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3 years ago

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In 3 seconds a car moving in a straight line increases its speed from 22.4 m/s to 29.1 m/s while a truck increases its speed fro

m 0 mph to 6.7m/s in the same amount of time. What is the acceleration of the truck in m/s? *
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1 answer:

ANTONII [103]3 years ago

6 0

Is there a picture?

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Barry is conducting an experiment and rolls a tennis ball down a ramp. Which best describes the motion of the tennis ball? It do

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A. It does not exhibit projectile motion and follows a straight path down the ramp.

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Maggie walks to a friends house which is exactly 1500 meters due South. It takes Maggie 45 minutes for the walk and Maggie has t

Sindrei [870]

Answer:

<em>Explanation below</em>

Explanation:

<u>Speed vs Velocity </u>

These are two similar physical concepts. They only differ in the fact that the velocity is vectorial, i.e. having magnitude and direction, and the speed is scalar, just the magnitude regardless of the direction. They are strongly related to the concepts of displacement and distance, which are the vectorial and scalar versions of the space traveled by a moving object. The velocity can be computed as

$\displaystyle \vec v=\frac{\vec r}{t}$

Where $\vec r$ is the position vector and t is the time. The speed is

$\displaystyle v=\frac{d}{t}$

To compute $\vec r$ , we only need to know the initial and final positions and subtract them. To compute d, we need to add all the distances traveled by the object, regardless of their directions.

Maggie walks to a friend's house, located 1500 meters from her place. The initial position is 0 and the final position is 1500 m. The displacement is

$\vec r=1500\ m \text{ to the south}$

and the velocity is

$\displaystyle \vec v=\frac{1500}{45}=33.33\ m/s\text{ to the south}$

Now, we know Maggie had to make three different turns of direction to finally get there. This means her distance is more than 1500 m. Let's say she walked 500 m in all the turns, then the distance is

$d=1500+500=2000\ m$

If she took the same time to reach her destiny, she would have to run faster, because her average speed is

$\displaystyle v=\frac{2000}{45}=44.44\ m/s$

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For those seeking for the answer, its a source of electrical energy.

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Can someone solve this problem and explain to me how you got it

evablogger [386]

Answer:

question5: F=74312.5N

question6: charge at the end of antenna=0.37N

Explanation:

Coulomb's law: the magnitude of the force of attraction or repulsion due to two charges is proportional to the product of the magnitude of the charges and inversely proportional to the square of distance between the charges.

⇒ $F\alpha\frac{q1*q2}{r^{2}}$

∴ $F=k\frac{q1*q2}{r^{2}}$

where $F$ is the force of attraction or repulsion

$k$ is Coulumb's constant= $9*10^{9}Nm^{2}C^{-2}$

$q1$ and $q2$ are the magnitude of the charges

$r$ is the distance between two charges

The force between the two charges is attractive if they are of different polarity

The force between the two charges is repulsive if they are of same polarity

Question5:

Given: q1=0.041 C, q2=0.029 C, r=12 m

therefore by Coulumb's law,

$F=9*10^{9}*\frac{0.041*0.029}{12^{2}}$

$F=74312.5N$

Question6:

Given: q1= $3*10^{-18}C$ , r=5 m, F= $4*10^{-11}N$

therefore by Coulumb's law,

$4*10^{-11}=9*10^{9}*\frac{3*10^{-18}*q2}{5^{2}}$

⇒ $q2=\frac{4*10^{-11}*25}{9*10^{9}*3*10^{-18}} \\=0.37C$

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3 years ago