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victus00 [196]
2 years ago
13

Blood can carry excess energy from the interior to the surface of the body, where the energy is dispersed in a number of ways. W

hile a person is exercising, 0.6 kg of blood flows to the body’s surface and releases 2000 J of energy. The blood arriving at the surface has the temperature of the body’s interior, 37.0 °C. Assuming that blood has the same specific heat capacity as water, determine the temperature of the blood that leaves the surface and returns to the interior.
Physics
1 answer:
stepladder [879]2 years ago
4 0

Answer:

T = 36.2 ^oC

Explanation:

As we know that heat released is given as

Q = ms \Delta T

here we know that

Q = 2000 J

m = 0.6 kg

s = 4186 J/kg C

now from above equation we have

2000 = 0.6(4186)(37 - T)

2000 = 2511.6 (37 - T)

T = 36.2 ^oC

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If you double the mass of an object while leaving the net force unchanged what is the result
valentinak56 [21]

Answer: If the net force acting on an object doubles, its acceleration is doubled. If the mass is doubled, then acceleration will be halved. If both the net force and the mass are doubled, the acceleration will be unchanged.

Explanation:

5 0
2 years ago
Which one of the following statements is true concerning an object executing simple harmonic motion?
timurjin [86]

Answer:

D) The objects velocity is zero when its acceleration is a maximum

Explanation:

In a simple harmonic motion, the total energy is constant (if we neglect air resistance and friction), and it is equal to the sum of the elastic potential energy U and the kinetic energy K:

E=K+U=\frac{1}{2}mv^2+\frac{1}{2}kx^2 (1)

where

m is the mass

v is the velocity

k is the spring constant

x is the displacement

As a consequence, since E must remain constant, when K increases U decreases, and vice-versa.

Also, in a simple harmonic motion the acceleration of the system is proportional to the negative of the displacement:

a\propto - x (2)

So, combining (1) with (2), we have the following situations:

- When the displacement is zero (x=0), the acceleration is also zero (a=0), and so the velocity is maximum, because the kinetic energy is maximum

- When the displacement is maximum (x=max), the acceleration is also maximum, while the velocity is zero because the kinetic energy is zero

So, the correct statement is

D) The objects velocity is zero when its acceleration is a maximum

4 0
2 years ago
An experiment based at New Mexico’s Apache Point observatory uses a laser beam to measure the distance to the Moon with millimet
Anna007 [38]

Answer:

Sorry but i dont know

Explanation:

6 0
2 years ago
One component of a metal sculpture consists of a solid cube with an edge of length 38.9 cm. The alloy used to make the cube has
vovangra [49]

Answer:

The mass of the cube is 420.8 kg.

Explanation:

Given that,

Length of edge = 38.9 cm

Density \rho= 7.15 \times10^{3}\ kg/m^3

We need to calculate the volume of cube

Using formula of volume

V = 38.9^3

V=0.058863\ m^3

We need to calculate the mass of the cube

Using formula of density

\rho = \dfrac{m}{V}

m = V\times\rho

m =0.058863\times7.15 \times10^{3}

m=420.8\ kg

Hence, The mass of the cube is 420.8 kg.

7 0
2 years ago
An object is released from rest near and above Earth’s surface from a distance of 10m. After applying the appropriate kinematic
Damm [24]

Answer:

v_y = 12.54 m/s

Explanation:

Given:

- Initial vertical distance y_o = 10 m

- Initial velocity v_y,o  = 0 m/s

- The acceleration of object in air = a_y

- The actual time taken to reach ground t = 3.2 s

Find:

- Determine the actual speed of the object when it reaches the ground?

Solution:

- Use kinematic equation of motion to compute true value for acceleration of the ball as it reaches the ground:

                             y = y_o + v_y,o*t + 0.5*a_y*t^2

                             0 = 10 + 0 + 0.5*a_y*(3.2)^2

                             a_y = - 20 / (3.2)^2 = 1.953125 m/s^2

- Use the principle of conservation of total energy of system:

                             E_p - W_f = E_k

Where,                  E_p = m*g*y_o

                             W_f = m*a_y*(y_i - y_f)      ..... Effects of air resistance

                             E_k = 0.5*m*v_y^2

Hence,                  m*g*y_o - m*a_y*(y_i - y_f) = 0.5*m*v_y^2

                             g*(10) - (1.953125)*(10) = 0.5*v_y^2

                             v_y = sqrt (157.1375)

                            v_y = 12.54 m/s

4 0
2 years ago
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