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3 years ago

6

Use the drop-down menu to complete the statement. Based on the field lines, the electric charges indicated by the question marks

are .

2 answers:

velikii [3]3 years ago

8 0

Answer:

the same

Explanation: got it right on edge

zheka24 [161]3 years ago

5 0

Answer: The same

Explanation: I just did it on Edg

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Answer this question, please

pashok25 [27]

Answer:

the correct answer is the 60

$+ 20 + 60 \gamma \beta$

3 0

3 years ago

Understanding that if we say something unkind to someone else his or her

OlgaM077 [116]

Answer: Moral

Explanation:

6 0

3 years ago

A 1.55-kg object hangs in equilibrium at the end of a rope (taken as massless) while a wind pushes the object with a 13.3-n hori

yarga [219]

<span>To answer this problem, we use balancing of forces: x and y components to determine the tension of the rope.

First, the vertical component of tension (Tsin theta) is equal to the weight of the object.
T * sin θ = mg =</span> 1.55 * 9.81 <span>
T * sin θ = 15.2055

Second, the horizontal component of tension (t cos theta) is equal to the force of the wind.
T * cos θ = 13.3

Tan θ = sin </span>θ / cos θ = 15.2055/13.3 = 1.143
we can find θ that is equal to 48.82.

T then is equal to 20.20 N

4 0

3 years ago

Light of wavelength 630 nm falls on two slits and produces an interference pattern in which the third-order bright red fringe is

goblinko [34]

Answer:

d $\frac{x}{l}$ = m× λ⇒ d = λ ×m×l / x

= 630× $10^{-9}$ m × 3×3m/ 45× $10^{-3}$ m

= 1.26× $10^{-4}$ m

Explanation:

the above calculation is based on Young’s double slit experiment where the two slits provide two coherent light sources which results either constructive interference or destructive interference when passing through a double slit.

6 0

3 years ago

A 1560-kilogram truck moving with a speed of 28.0 m/s runs into the rear end of a 1070-kilogram stationary car. If the collision

Nimfa-mama [501]

Answer:

Δ KE = 249158.6 kJ

Explanation:

given data

Truck mass M = 1560 Kg

Truck initial speed, u = 28 m/s

mass of car m = 1070 Kg

initial speed of car u1 = 0 m/s

solution

first we get here final speed by using conservation of momentum that is express as

Mu = (M+m) V .......................1

put here value we get

1560 × 28 = (1560 + 1070 ) V

solve it we get

final speed V = 16.60 m/s

and

Change in kinetic energy will be here

Δ KE = $\frac{1}{2} Mu^2 - \frac{1}{2}(M+m)V^2$ .................2

put here value and we get

Δ KE = $\frac{1}{2}\times 1560\times 28^2 - \frac{1}{2}\times (1560 + 1070)\times 16.60^2$

solve it we get

Δ KE = 249158.6 kJ

6 0

3 years ago