GIVING THE BRAINEST AND 15 POINTS

block is attached to an oscillating spring. The function below shows its position (cm) vs. time (s). What is the angular frequen

faltersainse [42]

Answer:

Angular frequency is 20 rad/s.

Explanation:

Given that,

A block is attached to an oscillating spring. The function below shows its position (cm) vs. time (s) is given by :

$x(t)=1.5\cos(20\ t)$ .....(1)

The general equation of oscillating particle is given by :

$x(t)=A\cos(\omega t)$ .......(2)

Compare equation (1) and (2) we get :

$\omega=20\ rad/s$

So, the angular frequency of the oscillation is 20 rad/s.

7 0

3 years ago

Which of these properties is the best measure of a star's brightness? (1 point)

storchak [24]

Answer:

ABSOLUTE MAGNITUDE, C

Explanation:

Absolute magnitude is how bright the star is from a distance of 10 parsecs. Apparent magnitude is how bright the star appears from Earth, which can vary, because some stars may be farther or closer away. Age and size are not very accurate either. For example, at the end of their lifetimes, very big start may explode in a violent and bright supernova, and outshine the stars near them. As for size, as many stars several times smaller than the sun shine much brighter than it, making it inaccurate.

6 0

2 years ago

A 3.00-kg block starts from rest at the top of a 33.0° incline and slides 2.00 m down the incline in 1.80 s. (a) Find the accele

ElenaW [278]

(a) $1.23 m/s^2$

Let's analyze the motion along the direction of the incline. We have:

- distance covered: d = 2.00 m

- time taken: t = 1.80 s

- initial velocity: u = 0

- acceleration: a

We can use the following SUVAT equation:

$d = ut + \frac{1}{2}at^2$

Since u=0 (the block starts from rest), it becomes

$d=\frac{1}{2}at^2$

So by solving the equation for a, we find the acceleration:

$a=\frac{2d}{t^2}=\frac{2(2.00 m)}{(1.80 s)^2}=1.23 m/s^2$

(b) 0.50

There are two forces acting on the block along the direction of the incline:

- The component of the weight parallel to the surface of the incline:

$W_p = mg sin \theta$

where

m = 3.00 kg is the mass of the block

g = 9.8 m/s^2 is the acceleration due to gravity

$\theta=33.0^{\circ}$ is the angle of the incline

This force is directed down along the slope

- The frictional force, given by

$F_f = - \mu mg cos \theta$

where

$\mu$ is the coefficient of kinetic friction

According to Newton's second law, the resultant of the forces is equal to the product between mass and acceleration:

$W-F_f = ma\\mg sin \theta - \mu mg cos \theta = ma$

Solving for $\mu$ , we find

$\mu = \frac{g sin \theta - a}{g cos \theta}=\frac{(9.8 m/s^2)sin 33.0^{\circ} - 1.23 m/s^2}{(9.8 m/s^2) cos 33.0^{\circ}}=0.50$

(c) 12.3 N

The frictional force acting on the block is given by

$F_f = \mu mg cos \theta$

where

$\mu = 0.50$ is the coefficient of kinetic friction

m = 3.00 kg is the mass of the block

g = 9.8 m/s^2 is the acceleration of gravity

$\theta=33.0^{\circ}$ is the angle of the incline

Substituting, we find

$F_f = (0.50)(3.00 kg)(9.8 m/s^2) cos 33.0^{\circ} =12.3 N$

(d) 6.26 m/s

The motion along the surface of the incline is an accelerated motion, so we can use the following SUVAT equation

$v^2 - u^2 = 2ad$

where

v is the final speed of the block

u = 0 is the initial speed

a = 1.23 m/s^2 is the acceleration

d = 2.00 m is the distance covered

Solving the equation for v, we find the speed of the block after 2.00 m:

$v=\sqrt{u^2 + 2ad}=\sqrt{0^2+2(9.8 m/s^2)(2.00 m)}=6.26 m/s$

5 0

3 years ago

n the railroad accident, a boxcar weighting 200 kN and traveling at 3 m/s on horizontal track slams into a stationary caboose we

USPshnik [31]

Answer:

ΔK = -6 10⁴ J

Explanation:

This is a crash problem, let's start by defining a system formed by the two trucks, so that the forces during the crash have been internal and the moment is preserved

initial instant. Before the crash

p₀ = m v₁ + M 0

final instant. Right after the crash

p_f = (m + M) v

p₀ = p_f

mv₁ = (m + M) v

v = $\frac{m}{m+M} \ v_1$

we substitute

v = $\frac{20}{20+40}$ 3

v = 1.0 m / s

having the initial and final velocities, let's find the kinetic energy

K₀ = ½ m v₁² + 0

K₀ = ½ 20 10³ 3²

K₀ = 9 10⁴ J

K_f = ½ (m + M) v²

K_f = ½ (20 +40) 10³ 1²

K_f = 3 10⁴ J

the change in energy is

ΔK = K_f - K₀

ΔK = (3 - 9) 10⁴

ΔK = -6 10⁴ J

The negative sign indicates that the energy is ranked in another type of energy

7 0

3 years ago

In which number are the zeros not significant?<br> 100.0<br> O 0.0003<br> O 4.00005<br> O 1.0004

tensa zangetsu [6.8K]

Answer:

0.0003

Explanation:

In the rules of Sig Figs, all zeros before with decimals are not sigificant. I.E. 0.00000000000000009. Despite how many 0's there are, only the 9 is significant. Zeros before a number is not significant. In 100, only the one is signficant in 100. with a dot at the end, the one and the two zeros are significant. hope this helps.

7 0

3 years ago

Read 2 more answers