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3 years ago

A ball is fired from a cannon at point 1 and follows the trajectory shown in the figure below. Air

Physics

1 answer:

AveGali [126]3 years ago

3 0

In trajectory, acceleration vector is not the same with the velocity vector but it depends on the change in velocity. The acceleration path is perpendicular to the velocity vector path. The Correct answers are:

a.) D

b.) B

c.) C

d.) A

A Vector is a quantity that has both magnitude and direction.

Given that a ball is fired from a cannon at point 1 and follows the trajectory shown in the figure.

a.) The vector best represents the ball's velocity at position 2 is D

b.) The vector best represents the ball's acceleration at point 2 is B

c.) Which vector best represents the ball's velocity at position 3 = C

d.) Which vector best represents the ball's acceleration at point = A

In trajectory, acceleration vector is not the same with the velocity vector but depends on the change in velocity. The acceleration path is perpendicular to the velocity vector path.

At point 3, since the ball path is parabolic, the velocity at that point is not equal to zero but will tend to take a horizontal path.

Learn more here: brainly.com/question/25519989

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You launch a cannonball at an angle of 35° and an initial velocity of 36 m/s (assume y = y₁=

velikii [3]

Answer:

Approximately $4.2\; {\rm s}$ (assuming that the projectile was launched at angle of $35^{\circ}$ above the horizon.)

Explanation:

Initial vertical component of velocity:

$\begin{aligned}v_{y} &= v\, \sin(35^{\circ}) \\ &= (36\; {\rm m\cdot s^{-1}})\, (\sin(35^{\circ})) \\ &\approx 20.6\; {\rm m\cdot s^{-1}}\end{aligned}$ .

The question assumed that there is no drag on this projectile. Additionally, the altitude of this projectile just before landing $y_{1}$ is the same as the altitude $y_{0}$ at which this projectile was launched: $y_{0} = y_{1}$ .

Hence, the initial vertical velocity of this projectile would be the exact opposite of the vertical velocity of this projectile right before landing. Since the initial vertical velocity is $20.6\; {\rm m\cdot s^{-1}}$ (upwards,) the vertical velocity right before landing would be $(-20.6\; {\rm m\cdot s^{-1}})$ (downwards.) The change in vertical velocity is:

$\begin{aligned}\Delta v_{y} &= (-20.6\; {\rm m\cdot s^{-1}}) - (20.6\; {\rm m\cdot s^{-1}}) \\ &= -41.2\; {\rm m\cdot s^{-1}}\end{aligned}$ .

Since there is no drag on this projectile, the vertical acceleration of this projectile would be $g$ . In other words, $a = g = -9.81\; {\rm m\cdot s^{-2}}$ .

Hence, the time it takes to achieve a (vertical) velocity change of $\Delta v_{y}$ would be:

$\begin{aligned} t &= \frac{\Delta v_{y}}{a_{y}} \\ &= \frac{-41.2\; {\rm m\cdot s^{-1}}}{-9.81\; {\rm m\cdot s^{-2}}} \\ &\approx 4.2\; {\rm s} \end{aligned}$ .

Hence, this projectile would be in the air for approximately $4.2\; {\rm s}$ .

8 0

2 years ago

Read 2 more answers

A country is deciding what to do about pollution glven off by power plants.

DENIUS [597]

Answer:

option B is the correct answer

Explanation:

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3 years ago

A person of mass m is standing on the surface of the Earth, of mass M E . What is the acceleration that the Earth experiences du

Lana71 [14]

Answer:

$a_E=\dfrac{Gm}{r^2}$

Explanation:

$M_E$ = Mass of the Earth = 5.972 × 10²⁴ kg

G = Gravitational constant = 6.67 × 10⁻¹¹ m³/kgs²

r = Radius of Earth = 6371000 m

m = Mass of person

The force on the person will balance the gravitational force

$M_Ea_E=\dfrac{GmM_E}{r^2}\\\Rightarrow a_E=\dfrac{Gm}{r^2}$

The acceleration that the Earth will feel is $a_E=\dfrac{Gm}{r^2}$

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3 years ago

The combination of all forces that act on an object are (2 points)

qaws [65]

Answer:

A) The net force

Explanation:

If two forces of equal strength act on an object in opposite directions, the forces will cancel, resulting in a net force of zero and no movement.

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4 years ago

What renewable source of energy would be suited to chicago IL?

Kay [80]

Answer:

Wind is the primary renewable resource used for electric power generation in the state. In 2019, wind provided 97% of the state's renewable energy generation, and Illinois was sixth in the nation in utility-scale (1 megawatt or greater) wind capacity, with about 5,200 megawatts online.

Explanation:

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3 years ago