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ioda
3 years ago
5

My dog Ubu can run at 21 mi/h. How far can he travel in 40 minutes?

Physics
1 answer:
Anna007 [38]3 years ago
4 0

Answer:

14

Explanation:

21miles/3=7

7*2=14

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Answer:

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3 years ago
Two upwards forces act: one of 34N which is 3.5m to the left of the axle, the other is 25N and is 2.6m to the right of the axle.
KiRa [710]

Answer:

See explanation below

Explanation:

In this case, you want to know if you put an object between these forces, which direction would go.

To know this, we need to calculate the moment of an object, which is defined as the product of a force and it's distance. In other words:

M = F * d   (1)

And, in order to reach equilibrium the force will exert a direction in clockwise or anticlosewise, and these moments, should be even:

anticlockwise moment = clockwise moment.

The clockwise would be the forces to the right, and anticlock would the only force to the left of the axle.

Clockwise moment = (10 * 0.8) + (25 * 2.6) = 73 Ns

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As we can see, the moment in the anticlockwise is higher than the actual clockwise moment, therefore, we can assume that the object will move anticlockwise, or simply move to the left.

Hope this helps

5 0
3 years ago
Scenario
zepelin [54]

Answer:

 t = 23.255 s,   x = 2298.98 m,    v_y = - 227.90 m / s

Explanation:

After reading your extensive writing, we are going to solve the approach.

The initial speed of the plane is 250 miles / h and it is at an altitude of 2650 m; In general, planes fly horizontally for launch, therefore this is the initial horizontal speed.

As there is a mixture of units in different systems we are going to reduce everything to the SI system.

         v₀ₓ = 250 miles h (1609.34 m / 1 mile) (1 h / 3600 s) = 111.76 m / s

         y₀ = 2650 m

Let's set a reference system with the x-axis parallel to the ground, the y-axis is vertical. As time is a scalar it is the same for vertical and horizontal movement

Y axis  

       y = y₀ + v₀ t - ½ g t²

the initial vertical velocity when the cargo is dropped is zero and when it reaches the floor the height is zero

       0 = y₀ + 0 - ½ g t²

       t = \sqrt{  \frac{2 y_o}{g} }

       t = √(2 2650/ 9.8)

       t = 23.255 s

Therefore, for the cargo to reach the desired point, it must be launched from a distance of

       x = v₀ₓ t

       x = 111.76 23.255

       x = 2298.98 m

at the point and arrival the speed is

        vₓ = v₀ₓ = 111.76

     

vertical speed is

         v_y = v_{oy} - gt

          v_y = 0 - gt

          v_y = - 9.8 23.25 555

         v_y = - 227.90 m / s

the negative sign indicates that the speed is down

in the attachment we have a diagram of the movement

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4 years ago
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