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3 years ago

A penny rolls along the table for a distance of 1.3 meters. Jackie pushes it 40 centimeters further in the same direction.

1 answer:

7 0

40 meters times 1 meter over 100 centimeters equals 0.4 meters. 1.3 meters + 40 centimeters =. 1.3 m + 0.4 m = 1.7 m. The answer is 1.7 meters

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Calculate the gravitational attraction between two objects which masses are: mass A: 2.5kg, mass B: 5kg. The distance between th

jonny [76]

From the calculation, the gravitational force of attraction is 1.33 * 10^-14 N.

<h3>What is the gravitational force?</h3>

The gravitational force is an attractive force that acts between any two masses.

It is given by;

F = Gm1m2/r^2

F = 6.67 * × 10−11 * 2.5 * 5/(250)^2

F = 83.4 × 10−11 /62500

F= 1.33 * 10^-14 N

Learn more about gravitational force:brainly.com/question/12528243

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4 0

2 years ago

Does an increase in velocity necessarily mean an increase in acceleration?

Vinil7 [7]

We know, acceleration = final velocity - initial velocity / time
Here, if velocity is increasing, then,
Final velocity > initial velocity, in that case, acceleration is also increasing, as it is directly proportional to velocity

In short, Your Answer would be "Yes"

Hope this helps!

3 0

3 years ago

Read 2 more answers

Find an equation of the line which is parallel to 2x-3y=6 and passes through the point (6,9)

tatuchka [14]

Step 2: Use the slope to find the y-intercept. Line is parallel so use m = 2/5. 6. Findthe equation of a line passing through the point (8, –9) perpendicular to the line 3x + 8y = 4.

8 0

3 years ago

You can use any coordinate system you like in order to solve a projectile motion problem. To demonstrate the truth of this state

posledela

Answer:

a) y₂ = 49.1 m , t = 1.02 s , b) y = 49.1 m , t= 1.02 s

Explanation:

a) We will solve this problem with the missile launch kinematic equations, to find the maximum height, at this point the vertical speed is zero

$v_{y}$ ² = $v_{oy}$ ² - 2 g (y –yo)

The origin of the coordinate system is on the floor and the ball is thrown from a height

y-yo = $v_{oy}² /2 g
 y- 0 = 10.0²/2 9.8
 y - 0 = 5.10 m
 
The height from the ground is the height that rises from the reference system plus the depth of the ground from the reference system
 y₂ = 5.1 + 44
 y₂ = 49.1 m
Let's use the other equation to find the time
 [tex]v_{y}$ = $v_{oy}$ - g t

t = $v_{oy}$ / g

t = 10 / 9.8

t = 1.02 s

b) the maximum height

y- 44.0 = $v_{y}$ ² / 2 g

y - 44.0 = 5.1

y = 5.1 +44.0

y = 49.1 m

The time is the same because it does not depend on the initial height

t = 1.02 s

7 0

3 years ago

What dating technique could scientists use to find the age of a meteorite?

kvasek [131]

Radiometric dating?
Also, possibly radiocarbon dating

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3 years ago