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Luden [163]
3 years ago
13

Chemical energy is a form of _____ energy.

Physics
1 answer:
Zepler [3.9K]3 years ago
4 0
The correct answer is C potential energy
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On a trolley ride around an amusement park, a child travelled from one signpost to a second signpost at a constant speed of 125
Tomtit [17]

Answer:

Explanation:

Speed given = 125 m /min

125 /60 m /s

In 450 second it will travel

= 450 x 125 / 60

=937.5 m.  

As the distance  is covered in less than 450 seconds , The distance must be less than 937.5 m

In 400 seconds , it will travel

= 400 x 125 / 60

833.33 m

Since the distance is covered in more than 400 seconds , the distance must be more than ie 833.33 .

Hence the distance covered is more than .833 m but less than 937.5

In either case these distance are more than .8 km .

5 0
3 years ago
Guyz... Its my first question in this app... Pls do answer​
Basile [38]

Answer:

1.414

Explanation:

Snell's law states:

n₁ sin θ₁ = n₂ sin θ₂

where n is the index of refraction and θ is the angle of incidence (relative to the normal).

The index of refraction of air is approximately 1.  So:

1 sin 45° = n sin 30°

n = sin 45° / sin 30°

n = 1.414

Round as needed.

6 0
3 years ago
At the end of a race a runner decelerates from a velocity of 8.90 m/s at a rate of 1.70 m/s2. (a) How far in meters does she tra
Ad libitum [116K]

Answer:

x=22.33m

Explanation:

Kinematics equation for constant deceleration:

x =v_{o}*t - 1/2*at^{2}=8.9*6.3-1/2*1.70*6.3^{2}=22.33m

7 0
3 years ago
How does the law of conservation of energy apply to machines?
trasher [3.6K]
The answer is letter C
7 0
3 years ago
Read 2 more answers
Problem One: A beam of red light (656 nm) enters from air into the side of a glass and then into water. wavelength, c. and speed
Ivanshal [37]

Answer:

Part a)

f_w = f_g = 4.57 \times 10^{14} Hz

Part b)

\lambda_w = 492 nm

\lambda_g = 437.3 nm

Part c)

v_w = 2.25 \times 10^8 m/s

v_g = 2.0 \times 10^8 m/s

Explanation:

Part a)

frequency of light will not change with change in medium but it will depend on the source only

so here frequency of light will remain same in both water and glass and it will be same as that in air

f = \frac{v}{\lambda}

f = \frac{3 \times 10^8}{656 \times 10^{-9}}

f = 4.57 \times 10^{14} Hz

Part b)

As we know that the refractive index of water is given as

\mu_w = 4/3

so the wavelength in the water medium is given as

\lambda_w = \frac{\lambda}{\mu_w}

\lambda_w = \frac{656 nm}{4/3}

\lambda_w = 492 nm

Similarly the refractive index of glass is given as

\mu_w = 3/2

so the wavelength in the glass medium is given as

\lambda_g = \frac{\lambda}{\mu_g}

\lambda_g = \frac{656 nm}{3/2}

\lambda_g = 437.3 nm

Part c)

Speed of the wave in water is given as

v_w = \frac{c}{\mu_w}

v_w = \frac{3 \times 10^8}{4/3}

v_w = 2.25 \times 10^8 m/s

Speed of the wave in glass is given as

v_g = \frac{c}{\mu_g}

v_g = \frac{3 \times 10^8}{3/2}

v_g = 2 \times 10^8 m/s

4 0
3 years ago
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