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CaHeK987 [17]
3 years ago
15

Study the four transverse waves shown. Compare the properties of waves B, C, & D to that of wave A.

Physics
1 answer:
tresset_1 [31]3 years ago
4 0
Wave D has the same frequency as Wave A. The same number of waves pass a point per unit time. They have different amplitudes.
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1) Halving the distance (i.s., decreasing by a factor of two) between two charged objects will cause the electrical force betwee
Vedmedyk [2.9K]

Answer:

Explanation:

For an electric force, F the formula:

F = kQq/r^2

Given:

r2 = 1/2 × r1

F1 × r1 = k

F1 × r1 = F2 × r2

F2 = (F1 × r1^2)/(0.5 × r1)^2

= (F1 × r1^2)/0.25r1^2

= 4 × F1.

7 0
4 years ago
If root-mean-square voltage of a supply is 240 V, calculate the maximum voltage of
Levart [38]

Answer:

Vp= Vr sq. root 2

= 240 × sq. root 2

=339.4 V

5 0
4 years ago
While the change in blank will remain the same during a collision, the force needed to bring an object to a stop can be blank if
Ahat [919]

Explanation :

There are two types of collision i.e. elastic and elastic collision.

  • Elastic collision : In this type of collision, the total momentum and the kinetic energy of the particles remains constant.
  • Inelastic collision : In this type of collision, only the momentum remains constant while there is some loss of kinetic energy occurs.

From Newton's second law,

F = m a

a is the rate of change of velocity.

F=m\dfrac{v}{t}

There is a inverse relation between the force and the time of collision.

The change in <em><u>momentum</u></em> will remain the same during a collision, the force needed to bring an object to a stop can be <em><u>increased</u></em> if the time of the collision is <u><em>decreased</em></u>.

6 0
3 years ago
Read 2 more answers
In your experiment, you measure a total deflection of 4.12 cm when an electric field of 1.10×103V/m is established between the p
Bond [772]

Answer:

B_0 = 1.69 \times 10^{-4}\ T

Explanation:

given,

total deflection = 4.12 cm

Electric field = 1.1 ×10³ V/m

plate length = 6 cm

distance between them = 12 cm

using formula

v_0 = \sqrt{\dfrac{q\epsilon_0d}{ym}(\dfrac{d}{2}+L)}

q = 1.6 × 10⁻¹⁹ C

m = 9.11 x 10⁻³¹ kg

d = 0.06 m

L = 0.12 m

v_0 = \sqrt{\dfrac{1.6 \times 10^{-19}\times 1.1 \times 10^{3}\times 0.06}{0.0412\times 9.11 \times 10^{-31} }(\dfrac{0.06}{2}+0.12)}

v_0 = 6496355.63 m/s

v_0 = \dfrac{E}{B_0}

B_0 = \dfrac{E}{v_0}

B_0 = \dfrac{1.1\times 10^{3}}{6496355.63}

B_0 = 1.69 \times 10^{-4}\ T

5 0
3 years ago
What are 1A, 3B, and 7A examples of on the periodic table?
In-s [12.5K]

Answer:

groups

Explanation:

I got a 100 on my quiz

8 0
4 years ago
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