PHYSICS COLLEGE

You are traveling on an interstate highway at the posted speed limit of 70 mph when you see that the traffic in front of you has stopped due to an accident up ahead. You step on your brakes to slow down as quickly as possible. Assume that you to slow down to 30 mph in about 5 seconds. A) With this same average acceleration, how much longer would it take you to stop?B) What total distance would you travel from when you first apply the brakes until the car stops?

Answers

Answer 1

Answer:

A.The time taken for the car to stop is 8.75 s

B.The distance travelled when the brakes were applied till the car stops is 136.89 m

A. Determination of the time taken for the car to stop.

We'llbegin bycalculatingthedecelerationof thecar

Initial velocity (u) = 70 mph = 0.447 × 70 = 31.29 m/s

Final velocity (v) = 30 mph = 0.447 × 30 = 13.41 m/s

Time (t) = 5 s

Deceleration (a) =?

$a \: = (v \: - u)/(t) \n \n a = (13.41 - 31.29)/(5) \n \n a \: = ( - 17.88)/(5) \n \n$

a = –3.576 m/s²

Finally,we shall determine the time taken for the car to stop.

Initial velocity (u) = 31.29 m/s

Final velocity (v) = 0 m/s

Deceleration (a) = –3.576 m/s²

Time (t) =?

$v \: = u \: + at \n 0 \: = 31.29 \: + \: ( - 3.576 * t) \n 0 \: = 31.29 \: - 3.576 * t \n collet \: like \: terms \n 0 - 31.29 \: = - 3.576 * t \n - 31.29 \: = - 3.576 * t \n divide \: both \: side \: by \: - 3.576 \n t \: = (- 31.29)/(- 3.576) \n$

t = 8.75 s

Thus, the time taken for the car to stop is 8.75 s

B.Determination of the total distance travelled when the brakes were applied.

Initial velocity (u) = 31.29 m/s

Final velocity (v) = 0 m/s

Deceleration (a) = –3.576 m/s²

Distance (s) =?

${v}^(2) = {u}^(2) + 2as \n {0}^(2) = {31.29}^(2) + (2 * - 3.576 * s) \n 0 = 979.0641 - 7.152 s \n collect \: like \: terms \n 0 - 979.0641 = - 7.152 s \n - 979.0641 = - 7.152 s \n divide \: both \: side \: by \: - 7.152 \n s = (- 979.0641)/(- 7.152) \n \n$

s = 136.89 m

Therefore, the total distance travelled by the car when the brakes were applied is 136.89 m

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Answer 2

Answer:

Answer:8.75 s,

136.89 m

Explanation:

Given

Initial velocity $=70 mph\approx 31.29 m/s$

velocity after 5 s is $30 mph\approx 13.41 m/s$

Therefore acceleration during these 5 s

$a=(v-u)/(t)$

$a=(13.41-31.29)/(5)=-3.576 m/s^2$

therefore time required to stop

v=u+at

here v=final velocity =0 m/s

initial velocity =31.29 m/s

$0=31.29-3.576* t$

$t=(31.29)/(3.576)=8.75 s$

(b)total distance traveled before stoppage

$v^2-u^2=2as$

$0^2-31.29^2=2* (-3.576)\cdot s$

s=136.89 m

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In constructing a moral argument, the point is toA.
come to a conclusion that leads to reasonable action consistent with moral values.

B.
clarify your own stage of moral reasoning.

C.
identify more prescriptive than descriptive premises.

D.
make certain you are being guided by good intentions and a clear conscience.

Answers

Hello there.

In constructing a moral argument, the point is to

A.
come to a conclusion that leads to reasonable action consistent with moral values.

A.) In constructing a moral argument, the point is to "come to a conclusion that leads to reasonable action consistent with moral values"

Hope this helps!

A 0.26 kg rock is thrown vertically upward from the top of a cliff that is 32 m high. When it hits the ground at the base of the cliff, the rock has a speed of 29 m/s. Assuming that air resistance can be ignore and using conservation of mechanical energy, find (a) the initial speed of the rock and (b) the greatest height of the rock as measured from the base of the cliff.

Answers

Answer:

a) The initial speed of the rock is approximately 14.607 meters per second.

b) The greatest height of the rock from the base of the cliff is 42.878 meters.

Explanation:

a) The rock experiments a free-fall motion, that is a vertical uniform accelerated motion due to gravity, in which air friction and effects of Earth's rotation. By Principle of Energy Conservation we have the following model:

$U_(g,1)+K_(1) = U_(g,2)+K_(2)$ (Eq. 1)

Where:

$U_(g,1)$ , $U_(g,2)$ - Initial and final gravitational potential energies, measured in joules.

$K_(1)$ , $K_(2)$ - Initial and final translational kinetic energies, measured in joules.

By definitions of gravitational potential and translational kinetic energies we expand and simplify the equation above:

$m\cdot g\cdot (y_(1)-y_(2))= (1)/(2)\cdot m\cdot (v_(2)^(2)-v_(1)^(2))$

$g\cdot (y_(1)-y_(2)) = (1)/(2)\cdot (v_(2)^(2)-v_(1)^(2))$ (Eq. 2)

Where:

$g$ - Gravitational acceleration, measured in meters per square second.

$y_(1)$ , $y_(2)$ - Initial and final height, measured in meters.

$v_(1)$ , $v_(2)$ - Initial and final speed of the rock, measured in meters per second.

If we know that $g = 9.807\,(m)/(s^(2))$ , $y_(1) = 32\,m$ , $y_(2) = 0\,m$ and $v_(2) = 29\,(m)/(s)$ , then the equation is:

$\left(9.807\,(m)/(s^(2)) \right)\cdot (32\,m-0\,m) = (1)/(2)\cdot \left[\left(29\,(m)/(s) \right)^(2)-v_(1)^(2)\right]$

$313.824 = 420.5-0.5\cdot v_(1)^(2)$

$0.5\cdot v_(1)^(2) = 106.676$

$v_(1) \approx 14.607\,(m)/(s)$

The initial speed of the rock is approximately 14.607 meters per second.

b) We use (Eq. 1) once again and if we know that $g =9.807\,(m)/(s^(2))$ , $y_(1) = 32\,m$ , $v_(1) \approx 14.607\,(m)/(s)$ and $v_(2) = 0\,(m)/(s)$ , then the equation is:

$\left(9.807\,(m)/(s^(2)) \right)\cdot (32\,m-y_(2)) = (1)/(2)\cdot \left[\left(0\,(m)/(s) \right)^(2)-\left(14.607\,(m)/(s) \right)^(2)\right]$

$313.824-9.807\cdot y_(2) = -106.682$

$9.807\cdot y_(2) = 420.506$

$y_(2) = 42.878\,m$

The greatest height of the rock from the base of the cliff is 42.878 meters.

A barbell spins around a pivot at its center at A. The barbell consists of two small balls, each with mass 450 grams (0.45 kg), at the ends of a very low mass rod of length d = 20 cm (0.2 m; the radius of rotation is 0.1 m). The barbell spins clockwise with angular speed 120 radians/s.What is the speed of ball 1?

Answers

The linear speed of the ball for the circular motion is determined as 12 m/s.

The given parameters;

mass of each ball, m = 450 g = 0.45 kg
length of the rod, L = 0.2 m
radius of the rod, r = 0.1 m
angular speed of the ball, ω = 120 rad/s

The linear speed of the ball is calculated as follows;

v = ωr

where;

ω is the angular speed of the ball
r is the radius of circular motion of the ball

The linear speed of the ball is calculated as follows;

v = ωr

v = 120 x 0.1

v = 12 m/s

Thus, the linear speed of the ball for the circular motion is determined as 12 m/s.

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Answer:

The speed of ball is 12 $(m)/(s)$

Explanation:

Given:

Mass of ball $m = 0.45$ kg

Radius of rotation $r = 0.1$ m

Angular speed $\omega = 120 (rad)/(s)$

Here barbell spins around a pivot at its center and barbell consists of two small balls,

From the formula of speed in terms of angular speed,

$v = r \omega$

Where $v =$ speed of ball

$v = 120 * 0.1$

$v = 12 (m)/(s)$

Therefore, the speed of ball is 12 $(m)/(s)$

A CO2 gun shoots a 0.2 gram round pellet (bb) at 2800 ft/sec, and as the bb leaves the gun it gets charged by friction . If Earths magnetic field points South to North at an intensity of 20 uT, and the bb is shot W->E. Find the charge the bb would need to stay level by balancing out the force of gravity.

Answers

Answer:

So it would need a charge of 0.115C for a upward force to act and cancel the force of gravity.

Explanation:

Fb = Fg

qvb= mg ⇒ q = mg/vB = 0.2 *10∧-3 * 9.8/853.44 * 20 * 10∧-6

= 0.115C

note:2800ft/sec = 853.44m/s

So it would need a charge of 0.115C for a upward force to act and cancel the force of gravity.

g Adjacent rows in the first part of the experiment are found to have potentials of 3.66 V and 4.22 V. If the distance between rows is found to be 0.4 cm, what is the magnitude of the electric field at the location between the rows

Answers

Answer:

$E=140V/m$

Explanation:

If the electric field is uniform, the electric field between two points at potentials $V_1$ and $V_2$ which are separated by a distance d will be given by the formula:

$E=(\Delta V)/(d)$

So in our case, we have $E=(4.22V-3.66V)/(0.004m)=140V/m$

A man with a mass of 65.0 kg skis down a frictionless hill that is 5.00 m high. At the bottom of the hill the terrain levels out. As the man reaches the horizontal section, he grabs a 20.0-kg backpack and skis off a 2.00-m-high ledge. At what horizontal distance from the edge of the ledge does the man land (the man starts at rest)?

Answers

Answer:

The horizontal distance is 4.823 m

Solution:

As per the question:

Mass of man, m = 65.0 kg

Height of the hill, H = 5.00 m

Mass of the backpack, m' = 20.0 kg

Height of ledge, h = 2 m

Now,

To calculate the horizontal distance from the edge of the ledge:

Making use of the principle of conservation of energy both at the top and bottom of the hill (frictionless), the total mechanical energy will remain conserved.

Now,

$KE_(initial) + PE_(initial) = KE_(final) + PE_(final)$