Water flow will be reduced in a pipe if the pipe is constricted. True False

TEA [102]

7kinetic energy is decreasing in B

4 0

2 years ago

An insulated thermos contains 106.0 cm3 of hot coffee at a temperature of 80.0 °C. You put in 11.0 g of ice cube at its melting

Andrei [34K]

Answer:

the final temperature is T f = 64.977 ° C≈ 65°C

Explanation:

Since the thermus is insulated, the heat absorbed by the ice is the heat released by the coffee. Thus:

Q coffee + Q ice = Q surroundings =0 (insulated)

We also know that the ice at its melting point , that is 0 °C ( assuming that the thermus is at atmospheric pressure= 1 atm , and has an insignificant amount of impurities ).

The heat released by coffee is sensible heat : Q = m * c * (T final - T initial)

The heat absorbed by ice is latent heat and sensible heat : Q = m * L + m * c * (T final - T initial)

therefore

m co * c co * (T fco - T ico) + m ice * L + m ice * c wat * (T fwa - T iwa) = 0

assuming specific heat capacity of coffee is approximately the one of water c co = c wa = 4.186 J/g°C and the density of coffee is the same as water

d co = dw = 1 gr/cm³

therefore m co = d co * V co = 1 gr / cm³ * 106 cm³ = 106 gr

m co * c wat * (T f - T ico) + m ice * L + m ice * c wat * (T f - T iwa) = 0

m co * c wat * T f+ m ice * c wat * T f = m ice * c wat * T iwa + m co * c wat * Tico -m ice * L

T f = (m ice * c wat * T iwa + m co * c wat * Tico -m ice * L ) /( m co * c wat * + m ice * c wat )

replacing values

T f = (11 g * 4.186 J/g°C * 0°C + 106 g * 4.186 J/g°C*80°C - 11 g * 334 J/gr) / ( 11 g * 4.186 J/g°C + 106 g * 4.186 J/g°C* ) = 64,977 ° C

T f = 64.977 ° C

7 0

3 years ago

) Force F = − + ( 8.00 N i 6.00 N j ) ( ) acts on a particle with position vector r = + (3.00 m i 4.00 m j ) ( ) . What are (a)

natali 33 [55]

To develop this problem it is necessary to apply the concepts related to the Cross Product of two vectors as well as to obtain the angle through the magnitude of the angles.

The vector product between the Force and the radius allows us to obtain the torque, in this way,

$\tau = \vec{F} \times \vec{r}$

$\tau = (8i+6j)\times(-3i+4j)$

$\tau = (8*4)(i\times j)+(6*-3)(j\times i)$

$\tau = 32k +18k$

$\tau = 50 k$

Therefore the torque on the particle about the origen is 50k

PART B) To find the angle between two vectors we apply the definition of the dot product based on the vector quantities, that is,

$cos\theta = \frac{r\cdot F}{|\vec{r}|*|\vec{F}|}$

$cos\theta = \frac{(8*-3)+(4*3)}{\sqrt{(-3)^2+4^2}*\sqrt{8^2+6^2}}$

$cos\theta = -0.24$

$\theta = cos^{-1} (-0.24)$

$\theta = 103.88\°$

Therefore the angle between the ratio and the force is 103.88°

5 0

3 years ago

One end of a string 4.32 m long is moved up and down with simple harmonic motion at a frequency of 75 Hz . The waves reach the o

max2010maxim [7]

To solve this problem, we will apply the concepts related to the kinematic equations of linear motion, which define speed as the distance traveled per unit of time. Subsequently, the wavelength is defined as the speed of a body at the rate of change of its frequency. Our values are given as,

$\text{Length of the string} = L = 4.32 m$