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r-ruslan
10 days 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 [1.1K]10 days 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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Use the formula t = (0.25) s1/2 to find the time t in seconds it will take a stone to drop a distance s of 200 feet. Round your
inna [987]

Answer:

The duration, t = 3.53 seconds

Explanation:

The following information is provided:

The equation to calculate the time t is expressed as:

t=(0.25)s^{1/2}...... (1)

Where

s denotes the distance in feet

We are to determine the duration taken by the stone to fall a distance of 200 feet, where s = 200 feet

Substituting the value of s into equation (1) yields:

t=(0.25)\times (200)^{1/2}

t = 3.53 seconds

Thus, the time taken by the object is 3.53 seconds, which provides the required answer.

4 0
4 days ago
A person is rowing across the river with a velocity of 4.5 km/hr northward. The river is flowing eastward at 3.5 km/hr (Figure 4
Yuliya22 [1153]

Answer: Her velocity magnitude (v) relative to the shore is 5.70 km/h.

Explanation:

Let Q be the speed of the boat, and P be the speed of the river flow.

R represents the resultant velocity combining boat velocity and river current.

According to vector addition using the law of triangles:

R=\sqrt{P^2+Q^2+2PQCos\theta}

From the diagram:

P = 3.5 km/h, Q = 4.5 km/h

\theta= 90^o

R=\sqrt{P^2+Q^2+2PQCos\theta}=\sqrt{(3.5)^2+(4.5)^2+3.5\times 4.5\times cos90^o}=5.70

(Cos90^o=0),(sin 90^o=1)

\alpha =tan^{-1}\frac{Qsin\theta}{P+Qcos\theta}=tan^{-1}\frac{4.5 sin 90^o}{3.5+4.5 cos90^o}=tan^{-1}\frac{4.5}{3.5}=52.12^o

Therefore, her velocity magnitude relative to the shore is 5.70 km/h.

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17 days ago
The spring in a retractable ballpoint pen is 1.8 cm long, with a 300 N/m spring constant. When the pen is retracted, the spring
Sav [1105]

Answer:

The pen requires 7.2 mJ of energy to extend.

Explanation:

Provided:

Length = 1.8 cm

Spring constant = 300 N/m

Initial compression = 1.0 mm

Additional compression = 6.0 mm

Total compression = 1.0 + 6.0 = 7.0 mm

We need to determine the energy needed

This energy is equivalent to the variation in spring potential energy

E=PE_{2}-PE_{1}

E=\dfrac{1}{2}kx_{2}^2-\dfrac{1}{2}kx_{1}^2

Substitute the values into the formula

E=\dfrac{1}{2}\times300\times(7.0\times10^{-3})^2-\dfrac{1}{2}\times300\times(1.0\times10^{-3})^2

E=0.0072\ J

E=7.2\ mJ

Therefore, a total of 7.2 mJ is needed to extend the pen.

7 0
12 days ago
Read 2 more answers
A Chevrolet Corvette convertible can brake to a stop from a speed of 60.0 mi/h in a distance of 123 ft on a level roadway. What
kicyunya [1025]

Response:

83.1946504051 m

Rationale:

u = Starting velocity = 60\ mph=\dfrac{60\times 1609.34}{3600}=26.82233\ m/s

s = Distance traveled = 123\ ft=\dfrac{123}{3.281}=37.4885705578\ m

\theta = Incline = 26^{\circ}

v^2-u^2=2as\\\Rightarrow a=\dfrac{v^2-u^2}{2s}\\\Rightarrow a=\dfrac{0^2-26.82233^2}{2\times 37.4885705578}\\\Rightarrow a=-9.5954230306\ m/s^2

Friction coefficient

\mu=-\dfrac{a}{g}\\\Rightarrow \mu=\dfrac{9.5954230306}{9.81}\\\Rightarrow \mu=0.978126710561

mg sin\theta - u mg cos\theta = ma\\\Rightarrow a=9.81(sin26-0.978126710561cos26)\\\Rightarrow a=-4.32382\ m/s^2

v^2-u^2=2as\\\Rightarrow s=\dfrac{v^2-u^2}{2a}\\\Rightarrow s=\dfrac{0^2-26.82233^2}{2\times -4.32382}\\\Rightarrow s=83.1946504051\ m

The calculated stopping distance is 83.1946504051 m

6 0
15 days ago
A 50-g cube of ice, initially at 0.0°C, is dropped into 200 g of water in an 80-g aluminum container, both initially at 30°C.
Ostrovityanka [942]

Answer:

b. 9.5°C

Explanation:

m_i = Ice mass = 50 g

T_i = Initial temperature of water and aluminum = 30°C

L_f = Latent heat of fusion = 3.33\times 10^5\ J/kg^{\circ}C

m_w = Water mass = 200 g

c_w = Water's specific heat = 4186 J/kg⋅°C

m_{Al} = Aluminum mass = 80 g

c_{Al} = Aluminum's specific heat = 900 J/kg⋅°C

The equation governing heat exchange in the system is presented by

m_i(L_f+c_wT)+m_wc_w(T-T_i)+m_{Al}c_{Al}=0\\\Rightarrow 0.05\times (3.33\times 10^5+4186\times T)+0.2\times 4186(T-30)+0.08\times 900(T-30)=0\\\Rightarrow 1118.5T-10626=0\\\Rightarrow T=\dfrac{10626}{1118.5}\\\Rightarrow T=9.50022\ ^{\circ}C

The final equilibrium temperature calculates to 9.50022°C

4 0
8 days ago
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