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1 year ago

A car travels 10 m/s east. Another car travels 10 m/s north. The relative speed of the first car with respect to the second is:

Physics

1 answer:

Thepotemich [5.8K]1 year ago

3 0

Answer:

d. less than 20m/s

Explanation:

To the 2nd car, the first car is travelling 10m/s east and 10m/s south. So the total velocity of the first car with respect to the 2nd car is

[tex]\sqrt{10^2 + 10^2} =10\sqrt{2}=14.14m/s

As 14.14m/s is less than 20m/s. d is the correct selection for this question.

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The speed of light of a ray of light traveling through a distance having an absolute index of refraction of 1.1 is?

kodGreya [7K]

<span>The speed of ray of light traveling through a substance whose index of refraction 1.1 is

(c/1.1) = (299,792,458 m/s) / 1.1 = </span>272,538,598 m/s

4 0

2 years ago

If a rock is thrown upward on the planet Mars with a velocity of 18 m/s, its height (in meters) after t seconds is given by H =

lukranit [14]

Answer:

a) The velocity of the rock after 1 s is 14.28 m/s.

b) The velocity of the rock after "a" seconds is 18 m/s - 3.72 m/s² · a

c) The rock will hit the ground after 9.7 s.

d) The velocity with which the rock hits the surface is -18.1 m/s

Explanation:

Hi there!

Let´s write the equation:

H(t) = 18 m/s · t - 1.86 m/s² · t²

The velocity is the variaiton of the height over time, in other words, it is the derivative of the height function with respect to time. Then:

v(t) = dH/dt = 18 m/s - 2 · 1.86 m/s² · t

v(t) = 18 m/s - 3.72 m/s² · t

a) When t = 1 s:

v(1 s) = 18 m/s - 3.72 m/s² · 1 s

v(1 s) = 14.28 m/s

b) When t = a:

v(a) = 18 m/s - 3.72 m/s² · a

c) When the rock hits the surface, the height of the rock is 0:

H(t) = 18 m/s · t - 1.86 m/s² · t²

0 = 18 m/s · t - 1.86 m/s² · t²

0 = t(18 m/s - 1.86 m/s² · t)

t = 0

and

18 m/s - 1.86 m/s² · t = 0

t = -18 m/s /- 1.86 m/s²

t = 9.7 s

d) Let´s use the equation of velocity to find the velocity at t = 9.7 s when the rock hits the ground.

v(t) = 18 m/s - 3.72 m/s² · t

v(9.7 s) = 18 m/s - 3.72 m/s² · 9.7 s

v(9.7 s) = -18.1 m/s

4 0

1 year ago

An object initially at rest experiences a constant horizontal acceleration due to the action of a resultant force applied for 10

Marianna [84]

Answer:

a = 18.28 ft/s²

Explanation:

given,

time of force application, t= 10 s

Work = 10 Btu

mass of the object = 15 lb

acceleration, a = ? ft/s²

1 btu = 778.15 ft.lbf

10 btu = 7781.5 ft.lbf

$m = \dfrac{15}{32.174}\ slug$

m = 0.466 slug

now,

work done is equal to change in kinetic energy

$W = \dfrac{1}{2} m (v_f^2-v_i^2)$

$7781.5 = \dfrac{1}{2}\times 0.466\times v_f^2$

$v_f = 182.75\ ft/s$

now, acceleration of object

$a = \dfrac{v_f-v_o}{t}$

$a = \dfrac{182.75-0}{10}$

a = 18.28 ft/s²

constant acceleration of the object is equal to 18.28 ft/s²

3 0

1 year ago

Type the correct answer in each box. Use numerals instead of words. If necessary, use / for the fraction bar. In a survey conduc

jarptica [38.1K]

Answer: The expected population that wants flexible working hours in ranging from 58% to 62 %.

Explanation:

Percentage of sample respondents agreed for flexible working hours = 60%

The margin of the error = ± 2%

The expected population that wants flexible working hours in ranging from:

$\text{percentage of agreed}-2\%$ lower range

$\text{percentage of agreed}+\2%$ upper range

$60\%-2\% to 60\%+2\%$ that is 58% to 62 %.

The expected population that wants flexible working hours in ranging from 58% to 62 %.

3 0

2 years ago

Our two intrepid relacar drivers are named Pam and Ned. We use these names to make it easy to remember: measurements made by Pam

zhannawk [14.2K]

The velocity of Ned as measured by Pam is the interpretation of v.

Answer: Option D

<u>Explanation:</u>

According to question, we know that this is an issue depending on the logical and translation of the factors. From the measured information taken what is gathered by the two people is communicated and we have given as:

The Ned reference framework : (x, t)

The Pam reference framework : $\left(x^{\prime}, t^{\prime}\right)$

From the reference framework, we realize that ν is the speed of Pam (the other reference framework) as estimation by Ned.

At that point, $v^{\prime}$ is the speed of Ned (from the other arrangement of the reference) as estimation by Pam.

3 0

2 years ago