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A stone is launched at an angle of 50 degree with initial velocity of 22m/s. Find out its initial and final velocity.

ss7ja [257]

(a) The initial vertical velocity of the stone is 16.85 m/s and the initial horizontal velocity is 14.14 m/s.

(b) The final vertical velocity is 0 and the final horizontal velocity is 14.14 m/s.

<h3>Initial vertical velocity</h3>

The initial vertical velocity of the stone is calculated as follows;

Vi = Vsinθ

Vi = 22 x sin(50)

Vi = 16.85 m/s

<h3>Initial horizontal velocity</h3>

Vxi = V cosθ

Vxi = 22 x cos(50)

Vxi = 14.14 m/s

<h3>Final vertical velocity of the stone</h3>

Vf = Vi - gt

where;

Vf is the final vertical velocity = 0 at maximum height

<h3 /><h3>Final horizontal velocity of the stone</h3>

Vfx = Vxi = 14.14 m/s

Thus, the initial vertical velocity of the stone is 16.85 m/s and the initial horizontal velocity is 14.14 m/s.

The final vertical velocity is 0 and the final horizontal velocity is 14.14 m/s.

Learn more about vertical velocity here: brainly.com/question/24949996

#SPJ1

3 0

2 years ago

In the compound below, what do the subscripts indicate?

mihalych1998 [28]

Answer:

b

that's how equations are

5 0

4 years ago

Read 2 more answers

<img src="https://tex.z-dn.net/?f=%5Chuge%20%5Cfrak%7B%E0%BC%86%20%20%5C%3A%20%20%5C%3A%20question%20%5C%3A%20%20%5C%3A%20%E0%BC

My name is Ann [436]

The crate is moving at constant velocity when the forces acting on it are

balanced.

The value of the force <em>F</em> required to pull the crate with constant velocity is; $\underline{F = \sqrt{2} \cdot \mu \cdot m\cdot g}$

Reasons:

Mass of the crate = m

Cross section of through = Right angled

Orientation (inclination) of the through to horizontal = 45°

Coefficient of kinetic friction = μ

Required:

The value of the required to pull the crate along the through at constant

velocity.

Solution:

When the through is moving at constant velocity, we have;

Friction force acting on crate = Force pulling the crate

Friction force = Normal reaction × Coefficient of kinetic friction

Normal reaction on an inclined plane = $\mathbf{F_N}$

Each side of the through gives a normal reaction.

The vertical component of the normal reaction on each side of the through

is therefore;

$F_N$ ·j = $\mathbf{F_N}$ × sin(θ)

The sum of the vertical component = $F_N$ ·j + $F_N$ ·j = 2· $F_N$ ·j = 2· $F_N$ ×sin(θ)

The sum of the vertical component of the normal reactions = The weight of the crate

Therefore;

2· $F_N$ ×sin(θ) = m·g

θ = 45°

Therefore;

2· $F_N$ ×sin(45°) = m·g

$\displaystyle sin(45^{\circ}) = \mathbf{\frac{\sqrt{2} }{2}}$

Therefore;

$\displaystyle 2 \cdot F_N \cdot sin(45^{\circ}) = 2 \cdot F_N \times \frac{\sqrt{2} }{2} = \sqrt{2} \cdot F_N$

$\displaystyle F_N = \mathbf{ \frac{m \cdot g}{\sqrt{2} }}$

Which gives;

$\displaystyle Force \ required, \ F = Sum \of \ friction \ forces \ = 2 \times F_N \times \mu = \mathbf{ 2 \times \frac{m \cdot g}{\sqrt{2} } \times \mu}$

$\displaystyle Force \ required, \ F = 2 \times \frac{m \cdot g}{\sqrt{2} } \times \mu = \mathbf{ \sqrt{2} \cdot \mu \cdot m \cdot g}$

Force required to pull the crate at constant velocity, <u>F = √2·μ·m·g</u>

Learn more about force of friction here:

brainly.com/question/6561298

8 0

2 years ago

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CAN YOU PLS CHECK IF ITS CORRECT I'LL MARK YOU BRAINLIST

Alexus [3.1K]

i think it looks good

yea it correct

BTW yw if it's right

5 0

3 years ago

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A capacitor stores 7.6 × 10–11 C when the electric potential difference between the plates is 5.8 V. What is the electric potent

N76 [4]

Energy stored in the capacitor= 2.2 x 10⁻¹⁰ J

Explanation:

The energy stored in the capacitor is given by

E= 1/2 q V

E= energy stored

q= charge= 7.6 x 10⁻¹¹ C

V= potential difference=5.8 V

so energy stored= E= 1/2 (7.6 x 10⁻¹¹) (5.8)

Energy stored= 2.2 x 10⁻¹⁰ J

5 0

4 years ago