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4 years ago

Why is the specific heat of salt solution lower than that of water?

1 answer:

3 0

When we dissolve NaCl in water, the ions are held in a rigid cage of water molecules. It takes less energy to activate these molecules, so the specific heat of the water decreases. The greater concentration of NaCl, the lower the specific heat capacity of the solution.

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A teacup on a spinning teacup ride takes 1.5 s to complete a revolution around the center of the platform. What is the centripet

arlik [135]

The correct answer is 53 meters per second squared

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3 years ago

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What is Circular Motion?

nevsk [136]

Circular motion is what an object has if it is moving around and around and around and around and around and around and around and around and around in a path that is a circle.

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4 years ago

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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

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3 years ago

On a 10 kg cart (shown below), the cart is brought up to speed with 50N of force for 7m, horizontally. At this point (A), the ca

pav-90 [236]

Answer: Letter B! Is your answer

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3 years ago

1.)Two objects, one of m=20,000 kg, and another of 12,500 kg, are placed at a distance of 5 meters apart. What is the force of g

Delvig [45]

1) $6.67\cdot 10^{-4} N$

The force of gravitation between the two objects is given by:

$F=G\frac{m_1 m_2}{r^2}$

where

$G=6.67\cdot 10^{-11} kg^{-1} m^{3} s^{-2}$ is the gravitational constant

m1 = 20,000 kg is the mass of the first object

m2 = 12,500 kg is the mass of the second object

r = 5 m is the distance between the two objects

Substituting the numbers inside the equation, we find

$F=(6.67\cdot 10^{-11})\frac{(20,000 kg)(12,500 kg)}{(5 m)^2}=6.67\cdot 10^{-4} N$

2) $2.7\cdot 10^{-3} N$

From the formula in exercise 1), we see that the force is inversely proportional to the square of the distance:

$F \sim \frac{1}{r^2}$

this means that if we cut in a half the distance without changing the masses, the magnitude of the forces changes by a factor

$F'\sim \frac{1}{(r/2)^2}=4 \frac{1}{r^2}=4F$

So, the gravitational force increases by a factor 4. Therefore, the new force will be

$F' = 4 F=4(6.67\cdot 10^{-4} N)=2.7\cdot 10^{-3} N$

3) 12.5 Nm

The torque is equal to the product between the magnitude of the perpendicular force and the distance between the point of application of the force and the centre of rotation:

$\tau=Fd$