Plzpzlpzlzplzplzplzpz this one also<br> all questions

Which stars have the lowest absolute brightness?

Finger [1]

Answer:

DWARFS

Explanation:

3 0

3 years ago

Read 2 more answers

A 1.0 kg mass is attached to the end of a vertical ideal spring with a force constant of 400 N/m. The mass is set in simple harm

Marina86 [1]

Answer:

$v(0)=2m/s$

Explanation:

The instantaneous velocity of a point mass that executes a simple harmonic movement is given by:

$v(t)=\omega *A*cos(\omega t + \phi)$

Where:

$\omega=Angular\hspace{3}frequency\\A=Amplitude\\\phi=Initial\hspace{3}phase$

Express the amplitude in meters:

$10cm*\frac{1m}{100cm} =0.1m$

The angular frequency can be found using the next equation:

$\omega=\sqrt{\frac{k}{m} }$

Using the data provided:

$\omega=\sqrt{\frac{400}{1} } =20$

At the equilibrium position:

$\phi=0$

$v(0)=20*(0.1)cos(20*0+0)=2*cos(0)=2*1=2m/s$

6 0

3 years ago

A convex mirror has a focal length of -10.8 cm. An object is placed 32.7 cm from the mirror's surface. Determine the image dista

KonstantinChe [14]

Answer:

-353.16

Explanation:

4 0

3 years ago

A 6 kilogram block in outer space is moving at -100 m/s (to the left). It suddenly experiences three forces as shown below.

Alika [10]

Newton's second law and the kinematic relations allow to find the results for the questions about forces and the movement of the block are:

B) the force applied to maintain the system is equilibrium is: F = 0.39 N with an angle of tea = 180º

C) The maximum force is: F = 24 N

D) The time to stop the block is: t = 25 s

Newton's second law establishes a relationship between the net force, the mass, and the acceleration of the body. In the special case that the acceleration is zero it is called the equilibrium condition.

B) They indicate a diagram of forces on the block, let's look for the components of the force that the block maintains with zero acceleration, in the attached we have a free-body diagram including the force applied to keep the system in equilibrium.

x-axis

-10 + 12 sin 60 + Fₓ = 0

Fₓ = 10- 12 sin 60 = -0.39 N

y-axis

12 cos 60 - 6 + F_y = 0

F_y = 6 - 12 cos 60 = 0 N

We can give the result of the force in two ways:

Form of coordinates F = -0.39 i ^ N

Form of module and angle.

Let's use Pythagoras' theorem to find the modulus.

$F = \sqrt{F_x^2 + F_y^2 } \\F = \sqrt{0.39^2 +0^2}$

F = 0.39N

We use trigonometry for the angle.

$tan \theta = \frac{F_y}{F_x}$

tan θ= 0º

The component of the force is negative therefore this angle is in the second quadrant, to measure the angle from the positive side of the x axis in a counterclockwise direction.

θ = 180 + θ'

θ = 180 + 0

θ = 180º

C) if the three forces can be moved and the maximum force occurs when they are all linear.

10+ 6 + 6 + F = 0

F = -24 N

D) if we maintain this force and eliminate the other three, the block stops, let's look for its acceleration.

$a = \frac{F}{m}$

a = $\frac{24}{6}$

a = 4 m / s²

The acceleration is in the opposite direction of the initial velocity of the block v₀ = -100 m / s

If we use kinematic relations.

v = v₀ - a t

Final velocity when stopped is zero

t = $\frac{0-v_o}{a}$

t = 100/4

t = 25 s

In conclusion using Newton's second law and the kinematics relations we can find the results for the questions about the forces and the motion of the block are:

B) the force applied to maintain the system is equilibrium is: F = 0.39 N with an angle of tea = 180º

C) The maximum force is: F = 24 N

D) The time to stop the block is: t = 25 s

Learn more about Newton's second law here: brainly.com/question/25545050