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4 years ago

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When a moving car is brought to a stop with the brakes, its acceleration vector is 29. (a) in the same direction as its velocity

vector b) in the opposite direction as its velocity vector c) equal to zero (d) directed downward

1 answer:

professor190 [17]4 years ago

7 0

To develop this problem we will apply the concepts related to the kinematic equations of motion, specifically that of acceleration. Acceleration can be defined as the change of speed in an instant of time, mathematically this is

$\vec{a} = \frac{\vec{v_2}-\vec{v_1}}{\Delta t}$

If a mobile is decreasing its speed (it is slowing down), then its acceleration is in the opposite direction to the movement. This would imply that the acceleration vector is opposite to the velocity vector.

Therefore the correct answer is B.

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horizontal force pushes block up a 20.0 incline with an initial speed 12.0 m//s. a) how high up the plane does slide before comi

Annette [7]

Answer:

Explanation:

We shall apply conservation of mechanical energy .

initial kinetic energy = 1/2 m v²

= .5 x m x 12 x 12

= 72 m

This energy will be spent to store potential energy . if h be the height attained

potential energy = mgh , h is vertical height attined by block

= mg l sin20 where l is length up the inclined plane

for conservation of mechanical energy

initial kinetic energy = potential energy

72 m = mg l sin20

l = 72 / g sin20

= 21.5 m

deceleration on inclined plane = g sin20

= 3.35 m /s²

v = u - at

t = v - u / a

= (12 - 0) / 3.35

= 3.58 s

it will take the same time to come back . total time taken to reach original point = 2 x 3.58

= 7.16 s

7 0

3 years ago

(a) Find the position vector of a particle that has the given acceleration and the specified initial velocity and position. a(t)

dolphi86 [110]

Answer:

$r(t)=(\frac{10t^3}{3}+t)i+(-sint+t+1)j+(\frac{-cos2t}{4}+\frac{1}{4})k$

Explanation:

a(t)=20t i+sin(t) j +cos(2t) k

v(t)= $\int\limits^a_b {a(t)} \, dt$

= $(10t^2+c_1)i+(-cost+c_2)j+\frac{sin2t}{2}k$ ---------------eqn 1

given v(0)=i

i= $c_1i+(-1+c_2)j+(0+c_3)k$

$c_1=1$ $c_2=1$ $c_3=0$

from equation 1

V(t)= $(10t^2+c_1)i+(-cost+c_2)j+\frac{sin2t}{2}k$ ----------eqn 2

now r(t)= $\int\limits^a_b {v (t)} \, dt$

$(\frac{10t^3}{3}+t+c_1)i+(-sint+t+c_2)j+(\frac{-cos2t}{4}+c_3)k$

given r(0)=j

0i+1j+ok = $c_1i+c_2j+(\frac{-1}{4}+c_3)k$

$c_1=0$ $c_2=0$ $c_3 = \frac{1}{4}$

$r(t)=(\frac{10t^3}{3}+t)i+(-sint+t+1)j+(\frac{-cos2t}{4}+\frac{1}{4})k$

8 0

3 years ago

Which principle states that sediments are deposited in sequence unless they are disturbed

Volgvan

The manner in which the sediments are being deposited unless the disturbance by tectonic processes makes it non sequential is theorized with the principle of superposition.

<u>Explanation: </u>

The principle of superposition lets us know that the oldest sediment layer is at the bottom with the newer at the top. For example, xenolith in igneous rock must be older than the rock it contains.

The formation of sedimentary rocks is done via sedimentation occurring in the sequential basis. Due to tectonic process or disruption or even say faults, the layers are sediment in non sequential manner. The entire manner of sequential layering gets turn upside down.

4 0

3 years ago

The resistanceless inductor is connected across the ac source whose voltage amplitude is 24.5 V and angular frequency is 850 rad

timurjin [86]

Answer:

Part A 2.88 A, PART B 28.82 mA PART C 0.288 mA

Explanation:

We have given angular frequency $\omega =850rad/sec$

Voltage source has an amplitude of 24.5 V, So V= 24.5 volt

Part A

We have given inductance $L=10^{-2}H$

So inductive reactance $X_L=\omega L=850\times 10^{-2}=8.5ohm$

So current $i=\frac{V}{X_L}=\frac{24.5}{8.5}=2.8823A$

Part B

We have given inductance $L=1 H$

So inductive reactance $X_L=\omega L=850\times 1=850ohm$

So current $i=\frac{V}{X_L}=\frac{24.5}{850}=28.82mA$

Part C

We have given inductance $L=100 H$

So inductive reactance $X_L=\omega L=850\times 100=85000ohm$

So current $i=\frac{V}{X_L}=\frac{24.5}{85000}=0.288mA$

8 0

3 years ago

Why do disk stars bob up and down as they orbit the galaxy?

Pie

Answer:

When the stars get a little bit farther from the disk, the other stars that are still in the disk will pull them toward it as a consequence of the gravity between them.

Explanation:

Galaxies have different structures due to parameters as the orbit of the stars around the center of the galaxy or the metallicity¹ of the stars (which serves to find the age of the stars). For example, in the case of the Milky Way, it has:

The Halo

The Bulge

The Thick Disk

The Thin Disk

The Halo has old stars with low metallicities and random orbits, while the bulge has stars with the same random motions like the one in the Halo.

The Thick Disk has stars with orbits oriented in the same direction but with metallicities a little bit higher than the one from the Halo but lower than the stars in the Thin Disk.

The stars that are in the thin disk have orbits oriented in the same direction and are the ones with the higher metallicity levels, which means that they are the most younger in the galaxy.

In general the stars in the disk of the galaxy (thick disk and thin disk) goes up and down as they orbit the center of the galaxy, <u>the reason for this is that when they get a little bit farther from the disk, the stars that are still in the disk pull them back toward it as a consequence of the gravity between them. </u>

Remember the equation of the Universal gravitation law:

$F = G\frac{m1.m2}{R^{2}}$ (1)

Where F is the force of gravity, G is the gravitational constant, m1 and m2 are the masses of two objects and R is the distance between them.

Notice how equation 1 expresses that the force of gravity is inversely proportional to the square of the distance between the two objects, which means that the force of gravity will decrease as the square of the distance increase.

Key terms:

¹Metallicity: the abundance of heavier elements against the presence of Helium or Hydrogen.

7 0

3 years ago