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3 years ago

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A ship going from sea water to river water has to displace more water to (a) change the buoyant force (b) decrease the buoyant f

orce in the same buoyant force (d) none of these

1 answer:

grin007 [14]3 years ago

4 0

Answer:

Change, or increase, the buoyant force.

Explanation:

As the salt water of the sea is denser than the river's fresh water, the boat will rise when entering the sea from the river and sink (to increase the buoyant force) when entering the river from the sea.

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A tennis ball covers a distance of 12 meters in 0.4 seconds what is the velocity of the tennis ball?

Usimov [2.4K]

V=d/t
V=12/0.4
V=30 m/s

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What is the amount of thermal energy needed to make 5 kg of ice at - 10 °C to

agasfer [191]

Answer:

The amount of thermal energy needed is 15167500 joules.

Explanation:

By First Law of Thermodynamics, we see that amount of thermal energy ( $Q$ ), in joules, is equal to the change in internal energy. From statement we understand that change in internal energy consisting in two latent components ( $U_{l,ice}$ , $U_{l,steam}$ ), in joules, and two sensible component ( $U_{s,w}$ ), in joules, that is:

$Q = U_{l,ice} + U_{s, w} + U_{s,ice} + U_{l,steam}$ (1)

By definitions of Sensible and Latent Heat, we expanded the formula:

$Q = m\cdot (h_{f,w}+h_{v,w}+c_{ice}\cdot \Delta T_{ice}+c_{w}\cdot \Delta T_{w})$ (2)

Where:

$m$ - Mass, in kilograms.

$h_{f,w}$ - Latent heat of fussion of water, in joules per kilogram.

$h_{v,w}$ - Latent heat of vaporization of water, in joules per kilogram.

$c_{ice}$ - Specific heat of ice, in joules per kilogram per degree Celsius.

$c_{w}$ - Specific heat of water, in joules per kilogram per degree Celsius.

$\Delta T_{ice}$ - Change in temperature of ice, measured in degrees Celsius.

$\Delta T_{w}$ - Change in temperature of water, measured in degrees Celsius.

If we know that $m = 5\,kg$ , $h_{f,w} = 3.34\times 10^{5}\,\frac{J}{kg}$ , $h_{v,w} = 2.26\times 10^{6}\,\frac{J}{kg}$ , $c_{ice} = 2.090\times 10^{3}\,\frac{J}{kg\cdot ^{\circ}C}$ , $c_{w} = 4.186\times 10^{3}\,\frac{J}{kg\cdot ^{\circ}C}$ , $\Delta T_{ice} = 10\,^{\circ}C$ and $\Delta T_{w} = 100\,^{\circ}C$ , then the amount of thermal energy is:

$Q = 15167500\,J$

The amount of thermal energy needed is 15167500 joules.

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3 years ago

Se pune<br>Numerical problems<br>Convert 100ºC to Kelvin scale

IgorC [24]

Answer:

373 K

Explanation:

To convert from °C to kelvin, you add the value to 273

Therefore,

100°C = (100+273)K

= 373K

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4 years ago

Tarzan is running with a horizontal velocity, along level ground. While running, he encounters a 2.21 m vine of negligible mass,

Lena [83]

Answer:

a) $a_c = 1.09m/s^2$

b) T = 720.85N

Explanation:

With a balance of energy from the lowest point to its maximum height:

$m*g*L(1-cos\theta)-1/2*m*V_o^2=0$

Solving for $V_o^2$ :

$V_o^2=2*g*L*(1-cos\theta)$

$V_o^2=2.408$

Centripetal acceleration is:

$a_c = V_o^2/L$

$a_c = 2.408/2.21$

$a_c = 1.09m/s^2$

To calculate the tension of the rope, we make a sum of forces:

$T - m*g = m*a_c$

Solving for T:

$T =m*(g+a_c)$

T = 720.85N

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4 years ago