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Luden [163]
3 years ago
7

Physical therapists know as you soak tired muscles in a hot tub, the water will cool down as you heat up. If a 67.9 kg person at

37.1 oC immerses in 50.2 kg of water at 40.5o C, the equilibrium temperature is 38.7 oC.
What is the specific heat of the person?
Use 4.186 kJ/kgoC for the specific heat of water.
Physics
1 answer:
Lubov Fominskaja [6]3 years ago
4 0

Answer:

c = 5032.8J/kg°C

Explanation:

the transfer of heat is given by

Q_1=-Q_2\\\\Q=cm(T_f-T_i)

c is the specific heat, m the mass and T the temperatures at equilibrium an at the beginning.

By replacing we obtain

c_1m_1(T_f-T_{1i})=-c_2m_2(T_f-T_{2i})\\\\c_2=-\frac{c_1m_1(T_f-T_{1i})}{m_2(T_f-T_{2i})}=-\frac{(67.9kg)(4186J/kg\°C)(38.7\°C-37.1\°C)}{(50.2kg)(38.7\°C-40.5\°C)}\\\\c_2=5032.8\frac{J}{kg\°C}

hope this helps!!

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If part of an electric circuit dissipates energy at 6 W when it draws a current of 3 A, what voltage is impressed across it?
Lena [83]

Answer:2 volts

Explanation:

Power=current x voltage

6=3 x voltage

Divide both sides by 3

6/3=(3 x voltage)/3

2=voltage

Voltage=2volts

5 0
3 years ago
Part C: Quantitative Problems when vf is not 0
Alina [70]

Answer:

(a)

\triangle v=-8\ m/s\\\triangle mv=-56\ kg.m/s

(b)

1120 N

Explanation:

Change in velocity, \triangle v is given by subtracting the initial velocity from the final velocity and expressed as \triangle v= v_f -v_i

Where v represent the velocity and subscripts f and i represent final and initial respectively. Since the ball finally comes to rest, its final velocity is zero. Substituting 0 for final velocity and the given figure of 8 m/s for initial velocity then the change in velocity is given by

\triangle v=0-8=-8\ m/s

To find m\triangle v then we substitute 7 kg for m and -8 m/s for \triangle v therefore \triangle\ v=7 Kg\times -8 m/s=-56\ Kg.m/s

(b)

The impact force, F is given as the product of mass and acceleration. Here, acceleration is given by dividing the change in velocity by time ie

a=\frac {\triangle v}{t}=\frac { v_f -v_i}{t}

Substituting t with 0.05 s then a=\frac {\triangle v}{t}=\frac { v_f -v_i}{t}=\frac {-8}{0.05}=-160 m/s^{2}

Since F=ma then substituting m with 7 Kg we get that F=7*-160=-1120 N

Therefore, the impact force is equivalent to 1120 N

3 0
4 years ago
Do all objects have inertia <br><br> a) true<br> b) false
Natasha2012 [34]

Answer:

The answer is A. Which is true

4 0
3 years ago
Interactive Solution 6.39 presents a model for solving this problem. A slingshot fires a pebble from the top of a building at a
mariarad [96]

(a) 29.8 m/s

To solve this problem, we start by analyze the vertical motion first. This is a free fall motion, so we can use the following suvat equation:

v_y^2 - u_y^2 = 2as

where, taking upward as positive direction:

v_y is the final vertical velocity

u_y = 0 is the initial vertical velocity (zero because the pebble is launched horizontally)

a=g=-9.8 m/s^2 is the acceleration of gravity

s = -25.0 m is the displacement

Solving for vy,

v_y = \sqrt{u^2+2as}=\sqrt{0+2(-9.8)(-25)}=-22.1 m/s (downward, so we take the negative solution)

The pebble also have a horizontal component of the velocity, which remains constant during the whole motion, so it is

v_x = 20.0 m/s

So, the final speed of the pebble as it strikes the ground is

v=\sqrt{v_x^2+v_y^2}=\sqrt{20.0^2+(-22.1)^2}=29.8 m/s

(b) 29.8 m/s

In this case, the pebble is launched straight up, so its initial vertical velocity is

u_y = 20.0 m/s

So we can find the final vertical velocity using the same suvat equation as before:

v_y^2 - u_y^2 = 2as

v_y = \sqrt{u^2+2as}=\sqrt{(20.0)^2+2(-9.8)(-25)}=-29.8 m/s (downward, so we take the negative solution)

The horizontal speed instead is zero, since the pebble is initially launched vertically, so the final speed is just equal to the magnitude of the vertical velocity:

v = 29.8 m/s

(c) 29.8 m/s

This case is similarly to the previous one: the only difference here is that the pebble is launched straight down instead than up, therefore

u_y = -20.0 m/s

Using again the same suvat equation:

v_y^2 - u_y^2 = 2as

v_y = \sqrt{u^2+2as}=\sqrt{(-20.0)^2+2(-9.8)(-25)}=-29.8 m/s (downward, so we take the negative solution)

As before, the horizontal speed instead is zero, since the pebble is initially launched vertically, so the final speed is just equal to the magnitude of the vertical velocity:

v = 29.8 m/s

We notice that the final value of the speed is always the same in all the three parts, so it does not depend on the direction of launching. This is due to the law of conservation of energy: in fact, the initial mechanical energy of the pebble (kinetic+potential) is the same in all three cases (because the height h does not change, and the speed v does not change either), and the kinetic energy gained during the fall is also the same (since the pebble falls the same distance in all 3 cases), therefore the final speed must also be the same.

7 0
3 years ago
Mrs. Smith can walk 1.4 m/s. If it takes her 8.5 seconds to get to the teacher lounge, how far is the teacher lounge from her ro
AlexFokin [52]

Answer:

6 meters away

Explanation:

6*1.4= 8.4 which is pretty close

5 0
3 years ago
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