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r-ruslan
3 months ago
6

Calculate the buoyant force in air on a kilogram of titanium (whose density is about 4.5 grams per cubic centimeter). compare wi

th the weight mg of this much titanium.
Physics
1 answer:
ValentinkaMS [3.4K]3 months ago
7 0
1) The buoyant force acting on an object submerged in a fluid can be described as:
B=d_f V_d g
where d_f indicates the fluid's density, V_d represents the volume of the fluid displaced, and g=9.81~m/s^2 signifies the gravitational acceleration.

2) To determine the volume of the displaced fluid, we note that the titanium object is entirely submerged in the fluid (air), thus this volume matches the volume of 1 Kg of titanium, which has a density of d=4.5~g/cm^3 = 4.5\cdot10^3~Kg/m^3. Using the correlation between density, volume, and mass, we derive
V_d= \frac{m}{d}= \frac{1~Kg}{4.5\cdot10^3Kg/m^3}=2.22\cdot10^{-4}~m^3

3) We can now revisit the equation in step 1) to compute the buoyant force. Given that the air density is d_f = 1~Kg/m^3, this provides us with
B=d_f V_d g=1~Kg/m^3 \cdot 2.22\cdot10^{-4}~m^3 \cdot 9.81~m/s^2=2.22\cdot10^{-3}~N

4) The weight of 1 Kg of titanium is:
W=mg=1~Kg \cdot 9.81~m/s^2=9.81~N
Therefore, the buoyant force is negligible when compared to the weight.
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Answer: small barrel gun

Explanation:

It is noted that short barrel guns have a higher muzzle velocity for bullets compared to longer barrel guns.

Acceleration is determined by the change in velocity with respect to time.

a=\dfrac{\Delta v}{\Delta t}

For short barrel guns, the bullet reaches its muzzle velocity more quickly, leading to greater acceleration than that of bullets from long barrel guns.

7 0
3 months ago
A 50-kg load is suspended from a steel wire of diameter 1.0 mm and length 11.2 m. By what distance will the wire stretch? Young'
Ostrovityanka [3204]

Answer:

3.5 cm

Explanation:

mass, m = 50 kg

diameter = 1 mm

radius, r = half the diameter = 0.5 mm = 0.5 x 10^-3 m

L = 11.2 m

Y = 2 x 10^11 Pa

Cross-sectional area of the wire = π r² = 3.14 x 0.5 x 10^-3 x 0.5 x 10^-3

= 7.85 x 10^-7 m^2

Let the change in length of the wire be ΔL.

The equation for Young's modulus is given by

Y =\frac{mgL}{A\Delta L}

\Delta L =\frac{mgL}{A\times Y}

ΔL = 0.035 m = 3.5 cm

Thus, the wire stretches by 3.5 cm.

5 0
3 months ago
Suppose you are myopic (nearsighted). You can clearly focus on objects that are as far away as 52.5 cm away. You can clearly foc
ValentinkaMS [3465]
La imagen del objeto distante se formará en el punto lejano o a 52.5 cm, así que la distancia del objeto u = infinito, la distancia de la imagen v = -52.5 cm, la longitud focal requerida = f. Usando la fórmula de la lente 1 / v - 1 / u = 1 / f, obtenemos f = -52.5 cm = -0.525 m. La potencia P = 1 / f = -1 / 0.525 = -1.90. Ahora, para el ojo con gafas, debemos encontrar el nuevo punto cercano.
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2 months ago
Water drops fall from the edge of a roof at a steady rate. a fifth drop starts to fall just as the first drop hits the ground. a
Keith_Richards [3271]

The height from which the drops fall is 3.57m

Assuming the drops fall at a rate of 1 drop every t seconds. The initial drop takes 5t seconds to hit the ground. The second drop reaches the bottom of the 1.00 m window in 4t seconds, while the third drop reaches the top of the window in 3t seconds.

Calculate the distances covered by the second and third drops s₂ and s₃, which begin from rest at the building's roof.

s_2=\frac{1}{2} g(4t)^2=8gt^2\\ s_3=\frac{1}{2} g(3t)^2=(4.5)gt^2

The height of the window s is described by,

s=s_2-s_3\\ (1.00 m)=8gt^2-4.5gt^2=3.5gt^2\\ t^2=\frac{1.00 m}{(3.5)(9.8m/s^2)} =0.02915s^2

The first drop has reached the base after taking 5t seconds.

The roof height h represents the distance covered by the first drop and is expressed as,

h=\frac{1}{2} g(5t)^2=\frac{25t^2}{2g} =\frac{25(0.02915s^2)}{2(9.8m/s^2)} =3.57 m

the height of the roof remains at 3.57 m



8 0
3 months ago
Read 2 more answers
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