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3 years ago

1. The speed limit on a particular freeway is 28.0 m/s (about 101 km/hour). A car

Physics

1 answer:

zavuch27 [327]3 years ago

3 0

Answer:

the speed limit 6 seconds and the car will travel in 90m

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Which of these is a chemical control used to help control invasive species?

astraxan [27]

Answer:

Explanation:

Using fences.................................

4 0

2 years ago

Part A (4 pts) Consider light of wavelength λ = 670nm traveling in air. The light is incident at normal incidence upon a thin fi

s344n2d4d5 [400]

Answer:

Explanation:

On both sides of the film , the mediums have lower refractive index.

for interfering pattern from above , for constructive interference of reflected wave from both sides of the film , the condition is

2μt = ( 2n +1 ) λ / 2

μ is refractive index of film ,t is thickness of film λ is wavelength of light

n is order of fringe

for minimum thickness

n = 0

2μt = λ / 2

t = λ / 4μ

= 670 / 1.75 x 4

= 95.71 nm .

6 0

3 years ago

A spiracle polish metallic ball is uses a diverging mirror with a focal length of -20 cm a bird 25 cm tall standing 50 cm away L

Aleksandr-060686 [28]

Answer:

1/f = 1/i + 1/o thin lens equation

1/i = 1/f - 1/o

i = o * f / ( o - f) = 50 * (-20) / (50 - (-20)) = -14.3 cm

The final image is erect and 14.3 cm behind the curved surface

M = -o / i = 14.3 / 50 = .29 magnificaton of object

S = .29 * 25 cm = 7.1 cm appearance of bird in mirror (height)

5 0

2 years ago

A 0.20-kg mass is oscillating on a spring over a horizontal frictionless surface. When it is at a displacement of 2.6 cm for equ

valentinak56 [21]

Explanation:

The given data is as follows.

mass = 0.20 kg

displacement = 2.6 cm

Kinetic energy = 1.4 J

Spring potential energy = 2.2 J

Now, we will calculate the total energy present present as follows.

Total energy = Kinetic energy + spring potential energy

= 1.4 J + 2.2 J

= 3.6 Joules

As maximum kinetic energy of the object will be equal to the total energy.

So, K.E = Total energy

= 3.6 J

Also, we know that

K.E = $\frac{1}{2}mv^{2}_{m}$

or, v = $\sqrt{\frac{2K.E}{m}}$

= $\sqrt{2 \times 3.6 J}{0.2 kg}$

= $\sqrt{36}$

= 6 m/s

thus, we can conclude that maximum speed of the mass during its oscillation is 6 m/s.

4 0

3 years ago

Select the correct answer.

Gelneren [198K]

Answer: It's b and c I got it right

Explanation:

Hope this helped!!!! :)

8 0

2 years ago