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3 years ago

Give a graph of velocity v, time does a horizontal line represent ?

Physics

1 answer:

riadik2000 [5.3K]3 years ago

8 0

The graph represents an object moving at constant velocity

Explanation:

A graph of velocity (v) versus time (t) of an object is generally used to give information about the motion of an object. In fact, from this graph, it is possible to infer several pieces of information:

The slope of the graph corresponds to the acceleration of the object. In fact, the acceleration of an object is defined as the rate of change of the velocity: $a=\frac{\Delta v}{\Delta t}$ , which also corresponds to the slope of the graph
The area under the curve between two times $t_1, t_2$ corresponds to the distance travelled by the object in that time interval.

In this problem, we have an object whose velocity-time graph corresponds to a horizontal line. We said that the slope of the graph corresponds to the acceleration of the object: since the slope of an horizontal line is zero, this means that the acceleration of this object is zero, and therefore the object is travelling at constant velocity.

Learn more about velocity:

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3 years ago

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A 0.500-kg potato is fired at an angle of 80.0° above the horizontal from a PVC pipe used as a "potato gun" and reaches a height

lions [1.4K]

Answer:

(a) $47.15ms^{-1}$

(b) $2470.13ms^{-2}$

Explanation:

From Newton's second law of motion

F=ma where m is mass, a is acceleration and F is net force

Also from kinetic equation of motion, velocity and displacement are related using equation

$v^{2}=u^{2}+2sa$

Where v is final velocity, u is initial velocity, a is acceleration and s is displacement

From the free body diagram attached, final velocity at maximum height is 0 and initial velocity is $usin80^{o}$

Also, the vertical component can be written as

$v_{y}^{2} }=u_{y}^{2} } -2gs$ The negative sign before 2gs means displacement is opposite the gravitational force

Where g is acceleration due to gravity, $u_{y}$ is vertical component of initial velocity and $v_{y}$ is vertical component of final velocity

Since $v_{y}$ is 0

$u_{y}^{2} } =2gs$

s=110 and g is taken as 9.8

$u_{y}=\sqrt{2*9.8*110}=46.43275$

$u_{y}=46.43ms^{-1}$

Also, it's evident that the vertical component of initial velocity is $u_{y}=u_{i}sin \theta$ where $\theta$ is angle of projection and $u_{i}$ is resultant velocity

Making $u_{i}$ the subject we obtain $u_{i}=\frac {u_{y}}{sin \theta}$

Since $u_{y}$ and $\theta$ are known as $46.43ms^{-1}$ and $80^{o}$ respectively, then $u_{i}=\frac {46.43ms^{-1}}{sin 80^{o}}=47.15ms^{-1}$

Therefore, the velocity of potato is $47.15ms^{-1}$

(b)

Displacement depends on length of tube hence s=0.450m hence going back to kinetic equation $v^{2}=u^{2}+2sa$

The final velocity v is answer in part a which is $47.15ms^{-1}$ , initial velocity u is $0ms^{-1}$ hence the equation is re-written as

$v^{2}=2sa$ and making a the subject we obtain

$a=\frac {v^{2}}{2s}$

$a=\frac {47.15^{2}}{2*0.450}=2470.13ms^{-2}$

Therefore, average acceleration is $2470.13ms^{-2}$

(c)

From Newton's second law of motion, F=ma where m=0.500kg and a is $2470.13ms^{-2}$

Therefore, the average force of potato is

F=0.5*2470.13=1235.06N

F=1235.06N

The weight, W of potato is given by W=mg

Taking R as ratio of average force and weight of potato

$R=\frac {F}{W}=\frac {F}{mg}$ and since F=1235.06, m=0.500kg and g=9.8

$R=\frac {1235.06}{0.500*9.8}$ =252.05

Therefore, ratio of average force to weight is 252.05

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4 years ago

a rocket has a mass 250(10^3) slugs on earth. Specify its mass in si units and its weight in si unites. if the rocket is on the

Katena32 [7]

Answer:

$W_{earth} = 35.74 * 10^6 N$ : Rocket weight on earth

$W_{moon} = 5.91 * 10^6 N$ : Rocket weight on moon

Explanation:

Conceptual analysis

Weight is the force with which a body is attracted due to the action of gravity and is calculated using the following formula:

W = m × g Formula (1)

W: weight

m: mass

g: acceleration due to gravity

The mass of a body on the moon is equal to the mass of a body on the earth

The acceleration due to gravity on a body is different on the moon and on the earth

Equivalences

1 slug = 14.59 kg

Known data

$m_{earth} = m_{moon} = 250 * 10^3 slug = 250* 10^3slug * \frac{14.59kg}{1slug} = 3647.5* 10^3 kg$

$g_{moon}= 1.62 \frac{m}{s^2}$

$g_{earth}= 9.8 \frac{m}{s^2}$

Problem development

To calculate the weight of the rocket on the moon and on earth we replace the data in formula (1):

$W_{earth} = 3647.5* 10^3 kg * 9.8 \frac{m}{s^2} = 35.74 * 10^6 N$ : Rocket weight on earth

$W_{moon} = 3647.5* 10^3 kg * 1.62 \frac{m}{s^2} = 5.91 * 10^6 N$ : Rocket weight on moon

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4 years ago

An object that is dropped from a height H falls with a constant acceleration of g. The final

GuDViN [60]

The height need to change by 4 to double the final velocity.

<h3>Final velocity of the object</h3>

The final velocity of an object during a free fall is related to maximum height of fall as given the equation below.

v = √2gh

v² = 2gh

v²/h = 2g

v₁²/h₁ = v₂²/h₂

when v₂ = 2v₁, change in height is calculated as;

h₂ = h₁v₂²/v₁²

h₂ = (h₁ (2v₁)²) / (v₁)²

h₂ = 4h₁v₁² / v₁²

h₂ = 4h₁

Thus, the height need to change by 4 to double the final velocity.

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4 0

2 years ago

Give a graph of velocity v, time does a horizontal line represent ?​

Give a graph of velocity v, time does a horizontal line represent ?