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Marina86 [1]
4 years ago
15

A person who weighs 864 N is riding a 85-N mountain bike. Suppose the entire weight of the rider plus bike is supported equally

by the two tires. If the gauge pressure in each tire is 7.90 x 105 Pa, what is the area of contact between each tire and the ground?
Physics
1 answer:
svetoff [14.1K]4 years ago
4 0

Answer:

A=60.06 x 10⁻⁵ m²

Explanation:

Given that

Weight of the person = 864 N

Weight of the bike = 85 N

The total weight ,car + bike ,Wt= 864 + 85 N

 Wt = 949 N

2 p = 949 N

The weight on the each tire(F) = 474.5  N

P = 7.9 x 10 ⁵  Pa

We know that

Force = Pressure x Area

F= P A

By putting the values

474.5 =  7.9 x 10 ⁵  A

A=60.06 x 10⁻⁵ m²

Therefore area of contact A

A=60.06 x 10⁻⁵ m²

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It is 92.96 millions miles away

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3 0
3 years ago
Do longitude lines run horizontally (east-west) or vertically (north-south)?
nlexa [21]
Longitude- Horizontal (East West)
Latitude- Vertical (North South)
3 0
3 years ago
A 72.8-kg swimmer is standing on a stationary 265-kg floating raft. The swimmer then runs off the raft horizontally with a veloc
nalin [4]

Answer:

-1.43 m/s relative to the shore

Explanation:

Total momentum must be conserved before and after the run. Since they were both stationary before, their total speed, and momentum, is 0, so is the total momentum after the run off:

m_sv_s + m_rv_r = 0

where m_s = 72.8, m_r = 265 are the mass of the swimmer and raft, respectively. v_s = 5.21 m/s, v_r are the velocities of the swimmer and the raft after the run, respectively. We can solve for v_r

265v_r + 72.8*5.21 = 0

v_b = -72.8*5.21/265 = -1.43 m/s

So the recoil velocity that the raft would have is -1.43 m/s after the swimmer runs off, relative to the shore

7 0
3 years ago
the pygmy shrew has an average mass of 2.0 g if 49 of these shrew are placed on a spring scale with a spring constant of 24 N/m
olga_2 [115]

Answer:

Spring's displacement, x = -0.04 meters.

Explanation:

Let the spring's displacement be x.

Given the following data;

Mass of each shrew, m = 2.0 g to kilograms = 2/1000 = 0.002 kg

Number of shrews, n = 49

Spring constant, k = 24 N/m

We know that acceleration due to gravity, g is equal to 9.8 m/s².

To find the spring's displacement;

At equilibrium position:

Fnet = Felastic + Fg = 0

But, Felastic = -kx

Total mass, Mt = nm

Fg = -Mt = -nmg

-kx -nmg = 0

Rearranging, we have;

kx = -nmg

Making x the subject of formula, we have;

x = \frac {-nmg}{k}

Substituting into the formula, we have;

x = \frac {-49*0.002*9.8}{24}

x = \frac {-0.9604}{24}

x = -0.04 m

Therefore, the spring's displacement is -0.04 meters.

3 0
3 years ago
Two protons are 9.261 fm apart. (1 fm= 1 femtometer = 1 x 10^-15 m.) What is the ratio of the electric force to the gravitationa
Strike441 [17]

Answer:

1.24\times 10^{36}

Explanation:

q = magnitude of charge on each proton = 1.6 x 10⁻¹⁹ C

m = mass of each proton = 1.67 x 10⁻²⁷ kg

r = distance between the two protons = 1 x 10⁻¹⁵ m

Electric force between the two protons is given as

F_{e} = \frac{kq^{2}}{r^{2}}

F_{e} = \frac{(9\times 10^{9})(1.6\times 10^{-19})^{2}}{(1\times 10^{-15})^{2}}

F_{e} = 230.4 N

Gravitational force between the two protons is given as

F_{g} = \frac{Gm^{2}}{r^{2}}

F_{g} = \frac{(6.67\times 10^{-11})(1.67\times 10^{-27})^{2}}{(1\times 10^{-15})^{2}}

F_{g} = 1.86\times 10^{-34} N

Ratio is given as

Ratio =\frac{F_{e}}{F_{g}}

Ratio =\frac{230.4}{1.86\times 10^{-34}}

Ratio = 1.24\times 10^{36}

3 0
3 years ago
Read 2 more answers
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