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Anastaziya [24]
3 years ago
13

Mr. Beall paddles his canoe at 8.0 m/s North in a river that flows at 6.0 m/s to the South. What is the magnitude and direction

of the resulting velocity?
a)

14 m/s South


b)

14 m/s North


c)

10 m/s North


d)

2 m/s North
Physics
1 answer:
kifflom [539]3 years ago
3 0

Answer:

2 /s north

Explanation:

Given that,

Velocity due North is 8 m/s and due south is 6 m/s

We need to find the magnitude and the direction of the resulting velocity.

Let North is positive and South is negative. When two velocities are in opposite direction, they adds up. So,

v=8+(-6)\\\\v=2\ m/s

It is positive. So, it is in North direction.

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How are the climates of coastal regions affected by the specific heat capacity of water?
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Answer:

The specific heat capacity can be defined as the amount of heat required to raise the temperature of 1 unit of mass by 1 unit temperature. The specific heat capacity of water is 4.186 joule/gram °C which is higher than common substances. The land has lower specific heat capacity. Thus, the land gets hot quickly than water.

This results in warming up air near the land which creates a difference in pressure across the coastal region. Sea breeze blows from sea towards landmass. Opposite happens at night, when water is still warm and land gets cooled down quickly. Then land breeze blows  from landmass towards the sea. This breeze maintains a moderate temperature and windy and humid weather in the coastal regions.

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A successful marriage is based on:
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Answer:

a and d

Explanation:

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3 years ago
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Where will the spacecraft be when the gravitational forces acting on it are equal?
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It would be not be able to move yet it would be in the air

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3 years ago
A tennis ball is served horizontally from 2.4m above the ground at net is 12m away and point 0.9 high will be ball clear the net
Schach [20]

Explanation:

Let us first calculate  long does it take to go 12m at 30m/s( assumed speed)

12/30 = 0.4 seconds

horizontal distance the ball drop in that time

H= (0)(0.4)+1/2(-9.8)(0.4)2

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H_1= V0y t’ + ½ g t’^2

-2.4= (0)t’ + ½ (-9.8) t’^2

t’= 0.69s

X’=V0x t’

X’=(30)(0.96)

X’= 20.7m

3 0
3 years ago
A space vehicle is traveling at 3760 km/h relative to Earth when the exhausted rocket motor is disengaged and sent backward. The
almond37 [142]

Answer:

3688 km/h

Explanation:

Given:-

- The speed of vehicle relative to earth, vs_e = 3760 km/h

- The relative speed of command and motor, v_c/m = 90 km/h

- The mass of command = m

- The mass of motor = 4m

Find:-

What is the speed of the command module relative to Earth just after the separation?

Solution:-

- Consider the space vehicle as a system that detaches itself into two parts ( command and motor ). We will assume that the gravitational pull due to Earth on the space vehicle is negligible. With that assumption we have our system in isolation. We will apply the principle of conservation of linear momentum on the system as follows:

             Initial momentum = Final momentum

                                       Pi = Pf

                  M*vs_e = m*vc_e + 4m*vm_e

Where,

                  M = m + 4m = 5m

                  vc_e = Velocity of command relative to earth

                  vm_e = Velocity of motor relative to earth  

- We will develop a relation of velocities of command and motor in the frame of earth as follows:

                  vm_e =  v_c/m + vc_e        

- Substituting (vm_e) from Equation 2 into Equation 1, we have:

                  5m*vs_e = m*vc_e + 4m*(v_c/m + vc_e)

                  5m*vs_e = 5m*vc_e + 4m*(v_c/m)

- Solve for vc_e:

                  5m*vs_e -  4m*(v_c/m) = 5m*vc_e

                   vs_e - 0.8*(v_c/m) = vc_e

- Plug in values and evaluate vc_e:

                  vc_e = 3760 - 0.8*(90)

                  vc_e = 3,688 km/h

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