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WINSTONCH [101]
2 years ago
15

A light string is wrapped around the edge of the smaller disk, and a 1.50 kgkg block is suspended from the free end of the strin

g. If the block is released from rest at a distance of 2.40 mm above the floor, what is its speed just before it strikes the floor
Physics
1 answer:
Mandarinka [93]2 years ago
6 0

Answer:

The speed of the block before it strikes the floor is 0.217 m/s.

Explanation:

Given;

mass of the block, m = 1.5 kg

height above the ground through the block was released, h = 2.4 mm = 2.4 x 10⁻³ m

The speed of the block before it strikes the floor will be maximum.

Let the speed of the block before it strikes the floor = v

Apply the principle of conservation of mechanical energy to determine the speed of the block.

K.E = P.E

¹/₂mv² = mgh

¹/₂v² = gh

v² = 2gh

v = √2gh

v = √(2 x 9.8 x 2.4 x 10⁻³)

v = 0.217 m/s

Therefore, the speed of the block before it strikes the floor is 0.217 m/s.

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A hot-water bottle contains 787 g of water at 75∘C. If the liquid water cools to body temperature (37 ∘C), how many kilojoules o
IgorC [24]

Answer:

Q = 787 gr * 1 \frac{cal}{gr C} *(37-75)C = -29906 cal

So then the answer for this case would be 29906 cal but we need to convert this into KJ and we know that 1 cal = 4.184 J and if we convert we got:

29906 cal *\frac{4.184 J}{1 cal}* \frac{1KJ}{1000 J}= 125.127 KJ

Explanation:

For this case we know the mass of the water given :

m = 787 gr

And we know that the initial temperature for this water is T_i =75 C.

We want to cool this water to the human body temperature T_f = 37 C

Since the temperatures given are not near to 0C (fusion point) or 100C (the boling point) we don't need to use latent heat, then the only heat involved for this case is the sensible heat given by:

Q= m c_p \Delta T

Where c_p represent the specific heat for the water and this value from tables we know that c_p =1 \frac{cal}{gr C} for the water.

So then we have everything in order to replace into the formula of sensible heat and we got:

Q = 787 gr * 1 \frac{cal}{gr C} *(37-75)C = -29906 cal

So then the answer for this case would be 29906 cal but we need to convert this into KJ and we know that 1 cal = 4.184 J and if we convert we got:

29906 cal *\frac{4.184 J}{1 cal}* \frac{1KJ}{1000 J}= 125.127 KJ

8 0
3 years ago
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5 0
3 years ago
What is the energy level of hydrogen?
barxatty [35]

Answer:

The energies corresponding to each of the allowed orbitals are called energy levels.

Explanation:

A scientist known as Niels Bohr put forward that electrons in an atom covers some permitted orbitals with a specific energy. In other words, the energy of an electron in an atom is not continuous, but 'quantized.' The energies corresponding to each of the allowed orbitals are called energy levels.

E = -\frac{E_0}{n^2} \\where \\E_0 = 13.6 eV (1 eV = 1.602\times 10^{-19}Joules)\\and\ n = 1,2,3...

8 0
3 years ago
A 300 gg ball on a 70-cmcm-long string is swung in a vertical circle about a point 200 cmcm above the floor. The string suddenly
Ratling [72]

Answer:

the   tension in the string an instant before it broke = 34 N

Explanation:

Given that :

mass of the ball m = 300 g = 0.300 kg

length of the string r = 70 cm = 0.7 m

At highest point, law of conservation of energy can be expressed as :

\frac{1}{2} mv^2 = mgh\\\\v = \sqrt{2gh}\\\\v = \sqrt{2*(9.8 \  m/s^2)*(6.00 \ m - 2.00 \ m)}\\\\

v = 8.854 \ m/s

The tension in the string is:

T = \frac{mv^2}{r}\\\\T = \frac{(0.300 \ kg)*(8.854 \ m/s^2)}{0.70 \ m}\\\\T = 33.59 N\\\\T = 34 \ N

Thus, the   tension in the string an instant before it broke = 34 N

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2 years ago
The mass of an electron is...
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<span>The mass of an electron is not significant to the overall mass of the atom. (B)

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3 years ago
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