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Liula [17]
3 years ago
7

Which cools faster, land or water? Give a suitable reason for your answer.

Physics
1 answer:
9966 [12]3 years ago
7 0

Answer: Land cools faster than water

Reason: Water has high specific heat capacity, so it absorbs huge amount of heat energy to increase comparatively small amount of temperature. Loosing a large amount of heat requires more time, so <em>water takes more time to cool down. </em>

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A plane is heading south with a velocity of 150 kilometers/hour. It experiences a tailwind with a velocity of 20 kilometers/hour
Tatiana [17]
The tailwind speed would actually add up to the speed of the plane.

It would blow from behind the tail of the plane.

Resultant Velocity = 20 km/h + 150 km/h = 170 km/h

6 0
3 years ago
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In 1656, the Burgmeister (mayor) of the town of Magdeburg, Germany, Otto Von Guericke, carried out a dramatic demonstration of t
sleet_krkn [62]

Answer:

Force required to pull the two hemisphere = 46622.72N

Explanation:

Complete question ( Note: 1 millibar=100 N/m2. One atmosphere is 1013 millibar = 1.013×105 N/m2 ] 

The contact area between the hemispheres is (pi x 0.400^2) = 0.5024m^2.

Pressure difference = (940 - 12) = 928 millibars.

(928 x 100) = 92,800N/m^2.

(92800 x 0.5024) = 46622.72N. force required to part the hemispheres.

6 0
3 years ago
Read 2 more answers
At 80 degrees celsius the vapor pressure of benzene is 753 torr and that of toluene is 290 torr. What is the total pressure and
nexus9112 [7]

Answer:

The total pressure of the solution is 598.66 torr

while the vapor pressure of the components is vapor pressure of benzene is =502 torr and that of toluene = 96.6 tor

Explanation:

To solve this question, we need to look at the known variables, the unknown variables, and then we select the appropriate relations between the known variables and the unknown

Here we have the temperature of the gases t = 80°C

the pressures of benzene = 753 torr

the pressures of toluene = 290 torr

mole fraction of benzene in solution = 2/3

mole fraction of toluene in solution = 1/3

the unknown variables = composition of the solution  and the total pressure of the solution

From here given the known variables and the required variables, the appropriate relation bstween the known and unknown variables is Route's Law

Raoult's relates the vapor pressure of a mixture of gases to the mole fraction of solute gases introduced in the  gas solution. Raoult's Law is expressed by the formula: Psolution = Χsolvent1×Psolvent1 + Χsolvent2×Psolvent2+ ... .

where. Psolution = vapor pressure of the solution and

Χsolvent1 = mole fraction of solvent 1

Psolvent1 = vapor pressure of solvent1

Thus we have by Raoult's Law

P_{solution} = X_{benzene} × P_{benzene}+ X_{toluene}×P_{toluene}

which is =2/3×753 + 1/3×290 = 502 + 96.6 = 598.66 torr

and composition vapor pressure of benzene is 502 torr

while the composition vapor pressure of toluene is 96.6 torr

6 0
3 years ago
Once a falling object has reached a constant velocity, the object ___.
Rashid [163]

'a',  'b',  and  'c'  are all reasonable statements.

7 0
3 years ago
Read 2 more answers
A toy car having mass m = 1.10 kg collides inelastically with a toy train of mass M = 3.55 kg. Before the collision, the toy tra
kkurt [141]

Answer:

V_{ft}= 317 cm/s

ΔK = 2.45 J

Explanation:

a) Using the law of the conservation of the linear momentum:

P_i = P_f

Where:

P_i=M_cV_{ic} + M_tV_{it}

P_f = M_cV_{fc} + M_tV_{ft}

Now:

M_cV_{ic} + M_tV_{it} = M_cV_{fc} + M_tV_{ft}

Where M_c is the mass of the car, V_{ic} is the initial velocity of the car, M_t is the mass of train, V_{fc} is the final velocity of the car and V_{ft} is the final velocity of the train.

Replacing data:

(1.1 kg)(4.95 m/s) + (3.55 kg)(2.2 m/s) = (1.1 kg)(1.8 m/s) + (3.55 kg)V_{ft}

Solving for V_{ft}:

V_{ft}= 3.17 m/s

Changed to cm/s, we get:

V_{ft}= 3.17*100 = 317 cm/s

b) The kinetic energy K is calculated as:

K = \frac{1}{2}MV^2

where M is the mass and V is the velocity.

So, the initial K is:

K_i = \frac{1}{2}M_cV_{ic}^2+\frac{1}{2}M_tV_{it}^2

K_i = \frac{1}{2}(1.1)(4.95)^2+\frac{1}{2}(3.55)(2.2)^2

K_i = 22.06 J

And the final K is:

K_f = \frac{1}{2}M_cV_{fc}^2+\frac{1}{2}M_tV_{ft}^2

K_f = \frac{1}{2}(1.1)(1.8)^2+\frac{1}{2}(3.55)(3.17)^2

K_f = \frac{1}{2}(1.1)(1.8)^2+\frac{1}{2}(3.55)(3.17)^2

K_f = 19.61 J

Finally, the change in the total kinetic energy is:

ΔK = Kf - Ki = 22.06 - 19.61 = 2.45 J

4 0
3 years ago
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