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Marta_Voda [28]
4 years ago
8

A bicycle with an initial velocity of +6 m/s accelerates at a

Physics
1 answer:
patriot [66]4 years ago
5 0
Answer :
It covers 27 meters
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The stoplights on a street are designed to keep traffic moving at 26 mi/h. the average length of a street block between traffic ligh
bonufazy [111]

We use the formula,

v = \frac{d}{t}.

Here, v is velocity and its value given 26 mi/h ( in m/s,  \frac{26 \times 1.609 \times 10^{3} m}{60 \times 60 s} =11.62 \ m/s ) and d is distance and its value is given 80 m.

Substituting these values in above formula we get,

t = \frac{80 m}{11.62m/s } = 6.88 \ s

Thus, the time delay between green lights on successive blocks to keep the traffic moving continuously is 6.88 s


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3 years ago
In the vacuum of outer space, there is no
natta225 [31]
Atmospheric pressure
7 0
4 years ago
What is gibbs paradox?
mr_godi [17]
This leads to a paradox known as the Gibbs paradox, after Josiah Willard Gibbs. The paradox allows for the entropy of closed systems to decrease, violating the second law of thermodynamics. A related paradox is the "mixing paradox".
7 0
3 years ago
Suppose you are an astronaut and you have been stationed on a distant planet. You would like to find the acceleration due to the
Maksim231197 [3]

Answer:

<h3> 1.40625m/s²</h3>

Explanation:

Using the equation of motion expressed as v = u+gt where;

v is the  final velocity of the ball

u is the initial velocity

g is the acceleration due to gravity

t is the time taken

Given

u = 9m/s

v = 0m/s

t = 6.4s

Required

acceleration due to gravity g

Since the rock is thrown up, g will be a negative value.

v = u+(-g)t

0 = 9-6.4g

-9 = -6.4g

6.4g = 9

divide both sides by 6.4

6.4g/6.4 = 9/6.4

g = 1.40625m/s²

Hence the acceleration due to gravity on the planet is  1.40625m/s²

6 0
3 years ago
a. When the electric field between the plates is 75% of the dielectric strength and energy density of the stored energy is 2800
Olegator [25]

Answer: The value of the dielectric constant k = 1.8

Explanation:

If C= ε A/d and

Electrostatic energy W = 1/2CV^2

Substitutes C in the first formula into the energy formula.

W = 1/2 ε A/d × V^2

Let us remember that electric field strength E is the ratio of potential V to the distance d. Where V = Ed

Substitute V = Ed into the energy W.

W = 1/2 × ε A/d ×( Ed )^2

W = 1/2 × ε A/d × E^2 × d^2

d will cancel one of the ds

W = 1/2 × ε Ad × E^2

W/Ad = 1/2 × ε × E^2

W/V = 1/2 × ε E^2

Where Ad = volume V

E = dielectric strength

εo = permittivity of free space = 8.84 x 10^-12 F/m

W/V = 2800 J/m^3

Let first calculate the dielectric strength

2800 = 1/2 × 8.84×10^-12 × E^2

5600 = 8.84×10^-12E^2

E^2 = 5600/8.84×10^-12

E = sqrt( 6.3 × 10^14)

E = 25 × 10^7

75% of E = 18.9 × 10^6Jm

The permittivity of the material will be achieved by using the same formula

2800 = 1/2 × ε E^2

2800 = 0.5 × ε × (18.9×10^6)^2

2800 = ε × 1.78 × 10^14

ε = 2800/1.78×10^14

ε = 1.57 × 10^-11

Dielectric constant k = relative permittivity

Relative permittivity is the ratio of the permittivity of the material to the permittivity of the vacuum in a free space. That is

k = 1.57×10^-11/8.84×10^-12

k = 1.776

k = 1.8 approximately

Therefore, the value of the dielectric constant k is 1.8

3 0
3 years ago
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