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3 years ago

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A uniform diameter rod is weighted at end and is floating in a liquid. Determine if the liquid is (a) lighter than water, (b) wa

ter, or (c) heavier than water. Justify your answer.

1 answer:

Yakvenalex [24]3 years ago

4 0

(C) Heavier than water

<u>Explanation:</u>

Given

Diameter of the rod is uniform

The road is floating in a liquid.

Let the area of cross-section of the rod = A

Let the weight density of water = γw

Let the weight density of liquid = γl

Let the density of water = ρw

g is the gravitational force

So, weight of rod = density of the rod X volume

Wrod = ρw X g X ( A X 2L) + 2 ρwg (A X L)

Wrod = 4ρwgAL

Wrod = 4γwAL

We can see that the weight of the rod is more than the density of water. So, in water, the rod will sink.

If the rod is floating on the liquid then the liquid is heavier than the water.

Therefore, The liquid is heavier than the water.

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John ( body mass= 65 kg) is taking off for a long jump . Horizontal accerleration ax is 5m/s^2 and vertical acceleration ay is 0

cestrela7 [59]

Answer:

(a) The horizontal ground reaction force $F_{g,x}=325\, N$

(b) The vertical ground reaction force $F_{g,y}=696\, N$

(c) The resultant ground reaction force $F_g=768\, N$

Explanation:

Given

John mass , m = 65 kg

Horizontal acceleration , $a_x= 5.0 \frac{m}{s^{2}}$

Vertical acceleration , $a_y=0.9 \frac{m}{s^{2}}$

(a) Using Newton's 2nd law in horizontal direction

$F_{g,x}=ma_x$

=> $F_{g,x}=65\times 5\, N=325\, N$

Thus the horizontal ground reaction force $F_{g,x}=325\, N$

(b) Using Newton's 2nd law in vertical direction

$F_{g,y}-mg=ma_y$

=> $F_{g,y}=mg+ma_y$

=> $F_{g,y}=65\times (9.81+0.9)\, N=696\, N$

Thus the vertical ground reaction force $F_{g,y}=696\, N$

(c) Resultant ground reaction force is

$F_g=(F_{g,x}^{2}+F_{g,y}^{2})^{\frac{1}{2}}$

=> $F_g=(325^{2}+696^{2})^{\frac{1}{2}}\, N=768\, N$

=> $F_g=768\, N$

Thus the resultant ground reaction force $F_g=768\, N$

5 0

3 years ago

You sit "at rest" in front of your computer to answer this question. But you sit on the surface of a planet that spins, so even

igomit [66]

Answer: Linear speed is 1,670 Kph.

Explanation:

If we assume that the earth is a perfect sphere, and that is spinning itself once every roughly 24 hr, we can get the angular velocity of the Earth, in magnitude, as follows:

ω = 2π / 24 Hr

Now, by definition, an angle is the relationship between the arc s, and the radius r, so we can replace these values in the angular velocity expression, as follows:

ω = (Δs / r) . 1/Δt ⇒ ω = (Δs/Δt). 1/r

But, by definition, Δs/At, is just the linear velocity, v, so we can conclude the following;

ω = v/r ⇒ v = ω. r

So, we can get v, as follows:

v = 2π /24 hr . 6378 Km = 1,670 Km/hr.

4 0

3 years ago

With mechanical waves, what is moving and what stays in roughly the same place?

Zielflug [23.3K]

The part that moves are called anti-nodes. The stationary pars are nodes

6 0

3 years ago

(a) According to Hooke's Law, the force required to hold any spring stretched x meters beyond its natural length is f(x)=kx. Sup

KengaRu [80]

Answer:

a) The work required to stretch the spring from 20 centimeters to 25 centimeters is 0.313 joules, b) The area of the region enclosed by one loop of the curve $r(\theta) = 2\cdot \sin 5\theta$ is $4\pi$ .

Explanation:

a) The work, measured in joules, is a physical variable represented by the following integral:

$W = \int\limits^{x_{f}}_{x_{o}} {F(x)} \, dx$

Where

$x_{o}$ , $x_{f}$ - Initial and final position, respectively, measured in meters.

$F(x)$ - Force as a function of position, measured in newtons.

Given that $F = k\cdot x$ and the fact that $F = 25\,N$ when $x = 0.3\,m - 0.2\,m$ , the spring constant ( $k$ ), measured in newtons per meter, is:

$k = \frac{F}{x}$

$k = \frac{25\,N}{0.3\,m-0.2\,m}$

$k = 250\,\frac{N}{m}$

Now, the work function is obtained:

$W = \left(250\,\frac{N}{m} \right)\int\limits^{0.05\,m}_{0\,m} {x} \, dx$

$W = \frac{1}{2}\cdot \left(250\,\frac{N}{m} \right)\cdot [(0.05\,m)^{2}-(0.00\,m)^{2}]$

$W = 0.313\,J$

The work required to stretch the spring from 20 centimeters to 25 centimeters is 0.313 joules.

b) Let be $r(\theta) = 2\cdot \sin 5\theta$ . The area of the region enclosed by one loop of the curve is given by the following integral:

$A = \int\limits^{2\pi}_0 {[r(\theta)]^{2}} \, d\theta$

$A = 4\int\limits^{2\pi}_{0} {\sin^{2}5\theta} \, d\theta$

By using trigonometrical identities, the integral is further simplified:

$A = 4\int\limits^{2\pi}_{0} {\frac{1-\cos 10\theta}{2} } \, d\theta$

$A = 2 \int\limits^{2\pi}_{0} {(1-\cos 10\theta)} \, d\theta$

$A = 2\int\limits^{2\pi}_{0}\, d\theta - 2\int\limits^{2\pi}_{0} {\cos10\theta} \, d\theta$

$A = 2\cdot (2\pi - 0) - \frac{1}{5}\cdot (\sin 20\pi-\sin 0)$

$A = 4\pi$

The area of the region enclosed by one loop of the curve $r(\theta) = 2\cdot \sin 5\theta$ is $4\pi$ .

5 0

3 years ago

A radioactive material has a count rate of 400 per minute. It has a half life of 40 years. How long will it take to decay to a r

cestrela7 [59]

Answer:

160 years.

Explanation:

From the question given above, the following data were obtained:

Initial count rate (Cᵢ) = 400 count/min

Half-life (t½) = 40 years

Final count rate (Cբ) = 25 count/min

Time (t) =?

Next, we shall determine the number of half-lives that has elapse. This can be obtained as follow:

Initial count rate (Cᵢ) = 400 count/min

Final count rate (Cբ) = 25 count/min

Number of half-lives (n) =?

Cբ = 1/2ⁿ × Cᵢ

25 = 1/2ⁿ × 400

Cross multiply

25 × 2ⁿ = 400

Divide both side by 25

2ⁿ = 400/25

2ⁿ = 16

Express 16 in index form with 2 as the base

2ⁿ = 2⁴

n = 4

Thus, 4 half-lives has elapsed.

Finally, we shall determine the time taken for the radioactive material to decay to the rate of 25 counts per minute. This can be obtained as follow:

Half-life (t½) = 40 years

Number of half-lives (n) = 4

Time (t) =?

n = t / t½

4 = t / 40

Cross multiply

t = 4 × 40

t = 160 years.

Thus, it will take 160 years for the radioactive material to decay to the rate of 25 counts per minute.

7 0

2 years ago