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AleksAgata
7 days ago
14

Stella is driving down a steep hill. She should keep her car __________ to help _________. a. light, it speed up in a higher b.

gear, slow her vehicle in a lower c. gear, slow her vehicle in a lower d. gear, speed up her vehicle
Physics
2 answers:
inna [2.7K]7 days ago
6 0
b. gear, slow her vehicle in a lower. c. The appropriate choice remains: When driving downhill, utilizing a lower gear enhances engine braking effect and mitigates the risk of brake overheating. Operating in a lower gear causes the engine to rev higher than typical, which is part of the engine braking system. This is crucial for handling steep inclines because excessive brake use can result in failures.
Ostrovityanka [2.8K]7 days ago
3 0
Both B and C are correct since they present identical options proposed in the question. Utilizing a lower gear in a vehicle aids in managing speed while descending a hill. It also helps preserve the brakes, which can overheat and lead to failures if overused on slopes. Shifting to a lower gear allows the engine to help reduce speed, although brakes may still be applied with less force. In this scenario, Stella should downshift to maintain control of her speed while descending the hill for safety.
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Use Wien’s Law to calculate the peak wavelength of Betelgeuse, based on the temperature found in Question #8. Note: 1 nanometer
inna [2732]

The peak wavelength for Betelgeuse is 828 nm

Explanation:

Wien's law describes how the surface temperature relates to a star’s peak wavelength:

\lambda=\frac{b}{T}

where

\lambda represents the peak wavelength

T is the surface temperature

b=2.898\cdot 10^{-3} m\cdot Kis Wien's constant

For Betelgeuse, the surface temperature is roughly

T = 3500 K

Consequently, its peak wavelength can be determined as:

\lambda=\frac{2.898\cdot 10^{-3}}{3500}=8.28\cdot 10^{-7} m = 828 nm

Learn more about wavelength:

8 0
16 days ago
A proton (mass = 1.67 10–27 kg, charge = 1.60 10–19 C) moves from point A to point B under the influence of an electrostatic for
serg [3228]

Answer:

20.353125 V

Explanation:

m = Mass of proton = 1.67\times 10^{-27}\ kg

q = Charge of proton = 1.6\times 10^{-19}\ C

v_A = Velocity of proton at point A = 50 km/s

v_B = Velocity of proton at point B = 80 km/s

The relationship derived from energy conservation is as follows:

\dfrac{1}{2}m(v_B^2-v_A^2)=q(V_B-V_A)\\\Rightarrow V_B-V_A=\dfrac{1}{2q}m(v_B^2-v_A^2)\\\Rightarrow V_B-V_A=\dfrac{1}{2\times 1.6\times 10^{-19}}\times 1.67\times 10^{-27}(80000^2-50000^2)\\\Rightarrow V_B-V_A=20.353125\ V

The determined potential difference is 20.353125 V

3 0
1 month ago
Water is made of two hydrogen atoms and one oxygen atom bonded together. Julia is describing how water undergoes a physical chan
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8 0
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Mosses don't spread by dispersing seeds; they disperse tiny spores. The spores are so small that they will stay aloft and move w
Keith_Richards [2897]

Solution:

Em_{f} / Em₀ = 0.30

Explanation:

In this problem, we apply the connection between mechanical energy, kinetic energy, and gravitational potential energy.

      K = ½ m v²

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We assess the mechanical energy at two positions:

Initial. Lower

    Em₀ = K = ½ m v²

At its highest point

    Em_{f} = U = mg and

Now let's compute

    Em₀ = ½ m 3.6²

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Thus the energy lost is given by:

    Em_{f} / Em₀ = m 1.96 / m 6.48

   Em_{f} / Em₀ = 0.30

This means that 30% of the sun's energy is transformed into potential energy.

There are various conversion possibilities.

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the term is hydrogen bridge.
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