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3 years ago

10

Your teacher's car can go from rest to 25 m/s (≈55 mph) in 10 seconds. The car's velocity changes at a uniform rate. Below is a

quantitative velocity vs time graph that represents the motion of the car. Determine the acceleration of the car.

1 answer:

Mice21 [21]3 years ago

5 0

If the speed of the car changes uniformly, as shown in the graph, this means that the acceleration is constant. As the acceleration is the derivative of the speed

$a = \frac{\delta v}{\delta t}$

Then, this means that the acceleration is the slope of the equation of the line shown in the graph.

Then, we find the acceleration using the following equation:

$a = \frac{v_f- v_i}{t_2-t_1}\\\\a = \frac{25-0}{10}\\\\a = 2.5\ m/s ^ 2$

Then, the acceleration of the car is $2.5\ m/s^2$

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An initially uncharged 3.53 × 10 − 6 F capacitor and a 7290 Ω resistor are connected in series to a 1.50 V battery that has negl

GarryVolchara [31]

Answer:

The initial current, Io = 2.06×10^-4A

Explanation:

The I itial current ,Io of a circuit is given by:

Io = Vo/R

Where Vo = initial votage

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Vo = 1.50V

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3 0

3 years ago

Help please, I don't get it

Sergio [31]

Answers:

a) $\hat F=(0.83,-0.55) N$

b) $\hat D=(-0.44,-0.89) m$

c) $\hat V=(-0.47,0.88) m/s$

Explanation:

A unit vector is a vector whose magnitude (length) is equal to 1. This kind of vector is identified as $\hat v$ and the way to calculate is as follows:

$\hat v=\frac{\vec v}{|v|}$

Where:

$\vec v=(x,y)$ is the vector

$|v|=\sqrt{x^{2}+y^{2}}$ is the magnitude of the vector

Having this information clarified, let's begin with the answers:

a) Force Vector

$\vec F=(9.0 \hat i - 6.0 \hat j) N$

Magnitude of $\vec F$ :

$|F|=\sqrt{(9.0 \hat i)^{2}+(-6.0 \hat j)^{2 }}N=10.81 N$

<u />

<u>Unit vector:</u>

$\hat F=\frac{\vec F}{|F|}$

$\hat F=\frac{(9.0 \hat i - 6.0 \hat j) N}{10.81 N}$

$\hat F=\frac{9.0}{10.81} N-\frac{6.0}{10.81}N$

$\hat F=(0.83,-0.55) N$

b) Displacement Vector

$\vec D=(-4.0 \hat i - 8.0 \hat j) m$

Magnitude of $\vec D$ :

$|D|=\sqrt{(-4.0 \hat i)^{2}+(-8.0 \hat j)^{2 }}m=8.94 m$

<u />

<u>Unit vector:</u>

$\hat D=\frac{\vec D}{|D|}$

$\hat D=\frac{(-4.0 \hat i - 8.0 \hat j) m}{8.94 m}$

$\hat D=\frac{-4.0}{8.94} Nm+\frac{-8.0}{8.94}m$

$\hat D=(-0.44,-0.89) m$

c) Velocity Vector

$\vec V=(-3.50 \hat i + 6.50 \hat j) m/s$

Magnitude of $\vec V$ :

$|V|=\sqrt{(-3.50 \hat i)^{2}+(6.50 \hat j)^{2}}m/s=7.38 m/s$

<u />

<u>Unit vector:</u>

$\hat V=\frac{\vec V}{|V|}$

$\hat V=\frac{(-3.50 \hat i +6.50 \hat j) m/s}{7.38 m/s}$

$\hat V=\frac{-3.50}{7.38} m/s+\frac{6.50}{7.38}m/s$

$\hat V=(-0.47,0.88) m/s$

8 0

4 years ago

If your speed slows down you will have what acceleration?

vitfil [10]

When speed slows down you have kinetic energy.

8 0

4 years ago

Read 2 more answers