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2 years ago

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

Physics

1 answer:

Alik [6]2 years 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 car is traveling with speed v0 when it begins to speed up at a rate of Δv every second. After t1 seconds, the car travels with

Rainbow [258]

Answer:

d = Δv(t2-t1)

Explanation:

Speed is defined as the change of displacement with respect to time. It is expressed as shown;

Speed = change in displacement/change in time

Δv = d/Δt

d = Δv*Δt

d = ΔvΔt

Δt = t2-t1

d = Δv(t2-t1)

Δv is the change in rate of speed

Δt = change in time

The correct expression for the displacement of the car during this motion is d = Δv(t2-t1)

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2 years ago

Why does frost bite occur in mountain climbers ?

san4es73 [151]

Answer:

The higher the altitude the colder

Explanation:

Mountains are very high up obviously higher then the ground. It is colder so it makes it easier for mountain climbers to get frost bite because of the high altitude.

Hope this helps:)

7 0

2 years ago

Read 2 more answers

An excited hydrogen atom releases an electromagnetic wave to return to its normal state. You use your futuristic dual electric/m

aniked [119]

Answer: The released electromagnetic wave will travel in +y direction

Explanation:

It should be noted that, in a situation, whereby an excited hydrogen atom releases an electromagnetic wave to return to its normal state. And it's also evident that the futuristic dual electric/magnetic field tester on the electromagnetic wave to find the directions of the electric field and magnetic field is used. Eventually, your device tells you that the electric field is pointing in the positive y direction and the magnetic field is pointing in the positive x direction. Therefore, the released electromagnetic wave will travel in +y direction.

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2 years ago

A block with mass m = 0.250 kg is attached to one end of an ideal spring and moves on a horizontal frictionless surface. The oth

ehidna [41]

Answer:

13.54 N/m

0.6 m

4.37 m/s

32.496 m/s²

Explanation:

m = Mass of block = 0.25 kg

k = Spring constant

A = Amplitude

x = Compression of spring = 0.24 m

a = Acceleration = -13 m/s²

v = Velocity = 4 m/s

The weight of the block and force on spring is equal

$ma=-kx\\\Rightarrow k=-\frac{ma}{x}\\\Rightarrow k=-\frac{0.25\times -13}{0.24}\\\Rightarrow k=13.54\ N/m$

The spring's force constant is 13.54 N/m

Total energy of the system is given by

$E=\frac{1}{2}(mv^2+kx^2)\\\Rightarrow E=\frac{1}{2}(0.25\times 4^2+13.54\times 0.24^2)\\\Rightarrow E=2.39\ J$

At maximum displacement v = 0

$E=\frac{1}{2}(mv^2+kA^2)\\\Rightarrow E=\frac{1}{2}(0+kA^2)$

$\\\Rightarrow E=\frac{1}{2}kA^2\\\Rightarrow A=\sqrt{\frac{E2}{k}}\\\Rightarrow A=\sqrt{\frac{2\times 2.39}{13.54}}\\\Rightarrow A=0.6\ m$

The amplitude of the motion is 0.6 m

Speed of the block

$E=\frac{1}{2}mv_m^2+0\\\Rightarrow v_m=\sqrt{\frac{2E}{m}}\\\Rightarrow v_m=\sqrt{\frac{2\times 2.39}{0.25}}\\\Rightarrow v_m=4.37\ m/s$

The maximum speed of the block during its motion is 4.37 m/s

Forces in the spring

$ma_m=kA\\\Rightarrow a_m=\frac{kA}{m}\\\Rightarrow a_m=\frac{13.54\times 0.6}{0.25}\\\Rightarrow a_m=32.496\ m/s^2$

Maximum magnitude of the block's acceleration during its motion is 32.496 m/s²

5 0

2 years ago

You have a remote-controlled car that has been programmed to have velocity v⃗ =(−3ti^+2t2j^)m/s, where t is in s. At t = 0 s, th

DiKsa [7]

Answer:

The y-component of the car's position vector is 670m/s.

The x-component of the acceleration vector is -3, and the y-component is 40.

Explanation:

The displacement vector of the car with velocity

$\boldsymbol{v}= (-3t\boldsymbol{i}+2t^2\boldsymbol{j})m/s$

is the integral of the velocity.

Integrating $\boldsymbol{v}$ we get the displacement vector $\boldsymbol{d}$ :

$\boldsymbol{d}=(-\dfrac{3}{2}t^2\boldsymbol{i}+\dfrac{2}{3}t^3\boldsymbol{j} )$

Now if the initial position if the car is

$\boldsymbol{r}= (3.0\boldsymbol{i}+2.0\boldsymbol{j})$

then the displacement of the car at time $t$ is

$\boldsymbol{d(t)}= \boldsymbol{r+d}$

$\boxed{\boldsymbol{d(t)}=(-\dfrac{3}{2}t^2+3.0\boldsymbol{i}+\dfrac{2}{3}t^3+2.0\boldsymbol{j} )}$

Now at $t=10s$ , we have

$\boxed{\boldsymbol{d(t)}=(-147\boldsymbol{i}+670\boldsymbol{j} )}m$

The y-component of the car's position vector is 670m/s.

The acceleration vector is the derivative of the velocity vector:

$\boldsymbol{a(t)}=\dfrac{d\boldsymbol{v(t)}}{dt} =(-3\boldsymbol{i}+4t\boldsymbol{j})$

and at $t=10s$ it is

$\boldsymbol{a(t)}=(-3\boldsymbol{i}+40\boldsymbol{j})m/s^2$

The x-component of the acceleration vector is -3, and the y-component is 40.

5 0

2 years ago