A car moved 120km to the north. what is its displacement?

A 2 kg mass is free falling in the negative Y direction when a 10 N force is exerted in the minus X direction. What is the accel

lara31 [8.8K]

Answer:

The mass's acceleration is 5 m/s^2 in the minus X direction and 9,8 m/s^2 in the minus Y direction.

Explanation:

By applying the second Newton's law in the X and Y direction we found that in the minus X direction an external force of 10 N is exerted, while in the minus Y direction the gravity acceleration is acting:

X-direction balance force: $-10 [N] = m.ax$

Y-direction balance force: $-m*9,8 \frac{m}{s^2} = m.ay$

Where ax and ay are the components of the respective acceleration and m is the mass. By solving for each acceleration:

$ax=(-10 [N]) / m$

$ay=-m*9,8\frac{m}{s^2} / m$

Note that for the second equation above the mass is cancelled and, the Y direction acceleration is minus the gravity acceleration:

$ay=-9,8\frac{m}{s^2}$

For the x component aceleration we must replace the Newton unit:

$N =\frac{kg.m}{s^2}$

$ax= -10 \frac{kg.m}{s^2} / (2 kg)$

$ax= - 5 \frac{m}{s^2}$

6 0

3 years ago

A 0.20-kg object is attached to the end of an ideal horizontal spring that has a spring constant of 120 N/m. The simple harmonic

Umnica [9.8K]

Answer:

A = 6.9 cm

Explanation:

Simple Harmonic Motion

A mass-spring system is a common example of a simple harmonic motion device since it keeps oscillating when the spring is stretched back and forth.

If a mass m is attached to a spring of constant k and they are set to oscillate, the angular frequency of the motion is

$\displaystyle w=\sqrt{\frac{k}{m}}$

The equation for the motion of the object is written as a sinusoid:

$\displaystyle X=A\ cos\ w\ t$

Where A is the amplitude.

The instantaneous speed is computed as the derivative of the distance

$\displaystyle X'=V=-A\ w\ sin\ w\ t$

And the maximum speed is

$\displaystyle V_{max}= A\ w$

Solving for the amplitude

$\displaystyle A= \frac{V_{max}}{w}$

Computing w

$\displaystyle w =\sqrt{\frac{120}{0.2}}=24.5\ rad/ s$

Calculating A

$\displaystyle A=\frac{1.7}{24.5}=0.069\ m$

$\displaystyle \boxed{A=6.9\ cm}$

7 0

3 years ago

Who is the founder of operant conditioning?

Gemiola [76]

Operant conditioning, sometimes called instrumental learning, was first extensively studied by Edward L. Thorndike, who observed the behavior of cats trying to escape from home-made puzzle boxes.

Hope this helps!

7 0

3 years ago

Which three quantities can be used to calculate acceleration?

PtichkaEL [24]

D is the correct answer, assuming that this is the special case of classical kinematics at constant acceleration. You can use the equation V = Vo + at, where Vo is the initial velocity, V is the final velocity, and t is the time elapsed. In D, all three of these values are given, so you simply solve for a, the acceleration.
A and C are clearly incorrect, as mass and force (in terms of projectile motion) have no effect on an object's motion. B is incorrect because it is not useful to know the position or distance traveled, unless it will help you find displacement. Even then, you would not have enough information to use a kinematics equation to find a.

4 0

3 years ago

A block of mass 4 kilograms is initially moving at 5m/s on a horizontal surface. There is friction between the block and the sur

emmasim [6.3K]

• Net vertical force on the block:

∑ F = n - w = 0

(n = magnitude of normal force, w = weight)

n = w = m g

(m = mass, g = 9.8 m/s²)

n = (4 kg) (9.8 m/s²) = 39.2 N

• Net horizontal force:

∑ F = -f = m a

(f = mag. of friction, a = acceleration)

We have f = µ n = 0.5 (39.2 N) = 19.6 N, so

-19.6 N = (4 kg) a

a = -4.9 m/s²

With this acceleration, the block comes to a rest from an initial speed of 5 m/s, so that it travels a distance ∆x in this time such that

0² - (5 m/s)² = 2 (-4.9 m/s²) ∆x

∆x = (25 m²/s²) / (9.8 m/s²) ≈ 2.55 m ≈ 2.6 m

8 0

2 years ago