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Cerrena [4.2K]
3 years ago
10

A car traveling west in a straight line on a highway decreases its speed from 30.0 meters per second to 23.0 meters per second i

n 2.00 seconds.The Cars average acceleration during the time interval. is
Physics
1 answer:
pychu [463]3 years ago
8 0
Acceleration, a =  (v - u) / t

Initial Velocity, u = 30 m/s
Final Velocity, v  =  23 m/s
time                 t  =  2.00 seconds

a = (23 - 30) / 2
a = -7 / 2 = -3.5 m/s2

So the acceleration  is negative, which means it is a deceleration of 3.5 m/s2.
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Two point charges, 3.4 μC and -2.0 μC , are placed 5.0 cm apart on the x axis. Assume that the negative charge is at the origin,
Elza [17]

The electric field is zero at x = -16.45cm

Data;

  • q1 = 3.4 μC
  • q2 = -2.0 μC
  • distance = 5cm

<h3>The Electric Field at point 0</h3>

As the 3μC is larger than -2.0μC  and the charges are opposite sign. The electric field will be zero at the negative axis.

Let the point be at x.

For an electric field to be equal to zero;

k(\frac{q_1}{d_1})^2 + k(\frac{q_2}{d_2})^2 = 0\\\frac{3.4}{(5-x)^2} - \frac{2}{x^2} = 0\\

Let's solve for x using mathematical methods.

\frac{3.4x^2 - 2(5-x)^2}{(5-x)^2x^2}= 0\\ 3.4x^2 - 2(5-x^2) = 0\\3.4x^2 - 50 - 2x^2 + 20x = 0\\1.4x^2 +20x - 50 = 0

Solving the above quadratic equation;

x = -16.45cm

The electric field is zero at x = -16.45cm

Learn more on electric field at a point here;

brainly.com/question/1592046

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8 0
3 years ago
Please help with this question ASAP!!!
Paladinen [302]
The answer is the second choice

7 0
4 years ago
A fan blade rotates with angular velocity given by ωz(t)= γ − β t2. part a calculate the angular acceleration as a function of t
Katyanochek1 [597]
Given:

w = z(t)= γ − β*t^2

Differentiating both sides with respect to t, we get:

α = z(t) = -2βt

Given: 
<span> γ = 5.35 rad/s and β = 0.810 rad/s3 
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so, For t = 3 sec,

angular acceleration = -2 * 0.810 * 3 = 
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8 0
4 years ago
Read 2 more answers
In the first stage of a two-stage Carnot engine, energy is absorbed as heat Q1 at temperature T1 = 500 K, work W1 is done, and e
Nataly [62]

Answer:

Efficiency = 52%

Explanation:

Given:

First stage

heat absorbed, Q₁ at temperature T₁ = 500 K

Heat released, Q₂ at temperature T₂ = 430 K

and the work done is W₁

Second stage

Heat released, Q₂ at temperature T₂ = 430 K

Heat released, Q₃ at temperature T₃ = 240 K

and the work done is W₂

Total work done, W = W₁ + W₂

Now,

The efficiency is given as:

\eta=\frac{\textup{Total\ work\ done}}{\textup{Energy\ provided}}

or

Work done = change in heat

thus,

W₁ = Q₁ - Q₂

W₂ = Q₂ - Q₃

Thus,

\eta=\frac{(Q_1-Q_2)\ +\ (Q_2-Q_3)}{Q_1}}

or

\eta=1-\frac{(Q_1-Q_3)}{Q_1}}

or

\eta=1-\frac{(Q_3)}{Q_1}}

also,

\frac{Q_1}{T_1}=\frac{Q_2}{T_2}=\frac{Q_3}{T_3}

or

\frac{T_3}{T_1}=\frac{Q_3}{Q_1}

thus,

\eta=1-\frac{(T_3)}{T_1}}

thus,

\eta=1-\frac{(240\ K)}{500\ K}}

or

\eta=0.52

or

Efficiency = 52%

8 0
4 years ago
A mass movement that involves the sudden movement of a block of material is called a
Margarita [4]
A mass movement that involves the sudden movement of a block of material is called a B.slide.

<span>♡♡Hope I helped!!! :)♡♡
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