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2 years ago

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A stone is thrown horizontally from the top of an inclined plane (angle of inclination θ). How would I find the initial speed of

the stone if it hit the plane a distance L down the slope from the point of launch?

1 answer:

Katen [24]2 years ago

7 0

Answer:

S = V t where S is the horizontal distance traveled

1/2 g t^2 = H where H is the vertical distance traveled

t^2 = 2 H / g

V^2 = S^2 / t^2 = S^2 g / (2 H) combining equations

tan theta = H / S

V^2 = S g / (2 tan theta)

Using S = L cos theta

V^2 = L g cos theta / (2 tan theta)

Giving V in terms of L and theta

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A 2.93 kg particle has a velocity of (2.98 i hat - 3.98 j) m/s.

cupoosta [38]

Answer:

a) The x and y components of the momentum are $8.731\,\frac{kg\cdot m}{s}$ and $-11.661\,\frac{kg\cdot m}{s}$ , respectively.

b) The magnitude and direction of its momentum are approximately 14.567 kilogram-meters per second and 306.823º.

Explanation:

a) The vectorial equation of momentum is represented by the following expression:

$\vec p = m\cdot \vec v$ (1)

Where:

$\vec p$ - Vector momentum, measured in kilogram-meters per second.

$m$ - Mass of the particle, measured in kilograms.

$\vec v$ - Vector velocity, measured in meters per second.

If we know that $m = 2.93\,kg$ and $\vec v = 2.98\,\hat{i}-3.98\,\hat{j}\,\,\,\left[\frac{m}{s} \right]$ , then the momentum is:

$\vec p = (2.93)\cdot (2.98\,\hat{i}-3.98\,\hat{j})\,\,\,\left[\frac{kg\cdot m}{s} \right]$

$\vec p = 8.731\,\hat{i}-11.661\,\hat{j}\,\,\,\left[\frac{kg\cdot m}{s} \right]$

The x and y components of the momentum are $8.731\,\frac{kg\cdot m}{s}$ and $-11.661\,\frac{kg\cdot m}{s}$ , respectively.

b) The magnitude and direction of momentum are represented by the following expressions:

$\|\vec p \| = \sqrt{p_{x}^{2}+p_{y}^{2}}$ (2)

$\theta = \tan^{-1}\left(\frac{p_{y}}{p_{x}} \right)$ (3)

Where:

$\|\vec p\|$ - Magnitude of momentum, measured in kilogram-meters per second.

$\theta$ - Direction of momentum, measured in sexagesimal degrees.

If we know that $p_{x} = 8.731\,\frac{kg\cdot m}{s}$ and $p_{y} = -11.661\,\frac{kg\cdot m}{s}$ , then the magnitude and direction of momentum are, respectively:

$\|\vec p\| = \sqrt{\left(8.731\,\frac{kg\cdot m}{s} \right)^{2}+\left(-11.661\,\frac{kg\cdot m}{s} \right)^{2}}$

$\|\vec p\| \approx 14.567\,\frac{kg\cdot m}{s}$

$\theta =\tan^{-1}\left(\frac{-11.661\,\frac{kg\cdot m}{s} }{8.731\,\frac{kg\cdot m}{s} } \right)$

$\theta \approx 306.823^{\circ}$

The magnitude and direction of its momentum are approximately 14.567 kilogram-meters per second and 306.823º.

6 0

3 years ago

Joffrey talks and moves slowly. When asked a question, he answers slowly in monotone monosyllables, if he answers at all. Joffre

lana [24]

Answer:

R.E.T.A.R.D.A.T.I.O.N

Explanation:

It won't let me spell it normal

7 0

3 years ago

What scale can I use for two force 0.8 Newtons and 0.9 Newtons? Someone please help

kirill [66]

Answer:

Explanation:

To plot the given forces of 0.8 N and 0.9 N. It is convenient to take

0.1 N as 1 unit in Graph

For example, 1 unit in the graph is 1 cm

then take 0.1 N equivalent to 1 cm

$0.8 N\equiv 8\ \text{units in graph}\\\\0.9\equiv 9\ \text{units in graph}$

5 0

3 years ago

The current in a circuit is tripled by connecting a 580 resistor in parallel with the resistance of the circuit. Determine the r

const2013 [10]

Answer:

1160 ohm

Explanation:

We are given that

R'=580 ohm

Current=3 I

We have to find the resistance of the circuit.

Let R be the resistance of circuit.

In parallel

$\frac{1}{R_{eq}}=\frac{1}{R_1}+\frac{1}{R_2}$

Using the formula

$\frac{1}{R_{eq}}=\frac{1}{580}+\frac{1}{R}=\frac{580+R}{580R}$

$R_{eq}=\frac{580R}{580+R}$

In parallel combination,Potential difference across each resistance remains same.

$V=IR$

Using the formula

$IR=3IR_{eq}$

$IR=3I\times \frac{580R}{580+R}$

$580+R=3\times 580=1740$

$R=1740-580=1160\Omega$

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3 years ago

ANSWER IN LESS THAN A MIN!! EASY!

Romashka [77]

Answer:

2 m/sec

Explanation:

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3 years ago