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hodyreva [135]
3 years ago
7

When you are ice skating, to get started, you push your skate backwards on the ice and, as a result, begin to move forward. Whic

h law of motion is being described in this scenario?
Physics
2 answers:
Artemon [7]3 years ago
7 0
<span>Newtons Third Law Of Motion </span>
Setler79 [48]3 years ago
3 0
I would say the third... force. 

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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}

7 0
3 years ago
2 Polnis
Marta_Voda [28]

Answer:

Oprion A

The length of time over which the conditions are measured

Explanation:

Weather captures the daily atmospheric conditions over a short duration hence it is short-term. Climate is the average of weather and covers a longer duration hence long-term. Therefore, what differentiatea these two is the length of time over which the conditions are measured.

7 0
3 years ago
Match the type of heat transfer with its description
VikaD [51]

Answer:

1. Convection

2. Radiation

3. Conduction

Hope this helps!

Explanation:

5 0
3 years ago
A baseball has a mass of 145 g. A bat exerts a force of 18,400 N on the ball. What is the acceleration of the ball?
amid [387]

The correct formula to use for the situation given above is: F = MA, where F is the applied force, M is the mass of the object and A is the acceleration.

From the details given in the question, we are told that:

F = 18, 400N

M = 145 g = 145 / 1000 = 0.145 kg

A = ?

From the equation F = MA

A = F / M

A = 18,400 / 0.145 = 126,896.55 = 1.27 *10^5.

Therefore, the correct option is C.

3 0
3 years ago
Read 2 more answers
Bonus: (It's not that hard, you just have to pay attention to units.) The Saturn V rocket first stage
agasfer [191]

v = 2.45×10^3\:\text{m/s}

Explanation:

Newton's 2nd Law can be expressed in terms of the object's momentum, in this case the expelled exhaust gases, as

F = \dfrac{d{p}}{d{t}} (1)

Assuming that the velocity remains constant then

F = \dfrac{d}{dt}(mv) = v\dfrac{dm}{dt}

Solving for v, we get

v = \dfrac{F}{\left(\frac{dm}{dt}\right)}\;\;\;\;\;\;\;(2)

Before we plug in the given values, we need to convert them first to their appropriate units:

The thrust <em>F</em><em> </em> is

F = 7.5×10^6\:\text{lbs}×\dfrac{4.45\:\text{N}}{1\:\text{lb}} = 3.34×10^7\:\text{N}

The exhaust rate dm/dt is

\dfrac{dm}{dt} = 15\dfrac{T}{s}×\dfrac{2000\:\text{lbs}}{1\:\text{T}}×\dfrac{1\:\text{kg}}{2.2\:\text{lbs}}

\;\;\;\;\;= 1.36×10^4\:\text{kg/s}

Therefore, the velocity at which the exhaust gases exit the engines is

v = \dfrac{F}{\left(\frac{dm}{dt}\right)} = \dfrac{3.34×10^7\:\text{N}}{1.36×10^4\:\text{kg/s}}

\;\;\;= 2.45×10^3\:\text{m/s}

6 0
2 years ago
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