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3 years ago

Can you help me??????????????????????????)?

Physics

1 answer:

garik1379 [7]3 years ago

5 0

Answer:

Velocity is a change in displacement over change in time and uses the units m/s.

Both are rates of change and can be positive or negative.

Acceleration is a change in velocity over change in time and uses the units m/s².

Explanation:

Velocity is the change in displacement over change in time, this makes it a rate of change. It can be positive or negative because it is a vector quantity. It uses the units m/s because that is a displacement unit over a time unit.

Acceleration is the change in velocity over change in time, this makes it a rate of change. It can be positive or negative because it is also a vector quantity. It uses the units m/s² (m/s/s) because that is a velocity unit over a time unit.

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Three point charges are arranged on a line. Charge q3 = 5 nC and is at the origin. Charge q2 = - 3 nC and is at x = 4 cm. Charge

Taya2010 [7]

Answer:

q₁ = + 1.25 nC

Explanation:

Theory of electrical forces

Because the particle q₃ is close to two other electrically charged particles, it will experience two electrical forces and the solution of the problem is of a vector nature.

Known data

q₃=5 nC

q₂=- 3 nC

d₁₃= 2 cm

d₂₃ = 4 cm

Graphic attached

The directions of the individual forces exerted by q1 and q₂ on q₃ are shown in the attached figure.

For the net force on q3 to be zero F₁₃ and F₂₃ must have the same magnitude and opposite direction, So, the charge q₁ must be positive(q₁+).

The force (F₁₃) of q₁ on q₃ is repulsive because the charges have equal signs ,then. F₁₃ is directed to the left (-x).

The force (F₂₃) of q₂ on q₃ is attractive because the charges have opposite signs. F₂₃ is directed to the right (+x)

Calculation of q1

F₁₃ = F₂₃

$\frac{k*q_{1}*q_3 }{(d_{13})^{2} } = \frac{k*q_{2}*q_3 }{(d_{23})^{2} }$

We divide by (k * q3) on both sides of the equation

$\frac{q_{1} }{(d_{13})^{2} } = \frac{q_{2} }{(d_{23})^{2} }$

$q_{1} = \frac{q_{2}*(d_{13})^{2} }{(d_{23} )^{2} }$

$q_{1} = \frac{5*(2)^{2} }{(4 )^{2} }$

q₁ = + 1.25 nC

3 0

3 years ago

Which statement best defines an electric field?

Shkiper50 [21]

A region around a charged particle or object within which a force would be exerted on other charged particles or objects

5 0

3 years ago

9. If everything in the universe – including atoms and particles stop moving, does time stop? Or does time continue even if ever

Vedmedyk [2.9K]

Time stops everything is made out of atoms so if atoms freeze everything freezes

5 0

3 years ago

In the lab downstairs physics majors use a rotating mirror to measure the speed of light within a few percent of the actual valu

iris [78.8K]

The number of complete cycles the rotating mirror goes through before the angular velocity gets to zero is approximately 1166.8 revs

<h3>What is angular velocity?</h3>

Angular velocity is the ratio of the angle turned to the time taken.

The kinematic equation for angular velocity are presented as follows;

ω = ω₀ + α·t

θ = θ₀ + ω₀·t + 0.5·α·t²

Where;

θ₀ = The initial angle turned = 0

ω₀ = The initial angular velocity of the mirrors = 115 rad/s clockwise

α = The angular acceleration = (115 - (-115))rad/s/(85 s) = -46/17 m/s²

t = The duration of the motion;

When the angular velocity, ω is zero, we get;

0 = 115 - 46/17·t

t = 85/2

Which indicates;

θ = 0 + 115× (85/2) + 0.5×(46/17) ×(85/2)² = 7331.25

θ = 7331.25 radians

θ = 7331.25/(2×π) ≈ 1166.8 rev

The mirrors would have turned through approximately 1166.8 revolutions when the angular gets to zero

Learn more about angular velocity and acceleration here:

brainly.com/question/13014974

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7 0

1 year ago

A certain spring stores 10.0J of potential energy when it isstretched by 2.00cm from its equilibrium position.How much potential

qaws [65]

Answer:

Answered

Explanation:

x= 0.02 m

E_p= 10.0 J

E_p= 0.5kx^2

10= 0.5k(0.02)^2

solving we get

K= 50.0 N/m

Now

E'_p= 0.5kx'^2

E'_p= 0.5×50×(0.04)^2

E'_p=40 J

b) potential energy is a scalar quantity and it only depends magnitude and not direction so it will remain same in compression and expansion both

c) 20 J = 0.5×50,000×x^2

solving

x= 0.028 m

d) k is 50.0 N/m from above calculation

3 0

3 years ago