A block slides down a rough ramp with a 30-degree incline as shown.

A car company wants to ensure its newest model can stop in less than 450 ft when traveling at 60 mph. If we assume constant dece

seraphim [82]

Answer:

The value of acceleration that accomplishes this is 8.61 ft/s² .

Explanation:

Given;

maximum distance to be traveled by the car when the brake is applied, d = 450 ft

initial velocity of the car, u = 60 mph = (1.467 x 60) = 88.02 ft/s

final velocity of the car when it stops, v = 0

Apply the following kinematic equation to solve for the deceleration of the car.

v² = u² + 2as

0 = 88.02² + (2 x 450)a

-900a = 7747.5204

a = -7747.5204 / 900

a = -8.61 ft/s²

|a| = 8.61 ft/s²

Therefore, the value of acceleration that accomplishes this is 8.61 ft/s² .

4 0

2 years ago

1. Two-point charges, QA = +8 μC and QB = -5 μC, are separated by a distance r = 10 cm. What is the magnitude and direction of t

tiny-mole [99]

Explanation:

Charges, $q_1=8\ \mu C=8\times 10^{-6}\ C$

$q_2=-5\ \mu C=-5\times 10^{-6}\ C$

The distance between charges, r = 10 cm = 0.1 m

We need to find the magnitude and direction of the electric force. It is given by :

$F=\dfrac{kq_1q_2}{r^2}\\\\F=\dfrac{9\times 10^9\times 8\times 10^{-6}\times 5\times 10^{-6}}{(0.1)^2}\\\\F=36\ N$

So, the required force between charges is 36 N and it is towards positive charge i.e. +8 μC.

6 0

3 years ago

Nucleus A decays into the stable nucleus B with a half-life of 22.07 s. At t=0 s there are 1,293 A nuclei and no B nuclei. At wh

Alexeev081 [22]

Answer:

29.38 seconds

Explanation:

Half life, T = 22.07 s

No = 1293

Let N be the number of atoms left after time t

N = 1293 - 779 = 514

By the use of law of radioactivity

$N=N_{0}e^{-\lambda t}$

Where, λ is the decay constant

λ = 0.6931 / T = 0.6931 / 22.07 = 0.0314 decay per second

so,

$514=1293e^{-0.0314t}$

$2.5155=e^{0.0314t}$

take natural log on both the sides

0.9225 = 0.0314 t

t = 29.38 seconds

5 0

3 years ago

In some circumstances, it is useful to look at the linear velocity of a point on the blade. The linear velocity of a point in un

mihalych1998 [28]

Answer:

$v=wr$

Explanation:

<u>Tangent and Angular Velocities</u>

In the uniform circular motion, an object describes the same angles in the same times. If $\theta$ is the angle formed by the trajectory of the object in a time t, then its angular velocity is

$\displaystyle w=\frac{\theta}{t}$

if $\theta$ is expressed in radians and t in seconds the units of w is rad/s. If the circular motion is uniform, the object forms an angle $2\theta$ in 2t, or $3\theta$ in 3t, etc. Thus the angular velocity is constant.