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3 years ago

If your body has a density of 955 kg/m3, what fraction of you will be submerged when floating gently in:

Physics

1 answer:

wel3 years ago

6 0

Answer: (a) 99.5℅ submerged

(B) 96.9℅ submerged

Explanation:

For 995 kilograms per cubic meter, It'll be the 995Kg/m3 density given, divided by the density of fresh water, which is one times ten to the three kilograms per cubic meter, and this works out to 99.5 percent submerged.

If the same person with the same density was floating in salt water, salt water has a slightly higher density of 1027 kilograms per cubic meter, this works out to 96.9 percent submerged.

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To initiate a nuclear reaction, an experimental nuclear physicist wants to shoot a proton into a 5.50-fm-diameter 12C nucleus. T

Fantom [35]

Answer:

$V_1= 3.4*10^7m/s$

Explanation:

From the question we are told that

Nucleus diameter $d=5.50-fm$

a 12C nucleus

Required kinetic energy $K=2.30 MeV$

Generally initial speed of proton must be determined,applying the law of conservation of energy we have

$K_2 +U_2=K_1+U_1$

where

$K_1$ =initial kinetic energy

$K_2$ =final kinetic energy

$U_1$ =initial electric potential

$U_2$ =final electric potential

mathematically

$U_2 = \frac{Kq_pq_c}{r_2}$

where

$r_f$ =distance b/w charges

$q_c$ =nucleus charge $=6(1.6*10^-^1^9C)$

$K$ =constant

$q_p$ =proton charge

Generally kinetic energy is know as

$K=\frac{1}{2} mv^2$

Therefore

$U_2 = \frac{Kq_pq_c}{r_2} + K_2=\frac{1}{2} mv_1^2 +U_1$

Generally equation for radius is $d/2$

Mathematically solving for radius of nucleus

$R=(\frac{5.50}{2}) (\frac{1*10^-^1^5m}{1fm})$

$R=2.75*10^-^1^5m$

Generally we can easily solving mathematically substitute into v_1

$q_p$ $=6(1.6*10^-^1^9C)$

$K_1=9.0*10^9 N-m^2/C^2$

$U_1= 0$

$R=2.75*10^-^1^5m$

$K=2.30 MeV$

$m= 1.67*10^-^2^7kg$

$V_1= (\frac{2}{1.67*10^-^2^7kg})^1^/^2 (\frac{(9.0*10^9 N-m^2/C^2)*(6(1.6*10^-^1^9C)(1.6*10^-^1^9C)}{2.75*10^-^1^5m+2.30 MeV(\frac{1.6*10^-^1^3 J}{1 MeV}) }$

$V_1= 3.4*10^7m/s$

Therefore the proton must be fired out with a speed of $V_1= 3.4*10^7m/s$

8 0

2 years ago

Neptune is approximately 4.5 billion kilometers from the sun. What is this distance in AU?

asambeis [7]

Well, one AU is 149,597,870 km. So, we would basically have to divide 4.5 billion km by 149,597,870, right?

4,500,000,000/149,597,870=30.080642 AU.

So, the correct answer would be 30 AU. Hoped this helped!

7 0

2 years ago

Susan, driving north at 53 mphmph , and Shawn, driving east at 63 mphmph , are approaching an intersection. Part A What is Shawn

mafiozo [28]

Answer:

Shawn's speed relative to Susan's speed = 10 mph

Resultant velocity = 82.32 mph

Explanation:

The given data :-

i) Susan driving in north and speed of Susan is ( v₁ ) = 53 mph.

ii) Shawn driving in east and speed of Shawn is ( v₂ ) = 63 mph.

iii) The speed of both Susan and Shawn is relative to earth.

iv) The angle between Susan in north and Shawn in east is 90°.

We have to find Shawn's speed relative to Susan's speed.

v₂₁ = v₂ - v₁ = 63 - 53 = 10 mph

Resultant velocity,

$v = \sqrt{v_{2} ^{2}+ v_{1} ^{2} } =\sqrt{63^{2} +53^{2} }$

v = 82.32 mph

5 0

3 years ago

Later, while taking your groceries back to the car, you accidentally let go of your cart. It rolls straight down a grassy slope

VladimirAG [237]

Car is moving on the glassy slope with constant speed

Now we know that

$a = \frac{dv}{dt}$

so acceleration is rate of change in velocity

as we know that velocity is constant here so acceleration is zero

so here

$a = 0$

now as we know by Newton's II law

$F = ma$

since a = 0

$F = 0$

so net force will be ZERO on it during this motion

6 0

3 years ago

A thin, rectangular sheet of metal has mass M and sides of length a and b. Find the moment of inertia of this sheet about an axi

slega [8]

We divide the thin rectangular sheet in small parts of height b and length dr. All these sheets are parallel to b. The infinitesimal moment of inertia of one of these small parts is
$dI =r^2*dm$
where $dm =M(b*dr)/(ab)$
Now we find the moment of inertia by integrating from $-a/2$ to $a/2$
The moment of inertia is
$I= \int\limits^{-a/2}_{a/2} {r^2*dm} = M \int\limits^{-a/2}_{a/2} r^2(b*dr)/(ab)=(M/a)(r^3/3)$ (from (-a/2) to $I=(M/3a)(a^3/8 +a^3/8)=(Ma^2)/12$ (a/2))

4 0

3 years ago