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Alina [70]
2 years ago
14

In a college homecoming competition, eighteen students lift a sports car. While holding the car off the ground, each student exe

rts an upward force of 400 N. (a) What is the mass of the car in kilograms? (b) What is its weight in pounds?
Physics
1 answer:
Nata [24]2 years ago
6 0

Answer:

Explanation:

Given

Each student exert a force of F=400 N

Let mass of car be m

there are 18 students who lifts the car

Total force by 18 students F=18\times 400=7200 N

therefore weight of car W=7200

mass of car m=\frac{W}{g}

m=\frac{7200}{9.8}=734.69 kg

(b)7200 N \approx 1618.624\ Pound-force

734.69 kg\approx 1619.71 Pounds                  

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A hot air balloon must be designed to support a basket, cords, and one person for a total payload weight of 1300 N plus the addi
RSB [31]

Answer:

r = 4.44 m

Explanation:

 

For this exercise we use the Archimedes principle, which states that the buoyant force is equal to the weight of the dislodged fluid

         B = ρ g V

Now let's use Newton's equilibrium relationship

         B - W = 0

         B = W

The weight of the system is the weight of the man and his accessories (W₁) plus the material weight of the ball (W)

         σ = W / A

         W = σ A

The area of ​​a sphere is

           A = 4π r²

       W = W₁ + σ 4π r²

The volume of a sphere is

           V = 4/3 π r³

Let's replace

     ρ g 4/3 π r³ = W₁ + σ 4π r²

If we use the ideal gas equation

     P V = n RT

    P = ρ RT

    ρ = P / RT

 

    P / RT g 4/3 π r³ - σ 4 π r² = W₁

    r² 4π (P/3RT  r - σ) = W₁

Let's replace the values

     r² 4π (1.01 10⁵ / (3 8.314 (70 + 273)) r - 0.060) = 13000

     r² (11.81 r -0.060) = 13000 / 4pi

     r² (11.81 r - 0.060) = 1034.51

As the independent term is very small we can despise it, to find the solution

       r = 4.44 m

3 0
2 years ago
A long-distance swimmer is able to swim through still water at 4.0 km/h. She wishes to try to swim from Port Angeles, Washington
Roman55 [17]

Let \theta be the direction the swimmer must swim relative to east. Then her velocity relative to the water is

\vec v_{S/W}=\left(4.0\dfrac{\rm km}{\rm h}\right)(\cos\theta\,\vec\imath+\sin\theta\,\vec\jmath)

The current has velocity vector (relative to the Earth)

\vec v_{W/E}=\left(3.0\dfrac{\rm km}{\rm h}\right)\,\vec\imath

The swimmer's resultant velocity (her velocity relative to the Earth) is then

\vec v_{S/E}=\vec v_{S/W}+\vec v_{W/E}

\vec v_{S/E}=\left(\left(4.0\dfrac{\rm km}{\rm h}\right)\cos\theta+3.0\dfrac{\rm km}{\rm h}\right)\,\vec\imath+\left(4.0\dfrac{\rm km}{\rm h}\right)\sin\theta\,\vec\jmath

We want the resultant vector to be pointing straight north, which means its horizontal component must be 0:

\left(4.0\dfrac{\rm km}{\rm h}\right)\cos\theta+3.0\dfrac{\rm km}{\rm h}=0\implies\cos\theta=-\dfrac{3.0}{4.0}\implies\theta\approx138.59^\circ

which is approximately 41º west of north.

6 0
2 years ago
An airplane travels horizontally at a constant velocity v. An object is dropped from the plane and one second later another obje
Delvig [45]

Answer:

the vertical distance between the two object will increase uniformly when they are dropped after a fixed interval of time

Explanation:

Since airplane is moving horizontally with constant speed v

so when object is dropped from the plane then the speed of the object will be same as that of the speed of the airplane

so we can say that two object when dropped after some interval of time then they always lie in same vertical line

now we know that they both have same acceleration in vertical line so the motion of two objects relative to each other in vertical direction is always uniform motion because they have no acceleration with respect to each other

So the vertical distance between the two object will increase uniformly when they are dropped after a fixed interval of time

8 0
2 years ago
A box of mass 8 kg slides across a frictionless surface at an initial speed 1.5 m/s into a relaxed spring of spring constant 69
Sav [38]

Answer:

1.1 sec

Explanation:

m = mass of the box = 8 kg

k = spring constant of the spring = 69 N/m

v = initial speed of the box = 1.5 m/s

t = time period of oscillation of box in contact with the spring

Time period is given as

t = \pi \sqrt{\frac{m}{k}}

Inserting the values

t = (3.14) \sqrt{\frac{8}{69}}

t = 1.1 sec

5 0
2 years ago
A tennis ball bounces on the floor three times. If each time it loses 22.0% of its energy due to heating, how high does it rise
lesya692 [45]

Answer:

H = 109.14 cm

Explanation:

given,                                                            

Assume ,                                                            

Total energy be equal to 1 unit                                

Balance of energy after first collision = 0.78 x 1 unit

                                                             = 0.78 unit

Balance after second collision = 0.78 ^2 unit

                                                   = 0.6084 unit

Balance after third collision = 0.78 ^3 unit

                                              = 0.475 unit

height achieved by the third collision will be equal to energy remained                                        

H be the height achieved after 3 collision

0.475 ( m g h) = m g H                  

H = 0.475 x h                                    

H = 0.475 x 2.3 m                          

H = 1.0914 m                      

H = 109.14 cm                      

6 0
2 years ago
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