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3 years ago

Many streets and parking lots are illuminated with ???

Physics

2 answers:

lora16 [44]3 years ago

4 0

The streets are illuminated with sodium vapor lights. hope this helps!!!!!!

UkoKoshka [18]3 years ago

3 0

Answer: C. sodium-vapor lights

Explanation:

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What happens to the resistance of a wire as it gets wider

Alchen [17]

Answer:

it snaps

Explanation:

the more force you put on it, the wired out it gets than it snaps. I think

5 0

3 years ago

Why is the top of a wave called a crest

Molodets [167]

Answer: A crest is the highest point the medium rises to and a trough is the lowest point the medium sinks to. It is also a point on the wave where the displacement of the medium is at a maximum.

4 0

4 years ago

A test charge of -1.4 x 10-7 coulombs experiences a force of 5.4 x 10-1 newtons. Calculate the magnitude of the electric field c

VARVARA [1.3K]

Answer:

3.86×10⁶ Newton/coulombs

Explaination:

Applying,

E = F/q....................... Equation 1

Where E = Electric Field, F = Force, q = charge.

From the question,

Given: F = 5.4×10⁻¹ N, q = -1.4×10⁻⁷ coulombs

Substitute these values into equation 1

E = 5.4×10⁻¹/ -1.4×10⁻⁷

E = -3.86×10⁶ Newtons/coulombs

Hence the magnitude of the electric field created by the

negative test charge is 3.86×10⁶ Newton/coulombs

5 0

3 years ago

Hello everyone. This is a question about Dimensional Analysis and I came across this question but I am unable to wrap my head ar

omeli [17]

Answer:

2. [B] = [L]/[T] and [C] = [L]/[T]

Explanation:

I assume you mean this:

A = B² + 2B⁴/C²

Since you can't add numbers with different units (for example, you can't add seconds to meters), each term in the sum must have the same units as A.

B² = [L]²/[T]²

B = [L]/[T]

B⁴/C² = [L]²/[T]²

C²/B⁴ = [T]²/[L]²

C² = B⁴ [T]²/[L]²

C² = ([L]/[T])⁴ [T]²/[L]²

C² = [L]²/[T]²

C = [L]/[T]

Notice we ignore the 2 coefficient, which is unitless.

7 0

3 years ago

block of mass 5kgriding on a horizontal frictionlessxy-plane surface is subjected tothree applied forces:→F1= 12√2N[ 45◦]→F2= (8

dsp73

Answer:

(i) See attached image for the drawing

(ii) net force given in component form: (20, 20)N with magnitude: $\sqrt{800} \,\,\,N$

Explanation:

First try to write all forces in vector component form:

The force F1 acting at 45 degrees would have multiplication factors of $\frac{\sqrt{2} }{2}$ on both axes, to take care of the sine and cosine projections. Therefore, the:

x-component of F1 is $F1_x=12\,\sqrt{2} \frac{\sqrt{2} }{2} =12\,\,N$

y-component of F1 is $F1_y=12\,\sqrt{2} \frac{\sqrt{2} }{2} =12\,\,N$

As far as force F2, it is given already in x and y components, then:

x-component of F2 = 8 N

y-component of F2 = -6 N (negative meaning pointing down the y-axis)

Force F3 has only component (upwards) in the y-direction

x-component of F3 = 0 N

y-component of F3 =14 N

The additions of all these component by component, gives the resultant force (R) acting on the 5 kg mass:

x-component of R = 12 + 8 = 20 N

y-component of R = 12 + 14 - 6 = 20 N

Therefore, the acceleration that the mass receives due to this force is given in component form as:

x-component of acceleration: 20 N / 5 kg = $4\,\,\,m/s^2$

y-component of acceleration: 20 N / 5 kg = $4\,\,\,m/s^2$

Now we can calculate the components of the velocity of this mass after 2 seconds of being accelerated by this force, using the formula of acceleration times time:

x-component of the velocity is: $v_x=4\,*\,2=8\,m/s$

y-component of the velocity is: $v_y=4\,*\,2=8\,m/s$

7 0

3 years ago