How many joules are utilized when warming 5g of steam from 100°C to 120°C?

A 72-kg boy is surfing and catches a wave which gives him an initial speed of 1.6 m/s. He then drops through a height of 1.61 m,

chubhunter [2.5K]

Answer:

2849.98 J

Explanation:

From the question,

Work done by the boy = change Potential energy of the boy + change in kinetic energy of the boy

W = ΔP + ΔK..................... Equation 1

Where W = work done by the boy, ΔP = change in potential energy of the boy, ΔK = Change in kinetic energy of the boy.

But,

ΔP = mgΔh.................... Equation 2

ΔK = 1/2mΔv²................. Equation 3

Where m = mass of the boy, Δh = change in height of the boy, Δv = change in velocity of the boy.

Substitute equation 2 and 3 into equation 1

W = mgΔh+1/2mΔv²................. Equation 4

Given: m = 72 kg, Δh = 1.61 m, Δv = 8.5-1.6 = 6.9 m/s, g = 9.8 m/s²

Substitute into equation 4

W = 72(9.8)(1.61)+1/2(72)(6.9²)

W = 1136.016+1713.96

W = 2849.98 J

8 0

3 years ago

If 36 grams of water is to be heated from 24.0°C to 48°C to make a cup of tea, how much heat must be added? The specific heat of

Vinvika [58]

We will have the following:

$\begin{gathered} Q=mc\Delta T\Rightarrow Q=(36)(4.18)(48-24) \\ \\ \Rightarrow Q=3611.52 \end{gathered}$

So, the heat to add is 3611.52 Joules.

3 0

1 year ago

A particle moves through an xyz coordinate system while a force acts on it. When the particle has the position vector r with arr

Paha777 [63]

Answer:

The question is incomplete, below is the complete question "A particle moves through an xyz coordinate system while a force acts on it. When the particle has the position vector r with arrow = (2.00 m)i hat − (3.00 m)j + (2.00 m)k, the force is F with arrow = Fxi hat + (7.00 N)j − (5.00 N)k and the corresponding torque about the origin is vector tau = (4 N · m)i hat + (10 N · m)j + (11N · m)k.

Determine Fx."

$F_{x}=-1N.m$

Explanation:

We asked to determine the "x" component of the applied force. To do this, we need to write out the expression for the torque in the in vector representation.

torque=cross product of force and position . mathematically this can be express as

$T=r*F$

Where

$F=F_{x}i+(7N)j-(5N)k$ and the position vector

$r=(2m)i-(3m)j+(2m)k$

using the determinant method to expand the cross product in order to determine the torque we have

$\left[\begin{array}{ccc}i&j&k\\2&-3&2\\ F_{x} &7&-5\end{array}\right]\\\\$

by expanding we arrive at

$T=(18-14)i-(-12-2F_{x})j+(12+3F_{x})k\\T=4i-(-12-2F_{x})j+(12+3F_{x})k\\\\$

since we have determine the vector value of the toque, we now compare with the torque value given in the question

$(4Nm)i+(10Nm)j+(11Nm)k=4i-(-12-2F_{x})j+(12+3F_{x})k\\$

if we directly compare the j coordinate we have

$10=-(-12-2F_{x})\\10=12+2F_{x}\\ 10-12=2F_{x}\\ F_{x}=-1N.m$

8 0

3 years ago

An automobile traveling 95 km/h overtakes a 1.30-km-long train traveling in the same direction on a track parallel to the road.

SOVA2 [1]

Answer:

Same direction: t=234s; d=6.175Km

Opposite direction: t=27.53s; d=0.73Km

Explanation:

If the automobile and the train are traveling in the same direction, then the automobile speed relative to the train will be $v_{AT}=v_A-v_T$ (the train must see the car advancing at a lower speed), where $v_A$ is the speed of the automobile and $v_T$ the speed of the train.

So we have $v_{AT}=(95km/h)-(75Km/h)=20Km/h$ .

So the train (anyone in fact) will watch the automobile trying to cover the lenght of the train L at that relative speed. The time required to do this will be:

$t = \frac{L}{v_{AT}} = \frac{1.3Km}{20Km/h} = 0.065h=234s$

And in that time the car would have traveled (relative to the ground):

$d=v_At=(95Km/h)(0.065h)=6.175Km$

If they are traveling in opposite directions, we have to do all the same but using $v_{AT}=v_A+v_T$ (the train must see the car advancing at a faster speed), so repeating the process:

$v_{AT}=(95km/h)+(75Km/h)=170Km/h$

$t = \frac{L}{v_{AT}} = \frac{1.3Km}{170Km/h} = 0.00765h=27.53s$

$d=v_At=(95Km/h)(0.00765h)=0.73Km$

5 0

3 years ago

HELPPP PLEASEEE!!!!!

devlian [24]

Answer:

1) The speed of sound increases

2) 440 Hz

3) 29°C

4) 17°C

5) 434 Hz

6) 12 m/s

7) 17.3 m

Explanation:

1) The speed of sound increases

2) V = f×λ

f = V/λ = 343/0.78 = 439.744 ≈ 440 Hz

3) V = f×λ

512 × 0.68 = 348.16 m/s

348.16 - 331 = 17.16

T = 17.16/0.6 = 28.6 ≈ 29°C

4) Increase in speed = 350 - 340 = 10

Increase in temperature = 10/0.6 = 16.67° ≈ 17°C

5) f = V/λ = 343/0.79 = 434 Hz

6) 331 + 0.6×30 - (331 × 0.6 ×10) = 12 m/s

7) V = 331 + 0.6×25 = 346m/s

λ = 346/20 = 17.3 m

5 0

3 years ago