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Reil [10]
3 years ago
10

Has anyone done this i need answers

Physics
2 answers:
Rudik [331]3 years ago
6 0
You cannot see the image it shows up as a black screen
kicyunya [14]3 years ago
6 0

I just see this so I can't see the image

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it is maybe A but i'm not 100% sure

Explanation:

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A farmer lifts his hay bales into the top loft of his barn by walking his horse forward with a constant velocity of 1 ft/s. Dete
Lesechka [4]

Answer:

The velocity of the hay bale is - 0.5 ft/s and the acceleration is 6.25\times 10^{- 3} ft/s^{2}

Solution:

As per the question:

Constant velocity of the horse in the horizontal, v_{x} = 1 ft/s

Distance of the horse on the horizontal axis, x = 10 ft

Vertical distance, y = 20 ft

Now,

Apply Pythagoras theorem to find the length:

20^{2} + 10^{2} = l^{2}

l^{2}= 500

Now,

x^{2} + y^{2} = 500                            (1)

Differentiating equation (1) w.r.t 't':

2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 0

x\frac{dx}{dt} = - y\frac{dy}{dt}

where

\frac{dx}{dt} = Rate of change of displacement along the horizontal

\frac{dy}{dt} = Rate of change of displacement along the vertical

v_{x} = velocity along the x-axis.

v_{y} = velocity along the y-axis

xv_{x} = -yv_{y}

v_{y} = - 10\times \frac{1}{20} = - 0.5 ft/s

|v_{y}| = 0.5\ ft/s

Acceleration of the hay bale is given by the kinematic equation:

v_{y}^{2} = u_{y} + 2ay

(-0.5)^{2} =0 + 2ay

0.25 = 2ay

\frac{0.25}{2y} = a

a = \frac{0.25}{2\times 20} = 6.25\times 10^{- 3} ft/s^{2}

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3 years ago
A force of 30.0 N is applied to a 3.00 kg object for 3.00 seconds. Calculate the velocity experienced by the object.
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Explanation:

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We have F, we have m, but in order to solve for v, we need a.

30.0 = 3.00a so

a = 10.0 m/s/s. Plug that in for a in the second equation and solve for v:

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v = 10.0(3.00) so

v = 30.0 m/s

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Which of these processes are chemical reactions
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Explanation:

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