How many meters across is a road sign that has an angular size of 120 arcseconds and is 1 km away?

Sam is observing the velocity of a car at different times. After two hours, the velocity of the car is 54 km/h. After four hours

MariettaO [177]

With that information you can only suppose a uniformly accelerated motion. This is, acceleration is constant.

Then, acceleration = change in velocity / change in time = (58 -54)km/h / 2 h = 4km/h / 2 h = 2 km/h^2

Then the equation for velocity, V is

V = Vo + a*t = Vo + 2 (km/h^2) * t = Vo + 2t

Vo is the initial velocity, which you can find using V = 54km/h and t = -2

Vo = V after 2 hours - a*(2hours) = 54km/h - 2(km/h^2)*2h = 54km/k - 4km/h = 50km/h

Then, the equation is: V = 50 km/h + 2t

Valid for constant acceleration.

5 0

2 years ago

15 Points , Physics HW, can someone please show me step by step how to calculate percent difference for this problem. My numbers

timofeeve [1]

The percent difference between two numbers $x$ and $y$ is given by

$\dfrac{|y-x|}x \times 100\%$

The absolute value is there because we only care about the absolute percent difference, and not taking into account whether we go from $x$ to $y$ or vice versa. If we remove them, we have two possible interpretations of percent difference.

For example, the (absolute) percent difference between 3 and 6 is

$\dfrac{|6-3|}3 \times 100\% = 100\%$

In other words, we add 100% of 3 to 3 to end up with 6. This is the same as the percent difference going from 3 to 6. On the other hand, the percent difference going from 6 to 3 is

$\dfrac{3-6}3\times100\%=-50\%$

which is to say, we take away 50% of 6 away from 6 to end up with 3.

"Make comparisons to object measurements" tells us that the differences should be computed relative to "measurements for object". In other words, take $x$ from the left column and $y$ from the right column.

$\dfrac{|7.1-7.3|}{7.1} \times 100\% \approx 2.82\%$

$\dfrac{|4.8-5.0|}{4.8} \times 100\% \approx 4.17\%$

$\dfrac{|7.2-7.5|}{7.2} \times 100\% \approx 4.17\%$

3 0

2 years ago

:What will be the value of the refractive index of the medium? Critical angle of that medium is 30 degree

earnstyle [38]

Answer:

Let the second medium be air (n₁=1)

The refractive index n₂ of the medium where first medium is air is found (a)

(a) n₂ = 2

Explanation:

Critical angle can be defined as the angle of incidence that provides the angle of refraction of 90°.

Refractive index of a medium can be defined as a number that describes that how fast a light will travel through that medium.

Critical angle and Refractive index are related by:

$\theta_{critical}= sin^{-1}(\frac{n_1}{n_2})$

$sin \theta_{critical}=\frac{n_1}{n_2}$

To find refractive index of medium with respect to air, substitute n₁=1 (Refractive index of air is 1)

Also θ(critical)=30°

Find n₂ :

$sin30= \frac{1}{n_2}\\0.5=\frac{1}{n_2}\\n_2=\frac{1}{0.5}\\n_2=2$

8 0

2 years ago

If you carry out an experiment measuring the weight and mass of objects in one particular location on the earth, what relation w

mario62 [17]

<span>Weight is directly proportional to mass.</span>

4 0

3 years ago

Read 2 more answers

Write a simple rule that will tell a person how many water molecule will be lost while putting monosaccharides together to form

belka [17]

<span>For hydrolysis to monosaccharides, one molecule of a disaccharide needs only one molecule of water. C12H22O11 (sucrose) + H2O = C6H12O6 (glucose) + C6H12O6 (fructose) Structurally, a disaccharide molecule may be viewed as a product formed by the condensation of two molecules of monosaccharides with the elimination of a water molecule. So, only one H2O molecule is needed for the reverse process.</span>

3 0

2 years ago