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4 years ago

What is providing the centripetal force in the following examples?

Physics

2 answers:

umka2103 [35]4 years ago

8 0

A yo-yo swung in a circle.

worty [1.4K]4 years ago

3 0

The yo yo being swung around in a circle is an example of centripetal

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16g 2g/cm3 what is the Volume

lord [1]

Answer:

V = 1.25E-7 cubic meter

7 0

3 years ago

If the pressure of a gas sample is quadrupled and the absolute temperature is doubled. True or False

Vanyuwa [196]

The given statement"If the pressure of a gas sample is quadrupled and the absolute temperature is doubled" is false.

Answer: Option B

<u>Explanation:</u>

As we know the direct relationship between Pressure and Temperature by the Gay-Lussac’s Law,

$P_{1} T_{1}=P_{2} T_{2}$

From this, we get,

$\frac{P_{1}}{T_{2}}=\frac{P_{2}}{T_{1}}$

$T_{2}=\frac{P_{1} T_{1}}{P_{2}}$

So, according to given statement, we have $P_{2} = 4 P_{1}$

Then from the above expression, we can find out the value of $T_{2}$ when pressure increased by 4 times of initial pressure as,

$T_{2}=\frac{P_{1} T_{1}}{4 P_{1}}$

Hence, we get,

$T_{1}=4 T_{2}$

Hence, from the above expression we can say that as we increase the pressure four times, the temperature does not get doubled. So, the given statement in the question is false.

6 0

4 years ago

(HELP FAST PLEASE. ILL GIVE BRAINLIEST ANSWER)

lana [24]

The answer is C, since the velocity is not changing then neither is the acceleration

8 0

3 years ago

slab of ice floats on water with a large portion submerged beneath the water surface. The slab is in the shape of a rectangular

n200080 [17]

Answer:

a) $\%V = 87.36\,\%$ , b) $x = 1.248\,m$ , c) $F_{B} = 176488.341\,N$ , d) Six polar bears.

Explanation:

a) The slab of ice is modelled by the Archimedes' Principles and the Newton's Laws, whose equation of equilibrium is:

$\Sigma F =\rho_{w}\cdot g \cdot A \cdot x-\rho_{i}\cdot g\cdot V = 0$

The height of the ice submerged is:

$\rho_{w}\cdot A \cdot x = \rho_{i}\cdot V$

$x = \frac{\rho_{i}\cdot V}{\rho_{w}\cdot A}$

$x = \frac{\left(900\,\frac{kg}{m^{3}}\right)\cdot (20\,m^{3})}{\left(1030\,\frac{kg}{m^{3}} \right)\cdot (14\,m^{2})}$

$x = 1.248\,m$

The percentage of the volume of the ice that is submerged is:

$\%V = \frac{(1.248\,m)\cdot (14\,m^{2})}{20\,m^{3}} \times 100\,\%$

$\%V = 87.36\,\%$

b) The height of the portion of the ice that is submerged is:

$x = 1.248\,m$

c) The buoyant force acting on the ice is:

$F_{B} = \left(1030\,\frac{kg}{m^{3}} \right)\cdot (1.248\,m)\cdot (14\,m^{2})\cdot \left(9.807\,\frac{m}{s^{2}} \right)$

$F_{B} = 176488.341\,N$

d) The new system is modelled after the Archimedes' Principle and Newton's Laws:

$\Sigma F = -n\cdot m_{bear}\cdot g-\rho_{i}\cdot V \cdot g + \rho_{w}\cdot V\cdot g = 0$

The number of polar bear is cleared in the equation:

$n\cdot m_{bear} = (\rho_{w} - \rho_{i})\cdot V$

$n = \frac{(\rho_{w}-\rho_{i})\cdot V}{m_{bear}}$

$n = \frac{\left(1030\,\frac{kg}{m^{3}} - 900\,\frac{kg}{m^{3}} \right)\cdot (20\,m^{3})}{400\,kg}$

$n = 6.5$

The maximum number of polar bears that slab could support is 6.

8 0

3 years ago

We can reasonably model a 90-W incandescent lightbulb as a sphere 5.7 cm in diameter. Typically, only about 5% of the energy goe

o-na [289]

Answer:

See answer

Explanation:

Given quantities:

$\eta = 0.05\\ W=90[W]\\r=0.0285[m]$

where $\eta$ is the efficiency of the lightbulb (visible light is 5% of the total power), $W$ is the total power of the lightbulb, r is the radius of the lightbulb in meters.

Intensity is power divided by area:

$I =\frac{P}{A}$

a) Now the effective power is $\eta*W$ , therefore:

$I =\frac{\eta*W}{\pi r^2}=\frac{0.05*90}{4\pi (0.0285)^2}=440.87[W/m^2]$

b) Now the intensity is the average poynting vector is related to the magnitudes of the maximum electric field and magnetic field amplitudes, following:

$S_{average}= \frac{EB}{2\mu_{0}}[W/m]$