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Nastasia [14]
2 years ago
8

Zoe has 25 grams of water (c = 4.186 ) at 10°C, which she mixes with 12 grams of water at 30°C. Assume that no heat is lost to t

he surroundings. What is the final temperature of the two liquids? The answer is expressed to the nearest whole number.
Physics
2 answers:
Pepsi [2]2 years ago
6 0

T = 16.49 degrees Celsius

fenix001 [56]2 years ago
3 0
Since heat here is conserved that means that the heat out is equal to the heat in. We use the expression Q = mC(T2-T1). We caclulate as follows:

Q absorbed = Q released
m1 C (T-T1) = -m2 C (T-T1)

C can be cancelled since they are the same substance.

m1 (T-T1) = -m2 (T-T1)
25 (T-10) = -12 (T-30)
T = 16.49 degrees Celsius

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If you peel two strips of transparent tape off the same roll and immediately let them hang near each other, they will repel each
g100num [7]

Answer:

1.

Firstly removing off one strip and it leaves electrons behind, so the strip becomes positively charged.

2. The roll however is not negatively charged because it is "earthed " by the hand holding it, thus excess negatives repel each other away through the hand.

3.Tearing off the next strip and once more it leaves electrons behind, the new strip is also positively charged and will repel the first strip.

4. Then, tear two strips apart and one will leave electrons behind on the other. Meaning that one strip is positive and the other is negative and they will attract each other.

5 0
2 years ago
Cesium-137 undergoes beta decay and has a half-life of 30.0 years. How many beta particles are emitted by a 14.0-g sample of ces
Mandarinka [93]

Answer: 0.81\times 10^{16} beta particles

Explanation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}

Given mass = 14.0 g

Molar mass = 137 g/mol

\text{Number of moles of cesium}=\frac{14.0g}{137g/mol}=0.102moles

According to avogadro's law, 1 mole of every substance weighs equal to its molecular mass and contains avogadro's number 6.023\times 10^{23} of particles.

1 mole of cesium contains atoms =  6.023\times 10^{23}

0.102 moles of cesium contains atoms =  \frac{6.023\times 10^{23}}{1}\times 0.102=0.614\times 10^{23}

The relation of atoms with time for radioactivbe decay is:

N_t=N_0\times \frac{1}{2}^{\frac{t}{t_{\frac{1}{2}}}}

Where N_t =atoms left undecayed

N_0 = initial atoms

t = time taken for decay = 3 minutes

{t_{\frac{1}{2}}} = half life = 30.0 years = 1.577\times 10^7 minutes

The fraction that decays  :  1-(\frac{1}{2})^{\frac{3}{1.577\times 10^7}}=1.32\times 10^{-7}

Amount of particles that decay is  = 0.614\times 10^{23}\times 1.32\times 10^{-7}=0.81\times 10^{16}

Thus 0.81\times 10^{16} beta particles are emitted by a 14.0-g sample of cesium-137 in three minutes.

7 0
2 years ago
A crate slides down a ramp that makes a 20∘ angle with the ground. To keep the crate moving at a steady speed, Paige pushes back
VMariaS [17]

Answer:

Hence, work done= 287.54 J

Explanation:

Given data:

angle of ramp with the ground θ =20°

force applied = 76 N

work done on the crate to slide down 4 m down the ramp

W= F×d cosθ ( only the cos component of the force will slide the crate down)

W= 76×4×cos20= 287.54 J

4 0
2 years ago
The mass of Venus is 81.5% that of the earth, and its radius is 94.9% that of the earth. If a rock weighs 75.0 N on earth, compu
IgorLugansk [536]

Answer:

g = 0.905 gE

W  = 67.9 N

Explanation:

given data

mass of Venus mv =  81.5% = 0.815

radius Rv = 94.9% = 0.949

weighs W = 75.0 N

solution

we apply here acceleration due to gravity at earth surface that is

g = \frac{Gm}{R^2}   = 9.80 m/s²  ............1

so

g = \frac{G(0.815)}{0.949R^2}  

g = 0.905 gE

and

W = m gv

W = 0.905 m gE

W  = 0.905 × 75

W  = 67.9 N

7 0
2 years ago
An electron and a proton are held on an x axis, with the electron at x = + 1.000 m and the proton at x = - 1.000 m.how much work
aksik [14]
It is required an infinite work. The additional electron will never reach the origin.

In fact, assuming the additional electron is coming from the positive direction, as it approaches x=+1.00 m it will become closer and closer to the electron located at x=+1.00 m. However, the electrostatic force between the two electrons (which is repulsive) will become infinite when the second electron reaches x=+1.00 m, because the distance d between the two electrons is zero:
F=k_e  \frac{q_e q_e}{d^2}
So, in order for the additional electron to cross this point, it is required an infinite amount of work, which is impossible.
5 0
2 years ago
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