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kow [346]
4 years ago
9

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]

now E and B are related:

E=cB\\ B=\frac{E}{c} and c=\frac{1}{\sqrt{\epsilon_{0} \mu_{0}}}

replace in S_{average}

S_{average}=I= \frac{c \epsilon_{0}E^2}{2}[W/m]

we replace the values and we get:

E= \sqrt{\frac{2I}{\epsilon_{0}c}}

E = \sqrt{\frac{2(440.8)}{8.85*10^{-12}3*10^8}}=576.24[V/m]

therefore

B=\frac{E}{c}=\frac{576.24}{3*10^{8}}=1.92*10^{-6}[T]

8 0
4 years ago
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