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Papessa [141]
3 years ago
7

The highest waterfall in the world is the Salto Angel in Venezuela. Its longest single falls has a height of 807 m. If water at

the top of the falls is at 17.3°C, what is the maximum temperature of the water at the bottom of the falls? Assume all the kinetic energy of the water as it reaches the bottom goes into raising its temperature.
Physics
1 answer:
valentina_108 [34]3 years ago
5 0

Answer:

Temperature at the bottom will be 19.18°C

Explanation:

We have given height h = 807 m

Temperature at the top =17.3^{\circ}C

Specific heat of water c = 4200 J/kg/^{\circ}C

From energy conservation

Kinetic energy at the bottom = potential energy at the top

So mc\Delta T=mgh

\Delta T=\frac{gh}{c}=\frac{9.8\times 807}{4200}=1.88^{\circ}C

So temperature at the bottom = 17.3+1.88 = 19.18°C

   

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A 1500 kg car traveling at 15.0 m/s to the south collides with a 4500 kg truck that is at rest at a stopligt. The car comes to a
Burka [1]

Answer:

5 m/s, moving to the South.

Explanation:

Parameters given:

Mass of car, m = 1500 kg

Initial velocity of car, u = 15 m/s

Mass of truck, M = 4500 kg

Initial velocity of truck, v = 0 m/s (Truck is at rest)

Final velocity of car, U = 0 m/s (Car comes to a stop)

Final velocity of truck = V

Because the collision is elastic, we can apply the principle of conservation of momentum, we have that:

Total initial momentum = Total final momentum

m*u + M*U = m*v + M*V

(1500 * 15) + (4500 * 0) = (1500 * 0) + (4500 * V)

22500 + 0 = 0 + 00V

=> V = 22500/4500

V = 5 m/s

The velocity carries a positive sign, hence, it's moving in the same direction as the car was moving initially.

That is, it's moving to the South.

8 0
3 years ago
Read 2 more answers
A parcel of land is in the shape of an isosceles triangle. The base has a length of 425 ft.; the other sides, which are of equal
kogti [31]

Answer:

The answer to your question is 636.6 ft    

Explanation:

Data

base = 425 ft

angle = 39°

See the picture below

1.- Divide the triangle to get two right triangles.

    Now the superior angle will measure 19.5° and the opposite side will measure 212.5 ft

2.- Use the trigonometric function sine to find the hypotenuse

     sin 19.5 = 212.5/hyp

solve for hyp

    hyp = 212.5 / sin 19.5

Result

    hyp = 212.5/ 0.333

    hyp = 636.6 ft    

4 0
3 years ago
Rank the ten objects from loudest to softest.
frez [133]
Um what are the ten objects..?
5 0
3 years ago
Momentum is a vector quantity because it includes velocity, which is also a
zvonat [6]

Answer: A; True

Explanation: Momentum is known to be a vector quality, and thus has been proven by modern scientists and resulting in this answer being true.

Hope this helps <3

Stay safe, stay warm

-Carrie

Ps. it would mean a lot if you marked brainliest (=

7 0
3 years ago
A long, thin rod parallel to the y-axis is located at x = - 1 cm and carries a uniform positive charge density λ = 1 nC/m . A se
zheka24 [161]

Answer:

The electric field at origin is 3600 N/C

Solution:

As per the question:

Charge density of rod 1, \lambda = 1\ nC = 1\times 10^{- 9}\ C

Charge density of rod 2, \lambda = - 1\ nC = - 1\times 10^{- 9}\ C

Now,

To calculate the electric field at origin:

We know that the electric field due to a long rod is given by:

\vec{E} = \frac{\lambda }{2\pi \epsilon_{o}{R}

Also,

\vec{E} = \frac{2K\lambda }{R}                  (1)

where

K = electrostatic constant = \frac{1}{4\pi \epsilon_{o} R}

R = Distance

\lambda = linear charge density

Now,

In case, the charge is positive, the electric field is away from the rod and towards it if the charge is negative.

At x = - 1 cm = - 0.01 m:

Using eqn (1):

\vec{E} = \frac{2\times 9\times 10^{9}\times 1\times 10^{- 9}}{0.01} = 1800\ N/C

\vec{E} = 1800\ N/C     (towards)

Now, at x = 1 cm = 0.01 m :

Using eqn (1):

\vec{E'} = \frac{2\times 9\times 10^{9}\times - 1\times 10^{- 9}}{0.01} = - 1800\ N/C

\vec{E'} = 1800\ N/C     (towards)

Now, the total field at the origin is the sum of both the fields:

\vec{E_{net}} = 1800 + 1800 = 3600\ N/C

7 0
3 years ago
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