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1 year ago

(b) Figure 4 shows a car travelling on a motorway.

Physics

1 answer:

Alik [6]1 year ago

5 0

Answer:

To calculate anything - speed, acceleration, all that - we need data. The more data we have, and the more accurate that data is, the more accurate our calculations will be. To collect that data, we need to measure it somehow. To measure anything, we need tools and a method. Speed is a measure of distance over time, so we'll need tools for measuring time and distance, and a method for measuring each.

Conveniently, the lamp posts in this problem are equally spaced, and we can treat that spacing as our measuring stick. To measure speed, we'll need to bring time in somehow too, and that's where the stopwatch comes in. A good method might go like this:

Press start on the stopwatch right as you pass a lamp post
Each time you pass another lamp post, press the lap button on the stopwatch
Press stop after however many lamp posts you'd like, making sure to hit stop right as you pass the last lamp post
Record your data
Calculate the time intervals for passing each lamp post using the lap data
Calculate the average of all those invervals and divide by 40 m - this will give you an approximate average speed

Of course, you'll never find an *exact* amount, but the more data points you have, the better your approximation will become.

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A toy rocket is launched vertically from ground level (y = 0 m), at time t = 0.0 s. The rocket engine provides constant upward a

Dvinal [7]

Answer:

2.13 s

Explanation:

Hi!

At t = 0s the rocket is at rest in its platform, so the intial speed is zero. I f the acceleration is A, then the height Y, and the speed V are:

$Y=\frac{A}{2}t^2$

$V=At$

We nedd to find time T during which the rocket engine provides upward acceleration. We know that:

$64\;m=\frac{A}{2}T^2\\ 60\frac{m}{s} =AT\\$

With these 2 equations we can find A and T (dropping units for simplicity):

$A=\frac{60}{T} \\64 =\frac{30}{T} T^2=30T\\T=\frac{64}{30}\approx 2.13\;s$

4 0

2 years ago

(a) On the axes below, sketch the graphs of the horizontal and vertical components of the sphere’s velocity as a function of tim

jeka94

Answer:

Two identical spheres are released from a device at time t = 0 from the same ... Sphere A has no initial velocity and falls straight down. ... (b) On the axes below, sketch and label a graph of the horizontal component of the velocity of sphere A and of sphere B as a function of time. ... Which ball has the greater vertical velocity

Explanation:

4 0

1 year ago

The robot arm is elevating and extending simultaneously. At a given instant, θ = 30°, ˙ θ = 10 deg / s = constant θ˙=10 deg/s=co

motikmotik

Explanation:

The position vector r:

$\overrightarrow{r(t)}=lcos\theta\hat{i}+lsin\theta\hat{j}$

The velocity vector v:

$\overrightarrow{v(t)}=\overrightarrow{\frac{dr}{dt}}=\dot{l}cos\theta-lsin\theta\dot{\theta}\hat{i}+\dot{l}sin\theta+lcos\theta\dot{\theta}\hat{j}$

The acceleration vector a:

$\overrightarrow{a(t)}}=cos\theta(\ddot{l}-l\dot{\theta}^2)-sin\theta(2\dot{l}\dot{\theta}+l\ddot{\theta})\hat{i}+cos\theta(2\dot{l}\dot{\theta}+l\ddot{\theta})+sin\theta(\ddot{l}-l\dot{\theta}^2)\hat{j}$

$\overrightarrow{v(t)}=0.13\hat{i}+0.18\hat{j}$

$\overrightarrow{a(t)}}=-0.3\hat{i}-0.1\hat{j}$

5 0

1 year ago

An electron is in motion at 4.0 × 106 m/s horizontally when it enters a region of space between two parallel plates, as shown, s

max2010maxim [7]

Answer:

xmax = 9.5cm

Explanation:

In this case, the trajectory described by the electron, when it enters in the region between the parallel plates, is a semi parabolic trajectory.

In order to find the horizontal distance traveled by the electron you first calculate the vertical acceleration of the electron.

You use the Newton second law and the electric force on the electron:

$F_e=qE=ma$ (1)

q: charge of the electron = 1.6*10^-19 C

m: mass of the electron = 9.1*10-31 kg

E: magnitude of the electric field = 4.0*10^2N/C

You solve the equation (1) for a:

$a=\frac{qE}{m}=\frac{(1.6*10^{-19}C)(4.0*10^2N/C)}{9.1*10^{-31}kg}=7.03*10^{13}\frac{m}{s^2}$

Next, you use the following formula for the maximum horizontal distance reached by an object, with semi parabolic motion at a height of d:

$x_{max}=v_o\sqrt{\frac{2d}{a}}$ (2)

Here, the height d is the distance between the plates d = 2.0cm = 0.02m

vo: initial velocity of the electron = 4.0*10^6m/s

You replace the values of the parameters in the equation (2):

$x_{max}=(4.0*10^6m/s)\sqrt{\frac{2(0.02m)}{7.03*10^{13}m/s^2}}\\\\x_{max}=0.095m=9.5cm$

The horizontal distance traveled by the electron is 9.5cm

4 0

2 years ago

At time t, gives the position of a 3.0 kg particle relative to the origin of an xy coordinate system ( ModifyingAbove r With rig

Elena-2011 [213]

Complete Question

The complete Question is shown on the first uploaded image

Answer:

The torque acting on the particle is $\tau = 48t \r k$

The magnitude of the angular momentum increases relative to the origin

Explanation:

From the equation we are told that

The position of the particle is $\= r = 4.0 t^2 \r i - (2.0 t - 6.0 t^2 ) \r j$

The mass of the particle is $m = 3.0 kg$

The time is t

The torque acting on the particle is mathematically represented as

$\tau = \frac{ d \r l }{dt}$

where $\r l$ is change in angular momentum which is mathematically represented as

$\r l = m (\r r \ \ X \ \ \r v)$

Where X mean cross- product

$\r v$ is the velocity which is mathematically represented as

$\r v = \frac{d \r r }{dt}$

Substituting for $\r r$

$\r v = \frac{d }{dt} [ 4 t^2 \r i - (2t + 6t^2 ) \r j]$

$\r v = 8t \r i - (2 + 12 t) \r j$

Now the cross product of $\r r \ and \ \r v$ is mathematically evaluated as

$\r r \ \ X \ \ \r v = \left[\begin{array}{ccc}{\r i}&{\r j}&{\r k}\\{4t^2}&{-2t -6t^2}&0\\{8t}&{-2 -12t}&0\end{array}\right]$

$= 0 \r i + 0 \r j + (- 8t^2 -48t^3 + 16t^2 + 48t^3 ) \r k$

$\r r \ \ X \ \ \r v = 8t^2 \r k$

So the angular momentum becomes

$\r l = m (8t^2 \r k)$

Substituting for m

$\r l = 3 * (8t^2 \r k)$

$\r l =24t^2 \r k$

Substituting into equation for torque

$\tau = \frac{d}{dt} [24t^2 \r k]$

$\tau = 48t \r k$

The magnitude of the angular momentum can be evaluated mathematically as

$|\r l| = \sqrt{(24 t^2) ^2}$

$|\r l| = 24 t^2$

From the is equation we see that the magnitude of the angular momentum is varies directly with square of the time so it would relative to the origin because at the origin t= 0s and we move out from origin t increases hence angular momentum increases also

4 0

2 years ago