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disa [49]
3 years ago
14

You are driving to the grocery store at 14 m/s. You are 115 m from an intersection when the traffic light turns red. Assume that

your reaction time is 0.50 s and that your car brakes with constant acceleration. How far are you from the intersection (in m) when you begin to apply the brakes
Physics
1 answer:
kogti [31]3 years ago
6 0

Answer:

111.5 m

Explanation:

Given that You are driving to the grocery store at 14 m/s. You are 115 m from an intersection when the traffic light turns red. Assume that your reaction time is 0.50 s and that your car brakes with constant acceleration. 

Use first equation of motion

V = U - at

Since the car is going to rest, V = 0 and a = negative

0 = 14 - a × 0.5

0.5a = 14

a = 14 /0.5

a = 28 m/s^2

Let us use second equation of motion

S = Ut - 1/2at^2

S = 14 × 0.5 - 0.5 × 28 × 0.5^2

S = 7 - 3.5

S = 3.5 m

115 - 3.5 = 111.5

Therefore, you are 111.5 metres from the intersection (in m) when you begin to apply the brakes.

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A glass optical fiber is used to transport a light ray across a long distance. The fiber has an index of refraction of 1.550 and
devlian [24]

To solve this exercise it is necessary to apply the concepts related to the Snells law.

The law defines that,

n_1 sin\theta_1 = n_2 sin\theta_2

n_1 = Incident index

n_2 = Refracted index

\theta_1 = Incident angle

\theta_2 = Refracted angle

Our values are given by

n_1 = 1.550

n_2 = 1.361

\theta_2 =90\° \rightarrowRefractory angle generated when light passes through the fiber.

Replacing we have,

(1.55)sin \theta_1 = (1.361) sin90

sin \theta_1 = \frac{(1.361) sin90}{(1.55)}

\theta_1 =sin^{-1} \frac{(1.361) sin90}{(1.55)}

\theta_1 =61.4\°

Now for the calculation of the maximum angle we will subtract the minimum value previously found at the angle of 90 degrees which is the maximum. Then,

\theta_{max} = 90-\theta \\\theta_{max} =90-61.4\\\theta_{max}=28.6\°

Therefore the critical angle for the light ray to remain insider the fiber is 28.6°

6 0
3 years ago
Explain why, if you go to the moon, your weight would change but your mass wouldnt
Evgen [1.6K]
There's little gravity so your weight would change but not your mass
3 0
3 years ago
To suck lemonade of density 1040 kg/m3 up a straw to a maximum height of 4.94 cm, what minimum gauge pressure (in atmospheres) m
Lady bird [3.3K]

Answer:

The minimum gauge pressure is 0.4969 atm.

Explanation:

Given that,

Density = 1040 kg/m³

Height = 4.94 cm

We need to calculate the pressure

Using formula of pressure

P_{g}=\rho g h

Where, \rho=density

h = height

Put the value into the formula

P_{g}=1040\times9.8\times4.94

P_{g}=50348.48\ Pa

Pressure in atmospheres

1\ atm =101.3\ kPa

P_{g}=\dfrac{50348.48}{101325}

P_{g}=0.4969\ atm

Hence, The minimum gauge pressure is 0.4969 atm.

7 0
3 years ago
Read 2 more answers
What is the gravitational force between Mars and the sun? 7.43 × 1030 N 1.79 × 1026 N 1.65 × 1021 N 3.76 × 1032 N
VMariaS [17]

The gravitational force between Mars and the Sun is 1.65\cdot 10^{21} N

Explanation:

The magnitude of the gravitational force between two objects is given by  the equation:

F=G\frac{m_1 m_2}{r^2}

where

G=6.67\cdot 10^{-11} m^3 kg^{-1}s^{-2} is the gravitational constant

m1, m2 are the masses of the two objects

r is the separation between them

In this problem, we have:

m_1 = 1.99\cdot 10^{30} kg is the mass of the Sun

m_2 = 6.39\cdot 10^{23} kg is the mass of Mars

r=229\cdot 10^6 km = 229\cdot 10^9 m is the average distance Mars-Sun

Substituting into the equation, we find the gravitational force:

F=(6.67\cdot 10^{-11})\frac{(1.99\cdot 10^{30})(6.39\cdot 10^{23})}{(229\cdot 10^9)^2}=1.62\cdot 10^{21} N

So, the closest answer is

1.65\cdot 10^{21} N

Learn more about gravitational force:

brainly.com/question/1724648

brainly.com/question/12785992

#LearnwithBrainly

4 0
3 years ago
What is the change in velocity of a 22-kg object that experiences a force of 15 N for
vagabundo [1.1K]

Answer:

Force = mass × acceleration

Acceleration:

{ \tt{15 = (22 \times a)}} \\ { \tt{a =  \frac{15}{22}  \:  {ms}^{ - 2} }}

From first Newton's equation of motion:

{ \bf{v = u + at}}

Change = v - u:

{ \tt{v - u = (a \times t)}} \\ { \tt{v - u = ( \frac{15}{22} \times 1.2) }} \\ { \tt{v - u = 0.82 \:  {ms}^{ - 2} }}

3 0
3 years ago
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