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Annette [7]
3 years ago
10

Which form or radiation has the greater frequency? UV radiation or violet light?

Physics
2 answers:
arlik [135]3 years ago
8 0
Ultraviolet (UV) has higher frequency and shorter wavelength than violet light.
Klio2033 [76]3 years ago
3 0
Ultra violet has a greater frequency. 
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A vehicle moving along at 5m/s. What should be the constant deceleration in order to stop it within 15m?​
Oduvanchick [21]

Answer:

a = -5/6 m/s²

Explanation:

v² = u² + 2as

a = (v² - u²) / 2s

a = (0² - 5²) / 2(15)

a = -5/6 m/s²

This assumes the initial velocity was in the positive direction.

5 0
3 years ago
A wheelchair moves upward on a 7.1 degree ramp at a speed of 20km/h what is the horizontal velocity
Charra [1.4K]

Answer:

If the wheelchair is up 7.1 ft. In hight the time of flight should be 0.664 seconds and the distance should be 12.108 ft.

Explanation: I divided the displacement by the time and I used the equation Vx = 20 km/m

6 0
3 years ago
the young’s modulus of aluminum is 69gpa, of nylon is 3gpa, of tungsten is 400gpa, and of copper is 117gpa. if equal-size sample
olga_2 [115]

Answer:

Least to most elongated: tungsten, copper, aluminum, nylon.

Explanation:

Materials with high Young's modulus are difficult to stretch. σ = Yε and ε = ΔL/L so an object with a high Young's modulus (Y) subject to a certain tensile stress (σ) will have a smaller strain than an object with a smaller Young's 's modulus subject to the same tensile stress. If strain (ε) is smaller, then ΔL will also be smaller.

3 0
2 years ago
How is energy involved in physical changes and in chemical changes?
GaryK [48]

During phase changes, energy changes are usually involved. For example, when solid dry ice vaporizes (physical change), carbon dioxide molecules absorb energy. Meanwhile, when liquid water becomes ice energy is released.

hope this helps :)

4 0
3 years ago
The driver of a car traveling at 31.3 m/s applies the brakes and undergoes a constant deceleration of 1.6 m/s2.How many revoluti
lisov135 [29]

Answer:

R=156.99\operatorname{Re}vs

Explanation: The equations used are as follows:

\begin{gathered} x(t)=x_o+v_ot+\frac{1}{2}at^2\Rightarrow(1) \\ v(t)=v_o+at\Rightarrow(2) \end{gathered}

By using equation (2), the time needed for the car to come to rest is calculated as follows:

\begin{gathered} v(t)=(31.3ms^{-1})_{}+(-1.6ms^{-2})t=0 \\ t=\frac{31.3ms^{-1}}{1.6ms^{-2}}=19.56s \\ t=19.563s \end{gathered}

By using equation (1), The total distance traveled in that time would be as:

\begin{gathered} x(19.563s)=_{}(31.3ms^{-1})\cdot(19.563s)+\frac{1}{2}(-1.6ms^{-2})\cdot(19.563s)^2\Rightarrow(1) \\ x(19.563s)=612.31-306.17=306.14m \\ \therefore\Rightarrow \\ x(19.563s)=306.14m \end{gathered}

The revolutions taken by the tire before the car comes to rest would be:

\begin{gathered} C=2\pi\cdot(0.31m)=1.95m \\ R=\frac{x(19.563s)}{C}=\frac{306.14m}{1.95m}=156.99\operatorname{Re}v \\ R=156.99\operatorname{Re}vs \end{gathered}

3 0
2 years ago
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