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2 years ago

What is the main reason people and bears slip and fall on ice, be careful when outside in the snow/ice.

Physics

1 answer:

Ket [755]2 years ago

8 0

Answer:

Both ice and snow have a lower coefficient of friction in comparison with grass, wood or pavement.

Explanation:

The main reason why people and bears, slip and fall on ice/snow is that both ice and snow have a lower coefficient of friction in comparison with grass, wood or pavement. We present below a comparative chart of coefficients of friction associated with different materials:

Material Coefficient of Friction - Static

Ice/Snow 0.05 - 0.3

Wood 0.2 - 0.6

Asphalt 0.5 - 0.7

Grass 0.3 - 0.5

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Problem 21-40a:

Greeley [361]

Answer:

Part a)

$\frac{F_e}{F_g} = 2.74 \times 10^{12}$

Part b)

$q = 3.37 \times 10^{-4} C$

Explanation:

As we know that electric force on electric charge is given as

$F = qE$

here we have

$q = 1.6 \times 10^{-19}C$

E = 153 N/C

now force is given as

$F = (1.6 \times 10^{-19})(153) = 2.45 \times 10^{-17} N$

Gravitational force on electric charge near surface of earth is given as

$F_g = mg$

$F_g = (9.1 \times 10^{-31})(9.81) = 8.93 \times 10^[-30} N$

now the ratio of two forces is given as

$\frac{F_e}{F_g} = \frac{2.45 \times 10^{-17}}{8.93 \times 10^{-30}}$

$\frac{F_e}{F_g} = 2.74 \times 10^{12}$

Part b)

Now the ball is balanced by the electric force and the force of gravity on it

so here we have

$F_g = qE$

$mg = qE$

$(5.25 \times 10^{-3})(9.81) = q(153)$

here we have

$q = 3.37 \times 10^{-4} C$

8 0

3 years ago

What is the relationship between atmospheric pressure and the density of gas particles in an area of increasing pressure

mezya [45]

Answer:

this is a no brainer

Explanation:

As air pressure in an area increases, the density of the gas particles in that area increases.

8 0

3 years ago

Help me, i don't know the answer

8_murik_8 [283]

Explanation:

A) Bb BB

B) 50%

A) 50%

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4 0

2 years ago

Suppose astronomers discover a radio message from a civilization whose planet orbits a star 35 light-years away. Their message e

Grace [21]

Answer:

The duration is $T =72 \ years /tex]Explanation:From the question we are told that The distance is [tex]D = 35 \ light-years = 35 * 9.46 *10^{15} = 3.311 *10^{17} \ m$