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3 years ago

6

How does Connecticut energy affect the stopping distance of a vehicle traveling at 30 mph compared to the same vehicle traveling

at 60 mph?

1 answer:

kotegsom [21]3 years ago

6 0

Formula for stopping distance can be calculated by

$v_f^2 - v_i^2 = 2 a d$

here

$v_f = 0$

$d = \frac{v_f^2}{2a}$

now we can see from above equation that stopping distance depends on square of the speed.

So as we can see that here 30 mph and 60 mph, if the speed is double then the stopping distance will become four times more.

You might be interested in

a projectile is shot horizontally from the edge of a cliff, 230m above the ground. the projectile lands 300m from base of the cl

bekas [8.4K]

Answer:

The time taken by the projectile to hit the ground is 6.85 sec.

Explanation:

Given that,

Vertical height of cliff = 230 m

Distance = 300 m

Suppose, determine the time taken by the projectile to hit the ground.

We need to calculate the time

Using second equation of motion

$s=ut+\dfrac{1}{2}gt^2$

Where, s = vertical height of cliff

u = initial vertical velocity

g = acceleration due to gravity

Put the value in the equation

$230=0+\dfrac{1}{2}\times9.8\times t^2$

$t=\sqrt{\dfrac{230\times2}{9.8}}$

$t=6.85 sec$

Hence, The time taken by the projectile to hit the ground is 6.85 sec.

7 0

3 years ago

If a soap bubble is 115 nm thick, what wavelength is most strongly reflected at the center of the outer surface when illuminated

sp2606 [1]

Answer:

$611.8$ nm

Explanation:

$n_{oil}$ = Index of refraction of soap bubble = 1.33

$t$ = thickness of the soap bubble = 115 nm = 115 x 10⁻⁹ m

$\lambda$ = wavelength of light = ?

$m$ = order = 0

For reflection , the necessary condition is

$2 n_{oil} t = (m + 0.5) \lambda$

$2 (1.33)(115\times 10^{-9})= (0 + 0.5) \lambda$

$\lambda = 6.118\times 10^{-7}$

$\lambda = 611.8$ nm

7 0

3 years ago

The atmosphere of Jupiter is essentially made up of hydrogen, H2. For H2, the specific gas constant is 4157 J/(kg K). The accele

Alenkinab [10]

Answer:

h=17357.9m

Explanation:

The atmospheric pressure is just related to the weight of an arbitrary column of gas in the atmosphere above a given area. So, if you are higher in the atmosphere less gass will be over you, which means you are bearing less gas and the pressure is less.

To calculate this, you need to use the barometric formula:

$P=P_0e^{-\frac{Mg}{RT}h}$

Where R is the gas constant, M the molar mass of the gas, g the acceleration of gravity, T the temperature and h the height.

Furthermore, the specific gas constant is defined by:

$R_{H_2}=\frac{R}{M}$

Therefore yo can write the barometric formula as:

$P=P_0e^{-\frac{g}{R_{H_2}T}h}$

at the surface of the planet (h =0) the pressure is $P_0[\tex]. The pressure at the height requested is half of that:[tex]P=\frac{P_0}{2}$

applying to the previuos equation:

$\frac{P_0}{2} =P_0e^{-\frac{g}{R_{H_2}T}h}$

solving for h:

h=17357.9m

3 0

4 years ago

How much charge can be added to each of the plates before a spark jumps between the two plates? For such flat electrodes, assume

Svetach [21]

Answer:

$9.56\cdot 10^{-7} C$

Explanation:

A parallel-plate capacitors consist of two parallel plates charged with opposite charge.

Since the distance between the plates (1 cm) is very small compared to the side of the plates (19 cm), we can consider these two plates as two infinite sheets of charge.

The electric field between two infinite sheets with opposite charge is:

$E=\frac{\sigma}{\epsilon_0}$

where

$\sigma=\frac{Q}{A}$ is the surface charge density, where

Q is the charge on the plate

A is the area of the plate

$\epsilon_0 = 8.85\cdot 10^{-12}F/m$ is the vacuum permittivity

In this problem:

- The side of one plate is

L = 19 cm = 0.19 m

So the area is

$A=L^2=(0.19)^2=0.036m^2$

Here we want to find the maximum charge that can be stored on the plates such that the value of the electric field does not overcome:

$E=3\cdot 10^6 N/C$

Substituting this value into the previous formula and re-arranging it for Q, we find the charge:

$E=\frac{Q}{A\epsilon_0}\\Q=EA\epsilon_0 = (3\cdot 10^6)(0.036)(8.85\cdot 10^{-12})=9.56\cdot 10^{-7} C$

7 0

4 years ago

A group of astrologers have written a book about how to use the stars to help you choose the right friends. This could be consid

Mrac [35]

Answer:b- Psuedoscience

Explanation:

It is a pseudoscience because it involves beliefs and practices which is mistakenly regarded as being based on scientific grounds.

No solid evidence is there which can prove that astrology is authentic and true that's why it is a Psuedo science

8 0

3 years ago