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vagabundo [1.1K]
3 years ago
11

Pls help look at photo

Physics
1 answer:
tatiyna3 years ago
6 0

Answer:

A

Explanation:

Intensity is proportional to amplitude

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A 1200-kg car initially at rest undergoes constant acceleration for 9.4 s, reaching a speed of 11 m/s. It then collides with a s
Natalka [10]

To solve this problem we will apply the principle of conservation of energy and the definition of kinematic energy as half the product between mass and squared velocity. So,

KE_i = KE_f

KE_f = \frac{1}{2} mv^2

Here,

m = Mass

V = Velocity

Replacing,

KE_f = \frac{1}{2} (12000)(11)^2

KE_f = 72600J

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8 0
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Which statement best describes electrons?
crimeas [40]
They are positive and remain inside the nucleus.
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2 years ago
Not in book
umka2103 [35]

Answer:

x=2.4365\ m

and

x=-1.4365\ m

Explanation:

Given:

  • first charge, q_1=5\times 10^{-3}\ C
  • second charge, q_2=3\times 10^{-3}\ C
  • position of first charge, x_1=-2\ m
  • position of second charge, x_2=-1\ m

Now since there are only 2 charges and of the same sign so they repel each other. This repulsion will be zero at some point on the line joining the charges.

<u>Now, according to the condition, electric field will be zero where the effects of field due to both the charges is equal.</u>

E_1=E_2

  • since first charge is greater than the second charge so we may get a point to the right of the second charge and the distance between the two charges is 1 meter.

\frac{1}{4\pi.\epsilon_0} \frac{q_1}{(r+1)^2} =\frac{1}{4\pi.\epsilon_0} \frac{q_2}{(r)^2}

\frac{5\times 10^{-3}}{(r+1)^2} = \frac{3\times 10^{-3}}{(r)^2}

3(r^2+1+2r)=5r^2

2r^2-6r-3=0

r=3.4365 \&\ r=-0.4365

Since we have assumed that the we may get a point to the right of second charge so we calculate with respect to the origin.

x=-1+3.4365=2.4365\ m

and

x=-1-0.4365=-1.4365\ m

6 0
3 years ago
A car travels 50km in 2 hours.
ser-zykov [4K]

Answer: it will travle 25km pr hour

Explanation:

sevide both by 2

3 0
3 years ago
An object with kinetic energy k explodes into two pieces, each of which moves with twice the speed of the original object.
zlopas [31]
<span>Assuming that the momenta of the two pieces are equal: when they have equal velocities, then the masses of the two pieces are also equal. Since there is no force from outside of the system, the center of mass moves on with the same velocity as before the equation. So the two pieces must fly at the side side of the mass center, i.e., they must always be at 90° to the side of the mass center. Otherwise it would not be the mass center, respectively the pieces would not have equal velocities. This is only possible, when the angle of their velocity with the initial direction is 60°. Because, cos (60°) = 1/2 = v/(2v).</span>
6 0
4 years ago
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