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2 years ago

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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Which explains why a gas does not have a fixed shape?

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A gas doesn't have a fixed shape because the amount of space between the molecules in gas can change easily. B. the particles are not compressible

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3 years ago

A loop of area 0.250 m^2 is in a uniform 0.020 0-T magnetic field. If the flux through the loop is 3.83 × 10-3T· m2, what angle

dangina [55]

Answer:

40.0⁰

Explanation:

The formula for calculating the magnetic flux is expressed as:

$\phi = BAcos\theta$ where:

$\phi$ is the magnetic flux

B is the magnetic field

A is the cross sectional area

$\theta$ is the angle that the normal to the plane of the loop make with the direction of the magnetic field.

Given

A = 0.250m²

B = 0.020T

$\phi$ = 3.83 × 10⁻³T· m²

3.83 × 10⁻³ = 0.020*0.250cosθ

3.83 × 10⁻³ = 0.005cosθ

cosθ = 0.00383/0.005

cosθ = 0.766

θ = cos⁻¹0.766

θ = 40.0⁰

<em>Hence the angle normal to the plane of the loop make with the direction of the magnetic field is 40.0⁰</em>

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3 years ago

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3 years ago

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3 years ago

6) Find the speed a spherical raindrop would attain by falling from 4.00 km. Do this:a) In the absence of air dragb) In the pres

sleet_krkn [62]

We are asked to determine the velocity of a rain drop if it falls from 4 km.

To do that we will use the following formula:

$2ah=v_f^2-v_0^2$

Where:

$\begin{gathered} a=\text{ acceleration} \\ h=\text{ height} \\ v_f,v_0=\text{ final and initial velocity} \end{gathered}$

If we assume the initial velocity to be 0 we get:

$2ah=v_f^2$

The acceleration is the acceleration due to gravity:

$2gh=v_f^2$

Now, we take the square root to both sides:

$\sqrt{2gh}=v_f$

Now, we substitute the values:

$\sqrt{2(9.8\frac{m}{s^2})(4000m)}=v_f$

solving the operations:

$280\frac{m}{s}=v$

Therefore, the velocity without air drag is 280 m/s.

Part B. we are asked to determine the velocity if there is air drag. To do that we will use the following formula:

$F_d=\frac{1}{2}C\rho_{air}Av^2$

Where:

$\begin{gathered} F_d=drag\text{ force} \\ C=\text{ constant} \\ \rho_{air}=\text{ density of air} \\ A=\text{ area} \\ v=\text{ velocity} \end{gathered}$

We need to determine the drag force. To do that we will use the following free-body diagram:

Since the velocity that the raindrop reaches is the terminal velocity and its a constant velocity this means that the acceleration is zero and therefore the forces are balanced: