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almond37 [142]
3 years ago
6

HELP ASAP!!! So stressed! :(

Physics
1 answer:
nasty-shy [4]3 years ago
7 0
Question #16:. last choice.
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Which of the following statements is TRUE for high-visibility clothing? A. High-visibility clothing helps to reduce insect probl
Vlad1618 [11]

Answer:

The answer for the above statement is:

C. High-visibility clothing is important to wear in areas with moving vehicles.

because in bright clothes you are easier to see, so people driving can see you.

Explanation:

3 0
3 years ago
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An atom that has 117 protons in its nucleus has not yet been made. Once this atom is made, to which group will element 117 belon
Pavel [41]
In nomine patris, et filii, et spiritus sancti. 
3 0
3 years ago
A professor's office door is 0.91 m wide, 2.0 m high, and 4.0 cm thick; has a mass of 25 kg; and pivots on frictionless hinges.
olasank [31]

Answer:

F= 5.71 N

Explanation:

width of door= 0.91 m

door closer torque on door= 5.2 Nm

In order to hold the door in open position we need to exert an equal and opposite torque, to the door closer torque, on the door.

so wee need to exert 5.2 Nm torque on the door.

If we want to apply minimum force to exert the required torque we need to apply force perpendicularly on the door knob (end of door) so that to to greater moment arm.

T= r x F

T= r F sin∅

F= T/ (r * sin∅)

F= 5.2/ (0.91 * 1)

F= 5.71 N

5 0
3 years ago
Find the magnitude of the electric field due to a charged ring of radius "a" and total charge "Q", at a point on the ring axis a
34kurt

Answer:

E=\frac{KQ}{2\sqrt 2a^2}

Explanation:

We are given that

Charge on ring= Q

Radius of ring=a

We have to find the magnitude of electric filed on the axis at distance a from the ring's center.

We know that the electric field at distance x from the center of ring of radius R is given by

E=\frac{kQx}{(R^2+x^2)^{\frac{3}{2}}}

Substitute x=a and R=a

Then, we get

E=\frac{KQa}{(a^2+a^2)^{\frac{3}{2}}}

E=\frac{KQa}{(2a^2)^{\frac{3}{2}}}

E=\frac{KQa}{2\sqrt 2a^3}

E=\frac{KQ}{2\sqrt 2a^2}

Where K=9\times 10^9 Nm^2/C^2

Hence, the magnitude of the electric filed due to charged ring on the axis of ring at distance a from the ring's center=\frac{KQ}{2\sqrt 2a^2}

4 0
3 years ago
A double-slit experiment yields an interference pattern due to the path length difference from light traveling through one slit
Lapatulllka [165]

Answer:

Therefore the correct statement is B.

Explanation:

In the interference and diffraction phenomena, the natural wave of electromagnetic radiation must be taken into account, the wave front that advances towards the slit can be considered as when it reaches it behaves like a series of wave emitters, each slightly out of phase from the previous one, following the Huygens principle that states that each point is compiled as a source of secondary waves.

The sum of all these waves results in the diffraction curve of the slit that has the shape

      I = Io sin² θ /θ²

Where the angle is a function of the wavelength and the width of the slit.

From the above, the interference phenomenon can be treated as the sum of two diffraction phenomena displaced a distance equal to the separation of the slits (d)

Therefore the correct statement is B

6 0
3 years ago
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