\omega_{BD}\cdot r_{BD}\cdot \cos \gamma = -\omega_{AB}\cdot r_{AB}\cdot \cos \theta   (2)

<h3>Accelerations</h3>

a_{D}+\alpha_{BD}\cdot \sin \gamma = -\omega_{AB}^{2}\cdot r_{AB}\cdot \cos \theta -\alpha_{AB}\cdot r_{AB}\cdot \sin \theta - \omega_{BD}^{2}\cdot r_{BD}\cdot \cos \gamma   (3)

-\alpha_{BD}\cdot r_{BD}\cdot \cos \gamma = - \omega_{AB}^{2}\cdot r_{AB}\cdot \sin \theta + \alpha_{AB}\cdot r_{AB}\cdot \cos \theta - \omega_{BD}^{2}\cdot r_{BD}\cdot \sin \gamma   (4)

If we know that \theta = 60^{\circ}, \gamma = 19.889^{\circ}, r_{BD} = 10\,in, \omega_{AB} = 16\,\frac{rad}{s}, r_{AB} = 3\,in and \alpha_{AB} = 0\,\frac{rad}{s^{2}}, then the solution of the systems of linear equations are, respectively:

<h3>Velocities</h3>

v_{D}+3.402\cdot \omega_{BD} = -41.569   (1)

9.404\cdot \omega_{BD} = -24   (2)

v_{D} = -32.887\,\frac{in}{s}, \omega_{BD} = -2.552\,\frac{rad}{s}

<h3>Accelerations</h3>

a_{D}+3.402\cdot \alpha_{BD} = -445.242   (3)

-9.404\cdot \alpha_{BD} = -687.264   (4)

a_{D} = -693.867\,\frac{in}{s^{2}}, \alpha_{BD} = 73.082\,\frac{rad}{s^{2}}

The <em>linear</em> acceleration of collar D when <em>θ = 60°</em> is - 693.867 inches per square second. \blacksquare

<h3>Remark</h3>

The statement is incomplete and figure is missing, complete form is introduced below:

<em>Arm AB has a constant angular velocity of 16 radians per second counterclockwise. At the instant when θ = 60°, determine the acceleration of collar D.</em>

To learn more on kinematics, we kindly invite to check this verified question: brainly.com/question/27126557

5 0
1 year ago
How to tackle questions on resultant force​
Solnce55 [7]

Answer:

February. we will be in the future types, I

Explanation:

Collins the best year fixed come in handy when but it at a very nice relaxing. it will not have the option?

3 0
3 years ago
A projectile is fired vertically from Earth's surface with
scoray [572]

Answer:

h=25.52\times10^6 m

Explanation:

Initial speed, v = 10 x 10^3 m/s

Mass of the earth, M = 6 x 10^24 kg

Radius of the earth, R = 6.4 x 10^6 m

Maximum from the surface of earth, h = ?

Let  m = Mass of the projectile

Solution:

Potential energy at maximum height =  ( Potential + Kinetic energy ) at the surface

-G M m / ( R + h )=- G M m / R + (1/2) m v^2

- G M / ( R + h ) = - G M / R + (1/2) v^2

-2\times G M / ( R + h ) = ( - 2 G M / R ) + v^2

-2\times6.67\times10^{-11}\times6\times10^{24}/ ( R + h )

=( (- 2\times 6.67\times10^{-11}\times6\times10^{24}) /(6.4\times10^6)} +10000^2

=-2.50625\times10^7 J

=- 8\times10^{14} / ( R + h )=-2.50625\times 10^7

R+h=31.92\times10^{6}

h=31.92\times10^{6}-6.4\times10^6

h=25.52\times10^6 m

5 0
2 years ago
Two boxers are fighting. Boxer 1 throws his 5 kg fist at boxer 2 with a speed of 9 m/s.
Sladkaya [172]

Answer:

0.001 s

Explanation:

The force applied on an object is equal to the rate of change of momentum of the object:

F=\frac{\Delta p}{\Delta t}

where

F is the force applied

\Delta p is the change in momentum

\Delta t is the time interval

The change in momentum can be written as

\Delta p=m(v-u)

where

m is the mass

v is the final velocity

u is the initial velocity

So the original equation can be written as

F=\frac{m(v-u)}{\Delta t}

In this problem:

m = 5 kg is the mass of the fist

u = 9 m/s is the initial velocity

v = 0 is the final velocity

F = -45,000 N is the force applied (negative because its direction is opposite to the motion)

Therefore, we can re-arrange the equation to solve for the time:

\Delta t=\frac{m(v-u)}{F}=\frac{(5)(0-9)}{-45,000}=0.001 s

4 0
3 years ago
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