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xenn [34]
3 years ago
5

Which of the following is an example of acceleration? I. A car speeds up. II. A car slows down. III. A car travels in a straight

line at a constant speed. IV. A car travels at a constant speed and turns left.
Physics
1 answer:
svetlana [45]3 years ago
6 0
I., II., and IV. are examples of acceleration. III. isn't.
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Why are there variations of traits within a population?
nydimaria [60]

Answer:

The variation and distribution of traits in a population depend on genetic and environmental factors. Genetic variation can result from mutations caused by environmental factors or errors in DNA replication, or from chromosomes swapping sections during meiosis.

Explanation:

Hope this helps!

3 0
3 years ago
Read 2 more answers
You are given an unknown solid substance of mass 500 g and you want to estimate its specific heat. unfortunately, you don’t have
Julli [10]

Answer:

Option C) 2,090 J/(mol K)

Explanation:

Data:

Volume in the beaker = 429 ml

temperature  = 20° C

Density = 789 kg/m³

Equilibrium reading = 429

volume change = 29 ml

                          = 0.029 L

Energy change = mcΔT

                       U + PΔV

7 0
2 years ago
An eagle flew for 3 hours at 115 km/h and 5 hours at 136 km/h. How far did the eagle fly?
Yanka [14]

Answer:

the eagle flew 1,025 km

Explanation:

since the eagle flew for 3 hours at 115 km/h it flew a total of 345 km during that time. During the 5 hours at 136 km/h the eagle flew a total of 680 km.

3 0
3 years ago
Coherent light with wavelength 610 nm passes through two very narrow slits, and theinterference pattern is observed on a screen
andriy [413]

Answer:

= 1220 nm

= 1.22 μm

Explanation:

given data:

wavelength \lambda = 610 nm = 610\times  10 ^{-9} m

distance of screen from slits D = 3 m

1st order bright fringe is 4.84 mm

condition for 1 st bright is

d sin \theta =\lambda     ---( 1)

andtan \theta = \frac{y}{ D}

\theta = tan^{-1}\frac{(y }{D})

= 0.0924 degrees

plug theta value in equation 1 we get

d sin ( 0.0924) = 610 \times 10 ^{-9}

d = 3.78\times 10^{-4} m

condition for 1 st dark fringe

d sin \theta =\frac{λ'}{2}

\lambda '= 2 d sin\theta

= 2λ    since from eq (1)

= 1220 nm

= 1.22 μm

7 0
3 years ago
A uniform thin rod of length 0.400 m and mass 4.40 kg can rotate in a horizontal plane about a vertical axis through its center.
elixir [45]
The rod has a mass of m = 4.4 kg and a length of L = 0.4 m.
Its polar moment of inertia is
J = (mL²)/12
   = (1/12) * [(4.4 kg)*(0.4 m)²]
   = 0.05867 kg-m²

The mass of the bullet is 0.3 g.
If its velocity is v m/s, then its linear momentum is
P = (0.3 x 10⁻³ kg)*(v m/s)
Its linear momentum perpendicular to the rod is
P*sin(60°) = 2.5981 x 10⁻⁴ v (kg-m)/s

The angular momentum about the center of the rod when the bullet strikes is
T = (2.5981 x 10⁻⁴ v (kg-m)/s)*(0.2 m) = 5.1962 x 10⁻⁵ v (kg-m²)/s

Because the bullet lodges into the end of the rod, the combined polar moment of inertia is
J + (0.3 x 10⁻³ kg)*(0.2 m)² = 0.05867 + 1.2 x 10⁻⁵ = 0.0587 kg-m²
The initial angular velocity is ω = 17 rad/s.

Because angular momentum is conserved, therefore
5.1962 x 10⁻⁵ v (kg-m²)/s = (0.0587 kg-m²)*(17 rad/s)
v = 19204 m/s

Answer:  19204 m/s

5 0
2 years ago
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