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3 years ago

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What happens to acceleration when mass is increased

1 answer:

Lubov Fominskaja [6]3 years ago

8 0

Answer:

Newtown's second law of motion

F= ma

a = F/m

if mass is increased then acceleration get decrease

because acceleration is inversely proportional to mass

You might be interested in

Hannah walks 0.30 km to class in 5.0 min. what is her average speed in m/s?

OverLord2011 [107]

Answer:

1.0 m/s

Explanation:

First, convert to SI units.

0.30 km × (1000 m / km) = 300 m

5.0 min × (60 s / min) = 300 s

Speed is distance divided by time:

300 m / 300 s = 1.0 m/s

3 0

3 years ago

A dragster in a race accelerated from stop to 60 m/s by the time it reached the finish line. The dragster moved in a straight li

Aleks [24]

Vf = final velocity
vo = initial velocity
a = acceleration
t = time

use the following equation

vf = vo + at

since vo = 0 m/s (stopped), that term drops out and you're left with . . .

vf = at

(60 m/s) = (8.0 m/s²)t

t = (60 m/s)/(8.0 m/s²) = 7.5 seconds

<u><em>t = 7.5 seconds</em></u>

7 0

4 years ago

The gravitational force between two asteroids is 2.59 × 10 (exponent)-6 N. The centers of mass are 2000 meters away and their ma

makkiz [27]

Answer:

$2.79 \times 10^5 \ \text{kg}$

Explanation:

Newton's Law of Universal Gravitation:

$$F= G\frac{m_1 m_2}{r^2}$
F = force of gravity (N)
G = gravitational constant $(6.67 \times 10^-^1^1 \ N\frac{m^2}{kg^2})$
$m_1$ = mass of Object 1 (kg)
$m_2$ = mass of Object 2 (kg)
r = distance between the center of mass (m)

Let's convert our given information to scientific notation:

$2000 \ m \rightarrow 2.0 \times 10^3 \ m$

Now using the gravitational force and the distance between centers of mass that are given, we can plug these into Newton's law:

$2.59 \times 10^-^6 $\ N = 6.67 \times 10^-^1^1 \ N \frac{m^2}{kg^2} \times \frac{m_1 m_2}{(2.0 \times 10^3 \ m)^2}$

Remove the units for better readability.

$2.59 \times 10^-^6=6.67 \times 10^-^1^1 \frac{m_1m_2}{(2.0 \times 10^3)^2}$

Divide both sides of the equation by the gravitational constant G.

$\frac{2.59 \times 10^-^6}{6.67 \times 10^-^1^1} =\frac{m_1m_2}{(2.0 \times 10^3)^2}$

Distribute the power of 2 inside the parentheses.

$\frac{2.59 \times 10^-^6}{6.67 \times 10^-^1^1} =\frac{m_1m_2}{2.0 \times 10^6}$

If we evaluate the left side of the equation, we get:

$3.88305847 \times 10^4 = \frac{m_1m_2}{2.0 \times 10^6}$

Multiply both sides of the equation by r.

$7.76611694 \times 10^1^0= m_1m_2$

In order to find the mass of one asteroid, we can use the fact that both asteroids have the same mass, therefore, we can rewrite $m_1m_2$ as $m^2$ .

$7.76611694 \times 10^1^0= m^2$

Square root both sides of the equation.

$m=\sqrt{7.76611694 \times 10^1^0}$

$m=2.78677536 \times 10^5$
$m=2.79 \times 10^5$

Since m is in units of kg, we can state that the mass of each asteroid is 2.79 * 10⁵ kg.

3 0

3 years ago

A gazelle leaps from a cliff 2.5 m high with a speed of 5.6m/s.

trapecia [35]

Initial speed of Gazelle is along x direction and its value will be

$v_x = 5.6 m/s$

also its initial height is given as

$y = 2.5 m$

Part a)

now from kinematics along Y direction

$\Delta y = v_y t + \frac{1}{2} at^2$

as we know that

$\Delta y = 0$

$v_y = 0$

$a = 9.8 m/s^2$

$2.5 = 0 + \frac{1}{2} (9.8) t^2$

$t = 0.714 s$

Part b)

distance moved horizontally

$\Delta x = v_x t$

as we know that

$v_x = 5.6 m/s$

now we will have

$v_x = 5.6 (0.714) = 4m$

so it will lend at distance of 4 m.

Part c)

final velocity in vertical direction

$v_{fy} = v_y + at$

$v_{fy} = 0 + (9.8)(0.714) = 7 m/s$

$v_x = 5.6 m/s$

so net speed will be

$v^2 = v_x^2 + v_y^2$

$v^2 = 7^2 + 5.6^2$

$v = 8.96 m/s$

7 0

4 years ago

15. Sandra decided to talk a walk at her neighborhood park. She walks 20 meters

FinnZ [79.3K]

Answer:

Long question good luck:)..............

Explanation:

8 0

3 years ago