What are the three layers of the sun's atmosphere, in order of increasing distance from the surface?

(3) A Ninja motorcycle is moving down I95 at 160 miles per hour (72 m/s) and a cop uses a radar gun to measure the speed. Using

Virty [35]

Answer:

it does use it at doppler

Explanation:

My dad told me the hes a cop so yw

5 0

2 years ago

HELP PLS.<br><br><br>What are the characteristics of the three states of matter?

vredina [299]

Matter can exist in one of three main states: solid, liquid, or gas. Solid matter is composed of tightly packed particles. A solid will retain its shape; the particles are not free to move around. Liquid matter is made of more loosely packed particles. Hopefully this helps:)

5 0

3 years ago

Read 2 more answers

Finally, you are ready to answer the main question. Cheetahs, the fastest of the great cats, can reach 50.0 miles/hour in 2.22 s

muminat

Answer:

Acceleration = 10.06 m/s²

Explanation:

1 mile = 1.6093km

1609.3m = 1 mile

1 m = $\frac{1}{1609}$ mile

50.0 miles/hour = $\frac{50 * 1609.3}{60 * 60}$ m/s

= 22.35m/s

from equation

S = Ut + 1/2 at²

v = U + at

22.35 = 0 + a * 2.22

a = 22.35 ÷ 2.22

= 10.06 m/s²

4 0

3 years ago

Which of these is NOT a possible type of energy transformation?

Orlov [11]

to be franc i really think the answer is B

6 0

3 years ago

FIGURE 2 shows a 1.5 kg block is hung by a light string which is wound around a smooth pulley of radius 20 cm. The moment of ine

Sindrei [870]

Answer:

At t = 4.2 s

Angular velocity: 6. 17 rad /s

The number of revolutions: 2.06

Explanation:

First, we consider all the forces acting on the pulley.

There is only one force acting on the pulley, and that is due to the 1.5 kg mass attached to it.

Therefore, the torque on the pulley is

$\tau=Fd=mg\cdot R$

where m is the mass of the block, g is the acceleration due to gravity, and R is the radius of the pulley.

Now we also know that the torque is related to angular acceleration α by

$\tau=I\alpha$

therefore, equating this to the above equation gives

$mg\cdot R=I\alpha$

solving for alpha gives

$\alpha=\frac{mgR}{I}$

Now putting in m = 1.5 kg, g = 9.8 m/s^2, R = 20 cm = 0.20 m, and I = 2 kg m^2 gives

$\alpha=\frac{1.5\cdot9.8\cdot0.20}{2}$ $\boxed{\alpha=1.47s^{-2}}$

Now that we have the value of the angular acceleration in hand, we can use the kinematics equations for the rotational motion to find the angular velocity and the number of revolutions at t = 4.2 s.

The first kinematic equation we use is

$\theta=\theta_0+\omega_0t+\frac{1}{2}\alpha t^2$

since the pulley starts from rest ω0 = 0 and theta = 0; therefore, we have

$\theta=\frac{1}{2}\alpha t^2$

Therefore, ar t = 4.2 s, the above gives

$\theta=\frac{1}{2}(1.47)(4.2)^2$

$\boxed{\theta=12.97}$

So how many revolutions is this?

To find out we just divide by 2 pi:

$\#\text{rev}=\frac{\theta}{2\pi}=\frac{12.97}{2\pi}$ $\boxed{\#\text{rev}=2.06}$

Or about 2 revolutions.

Now to find the angular velocity at t = 4.2 s, we use another rotational kinematics equation:

$\omega^2=w^2_0+2\alpha(\Delta\theta)_{}$

Since the pulley starts from rest, ω0 = 0. The change in angle Δθ we calculated above is 12.97. The value of alpha we already know to be 1.47; therefore, the above becomes:

$\omega^2=0+2(1.47)(12.97)$ $w^2=38.12$ $\boxed{\omega=6.17.}$

Hence, the angular velocity at t = 4.2 w is 6. 17 rad / s

To summerise:

at t = 4.2 s

Angular velocity: 6. 17 rad /s

The number of revolutions: 2.06

3 0

1 year ago