A 3.6kg mass is accelerated at 2.5m/s. Calculate the resultant force<br> acting on it.

The following situation will be used for the next three problems: A rock is projected upward from the surface of the moon, at ti

chubhunter [2.5K]

Explanation:

Given that,

Initial speed of the rock, u = 30 m/s

The acceleration due to gravity at the surface of the moon is 1.62 m/s².

We need to find the time when the rock is ascending at a height of 180 m.

The rock is projected from the surface of the moon. The equation of motion in this case is given by :

$h=ut-\dfrac{1}{2}gt^2\\\\180=30t-\dfrac{1}{2}\times 1.62t^2$

It is a quadratic equation, after solving whose solution is given by:

t = 7.53 s

or

t = 8 seconds

(e)If it is decending, v = -20 m/s

Now t' is the time of descending. So,

$v=-u+gt\\\\t=\dfrac{v+u}{g}\\\\t=\dfrac{20+30}{1.62}\\\\t=30.86\ s$

Let h' is the height of the rock at this time. So,

$h'=ut-\dfrac{1}{2}gt^2\\\\h'=30\times 30.86-\dfrac{1}{2}\times 1.62\times 30.86^2\\\\h'=154.40\ m$

or

h' = 155 m

6 0

3 years ago

Which measurement is equivalent to 350 milliliters?

tensa zangetsu [6.8K]

350 milliliters equals to .35 liters
Therefore, 3.5 times 10^-1 would be the only correct answer since that is the only one that matches .35 liters

4 0

3 years ago

A conductor of radius r, length and resistivity ρ has resistance r. what is the new resistance if it is stretched to 4 times its

max2010maxim [7]

The resistance of a conductor is given by
$R= \frac{\rho L}{A}$
where L is the length of the wire, $\rho$ the resistivity of the material and A the cross-sectional area.

We can see that if all the other quantities do not change, if the new length of the conductor is 4 times the original length: $L'=4 L$ , then the new resistance is also 4 times the original value:
$R'= \frac{\rho L'}{A}= \frac{\rho 4 L}{A}=4 \frac{\rho L}{A}=4 R$

7 0

3 years ago

What is the acceleration of a 10 kg mass pushed by a 5 N force?

yanalaym [24]

$g-\ gravitational \ acceleration \\ g= \frac{G}{m} = \frac{5N}{10kg}= \\ g=0,5 \frac{N}{kg}$

8 0

3 years ago

Read 2 more answers

8. A turtle crawls along a straight line, which we will call the x-axis with the positive direction to the right. The equation f

pychu [463]

(a) The turtle's initial velocity is 4 m/s, initial position of the turtle is 5 cm, and initial acceleration is -1.25 m/s².

(b) The time when the velocity of the turtle is zero is 3.2 s.

(c) The time taken for the turtle to return to its starting point is 6.4 s.

(d) The time taken for the turtle to travel 30 cm is 0.08 s.

<h3>Initial velocity of the turtle</h3>

The initial velocity of the turtle is calculated as follows;

$v = \frac{dx}{dt} \\\\x = 5 + 4t -0.625t^2\\\\v(t) = 4 - 1.25t\\\\v(0) = 4-0\\\\v(0) = 4 \ m/s$

<h3>Initial acceleration of the turtle</h3>

The initial acceleration of the turtle is calculated as follows;

$a = \frac{dv}{dt} \\\\v(t) = 4 - 1.25t\\\\a = -1.25\ m/s^2$

<h3>Initial position of the turtle</h3>

x(t) = 5 + 4t - 0.625t²

x(0) = 5 cm

<h3>Time when the velocity becomes zero</h3>

v(t) = 4 - 1.25t

0 = 4 - 1.25t

1.25t = 4

t = 4/1.25

t = 3.2 s

<h3>Time taken to return to starting point</h3>

The total distance traveled is calculated as follows

v² = u² + 2ad

0 = (4)² + 2(-1.25)d

0 = 16 - 2.5d

2.5d = 16

d = 16/2.5

d = 6.4 m

Time to travel the given distance;

d = ut + ¹/₂at²

6.4 = (4)t + ¹/₂(-1.25)t²

6.4 = 4t - 0.625t²

0.625t² - 4t + 6.4 = 0

solve the quadratic equation using formula method;

t = 3.2 s

The time travel the distance two times, = 2 x 3.2 s = 6.4 s

<h3>Time taken for the turtle to travel 30 cm</h3>

d = ut + ¹/₂at²

0.3 = (4)t + ¹/₂(-1.25)t²

0.3 = 4t - 0.625t²

0.625t² - 4t + 0.3 = 0

solve the quadratic equation using formula method;

t = 0.08 s

Learn more about velocity here: brainly.com/question/6504879

7 0

2 years ago

A 3.6kg mass is accelerated at 2.5m/s. Calculate the resultant force acting on it.