If the Net Force of the object is 30 N to the left, and mass is 3 kg what is the objects acceleration?

.A hard rubber ball, released at chest height, falls to the pavement and bounces back to nearly the same height. When it is in c

ohaa [14]

Answer:

a = 1.1 10⁵ m / s²

Explanation:

This is a momentum exercise, where we use the relationship between momentum and momentum

I = ∫ F dt = Δp

= p_f - p₀

as they indicate that the ball bounces at the same height, we can assume that the moment when it reaches the ground is equal to the moment when it bounces, but in the opposite direction

F t = 2 (m v)

therefore the average force is

F = 2 m v / t

where in general the mass of the ball unknown, the velocity of the ball can be calculated using the conservation of energy

starting point. Done the ball is released with zero initial velocity

Em₀ = U = mgh

final point. Upon reaching the ground, just before the deformation begins

Em_f = K = ½ m v²

energy is conserved in this system

Em₀ = Em_f

m g h = ½ m v²

v = √ (2gh)

This is the velocity of the body when it reaches the ground, so the force remains

F = 2m √(2gh) /t

where the height of the person's chest is known and the time that the impact with the floor lasts must be estimated in general is of the order of milli seconds

knowing this force let's use Newton's second law

F = m a

a = F / m

a = 2 √(2gh) / t

We can estimate the order of magnitude of this acceleration, assuming the person's chest height of h = 1.5 m and a collision time of t = 1 10⁻³ s

a = 2 √ (2 9.8 1.5) / 10⁻³

a = 1.1 10⁵ m / s²

6 0

2 years ago

What is the top of a wave called?

ohaa [14]

Answer:

The highest part of the wave is called the crest.

Explanation:) hope this helps

The highest part of the wave is called the crest. The lowest part is called the trough. The wave height is the overall vertical change in height between the crest and the trough and distance between two successive crests (or troughs) is the length of the wave or wavelength.

7 0

1 year ago

A hydrogen atom has a diameter of about 10 nm. Express this diameter in meters. Express this diameter in millimeters. Express th

Fynjy0 [20]

So, the first question is: how many meters are 10 nm?

1nm =<span>0.000000001 m.

So 10 nanometers are </span><span>0.00000001 m!

Now, how many milimeter are those?
let's start with meters, 1 meter are 1000 milimeters.
so </span>
0.00000001*1000=0.<span><span>00001</span> m!

now, micrometers .1 micrometer are 1000 nanometers.
so 10 nanometers are 0.01 micrometers! (1 nanometer is 0.001 micrometers)
</span>

8 0

3 years ago

20 Accelerated motion is represented by a<br> line on a nonlinear distance time graph.

Varvara68 [4.7K]

Curved line

Explanation:

Acceleration of motion is represented by a curved line on a non-linear distance-time graph.

The acceleration of a non-linear motion is depicted using a parabola which is a curve. This implies that the velocity is constantly changing and the distance covered by the body is also changing with equal amount of time.

A plot of this will give a parabola. This can be further established using one of the equations of motion below:

x = u + $\frac{1}{2}$ at ²

This is a quadratic function where:

x is the distance

u is the initial velocity

t is the time

a is acceleration

A quadratic function gives a curved line which is a parabola.

Learn more:

Acceleration brainly.com/question/10932946

#learnwithBrainly

3 0

3 years ago

A wad of clay of mass m1 = 0.49 kg with an initial horizontal velocity v1 = 1.89 m/s hits and adheres to the massless rigid bar

notka56 [123]

Answer:

<h2>The angular velocity just after collision is given as</h2><h2> $\omega = 0.23 rad/s$ </h2><h2>At the time of collision the hinge point will exert net external force on it so linear momentum is not conserved</h2>

Explanation:

As per given figure we know that there is no external torque about hinge point on the system of given mass

So here we will have

$L_i = L_f$

now we can say

$m_1v_1\frac{L}{2} = (m_2L^2 + m_1(\frac{L}{2})^2)\omega$

so we will have

$0.49(1.89)(0.45) = (2.13(0.90)^2 + 0.49(0.45)^2)\omega$

$\omega = 0.23 rad/s$

Linear momentum of the system is not conserved because at the time of collision the hinge point will exert net external force on the system of mass

So we can use angular momentum conservation about the hinge point

6 0

3 years ago