What is the initial vertical velocity?

Physics

1 answer:

Marat540 [252]2 years ago

3 0

Answer:

Thus, any projectile that has an initial vertical velocity of 21.2 m/s and lands 10.0 m below its starting altitude spends 3.79 s in the air.The initial vertical velocity is the vertical component of the initial velocity: v 0 y = v 0 sin θ 0 = ( 30.0 m / s ) sin 45 ° = 21.2 m / s .

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Determine the magnitude and direction of the force between two parallel wires 25 m long and 6.0 cm apart, each carrying 25 a in

mr Goodwill [35]

The magnetic force between two wires is 0.052 N which is attract each other.

We need to know about magnetic force on a current-carrying wire formula to solve this problem. The magnetic force on two wires with same direction of current is

F = μ₀ . I1 . I2 . L / ( 2π . r )

where μ₀ is vacuum permeability (4π×10‾⁷ H/m) F is the magnetic force, I is current, L is the length of wire, r is distance of 2 wires.

From the question above, we know that:

L = 25 m

r = 6 cm

I1 = I2 = 25 A

By substituting the parameter, we get

F = μ₀ . I1 . I2 . L / ( 2π . r )

F = 4π×10‾⁷ . 25 . 25 . 25 / (2π . 0.06)

F = 0.052 N

Hence, the force between two wires is 0.052 N which is attract each other.

Find more on magnetic force at: brainly.com/question/13277365

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7 0

2 years ago

Suppose you walk 11 m in a direction exactly 24° south of west then you walk 21 m in a direction exactly 39° west of north. 1) H

DENIUS [597]

Answer:

(1) 42.94 m

(2) $16.02^\circ$

Explanation:

Let us first draw a figure, for the given question as below:

In the figure, we assume that the person starts walking from point A to travel 11 m exactly $24^\circ$ south of west to point B and from there, it walks 21 m exactly $39^\circ$ west of north to reach point C.

Let us first write the two displacements in the vector form:

$\vec{AB} = (-11\cos 24^\circ\ \hat{i}-11\sin 24^\circ\ \hat{j})\ m =(-10.05\ \hat{i}-4.47\ \hat{j})\ m\\\vec{BC} = (-21\sin 39^\circ\ \hat{i}+21\cos 39^\circ\ \hat{j})\ m =(-31.22\ \hat{i}+16.32\ \hat{j})\ m$

Now, the vector sum of both these vectors will give us displacement vector from point A to point C.

$\vec{AC}=\vec{AB}+\vec{BC}\\\Rightarrow \vec{AC}=(-10.05\ \hat{i}-4.47\ \hat{j})\ m+(-31.22\ \hat{i}+16.32\ \hat{j})\ m\\\Rightarrow \vec{AC}=(-41.25\ \hat{i}+11.85\ \hat{j})\ m$

Part (1):

the magnitude of the shortest displacement from the starting point A to point the final position C is given by:

$AC=\sqrt{(-41.25)^2+(11.85)^2}\ m= 42.94\ m$

Part (2):

As the vector AC is coordinates lie in the third quadrant of the cartesian vector plane whose angle with the west will be positive in the north direction.

The angle of the shortest line connecting the starting point and the final position measured north of west is given by:

$\theta = \tan^{-1}(\dfrac{11.85}{41.27})\\\Rightarrow \theta = 16.02^\circ$