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Ulleksa [173]
3 years ago
14

What are the kinematic formulas?​

Physics
1 answer:
12345 [234]3 years ago
7 0

Here's a picture of Kinematic formulas

The kinematic formulas are a set of formulas that relate the five kinematic variables listed below.

Δx - Displacement

t - Time interval

v0 - Initial velocity

v- Final velocity

a - Constant acceleration

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shutvik [7]

c ... max speed zero accel

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It has made us think we have to live perfect lives?
Citrus2011 [14]

Answer: Is this a question? I dont understand.

Explanation:

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3 years ago
An airplane is flying at a speed of 45 m/s when it drops a 40 kg food package to the Polar exploration team. If the plane drops
Alina [70]

To solve this problem we will apply the concepts related to energy conservation, so the potential energy in the package must be equivalent to its kinetic energy. From there we will find the speed of the package in the vertical component. The horizontal component is given, as it is the same as the one the plane is traveling to. Vectorially we will end up finding its magnitude. So,

PE = KE

mgh = \frac{1}{2}mv^2

Here,

m = Mass

g = Gravity

h = Height

v = Velocity

Rearranging to find the velocity

v = \sqrt{2gh}

Replacing,

v = \sqrt{2(9.8)(120)}

v = 48.49m/s

Using the vector properties the magnitude of the velocity vector would be given by,

|V| = \sqrt{v_x^2+v_y^2}

|V| = \sqrt{45^2+48.42^2}

|V| = 66.2m/s

Therefore the package is moving to 66.2m/s

3 0
3 years ago
Two horizontal rods are each held up by vertical strings tied to their ends. Rod 1 has length L and mass M; rod 2 has length 2L
antiseptic1488 [7]

Answer:

Rod 1 has greater initial angular acceleration; The initial angular acceleration for rod 1 is greater than for rod 2.

Explanation:

For the rod 1 the angular acceleration is

\tau_1 = I_1\alpha _1 \\\\\alpha_1 = \dfrac{\tau_1}{I_1}

Similarly, for rod 2

\alpha_2 = \dfrac{\tau_2}{I_2}.

Now, the moment of inertia for rod 1 is

I_1 = \dfrac{1}{3}ML^2,

and the torque acting on it is (about the center of mass)

\tau_1 = Mg\dfrac{L}{2};

therefore, the angular acceleration of rod 1 is  

\alpha_1 = \dfrac{Mg\dfrac{L}{2}}{\dfrac{1}{3}ML^2},

\boxed{\alpha_1 = \dfrac{3g}{2L} }

Now, for rod 2 the moment of inertia is

I_2 = \dfrac{1}{3}(2M)(2L)^2

I_2 = \dfrac{8}{3} ML^2,

and the torque acting is (about the center of mass)

\tau _2 = (2M)g \dfrac{(2L)}{2}

\tau _2 = 2MgL;

therefore, the angular acceleration \alpha_2 is

\alpha_2 = \dfrac{2MgL;}{\dfrac{8}{3} ML^2,}.

\boxed{\alpha_2 = \dfrac{3g}{4L}}

We see here that

\dfrac{3g}{2L} > \dfrac{3g}{4L}

therefore

\boxed{\alpha_1 > \alpha_2.}

In other words , the initial angular acceleration for rod 1 is greater than for rod 2.

7 0
2 years ago
Define 1ohm resistance​
algol13

Answer:

1 ohm is equal to one volt (V)/ one ampere (1A) 1 Ohm is defined as the resistance of a conductor with a potential difference of 1 volt applied to the ends through which 1-ampere current flows. Ohms is the SI unit of electrical resistance.

Explanation:

I hope it's help u

7 0
2 years ago
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