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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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The ______ length of a lens is the distance from the center of the lens to<br> its principal focus.
Marizza181 [45]

Answer:

<h2>Focal length</h2>

Explanation:

The Focal length of a lens is the distance from the center of the lens to

its principal focus.

3 0
3 years ago
A farm tractor tows a 3300-kg trailer up a 14" incline with a steady speed of 2.8 m/s. what force does the tractor exert on the
gladu [14]

The force exerted by gravity is:

F = m g

F = 3300 kg * 9.8 m/s^2

F = 32,430 N

 

The force exerted due to the inclined plane is:

F tractor = 32,430 N * sin 14

<span>F tractor = 7,823.75 N      (answer)</span>

7 0
2 years ago
What impulse occurs when an average force of 7.0 N is exerted on a cart for 2.5 s ?
makkiz [27]

Answer:1.7\times 10

Explanation:

Given

Force F=7 N

time interval t=2.5 s

Impulse is given by =Force \times time\ interval\ for\ applied\ force

Impulse=7\times 2.5=17.5

For two significant Figure

Impulse=1.7\times 10

                       

7 0
3 years ago
At what rate must a cylindrical spaceship rotate if occupants are to experience simulated gravity of 0.50 gg? Assume the spacesh
Svetradugi [14.3K]

Answer:

The time needed is T  = 16.8 s

Explanation:

From the question we are told that

      The magnitude of the stimulated acceleration due gravity is  a  =  0.5 g

        The diameter of the spaceship is  d =  35m

       

Generally the force acting on the spaceship is  

       F  =  ma

Given that the spaceship is rotating it implies that the force experienced by the occupant is a centripetal force so

      F  = \frac{mv^2}{r}

Thus  

       ma  =  \frac{mv^2}{r}

=>    \frac{v^2}{r}  =  a

      Generally the speed of this spaceship is mathematically represented as

      v =  \frac{2 \pi}{T}

=>    v^2  =   [\frac{2\pi}{T}] ^2

=>     \frac{\frac{4\pi^2 r^2}{T^2} }{r}  = 0.5g

=>       \frac{4 \pi^2 r }{T^2} =  0.5 g

=>         T  = \sqrt{ \frac{4\pi^2 r}{0.5g}}

substituting values

          T  = \sqrt{ \frac{4* (3.142)^2 *(35)}{0.5 * 9.8}}

         T  = 16.8 s

4 0
2 years ago
Read 2 more answers
Martha was leaning out of the window on the second floor of her house and speaking to Steve. Suddenly, her glasses slipped from
DiKsa [7]

Answer:

s = 23.72 m

v = 21.56 m/s²  

Explanation:

given

time to reach the ground (t) = 2.2 second

we know that

a) s = u t + 0.5 g t²

   u = 0 m/s

   g = 9.8 m/s²

   s = 0 + 0.5 × 9.8 × 2.2²

  s = 23.72 m

b) impact velocity

      v = √(2gh)

      v = √(2× 9.8 × 23.72)    

      v =  √464.912

      v = 21.56 m/s²  

6 0
3 years ago
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