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4 days ago

Look at the diagram of the moon and three different paths that it could take. A planet with a moon in orbit. The moon has arrows

labeled 1 through 3. 1: Curves up and away from the planet. 2: Moves straight forward. 3: Curves down to follow planet's orbit. Which description applies best to pathway 2? the path of the moon toward the Sun the path of the moon’s orbit around Earth the path the moon would take if it had a larger mass than the Sun the path the moon would take if there was no force of Earth’s gravity

Physics

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A 1.7-kg block of wood rests on a rough surface. A 0.011-kg bullet strikes the block with a speed of 670 m/s and embeds itself.

ValentinkaMS [3465]

What is the coefficient of kinetic friction between the block and the surface? Provide your answer using two significant figures.

Response:

0.39

To elaborate:

The provided data includes

m1=1.7 Kg

m2=0.011 Kg

v2=670 m/s

d=2.4 m

The relationship m2v2=(m1+m2)v applies here hence $v=\frac {m2v2}{m1+m2}$ also $a=\frac {v^{2}}{2d}$

Friction's deceleration can be calculated using

$F=\mu_k N=\mu_k W=\mu_k (m1+m2)g$

$F=(m1+m2)a=\mu_k (m1+m2)g$ thus $a=\mu_k g$

$a=\frac {v^{2}}{2d}=\mu_k g$

$\mu_k=\frac {1}{2dg}(\frac {m2v}{m1+m2})^{2}=\frac {1}{2*2.4*9.81}\times (\frac {0.011*670}{1.7+0.011})^{2}$

$\mu_k=0.39$

7 0

1 month ago

Two resistors ( 3 ohms & 6 ohms) in a series circuit with a power supply = 12 volts. The current through resistor 6 ohms is

Ostrovityanka [3204]

In a series circuit...

-- The overall resistance equals the sum of the individual resistances.

-- The current remains identical throughout the circuit.

The total resistance in this circuit is (3Ω + 6Ω ) = 9Ω

<pThe current at every point measures (V/R) = (12v / 9Ω ) = 1.33 A.

Select choice (a).

6 0

3 months ago

An ore sample weighs 17.50 N in air. When the sample is suspended by a light cord and totally immersed in water, the tension in

kicyunya [3294]

Answer:

Estimate of the sample's volume: approximately $\rm 0.6422 \; L = 6.422 \times 10^{-4} \; m^{3}$ .

Mean density of the sample: approximately $\rm 2.77\; g \cdot cm^{3} = 2.778 \times 10^{3}\; kg \cdot m^{3}$ .

Assumption:

$\rm g = 9.81\; N \cdot kg^{-1}$ .
$\rho(\text{water}) = \rm 1.000\times 10^{3}\; kg \cdot m^{-3}$ .
The volume of the cord is considered negligible.

Explanation:

Overall volume of the sample

The magnitude of the buoyant force equals $\rm 17.50 - 11.20 = 6.30\; N$ .

This also corresponds to the weight (weight, $m \cdot g$ ) of the water displaced by the object. To determine the mass of the displaced water from its weight, apply the formula: divide weight by $g$ .

$\displaystyle m = \frac{m\cdot g}{g} = \rm \frac{6.30\; N}{9.81\; N \cdot kg^{-1}} \approx 0.642\; kg$ .

Assuming the density of water is $\rho(\text{water}) = \rm 1.000\times 10^{3}\; kg \cdot m^{-3}$ . To find the volume of the displaced water, use the formula: divide mass by density $\rho(\text{water})$ .

$\displaystyle V(\text{water displaced}) = \frac{m}{\rho} = \rm \frac{0.642\; kg}{1.000\times 10^{3}\; kg \cdot m^{-3}} \approx 6.42201 \times 10^{-4}\; m^{3}$ .

Assuming the cord's volume is negligible, since the sample is completely submerged in water, its volume should equal the volume of the displaced water.

$V(\text{sample}) = V(\text{water displaced}) \approx \rm 6.422\times 10^{-4}\; m^{3}$ .

Mean Density of the sample

Average density can be calculated by the mass divided by volume.

To compute the mass of the sample from its weight, utilize the formula: divide by $g$ .

$\displaystyle m = \frac{m \cdot g}{g} = \rm \frac{17.50\; N}{9.81\; N \cdot kg^{-1}} \approx 1.78389 \; kg$ .

The volume from the previous section can be utilized.

Lastly, divide mass by volume to find the average density.

$\displaystyle \rho(\text{sample, average}) = \frac{m}{V} = \rm \frac{1.78389\; kg}{6.42201 \times 10^{-4}\; m^{3}} \approx 2.778\; kg \cdot m^{-3}$ .

3 0

1 month ago

An object of mass 8.0 kg is attached to an ideal massless spring and allowed to hang in the Earth's gravitational field. The spr

inna [3103]

The derived frequency equals 2.63 Hz. Explanation: For an object weighing 8.0 kg with a spring stretching 3.6 cm, calculations involving the spring constant and oscillation frequency lead to this specific oscillation rate.

4 0

2 months ago

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