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-BARSIC- [3]
2 years ago
15

In reaching her destination, a backpacker walks with an average velocity of 1.32 m/s, due west. This average velocity results, b

ecause she hikes for 5.21 km with an average velocity of 3.49 m/s due west, turns around, and hikes with an average velocity of 0.687 m/s due east. How far east did she walk (in kilometers)?
Physics
1 answer:
Svetach [21]2 years ago
8 0

Answer:

distance in east is 1273.78 m

Explanation:

given data

average velocity = 1.32 m/s west =

hike = 5.21 km = 5.21 × 10³ m

average velocity = 3.49 m/s west

average velocity = 0.687 m/s east

to find out

distance in east

solution

we consider here distance in east  is = x

so distance from starting point = 5.21 × 10³ - x        ...................1

and we can say time required to reach end

time required = distance / speed

time required = \frac{5.21 *10^3 - x}{1.32}    ................2

and

time required for 6.44 km west

time required = \frac{5.21 *10^3 - x}{3.49}    ................3

and time required for distance x

time required = \frac{x}{0.687}    ................4

so from equation 2 , 3 and 4

\frac{5.21 *10^3 - x}{1.32} = \frac{5.21 *10^3 - x}{3.49}  + \frac{x}{0.687}

x = 1273.78 m

so distance in east is 1273.78 m

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Assume: Moving to the right is positive. A(n) 7.7 g object moving to the right at 22 cm/s makes an elastic head-on collision wit
krek1111 [17]

Answer:

Explanation:

We shall apply the formula for velocity in case of elastic collision which is given below

v₁ = (m₁ - m₂)u₁ /  (m₁ +  m₂)  + 2m₂u₂ / (m₁ +  m₂)

m₁ and u₁ is mass and velocity of first object , m₂ and  u₂ is mass and velocity of second object before collision and v₁ is velocity of first velocity after collision.

Here u₁ = 22 cm /s , u₂ = - 14 cm /s . m₁ = 7.7 gm , m₂ = 18 gm

v₁ = ( 7.7 - 18 ) x 22 / ( 7.7 + 18 )  + 2 x 18 x - 14 / ( 7.7 + 18 )

= - 8.817 - 19.6

= - 28.4 cm / s

5 0
3 years ago
A loaded barge has a mass of 1 500 000 kg and is traveling at 3 m/s. If a tugboat applies an opposing force of 12 000 N for 10 s
yan [13]

Answer:

Explanation:

Initial momentum is 1.5e6(3) = 4.5e6 kg•m/s

An impulse results in a change of momentum

The tug applied impulse is 12000(10) = 120000 N•s or 0.12e6 kg•m/s

The remaining momentum is 4.5e6 - 0.12e6 =  4.38e6 kg•m/s

The barge velocity is now 4.38e6 / 1.5e6 = 2.92 m/s

The tug applies 0.012e6 N•s of impulse each second.

The initial barge momentum will be zero in

t = 4.5e6 / 0.012e6 = 375 s or 6 minutes and 15 seconds

To stop the barge in one minute(60 s), the tug would have to apply

4.5e6 / 60 = 75000 N•s /s or 75 000 N

5 0
2 years ago
A wattage rating of a lightbulb is the power it consumers when it is connected across a 120 V potential difference. How does the
navik [9.2K]

Answer:

The answer is C.

120 V with 60 W light bulb is 240 ohms.

120 V with 100 W light bulb is 144 ohms.

The 100 W bulb has less resistance :)

7 0
3 years ago
A student drops two metallic objects into a 120-g steel container holding 150 g of water at 25°C. One object is a 206-g cube of
spayn [35]

Answer:

Mass of the aluminium chunk = 278.51 g

Explanation:

For an isolated system as given the energy lost and gains in the system will be zero therefore sum of all transfer of energy will be zero,as the temperature will also remain same

A specific heat formula is given as                  

Energy Change = Mass of liquid x Specific Heat Capacity x Change in temperature

                                       Q =  m×c×ΔT

                        Heat gain by aluminium + heat lost by copper  = 0    (1)

For Aluminium:

      Q = m\times0.897\frac{J}{g.k}\times(25-5)

      Q = m x 17.94 joule

For Copper:

Q= 206g\times0.385\frac{J}{g.k} \times(88-25)

       Q= 4996.53 Joule

from eq 1

     m x 17.94 = 4996.53

     mass of aluminium = \frac{4996.53}{17.94} g

    Mass of the aluminium chunk = 278.51 g

                         

3 0
3 years ago
A 10.0 kg ball weighs 98.0 N in air and weighs 65.0 N when submerged in water. The volume of the ball is:_________.A) 0.00245 m3
Kamila [148]

Answer: B) 0.00337 m3.

Explanation:

Given data:

Mass of the ball = 10kg

Weight of the ball in air = 98N

Weight of the ball in water = 65N

Solution:

To get the Volume of the ball when submerged in water, we divide the weight of the ball in water with the difference in apparent weight by 9.8m/s^2.

= 98 - 65 / 9.8

= 33 / 9.8

= 3.37kg

The volume of the ball is 3.37kg

The density of water is 1kg per Liter.

So 3.37 kg of water would have a volume of 3.37 Liters.

Therefore the ball would have a volume of 3.37 Liters (or 0.00337 cubic meters).

7 0
2 years ago
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