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Bezzdna [24]
3 years ago
13

A person who weighs 625 N is riding a 98-N mountain bike. Suppose that the entire weight of the rider and bike is supported equa

lly by the two tires. If the pressure in each tire is 7.60 x 10^5 Pa, what is the area of contact between each tire and the ground?
Physics
1 answer:
kozerog [31]3 years ago
7 0

Answer:

A = 4.76 x 10⁻⁴ m²

Explanation:

given,                                  

weight of the person = 625 N

weight of the bike = 98 N        

Pressure on each Tyre = 7.60 x 10⁵ Pa

Area of contact on each Tyre = ?          

total weight of the system = 625 + 98

                                             = 723 N

Let F be the force on both the Tyre

F + F = W                                    

2 F  = 723                                    

F = 361.5 N                                

F = P A                                            

A = \dfrac{F}{P}                          

A = \dfrac{361.5}{7.60 \times 10^5}

A = 4.76 x 10⁻⁴ m²

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A space shuttle orbits Earth at a speed of 21,000 km/hr. How far will it travel in 5 hours?
lianna [129]

Answer:

105000km

Explanation:

Given parameters:

Speed of shuttle = 21000km/hr

Time of travel  = 5hrs

Unknown:

The distance covered by the space shuttle;

Solution:

Distance is the length of the path traveled.

   Distance  = speed x time

Input the parameters and solve;

     Distance = 21000km/hr x 5hrs

     Distance  = 105000km

6 0
3 years ago
What distance will a vehicle travel before coming to a complete stop from a speed of 70 mph, (a) When the vehicle is traveling o
OLga [1]

Answer:

(a), The SSD will be 723.9 ft.

(b-1), The SSD will be 620.2 ft.

(b-2), The SSD will be 723.91>SSD>620.2

(c), The SSD will be 910.5 ft.

Explanation:

Given that,

Speed = 70 mph

Suppose, a perception reaction time of 2.5 sec and the coefficient of friction is 0.35

We need to calculate the stopping sight distance

Using formula of SSD

SSD=1.47\times v\times t+\dfrac{v^2}{30\times(f\pm g)}

Where, v = speed of vehicle

t = perception reaction time

f = coefficient of friction

g = gradient of road

(a). If the gradient of road is zero.

Then, the stopping sight distance will be

SSD=1.47\times 70\times 2.5+\dfrac{70^2}{30\times(0.35)}

SSD=723.9\ ft

(b-1). If the gradient of road is 0.1

Then, the stopping sight distance will be

SSD=1.47\times 70\times 2.5+\dfrac{70^2}{30\times(0.35+0.1)}

SSD=620.2\ ft

(b-2). If the grade continuously decrease then the SSD will be increase.

But if the grade is increase then the SSD will be decrease and for flat grade the SSD will be more.

So, The SSD will be 723.91>SSD>620.2

(c). When the vehicle is traveling downhill on a roadway of constant grade then the vehicle take will be more SSD

So, The SSD will be

SSD=1.47\times 70\times 2.5+\dfrac{70^2}{30\times(0.35-0.1)}

SSD=910.5\ ft

Hence, (a), The SSD will be 723.9 ft.

(b-1), The SSD will be 620.2 ft.

(b-2), The SSD will be 723.91>SSD>620.2

(c), The SSD will be 910.5 ft.

7 0
4 years ago
Most cars have a coolant reservoir to catch radiator fluid that may overflow when the engine is hot. A radiator is made of coppe
alexandr402 [8]

Answer:

0.699 L of the fluid will overflow

Explanation:

We know that the change in volume ΔV = V₀β(T₂ - T₁) where V₀ = volume of radiator = 21.1 L, β = coefficient of volume expansion of fluid = 400 × 10⁻⁶/°C

and T₁ = initial temperature of radiator = 12.2°C and T₂ = final temperature of radiator = 95.0°C

Substituting these values into the equation, we have

ΔV = V₀β(T₂ - T₁)

= 21.1 L × 400 × 10⁻⁶/°C × (95.0°C - 12.2°C)

= 21.1 L × 400 × 10⁻⁶/°C × 82.8°C = 698832 × 10⁻⁶ L

= 0.698832 L

≅ 0.699 L = 0.7 L to the nearest tenth litre

So, 0.699 L of the fluid will overflow

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8 0
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Wewaii [24]
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6 0
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