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4 years ago

During light activity, a 70-kg person may generate 200 kcal

Physics

1 answer:

Lerok [7]4 years ago

3 0

Answer:

$2.75387\ ^{\circ}$

Explanation:

m = Mass of person = 70 kg

c = Specific heat of human body = $0.83\ J/kg^{\circ}C$

$\Delta T$ = Change in temperature

Time taken is 1 hour

Heat is given by

$Q=mc\Delta T\\\Rightarrow \Delta T=\dfrac{Q}{mc}\\\Rightarrow \Delta T=\dfrac{0.8\times 200\times 1}{70\times 0.83}\\\Rightarrow \Delta T=2.75387\ ^{\circ}C$

The rise in temperature is $2.75387\ ^{\circ}C$

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Phoenix [80]

Answer: I believe it’s location D. Or whichever is the location of slow crystallization.

Explanation: Slow crystallization of magma forms granite. Hope this helps. :)

6 0

3 years ago

Read 2 more answers

A box of mass 20kg is pulled up an inclined plane by a force of 285N. Given that the value of the incline angle is 30 degrees an

docker41 [41]

Given :

Mass of box , m = 250 kg.

Force applied , F = 285 N.

The value of the incline angle is 30°.

the coefficient of dynamic friction is $\mu=0.72$ .

To Find :

The speed with which the box is moving with, assuming it takes 4 seconds to reach the top of the incline.

Solution :

Net force applied in box is :

$F=285 - mgsin\ \theta - \mu mg cos \ \theta\\ \\F=285-mg( sin \ \theta - \mu cos\ \theta)\\\\F=285 - 20\times 10( \dfrac{1}{2}+0.72\times \dfrac{\sqrt{3}}{2})\\\\F=60.29\ N$

Acceleration , $a=\dfrac{F}{m}=\dfrac{60.29}{20}=3.01\ m/s^2$ .

By equation of motion :

$v=u+at\\\\v=0+3.01\times 4\\\\v=12.04\ m/s$

Therefore, the speed of box is 12.04 m/s.

Hence, this is the required solution.

6 0

3 years ago

A uniform rod rotates in a horizontal plane about a vertical axis through one end. The rod is 3.46 m long, weighs 12.8 N, and ro

Mkey [24]

Answer:

a. Rotational inertia: 5.21kgm²

b. Magnitude of it's angular momentum: 123.32kgm²/s

Explanation:

Length of the rod = 3.46m

Weight of the rod = 12.8 N

Angular velocity of the rod= 226 rev/min

a. Rotational Inertia (I) about its axis

The formula for rotational inertia =

I = (1/12×m×L²) + m × ( L ÷ 2)²

Where L = length of the rod

m = mass of the rod

Mass of the rod is calculated by dividing the weight of the rod with the acceleration due to gravity.

Acceleration due to gravity = 9.81m/s²

Mass of the rod = 12.8N/ 9.81m/s²

Mass of the rod = 1.305kg

Rotational Inertia =

(1/12× 1.305 × 3.46²)+ 1.305 ( 3.46÷2)²

Rotational Inertia = 1.3019115 + 3.9057345

Rotational Inertia = 5.207646kgm²

Approximately = 5.21kgm²

b. The magnitude of the rod's angular momentum about the rotational axis is calculated as

Rotational Inertia about its axis × angular speed of the rod.

Angular speed of the rod is calculated as= (Angular velocity of the rod × 2π)/60

= (226×2π) /60

= 23.67 rad/s

Rotational Inertia = 5.21kgm²

The magnitude of the rod's angular momentum about the rotational axis

= 5.21kgm²× 23.67 rad/s

= 123.3207kgm²/s

Approximately = 123.32kgm²/s

7 0

4 years ago

A ball of mass 4.5 kg moving with speed of 2.2 m/s in the +x-direction hits a wall and bounces back with the same speed in the x

Levart [38]

Answer:

The change of the momentum of the ball is $-19.8\, \frac{mkg}{s}$

Explanation:

We should find $\varDelta\overrightarrow{p}=\overrightarrow{p_{f}}-\overrightarrow{p_{i}}$ (1)with $\overrightarrow{p_{i}}$ the initial momentum and $\overrightarrow{p_{f}}$ the final momentum. Linear momentum is defined as $\overrightarrow{p}=m\overrightarrow{v}$ , using that on (1):

$\varDelta\overrightarrow{p}=m \overrightarrow{v_{f}}-m \overrightarrow{v_{i}}$ (2)

It's important to note that momentum and velocity are vectors and direction matters, so if +x direction is the direction towards the wall and the -x direction away the wall $\overrightarrow{v_{i}}=+2.2\, \frac{m}{s}$ and $\overrightarrow{v_{f}}=-2.2\, \frac{m}{s}$ so (2) becomes: