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ryzh [129]
3 years ago
14

A car starts from rest at a stop sign. It accelerates at 3.8 m/s 2 m/s2 for 6.0 s, coasts for 1.6 ss , and then slows down at a

rate of 3.3 m/s 2 m/s2 for the next stop sign.
How far apart are the stop signs?
Physics
1 answer:
algol [13]3 years ago
7 0

To solve this problem we will start by calculating the distance traveled while relating the first acceleration in the given time. From that acceleration we will calculate its final speed with which we will calculate the distance traveled in the second segment. With this speed and the acceleration given, we will proceed to calculate the last leg of its route.

Expression for the first distance is

s_1 = ut +\frac{1}{2} at^2

s_1 = 0+\frac{1}{2} (3.8)(6)^2

s_1 = 68.4m

The expression for the final speed is

v = v_0 +at

v = 0+(3.8)(6)

v = 22.8m/s

Then the distance becomes as follows

s_2 = vt

s_2 = (22.8)(1.6)

s_2 = 36.48m

The expression for the distance at last sop is

v_1^2=v_0^2 +2as_3

22.8^2 = 0+2(3.3)s_3

s_3 =78.7636m

Therefore the required distance between the signs is,

S = s_1+s_2+s_3

S = 68.4+36.48+78.76

S = 183.64m

Therefore the total distance between signs is 183.54m

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If someone were monitoring your vital signs (like your heart rate, oxygen content in your blood, blood pressure, etc.) every hou
Lilit [14]

Answer:

Accuracy

Explanation:

I think accuracy is more important. When it comes to vital organs in the body, the exactness of getting the measurement is paramount. Accuracy deals with getting very close, almost exact you may say, to a known standard. Precision on the other hand, deals with how easy a measurement can be retaken, reproduced or remade, irrespective of how far or close they are from the accepted norm.

From this, we can agree that precision neglects the most important factor, closeness or say, exactness. Precision isn't bothered by it. And while that can be excused in a few instances, it certainly can not be permitted when it comes to life, or organs of the body

7 0
3 years ago
How fast is a car going if it accelerates at 10m/s/s for 4 seconds
lara31 [8.8K]

Answer:

40 ms-1

Explanation:

v=u+at

v=0+10×4

v=40

5 0
3 years ago
A projectile is fired vertically from Earth's surface with
scoray [572]

Answer:

h=25.52\times10^6 m

Explanation:

Initial speed, v = 10 x 10^3 m/s

Mass of the earth, M = 6 x 10^24 kg

Radius of the earth, R = 6.4 x 10^6 m

Maximum from the surface of earth, h = ?

Let  m = Mass of the projectile

Solution:

Potential energy at maximum height =  ( Potential + Kinetic energy ) at the surface

-G M m / ( R + h )=- G M m / R + (1/2) m v^2

- G M / ( R + h ) = - G M / R + (1/2) v^2

-2\times G M / ( R + h ) = ( - 2 G M / R ) + v^2

-2\times6.67\times10^{-11}\times6\times10^{24}/ ( R + h )

=( (- 2\times 6.67\times10^{-11}\times6\times10^{24}) /(6.4\times10^6)} +10000^2

=-2.50625\times10^7 J

=- 8\times10^{14} / ( R + h )=-2.50625\times 10^7

R+h=31.92\times10^{6}

h=31.92\times10^{6}-6.4\times10^6

h=25.52\times10^6 m

5 0
3 years ago
How do you measure the wavelength of wave?
Tasya [4]

Explanation:

by finding the distance between two successful crest

6 0
3 years ago
Read 2 more answers
An ideal spring hangs from the ceiling. A 2.15 kg mass is hung from the spring, stretching the spring a distance d = 0.0895 m fr
Igoryamba

Answer:

The kinetic energy of the mass at the instant it passes back through the equilibrium position is 0.06500 J.

Explanation:

Given that,

Mass = 2.15 kg

Distance = 0.0895 m

Amplitude = 0.0235 m

We need to calculate the spring constant

Using newton's second law

F= mg

Where, f = restoring force

kx=mg

k=\dfrac{mg}{x}

Put the value into the formula

k=\dfrac{2.15\times9.8}{0.0895}

k=235.41\ N/m

We need to calculate the kinetic energy of the mass

Using formula of kinetic energy

K.E=\dfrac{1}{2}mv^2

Here, v = A\omega

K.E=\dfrac{1}{2}m\times(A\omega)^2

Here, \omega=\sqrt{\dfrac{k}{m}}^2

K.E=\dfrac{1}{2}m\times A^2\sqrt{\dfrac{k}{m}}^2

K.E=\dfrac{1}{2}kA^2

Put the value into the formula

K.E=\dfrac{1}{2}\times235.41\times(0.0235)^2

K.E=0.06500\ J

Hence, The kinetic energy of the mass at the instant it passes back through the equilibrium position is 0.06500 J.

8 0
3 years ago
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