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3 years ago

Pls give me an example of newton's 1 ,2 ,3 law of motion pls this is due tomorrow

1 answer:

6 0

"He who lays up treasures on earth spends his life backing away from his treasures. To him, death is loss. He who lays up treasures in heaven looks forward to eternity; he’s moving daily toward his treasures. To him, death is gain." - Randy Alcorn

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An object is projected from the ground with an upward speed of ų m/s has a speed of 23m/s when it is at a height of 5m above the

vovikov84 [41]

Answer:

25.08m/s

Explanation:

mgh1 + 0.5mv1² = mgh2 + 0.5mv2²

h1 = 0m

v1 = u

h2 = 5m

v2 = 23m/s

putting the values into the formula above;

m(10)(0) + 0.5m(u²) = m(10)(5) + 0.5m(23²)

0 + 0.5mu² = 50m + 264.5m

0.5mu² = 314.5m

dividing through by m

0.5u² = 314.5

u² = 629

u = 25.08m/s

Therefore, the initial speed "u" = 25m/s

6 0

2 years ago

How does temperature,barometric pressure,humidity,wind speed and direction, and precipitation determine the weather in a particu

Deffense [45]

Temperature is how hot or cold something is,barometric is air pressure(when the pressures high,the weather is dry)humidity is how moist the air is and how many water particles are there, wind speed and direction is how the hot and cold air is moving and how fast,which makes wind (high to low, hot to cold) precipitation is how much it is raining at that point.

7 0

3 years ago

A rock weighing 98 newtons is pushed off the edge of a bridge 50 meters above the ground. What was the kinetic energy of the roc

Ahat [919]

2,450 Joules, kinetic energy is 1/2 mass x velocity squared.

3 0

3 years ago

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: