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vladimir2022 [97]
2 years ago
11

You are standing on a street corner with your friend. You then travel 14.0 m due west across the street and into your apartment

building. You travel in the elevator 22.0 m upward to your floor, walk 12.0 m north to the door of your apartment, and then walk 6.0 m due east to your balcony that overlooks the street. Your friend is standing where you left her.
Now how far are you from your friend?
Express your answer with the appropriate units.
Physics
1 answer:
Margarita [4]2 years ago
5 0

Answer:

Explanation:

We shall express each displacement vectorially , i for each unit displacement towards east , j for northward displacement and k for vertical displacement .

14 m due west = - 14 i

22.0 m upward in the elevator = 22 k

12 m north = 12 j

6.00 m east = 6 i

Total displacement = - 14 i + 22 k + 12 j + 6 i

D = - 8 i + 12 j + 22 k

magnitude = √ ( 8² + 12² + 22² )

= √ ( 64 + 144 + 484 )

= √ 692

= 26.3 m

Net displacement from starting point = 26.3 m .

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Two students race up a flight of steps together. "Student A" reaches the top of the steps 2 seconds ahead of
oksano4ka [1.4K]
Um student a because they were there a few seconds ahead
3 0
3 years ago
A well-insulated bucket of negligible heat capacity contains 129 g of ice at 0°C.
Luba_88 [7]

Answer:

The final equilibrium temperature of the system is T = 12.48^oC

For the ice it would melt completely the mass that would remain is Zero

Explanation:

In the following question we are provided with

Mass of the ice M_{i} = 129 g = 0.129 kg

Mass of the steam M_s = 19 g = 0.019 kg

Initial temperature is  T_i = 0°C

Temperature of  steam  T_s = 100°C

Following the change of state of water in the question

 The energy required by ice to change to water is mathematically given as

          Q_A = M_iL_f

Where L_f is a constant known as heat of fusion  and the value is 334*10^3 J/kg

           Q_A = 0.129 *334 *10^3  = 43086 J

The energy been released when the steam changes to water is mathematically given as

            Q_B = M_s * L_v

           Where L_v is a constant known as heat of vaporization and the value is 2256*10^3J/kg

           Q_B = 0.019 * 2256*10^3 = 42864J

         The energy released when the temperature of water decrease from 100°C to 0°C is

                 Q_C = M_s *C_water (100°C)

Where C_{water} is the specific heat of water which has a value 4186J/kg \cdot K

                  Q_C = 0.019 *4186*100 = 7953.4

Looking at the values we obtained we noticed that ]

             Q_B + Q_C > Q_A

What this means is that the ice will melt

bearing in mind the conservation of energy

     looking the way at which water at different temperature were mixed according to the question

     Heat lossed by the vapor   = heat gained by ice

        Q_B + M_s *C_{water}(100-T) = Q_A + M_i C_{water} T

                                               T = \frac{Q_B+M_s *C_{water}(100^oC)-Q_A}{(M_s *C_{water})+(M_i*C_{water})}

                                               T = \frac{42864+7953.4-43086}{(0.019+0.129)(4186)}

                                              T = 12.48^oC

       

3 0
3 years ago
A child walks due east on the deck of a ship at 2 miles per hour. the ship is moving north at a speed of 18 miles per hour. find
garik1379 [7]
Refer to the figure shown below.

The velocity of the child and the velocity of the ship should be added vectorially to find the speed and direction of the child relative to the water surface.

The magnitude of the child's velocity is
v = √(2² + 18²) = 18.11 mph

The direction of the child's speed is
θ = tan⁻¹ (18/2) = tan⁻¹ 9 = 83.7° north of east or counterclockwise from the eastern direction.

Answer:
The magnitude is 18.1 mph.
The direction is 84° north of east.

8 0
3 years ago
A ____________ galaxy has arms of gas and dust; where as _____________ galaxy is a category of galaxy that does not have a disti
Jet001 [13]

Answer:

spiral, an irregular

Explanation:

i took the test :)

3 0
3 years ago
Two equal forces are applied to a door. The first force is applied at the midpoint of the door, the second force is applied at t
Orlov [11]

Answer:

D) the second at the doorknob

Explanation:

The torque exerted by a force is given by:

\tau = Fdsin \theta

where

F is the magnitude of the force

d is the distance between the point of application of the force and the centre of rotation

\theta is the angle between the direction of the force and d

In this problem, we have:

- Two forces of equal magnitude F

- Both forces are perpendicular to the door, so \theta=90^{\circ}, sin \theta=1

- The first force is exerted at the midpoint of the door, while the 2nd force is applied at the doorknob. This means that d is the larger for the 2nd force

--> therefore, the 2nd force exerts a greater torque

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3 years ago
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