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2 years ago

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A scientist heated a tank containing 50 g of water. The specific heat of water is 4.18 J/gºC. The temperature of the water incre

ased from 25ºC to 37ºC. How much heat energy did the water absorb?
Your answer:

2,508 Joules

-2,508 Joules

5,225 Joules

7,733 Joules

1 answer:

Amanda [17]2 years ago

6 0

I got 5,225 by 50x4.18= 209(25)=5,225

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A 7.36 g sample of copper is contaminated with a additional 0.51 g sample of zinc. Suppose an atomic mass measurement was perfor

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The molar mass of the sample is equal to the summation of the molar mass of the elementas multiplied by the abundance of the elements by mole. In this case, copper has an abundance of 93.69 percent while zinc has 6.31 percent. In this case, the average molecular weight is 63.67 g/mol

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2 years ago

The wheels of an automobile are locked as it slides to a stop from an initial speed of 30.0 m/s. If the coefficient of kinetic f

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Answer:

x = 76.5 m

Explanation:

Let's use Newton's second law at the point of contact between the wheel and the floor.

fr = m a

fr = miy N

N-W = 0

N = W

μ mg = m a

a = miu g

a = 0.600 9.8

a = 5.88 m / s²

Having the acceleration we can use the kinematic relationships to find the distance

$v_{f}$ ² = v₀² + 2 a x

$v_{f}$ = 0

x = -v₀² / 2 a

Acceleration opposes the movement by which negative

x = - 30²/2 (-5.88)

x = 76.5 m

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2 years ago

At takeoff, a commercial jet has a speed of 72 m/s. Its tires have a diameter of 0.89 m. Part (a) At how many rev/min are the ti

34kurt

Answer:

a) Revolutions per minute = 2.33

b) Centripetal acceleration = 11649.44 m/s²

Explanation:

a) Angular velocity is the ratio of linear velocity and radius.

Here linear velocity = 72 m/s

Radius, r = 0.89 x 0. 5 = 0.445 m

Angular velocity

$\omega =\frac{72}{0.445}=161.8rad/s$

Frequency

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Revolutions per minute = 2.33

b) Centripetal acceleration

$a=\frac{v^2}{r}$

Here linear velocity = 72 m/s

Radius, r = 0.445 m

Substituting

$a=\frac{72^2}{0.445}=11649.44m/s^2$

Centripetal acceleration = 11649.44m/s²

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2 years ago

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2 years ago

The electric flux through a square-shaped area of side 5 cm near a very large, thin, uniformly-charged sheet is found to be 3\ti

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Answer:

Explanation:

Given

side of square shape $a=5\ cm$

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Flux is given by

$\phi =EA\cos \theta$

where E=electric field strength

A=area

$\theta$ =Angle between Electric field and area vector

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and Electric field by a uniformly charged sheet is given by

$E=\frac{\sigma }{2\epsilon_0}$

where $\sigma$ =charge density

$=\frac{\sigma }{\epsilon_0}$

$\sigma =0.012\times 8.85\times 10^{-12}$

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2 years ago