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3 years ago

There are 2 questions into 1

Physics

1 answer:

kiruha [24]3 years ago

3 0

As we know that magnitude of the vector is 670 m

and it is inclined 25 degree with x axis

now we know that component of vector along x direction is given as

$x = L cos\theta$

$x = 670 cos25$

$x = 610 m$

Part 2)

In order to find the other component of the vector we can use

$y = L sin\theta$

$y = 670 sin25$

$y = 280 m$

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If a small rock is dropped from a height of 3.1 m how fast will it be moving when it reaches the ground 0.80 seconds later

garik1379 [7]

X=1/2 at^2
3.1=1/2 a *0.64
a=9.68
v=at
v=0.8*9.6875=7.75

5 0

3 years ago

A motorcycle is following a car that is traveling at a constant speed on a straight highway. Initially, the car and the motorcyc

Artist 52 [7]

Answer:

(a) 3.807 s

(b) 145.581 m

Explanation:

Let Δt = t2 - t1 be the time it takes from the moment when the motorcycle starts to accelerate until it catches up with the car. We know that before the acceleration, both vehicles are travelling at a constant speed. So they would maintain a distance of 58 m prior to the acceleration.

The distance traveled by car after Δt (seconds) at $v_c = 23m/s$ speed is

$s_c = \Delta t v_c = 23\Delta t$

The distance traveled by the motorcycle after Δt (seconds) at $m_m = 23 m/s$ speed and acceleration of a = 8 m/s2 is

$s_m = \Delta t v_m + a\Delta t^2/2$

$s_m = 23\Delta t + 8\Delta t^2/2 = 23 \Delta t + 4 \Delta t^2$

We know that the motorcycle catches up to the car after Δt, so it must have covered the distance that the car travels, plus their initial distance:

$s_m = s_c + 58$

$23 \Delta t + 4 \Delta t^2 = 23\Delta t + 58$

$4 \Delta t^2 = 58$

$\Delta t^2 = 14.5$

$\Delta t = \sqrt{14.5} = 3.807s$

(b)

$s_m = 23 \Delta t + 4 \Delta t^2$

$s_m = 23*3.807 + 58 = 145.581 m$

5 0

3 years ago

Starting from rest, a disk rotates about its central axis with constant angular acceleration. in 6.00 s, it rotates 44.5 rad. du

Klio2033 [76]

a. The disk starts at rest, so its angular displacement at time $t$ is

$\theta=\dfrac\alpha2t^2$

It rotates 44.5 rad in this time, so we have

$44.5\,\mathrm{rad}=\dfrac\alpha2(6.00\,\mathrm s)^2\implies\alpha=2.47\dfrac{\rm rad}{\mathrm s^2}$

b. Since acceleration is constant, the average angular velocity is

$\omega_{\rm avg}=\dfrac{\omega_f+\omega_i}2=\dfrac{\omega_f}2$

where $\omega_f$ is the angular velocity achieved after 6.00 s. The velocity of the disk at time $t$ is

$\omega=\alpha t$

so we have

$\omega_f=\left(2.47\dfrac{\rm rad}{\mathrm s^2}\right)(6.00\,\mathrm s)=14.8\dfrac{\rm rad}{\rm s}$

making the average velocity

$\omega_{\rm avg}=\dfrac{14.8\frac{\rm rad}{\rm s}}2=7.42\dfrac{\rm rad}{\rm s}$

Another way to find the average velocity is to compute it directly via

$\omega_{\rm avg}=\dfrac{\Delta\theta}{\Delta t}=\dfrac{44.5\,\rm rad}{6.00\,\rm s}=7.42\dfrac{\rm rad}{\rm s}$

c. We already found this using the first method in part (b),

$\omega=14.8\dfrac{\rm rad}{\rm s}$

d. We already know

$\theta=\dfrac\alpha2t^2$

so this is just a matter of plugging in $t=12.0\,\mathrm s$ . We get

$\theta=179\,\mathrm{rad}$

Or to make things slightly more interesting, we could have taken the end of the first 6.00 s interval to be the start of the next 6.00 s interval, so that

$\theta=44.5\,\mathrm{rad}+\left(14.8\dfrac{\rm rad}{\rm s}\right)t+\dfrac\alpha2t^2$