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Likurg_2 [28]
2 years ago
5

A car traveling at 100km/hr .how many hours will it take to cover a distance of 750 km

Physics
2 answers:
grin007 [14]2 years ago
4 0

Answer:

7.5 right?

Explanation: if im wrong shoot me

DedPeter [7]2 years ago
4 0

Answer:

7.5 hours( 7 hours 30 minutes)

Explanation:

Speed = distance/time

time = distance/speed

750 km divided by 100 km/hr = 7.5 hours

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6 0
3 years ago
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3 0
2 years ago
(a) What is the entropy change of a 14.6 g ice cube that melts completely in a bucket of water whose temperature is just above t
Alex73 [517]

Answer:

a) 17.81 J/K

b) 33.325 J/K

Explanation:

The expression to use here is the following:

ΔS = Q/T

Where:

Q: heat released or absorbed

T: Temperature in K

Now, in order to do this, we need to gather the data. We know that the temperature in part a) is above the freezing temperature of water, which is 0 ° C or 273 K. and the mass of the ice cube is 14.6 g.

a) Using the water heat of fusion (Cause it's melting), we can calculated the heat released using the following expression:

Q = m * Lf

Lf = 333,000 J/kg

Solving for Q first we have:

Q = (14.6 / 1000) * 333,000

Q = 4,861.8 J

Now, the entropy change is:

ΔS = 4,861.8 / 273

ΔS = 17.81 J/K

b) In this part, we follow the same procedure than in part a) but using the water heat of boiling (Lv = 2,256,000 J/kg), the temperature of boiling which is 100 °C (or 373 K) and the mass of 5.51 g (0.00551 kg)

Calculating the heat:

Q = 0.00551 * 2,256,000 = 12,430.56 J

Now the entropy change:

ΔS = 12,430.56 / 373

ΔS = 33.325 J/K

8 0
2 years ago
How popsicles are made?
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6 0
3 years ago
A 20-cm-long, 190 g rod is pivoted at one end. A 19 g ball of clay is stuck on the other end.
MaRussiya [10]

Answer:

0.76 s

Explanation:

We are given that

Length of rod,L=20 cm=\frac{20}{100}=0.20m

1 m=100 cm

Mass of rod,M=190 g=\frac{190}{1000}=0.19kg

Mass of ball,m=19 g=\frac{19}{1000}=0.019 kg

Using 1 kg=1000g

We have to find the period if the rod and clay swing as a  pendulum.

Moment of inertia of rod-clay=Moment of inertia of rod+moment of inertia of clay

I_{rod-clay}=I_{rod}+I_{clay}

I_{rod-clay}=\frac{1}{3}ML^2+mL^2

Substitute the values then we get

I_[rod-clay}=\frac{1}{3}(0.19)(0.20)^2+(0.019)(0.20)^2

I_{rod-clay}=3.29\times 10^{-3} kgm^2

Now, the center of mass of the combination of the rod and clay is given by

y=\frac{Md_1+md_2}{M+m}

Substitute d_1=\frac{L}{2}=Distance between pivot and the center of the rod

d_2=L=The distance  between rod and clay

Using the formula

y=\frac{0.19\times \frac{0.20}{2}+0.019\times 0.20}{0.19+0.019}

y=0.1091 m

Time period of the oscillation of the system of the rod and the clay is given by

T=2\pi\sqrt{\frac{I_{rod-clay}}{(M+m)yg}}

g=9.8m/s^2

Using the formula

Time period=2\times 3.14\sqrt{\frac{3.29\times 10^{-3}}{(0.19+0.019)\times 9.8\times 0.1091}}

Time-period=0.76 s

Hence, the period =0.76 s

8 0
3 years ago
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