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4 years ago

How did the horizontal velocity vector component change during the flight of the cannonball in the simulation

Physics

1 answer:

arlik [135]4 years ago

7 0

Answer:

The horizontal velocity vector of the canonball does not change at all, but is constant throughout the flight.

Explanation:

First, I'll assume this is a projectile simulation, since no simulation is shown here. That been the case, in a projectile flight, there is only a vertical component force (gravity) acting on the body, and no horizontal component force on the body. The effect of this on the canonball is that the vertical velocity component on the canonball goes from maximum to zero at a deceleration of 9.81 m/s^2, in the first half of the flight. And then zero to maximum at an acceleration of 9.81 m/s^2 for the second half of the flight before hitting the ground. Since there is no force acting on the horizontal velocity vector of the canonball, there will be no acceleration or deceleration of the horizontal velocity component of the canonball. This means that the horizontal velocity component of the canonball is constant throughout the flight

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Which motor and body should Devon use to build the car with the greatest acceleration?

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Complete Question:

Devon has several toy car bodies and motors. The motors have the same mass, but they provide different amounts of force, as shown in this table.

The bodies have the masses shown in this table (refer attached figure).

Which motor and body should Devon use to build the car with the greatest acceleration?

motor 1, with body 1

motor 1, with body 2

motor 2, with body 1

motor 2, with body 2

Answer:

Devon should build the car with motor 2 and body 1 for having the greatest acceleration.

Explanation:

As per Newton's second law of motion, the acceleration of any object is directly proportional to the force on the object and inversely proportional to the mass of the object.

It can be seen that motor 2 has greater force than the force provided by motor 1. Similarly, the mass of body 1 is found to be lesser compared to mass of body 2. So,

$acceleration =\frac{\text { Force }}{\text { mass }}$

It gives, the system with motor 2 and body 1 the maximum acceleration. So the car should be built with motor 2 and body 1.

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3 years ago

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A reaction that releases energy as it occurs is classified as a(n ________.

cluponka [151]

The reaction that releases energy is called an EXOTHERMIC process. This is a word from the prefix exo- which means to exit and from the root word therm which is a unit of heat which in turn is form of energy. This is the opposite process of ENDOTHERMIC process.

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3 years ago

A rock is thrown off a cliff at an angle of 53° with respect to the horizontal. The cliff is 100 m high. The initial speed of th

Nadusha1986 [10]

(a) 129.3 m

The motion of the rock is a projectile motion, consisting of two indipendent motions along the x- direction and the y-direction. In particular, the motion along the x- (horizontal) direction is a uniform motion with constant speed, while the motion along the y- (vertical) direction is an accelerated motion with constant acceleration $g=-9.8 m/s^2$ downward.

The maximum height of the rock is reached when the vertical component of the velocity becomes zero. The vertical velocity at time t is given by

$v(t) = v_0 sin \theta +gt$

where

$v_0 = 30 m/s$ is the initial velocity of the rock

$\theta=53^{\circ}$ is the angle

t is the time

Requiring $v(t)=0$ , we find the time at which the heigth is maximum:

$0=v_0 sin \theta + gt\\t=\frac{-v_0 sin \theta}{g}=-\frac{(30)(sin 53^{\circ})}{(-9.8)}=2.44 s$

The heigth of the rock at time t is given by

$y(t) = h+(v_0 sin\theta) t + \frac{1}{2}gt^2$

Where h=100 m is the initial heigth. Substituting t = 2.44 s, we find the maximum height of the rock:

$y=100+(30)(sin 53^{\circ})(2.44)+\frac{1}{2}(-9.8)(2.44)^2=129.3 m$

(b) 44.1 m

For this part of the problem, we just need to consider the horizontal motion of the rock. The horizontal displacement of the rock at time t is given by

$x(t) = (v_0 cos \theta) t$

where

$v_0 cos \theta$ is the horizontal component of the velocity, which remains constant during the entire motion

t is the time

If we substitute

t = 2.44 s

Which is the time at which the rock reaches the maximum height, we find how far the rock has moved at that time:

$x=(30)(cos 53^{\circ})(2.44)=44.1 m$

(c) 7.58 s

For this part, we need to consider the vertical motion again.

We said that the vertical position of the rock at time t is

$y(t) = h+(v_0 sin\theta) t + \frac{1}{2}gt^2$

By substituting

y(t)=0

We find the time t at which the rock reaches the heigth y=0, so the time at which the rock reaches the ground:

$0=100+(30)(sin 53^{\circ})t+\frac{1}{2}(-9.8)t^2\\0=100+23.96t-4.9t^2$

which gives two solutions:

t = -2.69 s (negative, we discard it)

t = 7.58 s --> this is our solution

(d) 136.8 m

The range of the rock can be simply calculated by calculating the horizontal distance travelled by the rock when it reaches the ground, so when

t = 7.58 s

Since the horizontal position of the rock is given by

$x(t) = (v_0 cos \theta) t$

Substituting

$v_0 = 30 m/s\\\theta=53^{\circ}$

and t = 7.58 s we find:

$x=(30)(cos 53^{\circ})(7.58)=136.8 m$

(e) (36.1 m, 128.3 m), (72.2 m, 117.4 m), (108.3 m, 67.4 m)

Using the equations of motions along the two directions:

$x(t) = (v_0 cos \theta) t$

$y(t) = h+(v_0 sin\theta) t + \frac{1}{2}gt^2$

And substituting the different times, we find:

$x(2.0 s)=(30)(cos 53^{\circ})(2.0)=36.1 m$

$y(2.0 s)= 100+(23.96)(2.0)-4.9(2.0)^2=128.3 m$

$x(4.0 s)=(30)(cos 53^{\circ})(4.0)=72.2 m$

$y(4.0 s)= 100+(23.96)(4.0)-4.9(4.0)^2=117.4 m$

$x(6.0 s)=(30)(cos 53^{\circ})(6.0)=108.3 m$

$y(6.0 s)= 100+(23.96)(6.0)-4.9(6.0)^2=67.4 m$

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3 years ago

A gas is confined to a vertical cylinder by a piston of mass 2 kg and radius 1 cm. When 5J of heat are added, the piston rises b

Anarel [89]

The work done by gas is 0.753 J and change in internal energy is 4.247J

So we are given that mass is 2kg , radius 1 cm and the amount of heat is 5 cm

The piston raised by 2.4cm

As we know that Work done is PΔV

Where ΔV is change in volume

Therefore ΔV = πr^2 h = π x (.01)^2 x .024 =7.53×10^(-6)m^3

Here pressure is 10^5 pa

So W = $10^5\times7.53\times10^-(6)$

Therefore W = 0.753 J

Now coming to change in internal energy

Change in Internal Energy = Heat Added - Energy lost in work

∴ 5J - 0.753 J = 4.247J

Hence the change in internal energy is 4.247 J

Learn more about Work done here

brainly.com/question/16951089

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2 years ago

Object C has a mass of 3,600 kilograms. Object D has a mass of 900 kilograms. Both objects were placed on different planets so t

jek_recluse [69]

The answer is 1/4
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4 years ago

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