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aleksandrvk [35]
3 years ago
11

An insulated beaker with negligible mass contains 0.250 kg of water at 75.0C. How many kilograms of ice at -20.0C must be droppe

d into the water to make the final temperature of the system 40.0C?
Physics
1 answer:
kkurt [141]3 years ago
3 0

Answer:

The amount of kilograms of ice at -20.0°C that must be dropped into the water to make the final temperature of the system 40.0°C = 0.0674 kg

Explanation:

Heat gained by ice in taking the total temperature to 40°C = Heat lost by the water

Total Heat gained by ice = Heat used by ice to move from -20°C to 0°C + Heat used to melt at 0°C + Heat used to reach 40°C from 0°C

To do this, we require the specific heat capacity of ice, latent heat of ice and the specific heat capacity of water. All will be obtained from literature.

Specific heat capacity of ice = Cᵢ = 2108 J/kg.°C

Latent heat of ice = L = 334000 J/kg

Specific heat capacity of water = C = 4186 J/kg.°C

Heat gained by ice in taking the total temperature to 40°C = mCᵢ ΔT + mL + mC ΔT = m(2108)(0 - (-20)) + m(334000) + m(4186)(40 - 0) = 42160m + 334000m + 167440m = 543600 m

Heat lost by water = mC ΔT = 0.25 (4186)(75 - 40) = 36627.5 J

543600 m = 36627.5

m = 0.0674 kg = 67.4 g of ice.

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Marina86 [1]

Answer:

F=-88Ib

Explanation:

From the question we are told that:

Force P=88Ib

Mass of crate M_c=210Ib

Generally the equation for Frictional force F is mathematically given by

Friction\ force (f) = friction\ coefficient\ (u) * Normal\ reaction (N)

F=u*N

with \mu =0.47

F=98.7Ib

Therefore since Static Friction supersedes applied force body remains at rest.

Frictional force =88Ib (negative)

F=-88Ib

5 0
3 years ago
Calculate the momentum of a 4 kg Bolling ball being thrown at a speed of 3 m/s
bonufazy [111]

We are given –

  • Mass of boiling ball is, m = 4 kg
  • Speed is, v = 3 m/s
  • Momentum, P =?

As we know –

↠Momentum = Mass × Speed(Velocity)

↠Momentum = 4 × 3 kgm/s

↠Momentum = 12 kgm/s

  • Henceforth,Momentum will be 12 kgm/s.
7 0
2 years ago
A person travels by car from one city to an-
Mademuasel [1]
<h2>Answer:</h2>

<h2>Explanation:</h2>

First, let's refer to the distance formula:

d=v*t, where d is distance, v is velocity or speed and t is time.

Now, let's find the distance covered by each individual speed that the car had:

<h3>1. Speed 1.</h3>

In order to use the formula, we need to convert minutes into hours since the speed is given in km/h.

21.1 min/60= 0.35 h.

Now, apply the distance formula.

d=(0.35h)*(86.8km/h)= 30.38 km.

<h3>2. Speed 2.</h3>

Convert minutes to hours again and do the same calculations.

10.6min/60=0.18h

d=(0.18h)*(106km/h)= 19.08 km.

<h3>3. Speed 3.</h3>

36.5min/60= 0.61h

d=(0.61h)*(30.9km/h)= 18.85 km.

<h3>4. Obtain the total distance.</h3>

The total distance must be given by the addition of all individual distances traveled by the car on each speed:

Total distance= 30.38 km + 19.08 km + 18.85 km= 68.31 km.

5 0
1 year ago
Calculate V1 and V2 ( V = voltage )
Kitty [74]

FORMULA:

  • V = IR, where V = P.D; I = Current; R = Resistance.

ANSWER:

Total equivalent resistance for circuit:

R(eq) = R1 + R2 [It is in series]

  • 330Ω + 470Ω
  • 800Ω

Now, Current passing through whole circuit:

I = V/R

  • 16/800
  • 1/50 ampere

We know that, In series combination current passing through whole circuit is same.

So, V¹ = IR¹

V¹ = 1/50 × 330

  • V¹ = 6.6 volt

And V² = IR²

V² = 1/50 × 470

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7 0
3 years ago
Read 2 more answers
A 160cm long string has two adjacent resonance at 85hz frequencies . calculate : 1- the fundamental frequency , 2- the speed of
lara [203]

Length=l=160cm=1.6m

Frequency=85Hz

#1

Fundamental frequency be v

\\ \sf\bull\longmapsto v=\dfrac{\nu}{2\ell}

\\ \sf\bull\longmapsto v=\dfrac{85}{2(1.6)}

\\ \sf\bull\longmapsto v=\dfrac{85}{3.2}

\\ \sf\bull\longmapsto v=26.6Hz

#2

First we have to find wavelength

\\ \sf\bull\longmapsto \lambda=\dfrac{c}{v}

\\ \sf\bull\longmapsto \lambda=\dfrac{3\times 10^8ms^{-1}}{26.6}

\\ \sf\bull\longmapsto \lambda=0.112\times 10^8m

\\ \sf\bull\longmapsto \lambda=112\times 10^5m

\\ \sf\bull\longmapsto \lambda=1.12\times 10^6m

Now..

Velocity be v

\\ \sf\bull\longmapsto v=f\lambda

  • f is frequency

\\ \sf\bull\longmapsto v=26.5\times1.12\times 10^6m

\\ \sf\bull\longmapsto v=29.68\times 10^6

\\ \sf\bull\longmapsto v=2.9\times 10^7m/s

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3 years ago
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