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2 years ago

How much heat is required to raise the temperature of 0.25 kg of water from 20°C to 30°C

Physics

1 answer:

pentagon [3]2 years ago

3 0

Answer:

10500 J/kg/*C

Explanation:

Quantity of heat required=mass of substance x specific heat capacity x change in temperature

Quantity of heat required=0.25 x 4200 x [30-20]

Quantity of heat required=0.25 x 4200 x 10

Quantity of heat required=10500 J/kg/*C

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What force must the deltoid muscle provide to keep the arm in this position?

ruslelena [56]

Answer:

Deltoid Force, $F_{d} = \frac {r_{a}mgsin\alpha_{a}}{r_{d}sin\alpha_{d}}$

Additional Information:

Some numerical information are missing from the question. However, I will derive the formula to calculate the force of the deltoid muscle. All you need to do is insert the necessary information and calculate.

Explanation:

The deltoid muscle is the one keeping the hand arm in position. We have two torques that apply to the rotating of the arm.

1. The torque about the point in the shoulder for the deltoid muscle, $T_{Deltoid}$

2. The torque of the arm, $T_{arm}$

Assuming the arm is just being stretched and there is no rotation going on,

$T_{Deltoid}$ = 0

$T_{arm}$ = 0

⇒ $T_{Deltoid} = T_{arm}$

$r_{d}F_{d}sin\alpha_{d} = r_{a}F_{a}sin\alpha_{a}$

Where,

$r_{d}$ is radius of the deltoid

$F_{d}$ is the force of the deltiod

$\alpha_{d}$ is the angle of the deltiod

$r_{a}$ is the radius of the arm

$F_{a}$ is the force of the arm , $F_{a} = mg$ which is the mass of the arm and acceleration due to gravity

$\alpha_{a}$ is the angle of the arm

The force of the deltoid muscle is,

$F_{d} = \frac {r_{a}F_{a}sin\alpha_{a}}{r_{d}sin\alpha_{d}}$

but $F_{a} = mg$ ,

∴ $F_{d} = \frac {r_{a}mgsin\alpha_{a}}{r_{d}sin\alpha_{d}}$

7 0

2 years ago

A fluid in an aquifer is 23.6 m above a reference datum, the fluid pressure (in gage pressure) is 4390 n/m2 and the flow velocit

Phoenix [80]

As per Bernuolli's Theorem total energy per unit mass is given as

$\frac{P}{\rho} + \frac{1}{2}v^2 + gH = E$

now from above equation

$P = 4390 N/m^2$

$\rho = 0.999 * 10^3$

$v = 7.22 * 10^{-4} m/s$

$H = 23.6 m$

now by above equation

$\frac{4390}{0.999*10^3} + \frac{1}{2}*(7.22*10^{-4})^2 + 9.8*23.6 = E$

$E = 235.7 J/kg$

Part B)

Now energy per unit weight

$U = \frac{E}{g}$

$U = \frac{235.7}{9.8}$

$U = 24 m$

7 0

3 years ago

When can a high speed velocity cause damage?'

sweet-ann [11.9K]

Answer:

50 Mph.

Explanation:

According to the National Severe Storms Laboratory, winds can really begin to cause damage when they reach <em><u>50 mph</u></em>. But here’s what happens before and after they reach that threshold, according to the Beaufort Wind Scale (showing estimated wind speeds): - at 19 to 24 mph, smaller trees begin to sway.

7 0

2 years ago

Water flows from a large drainage pipe at a rate of 950 gal/min. What is this volume rate of flow in (a) m3/s , (b) liters/min,

Paladinen [302]

Answer:

$0.05997\ m^3/s$