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3 years ago

1. How would you expect an instantaneous acceleration vs. time graph to look for a cart moving with a constant

Physics

1 answer:

r-ruslan [8.4K]3 years ago

6 0

Answer:

a=0 v = v₀ + a t

a=0 line is horizontal

Explanation:

1, In a graph of acceleration vs. time, we have lines, when the line is horizontal it is zero, when the line has a positive slope the increasing accelerations and when the slope is negative the decreasing acceleration

2, speed and relationship of a car is given by

v = v₀ + a t

where vo is the initial velocity, a is the acceleration and tel time

in this case I will calcograph velocity vs. time the constant acceleration is a straight line.

In general from the graph we can find the initial velocity with the cut at that x and the acceleration of the car with the slope

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A 2.0-kilogram ball traveling north at 4.0 meters per second collides head on with a 1.0-kilogram ball traveling south at 8.0 me

enyata [817]

Answer:

We know the momentum after the collision MUST be equal to the momentum BEFORE the collision.

Momentum is a VECTOR quantity having both magnitude and direction. The first ball has momentum P =m*v = 2*4 = 8 at 90degrees. The second ball has momentum P = 1*8 = 8 at -90 or 270 degrees. They sum to zero when you perform vector addition.

Explanation:

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3 years ago

A baseball player leads off the game and hits a long home run. The ball leaves the bat at an angle of 70.0 from the horizontal w

professor190 [17]

Answer: 211.059 m

Explanation:

We have the following data:

$\theta=70\°$ The angle at which the ball leaves the bat

$V_{o}=55 m/s$ The initial velocity of the ball

$g=-9.8 m/s^{2}$ The acceleration due gravity

We need to find how far (horizontally) the ball travels in the air: $x$

Firstly we need to know this velocity has two components:

<u>Horizontally:</u>

$V_{ox}=V_{o}cos \theta$ (1)

$V_{ox}=55 m/s cos(70\°)=18.811 m/s$ (2)

<u>Vertically:</u>

$V_{oy}=V_{o}sin \theta$ (3)

$V_{oy}=55 m/s sin(70\°)=51.683 m/s$ (4)

On the other hand, when we talk about parabolic movement (as in this situation) the ball reaches its maximum height just in the middle of this parabola, when $V=0$ and the time $t$ is half the time it takes the complete parabolic path.

So, if we use the following equation, we will find $t$ :

$V=V_{o}+gt=0$ (5)

Isolating $t$ :

$t=\frac{-V_{o}}{g}$ (6)

$t=\frac{-55 m/s}{-9.8 m/s^{2}}$ (7)

$t=5.61 s$ (8)

Now that we have the time it takes to the ball to travel half of is path, we can find the total time $T$ it takes the complete parabolic path, which is twice $t$ :

$T=2t=2(5.61 s)=11.22 s$ (9)

With this result in mind, we can finally calculate how far the ball travels in the air:

$x=V_{ox}T$ (10)

Substituting (2) and (9) in (10):

$x=(18.811 m/s)(11.22 s)$ (11)

Finally:

$x=211.059 m$

8 0

3 years ago

What is the metric unit of force? kg kg/cm^3 kg/cm^2 N

Feliz [49]

Force is found by multiplying mass (kg) and acceleration (m/s^2), so the metric unit of force is kg*m/s^2 or N (newtons)

4 0

3 years ago

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Someone please help me

Lapatulllka [165]

False... I hope that helps ;)

6 0

3 years ago

Which best explains how the diffraction pattern observed in young’s experiment supports the wave theory of light?.

KIM [24]

Answer: the airy pattern can only arise from wave propagation

Explanation:if particles went in straight lines through a slit, they would progate linearly and not interfere. The airy pattern arises from diffraction as waves interfere, producing peaks (constructive interference where peaks of waves from each slit coincide) and troughs (destructive interference where peaks and troughs of waves from each slit cancel out). If intensity rather than field is measured nodes occur where 0 values line up instead of troughs

6 0

2 years ago