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3 years ago

15

The density of ice is 917 kg/m3, and the density of sea water is 1025 kg/m3. A swimming polar bear climbs onto a piece of floati

ng ice that has a volume of 6.22 m3. What is the weight of the heaviest bear that the ice can support without sinking completely beneath the water

1 answer:

Oksanka [162]3 years ago

3 0

Answer:

671.76 kg or 6590 N

Explanation:

So the buoyant force generated by the floating ice is equals to the mass of water displaced by the submerged ice. We also need to account for gravity of ice. The resulting additional mass that the ice sheet can support is the difference between the mass of water displaced by ice and the mass of ice submerged totally in water.

$m = m_w - m_i$

$m = V\rho_w - V\rho_i$

$m = V(\rho_w - \rho_i) = 6.22*(1025 - 917) = 671.76 kg$

So the ice piece can support an additional 671.76 kg of bear, or 671.76 * 9.81 = 6590 N

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In a physics lab, Asha is given a 10.7 kg uniform rectangular plate with edge lengths 67.3 cm by 53.5 cm . Her lab instructor re

Leya [2.2K]

Answer:

$I=2.6363\ kg.m^2$

Explanation:

Given:

dimension of uniform plate, $(0.673\times 0.535)\ m^2$

mass of plate, $m=10.7\ kg$

Now we find the moment of inertia about the center of mass of the rectangular plate is given as:

$I_{cm}=\frac{1}{12} \times m(L^2+B^2)$

where:

$L=$ length of the plate

$B=$ breadth of the plate

$I_{cm}=\frac{1}{12} \times 10.7\times(0.673^2+0.535^2)$

$I_{cm}=0.6591\ kg.m^2$

We know that the center of mass of the rectangular plane is at its geometric center which is parallel to the desired axis XX' .

Now we find the distance between the center of mass and the corner:

$s=\frac{\sqrt{ (0.673^2+0.535^2)}}{2}$

$s=0.4299\ m$

Now using parallel axis theorem:

$I=I_{cm}+m.s^2$

$I=0.6591+10.7\times 0.4299^2$

$I=2.6363\ kg.m^2$

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3 years ago

Is the frictional force the same as the applied force when the net force equals zero?

DerKrebs [107]

Answer:

Since the net force is to the right (in the direction of the applied force), then the applied force must be greater than the friction force. The friction force can be determined using an understanding of net force as the vector sum of all the forces.

Explanation:

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2 years ago

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3 years ago

Estimate the wavelength corresponding to maximum emission from each of the following surfaces: the sun, a tungsten filament at 2

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Answer

Applying Wein's displacement

$\lamda_{max}\ T = 2898 \mu_mK$

1) for sun T = 5800 K

$\lambda_{max} = \dfrac{2898}{5800}$

$\lambda_{max} = 0.5 \mu_m$

2) for tungsten T = 2500 K

$\lambda_{max} = \dfrac{2898}{2500}$

$\lambda_{max} = 1.16 \mu_m$

3) for heated metal T = 1500 K

$\lambda_{max} = \dfrac{2898}{1500}$

$\lambda_{max} = 1.93 \mu_m$

4) for human skin T = 305 K

$\lambda_{max} = \dfrac{2898}{305}$

$\lambda_{max} = 9.50 \mu_m$

5) for cryogenically cooled metal T = 60 K

$\lambda_{max} = \dfrac{2898}{60}$

$\lambda_{max} = 48.3 \mu_m$

range of different spectrum

UV ----0.01-0.4

visible----0.4-0.7

infrared------0.7-100

for sun T = 5800

λ 0.01 0.4 0.7 100

λT 58 2320 4060 5.8 x 10⁵

F 0 0.125 0.491 1

fractions

for UV = 0.125

for visible = 0.441-0.125 = 0.366

for infrared = 1 -0.491 = 0.509

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3 years ago

What is the speed of a cheetah if it takes 20 sesonds to run 300 m

natta225 [31]

the answer to your question is 15 :)

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3 years ago