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8090 [49]
3 years ago
9

Please help have due very soon?thank you

Physics
1 answer:
Andru [333]3 years ago
8 0
The answer is :78 I think
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180 cm3 of hot tea at 97 °C are poured into a very thin paper cup with 20 g of crushed ice at 0 °C. Calculate the final temperat
Zolol [24]

Answer : The final temperature of the mixture is 91.9^oC

Explanation :

First we have to calculate the mass of water.

Mass = Density × Volume

Density of water = 1.00 g/mL

Mass = 1.00 g/mL × 180 cm³ = 180 g

In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.

q_1=-q_2

m_1\times c_1\times (T_f-T_1)=-m_2\times c_2\times (T_f-T_2)

where,

c_1 = specific heat of hot water (liquid) = 4.18J/g^oC

c_2 = specific heat of ice (solid)= 2.10J/g^oC

m_1 = mass of hot water = 180 g

m_2 = mass of ice = 20 g

T_f = final temperature of mixture = ?

T_1 = initial temperature of hot water = 97^oC

T_2 = initial temperature of ice = 0^oC

Now put all the given values in the above formula, we get

(180g)\times (4.18J/g^oC)\times (T_f-97)^oC=-(20g)\times 2.10J/g^oC\times (T_f-0)^oC

T_f=91.9^oC

Therefore, the final temperature of the mixture is 91.9^oC

8 0
3 years ago
Three ropes A, B and C are tied together in one single knot K. (See figure.)
masha68 [24]

The tension in the rope B is determined as 10.9 N.

<h3>Vertical angle of cable B</h3>

tanθ = (6 - 4)/(5 - 0)

tan θ = (2)/(5)

tan θ = 0.4

θ = arc tan(0.4) = 21.8 ⁰

<h3>Angle between B and C</h3>

θ = 21.8 ⁰ + 21.8 ⁰ = 43.6⁰

Apply cosine rule to determine the tension in rope B;

A² = B² + C² - 2BC(cos A)

B = C

A² = B² + B² - (2B²)(cos A)

A² = 2B² - 2B²(cos 43.6)

A² = 0.55B²

B² = A²/0.55

B² = 65.3/0.55

B² = 118.73

B = √(118.73)

B = 10.9 N

Thus, the tension in the rope B is determined as 10.9 N.

Learn more about tension here: brainly.com/question/24994188

#SPJ1

6 0
2 years ago
Imagine that the earth and the-moon have positive charges of the same magnitud. How big isäºthe charge necesary to produce an el
lions [1.4K]

Answer:

5.7 x 10^12 C

Explanation:

Let the charge on earth and moon is q.

mass of earth, Me = 5.972 x 10^24 kg

mass of moon, Mm = 7.35 x 10^22 kg

Let d be the distance between earth and moon.

the gravitational force between them is

F_{g}=G\frac{M_{e} \times M_{m}}{d^{2}}

The electrostatic force between them is

F_{e}=\frac{Kq^{2}}{d^{2}}

According to the question

1 % of Fg = Fe

0.01 \times 6.67\times10^{-11}\frac{5.97 \times 10^{24}\times7.35 \times 10^{22}}{d^{2}}=9 \times 10^{9}\frac{q^{2}}{d^{2}}

2.927 \times 10^{35}=9 \times10^{9}q^{2}

3.25 \times 10^{25}=q^{2}

q = 5.7 x 10^12 C

Thus, the charge on earth and the moon is 5.7 x 10^12 C.

6 0
3 years ago
How much will a 200kg hippo accelerate if you push it with a force of 800 N?
Fantom [35]

Answer:

Explanation:

force = f, mass = m, acceleration = a

f = m a

m = 200 kg

f = 800 N

f = m a

800 = 200a

a = 800 / 200

<u><em>a = 4</em></u>

Hope this helps

plz mark as brainliest!!!!!!!

4 0
3 years ago
1. A sample of gas has a constant temperature and number of particles. As the volume of the gas sample is increased, the pressur
AURORKA [14]

Answer:

1.C

2.C

3.C

Explanation:

hope its help hehe:(

4 0
3 years ago
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