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3 years ago

A car goes from 4.47 m/s to 17.9 m/s in 3 seconds. calculate the acceleration of the car

Physics

2 answers:

Vanyuwa [196]3 years ago

5 0

Answer:

4.48 m/s²

Explanation:

Acceleration is change in velocity over time.

a = Δv / Δt

a = (17.9 m/s − 4.47 m/s) / 3 s

a = 4.48 m/s²

Lera25 [3.4K]3 years ago

4 0

Answer:

$\boxed {\tt a \approx4.48 \ m/s^2}$

Explanation:

Acceleration can be found by dividing the change in velocity over the time.

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

The change in velocity can be found by subtracting the final velocity by the initial velocity.

$\Delta v= final \ velocity - initial \ velocity$

The final velocity is 17.9 meters per second.

The initial velocity is 4.47 meters per second.

$\Delta v=17.9 \ m/s-4.47 \ m/s$

$\Delta v=13.43 \ m/s$

The change in velocity is 13.43 meters per second and the time is 3 seconds.

$a=\frac{13.43 \ m/s}{3 \ s}$

Divide.

$a=4.47666667 \ m/s^2$

$a \approx4.48 \ m/s^2$

The acceleration is about 4.48 meters per seconds squared.

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2 years ago

On the way to school, the bus speeds up from 20 m/s to 36 m/s in 4 seconds. What distance

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Answer B. 112 m

Step-by-Step Explanation

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2 years ago

Four charges with equal magnitudes of 10.6 × 10-12 C are placed at the corners of a rectangle. The lengths of the sides of the r

cricket20 [7]

Answer:

Figure a. E_net = 99.518 N/C

Figure b. E_net = 177.151 N / C

Explanation:

Given:

- Attachment for figures missing in the question.

- The dimensions for rectangle are = 7.79 x 3.99 cm

- All four charges have equal magnitude Q = 10.6*10^-12 C

Find:

Find the magnitude of the electric field at the center of the rectangle in Figures a and b.

Solution:

- The Electric field generated by an charged particle Q at a distance r is given by:

E = k*Q / r^2

- Where, k is the coulomb's constant = 8.99 * 10^9

Part a)

- First we see that the charges +Q_1 and +Q_3 produce and electric field equal but opposite in nature. So the sum of Electric fields:

E_1 + E_3 = 0

- For Charges -Q_2 and +Q_4, they are equal in nature but act in the same direction towards the negative charge -Q_2. Hence, the net Electric Field at center of the rectangle can be given as:

E_net = E_2 + E_4

E_2 = E_4

E_net = 2*E = 2*k*Q / r^2

- The distance r from each corner to mid-point of the rectangle is constant. It can be evaluated by Pythagoras Theorem as follows:

r = sqrt ( (7.79/200)^2 + (3.99/200)^2 )

r = sqrt ( 1.9151*10^-3 ) = 0.043762 m

- Plug the values in the E_net expression developed above:

E_net = 2*(8.99*10^9)*(10.6*10^-12) / 1.9151*10^-3

E_net = 99.518 N/C

Part b)

- Similarly for Figure b, for Charges -Q_2 and +Q_4, they are equal in nature but act in the same direction towards the negative charge -Q_2. Also, Charges -Q_1 and +Q_3, they are equal in nature but act in the same direction towards the negative charge -Q_1. These Electric fields are equal in magnitude to what we calculated in part a).

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Hence,

E_net = 2*E_part(a)*cos(Q)

- Where, Q is the angle between resultant, vertical in direction, and each of the electric field. We can calculate Q using trigonometry as follows:

Q = arctan ( 3.99 / 7.79 ) = 27.12 degrees.

- Now, compute the net electric field E_net:

E_net = 2*(99.518)*cos(27.12)

E_net = 177.151 N / C

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2 years ago

if a ball is thrown straight up into the air with an initial velocity of 8080 ft/s, its height in feet after tt second is given

yanalaym [24]

The average velocity for the time period beginning when t=1 and lasting

(i) 0.01 seconds = 63.84 ft/s

(ii) 0.001 seconds = 63.984 ft/s

Given that a ball is thrown with an initial velocity = 80 ft/s

Let 'y' be the height in feet after 't' seconds.

Given, $y=80t-16t^2$ gives the height in 't' seconds.

Average velocity = Rate of change of distance

= Change in distance/Change in time.

The initial time can be taken as 0 s.

When t =1 s, y = 80 - 16 = 64 ft

(1) t = 0.01 s

y = 80 x 0.01 - 16 x 0.01 x 0.01 = 0.7984 ft

Average velocity = (64 - 0.7984) / (1 -0.01) = 63.84 ft/s

(2) t = 0.001 s

y = 80 x 0.001 - 16 x 0.001 x 0.001 = 0.079984 ft

Average velocity = (64 - 0.079984) / (1 -0.001) = 63.984 ft/s

The question is incomplete. Find out the complete question below:

If a ball is thrown straight up into the air with an initial velocity of 80 ft/s, it height in feet after t second is given by $y=80t-16t^2$ .Find the average velocity for the time period beginning when t=1 and lasting

(i) 0.01 seconds

(ii) 0.001 seconds

Learn more about average velocity at brainly.com/question/6504879

#SPJ4

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2 years ago