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3 years ago

What is projectile. what is projectile motion

Physics

2 answers:

GenaCL600 [577]3 years ago

5 0

<h3> Projectile:</h3>

The projectile is any object thrown into space upon which the only acting force is gravity. In other words, the primary force acting on a projectile is gravity. This doesn’t necessarily mean that the other forces do not act on it, just that their effect is minimal compared to gravity. The path followed by a projectile is known as a trajectory. A baseball batted or thrown and the instant the bullet exits the barrel of a gun are all examples of the projectile.

<h3> Projectile Motion:</h3>

When a particle is thrown obliquely near the earth’s surface, it moves along a curved path under constant acceleration that is directed towards the center of the earth (we assume that the particle remains close to the surface of the earth). The path of such a particle is called a projectile and the motion is called projectile motion. Air resistance to the motion of the body is to be assumed absent in projectile motion.

In a Projectile Motion, there are two simultaneous independent rectilinear motions:

Along the x-axis: uniform velocity, responsible for the horizontal (forward) motion of the particle.

Along y-axis: uniform acceleration, responsible for the vertical (downwards) motion of the particle.

Accelerations in the horizontal projectile motion and vertical projectile motion of a particle: When a particle is projected in the air with some speed, the only force acting on it during its time in the air is the acceleration due to gravity (g). This acceleration acts vertically downward. There is no acceleration in the horizontal direction, which means that the velocity of the particle in the horizontal direction remains constant.

zhenek [66]3 years ago

3 0

Explanation:

If an object is given an initial velocity in any direction and then allowed to travel freely under gravity, it is called a projectile motion

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A solid disc with a radius of 5.00 m and a mass of 20.0 kg is initially at rests and lies on the plane of the paper. A smaller s

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Answer:

Explanation:

This problem is based on conservation of angular momentum.

moment of inertia of larger disc I₁ = 1/2 m r² , m is mass and r is radius of disc . I

I₁ = .5 x 20 x 5²

= 250 kgm²

moment of inertia of smaller disc I₂ = 1/2 m r² , m is mass and r is radius of disc . I

I₂ = .5 x 10 x 2.5²

= 31.25 kgm²

3500 rmp = 3500 / 60 rps

n = 58.33 rps

angular velocity of smaller disc ω₂ = 2πn

= 2π x 58.33

= 366.3124 rad /s

applying conservation of angular momentum

I₂ω₂ = ( I₁ +I₂) ω , ω is the common angular velocity

31.25 x 366.3124 = ( 250 +31.25) ω

ω = 40.7 rad / s .

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You attach a 3.10 kg weight to a horizontal spring that is fixed at one end. You pull the weight until the spring is stretched b

dolphi86 [110]

Answer:

0.785 m/s

Explanation:

Hi!

To solve this problem we will use the equation of motion of the harmonic oscillator, <em>i.e.</em>

$x(t) = A cos(\omega t )+B sin(\omega t)$ - (1)

$\frac{dx}{dt}(t) = \omega (B cos(\omega t )- A sin(\omega t)$ - (1)

The problem say us that the spring is released from rest when the spring is stretched by 0.100 m, this condition is given as:

$x(0) = 0.100$

$\frac{dx}{dt}(0) = 0$

Since cos(0)=1 and sin(0) = 0:

$x(0)=A$

$\frac{dx}{dt}(0) = -B\omega$

We get

$A =0.100\\B = 0$

Now it say that after 0.4s the weigth reaches zero speed. This will happen when the sping shrinks by 0.100. This condition is written as:

$x(0.4) = - 0.100$

Since

$x(t) = 0.100 cos(\omega t)\\ -0.100=x(0.4)=0.100cos(\omega 0.4)$

This is the same as:

$-1 = cos(0.4\omega)$

We know that cosine equals to -1 when its argument is equal to:

(2n+1)π

With n an integer

The first time should happen when n=0

Therefore:

π = 0.4ω

ω = π/0.4 -- (2)

Now, the maximum speed will be reached when the potential energy is zero, <em>i.e. </em>when the sping is not stretched, that is when x = 0

With this info we will know at what time it happens:

$0 = x(t) = 0.100cos(\omega t)$

The first time that the cosine is equal to zero is when its argument is equal to π/2

$t_{maxV}=\pi /(2\omega)$

And the velocity at that time is:

$\frac{dx}{dt}(t_{maxV} ) =- 0.100\omega sin(\omega t_{maxV})\\\frac{dx}{dt}(t_{maxV} ) =- 0.100\omega sin(\pi/2)\\$

But sin(π/2) = 1.

Therefore, using eq(2):

$\frac{dx}{dt}(t_{maxV} ) = 0.100*\omega = 0.100\frac{\pi}{0.400} = \pi/4$

And so:

$V_{max} = \pi / 4 =0.785$

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3 years ago

How much extra water does a 147-lb concrete canoe displace compared to an ultralightweight 38-lb kevlar canoe of the same size c

romanna [79]

We use the formula, to calculate the volume of water displaced by concrete canoe,

$V =\frac{W}{\gamma }$

Here, W is the weight of concrete canoe and $\gamma$ is the specific weight of water and its value is $62.4\ lb/ft^3$ .

So,

$V =\frac{ 147\ lb}{62.4\ lb/ft^3}=2.356\ ft^3$ .

Now the volume of water occupied in ultra lightweight kevlar canoe,

$v=\frac{w}{\gamma}$

Here, w is weight of kevlar canoe.

So,

$v=\frac{38\ lb}{62.4\ lb/ft^3} =0.6089\ ft^3$

Thus, the volume of water displaced,

$=V-v=2.356\ ft^3-0.6089\ ft^3=2.19\ ft^3$ .

Hence, the volume of water displaced canoe compared to an ultra-lightweight kevlar canoe is $2.19\ ft^3$

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3 years ago

What is projectile. what is projectile motion​

What is projectile. what is projectile motion