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3 years ago

An object at rest increases its velocity to 10m/s in 3 seconds. What is its acceleration

Physics

1 answer:

BabaBlast [244]3 years ago

5 0

Acceleration = (change in speed) / (time for the change)

Change in speed = (speed at the end) - (speed at the beginning).

Change in speed = (10 m/s) - (zero) = 10 m/s

Time for the change = 3 sec

Acceleration = (10 m/s) / (3 sec)

<em>Acceleration = (3 and 1/3) m/s²</em> or 3.333 m/s²

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On a certain nonstop trip, Marta averaged x miles per hour for 2 hours and y miles per hour for the remaining 3 hours. What was

Yuri [45]

Answer:

Average speed = (2x+3y)/5 miles/hour

Explanation:

Average speed is the total distance travelled per unit time.

Average speed = total distance/total time taken

Given;

1. Travelled x miles per hour for 2hours.

2 travelled y miles per hour for 3 hours.

Distance covered is;

1. Distance = speed × time

d1 = x × 2 = 2x

2. d2 = y × 3 = 3y

Total distance travelled = d1+d2 = 2x + 3y

Total time taken = t1+t2 = 2+3 = 5hours

Average speed = (2x+3y)/5 miles/hour

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4 years ago

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Answer:

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3 years ago

Mark creates a graphic organizer to review his notes about electrical force. Which labels belong in the regions marked X and Y?

sveta [45]

Answer:

The correct answer is A

Explanation:

The question requires as well the attached image, so please see that below.

Coulomb's Law.

The electrical force can be understood by remembering Coulomb's Law, that describes the electrostatic force between two charged particles. If the particles have charges $q_1$ and $q_2$ , are separated by a distance r and are at rest relative to each other, then its electrostatic force magnitude on particle 1 due particle 2 is given by:

$|F|=k \cfrac{q_1 q_2}{r^2}$

Thus if we decrease the distance by half we have

$r_1 =\cfrac r2$

So we get

$|F|=k \cfrac{q_1 q_2}{r_1^2}$

Replacing we get

$|F|=k \cfrac{q_1 q_2}{(r/2)^2}\\|F|=k \cfrac{q_1 q_2}{r^2/4}$

We can then multiply both numerator and denominator by 4 to get

$|F|=k \cfrac{4q_1 q_2}{r^2}$

So we have

$|F|=4 \left(k \cfrac{q_1 q_2}{r^2}\right)$

Thus if we decrease the distance by half we get four times the force.

Then we can replace the second condition

$q_{2new} =2q_2$

So we get

$|F|=k \cfrac{q_1 q_{2new}}{r_1^2}$

which give us

$|F|=k \cfrac{q_1 2q_2}{r_1^2}\\|F|=2\left(k \cfrac{q_1 q_2}{r_1^2}\right)$

Thus doubling one of the charges doubles the force.

So the answer is A.

8 0

3 years ago

Read 2 more answers

One of the best candidates for a black hole is found in the binary system called A0620-0090. The two objects in the binary syste

Darya [45]

Answer:

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Explanation:

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3 years ago

A fish is able to jump vertically out of the water with a speed of 4.45 m/s. What is the speed of the fish as it passes a point

rjkz [21]

Answer:

Explanation:

given

initial velocity u = 4.45m/s

Height = 0.6m

g = 9.8m/s²

Required

final velocity v

Using the equation of motion;

v² = u²-2gH (upward motion of the fish makes g to be negative)

v² = 4.45²-2(9.8)(0.6)

v² = 19.8025-11.76

v² = 8.0425

v = 2.84 m/s

Hence the speed of the fish as it passes a point 0.6 m above the water is 2.84m/s

To get the time, we will use the formula

v = u - gt

2.84 = 4.45 - 9.8t

2.84-4.45 = -9.8t

-1.61 = -9.8t

t = 1.61/9.8

t = 0.164secs

Hence the time taken is 0.164secs

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3 years ago