What’s the awnser to 1 and 2

light travels approximately 982,080,000 ft/s, and one year has approximately 32,000,000 seconds. A light year is the distance li

lapo4ka [179]

Answer:

The distance traveled in 1 year is: $3.143*10^{16}ft$

Explanation:

Given

$s = 982,080,000 ft/s$ --- speed

$t = 32,000,000 s$ --- time

Required

The distance traveled

This is calculated as:

$Speed = \frac{Distance}{Time}$

So, we have:

$Distance = Speed * Time$

This gives:

$Distance = 982,080,000 ft/s * 32,000,000 s$

$Distance = 982,080,000 * 32,000,000ft$

$Distance = 3.143*10^{16}ft$ -- approximated

5 0

3 years ago

A rope passes over a fixed sheave with both ends hanging straight down. The coefficient of friction between the rope and sheave

Oliga [24]

Answer:3.51

Explanation:

Given

Coefficient of Friction $\mu =0.4$

Consider a small element at an angle \theta having an angle of $d\theta$

Normal Force $=T\times \frac{d\theta }{2}+(T+dT)\cdot \frac{d\theta }{2}$

$N=T\cdot d\theta$

Friction $f=\mu \times Normal\ Reaction$

$f=\mu \cdot N$

and $T+dT-T=f=\mu Td\theta$

$dT=\mu Td\theta$

$\frac{dT}{T}=\mu d\theta$

$\int_{T_2}^{T_1}\frac{dT}{T}=\int_{0}^{\pi }\mu d\theta$

$\frac{T_2}{T_1}=e^{\mu \pi}$

$\frac{T_2}{T_1}=e^{0.4\times \pi }$

$\frac{T_2}{T_1}==e^{1.256}$

$\frac{T_2}{T_1}=3.51$

7 0

3 years ago

"In a Young’s double-slit experiment, the separation between slits is d and the screen is a distance D from the slits. D is much

san4es73 [151]

Answer:

The number of bright fringes per unit width on the screen is, $x=\dfrac{\lambda D}{d}$

Explanation:

If d is the separation between slits, D is the distance between the slit and the screen and $\lambda$ is the wavelength of the light. Let x is the number of bright fringes per unit width on the screen is given by :

$x=\dfrac{n\lambda D}{d}$

$\lambda$ is the wavelength

n is the order

If n = 1,

$x=\dfrac{\lambda D}{d}$

So, the the number of bright fringes per unit width on the screen is $\dfrac{\lambda D}{d}$ . Hence, the correct option is (B).

6 0

3 years ago

I started to solve this problem but I’m not sure on what to do next.

snow_tiger [21]

ANSWER:

3408.81 kg

STEP-BY-STEP EXPLANATION:

Given:

v = 111 m/s

Ek = 21000000 J

We have that the formula for kinetic energy is as follows:

$E_k=\frac{1}{2}\cdot m\cdot v^2$

We substitute the values given in the exercise and solve for m (mass)

$\begin{gathered} 21000000=\frac{1}{2}\cdot m\cdot111^2 \\ m=\frac{21000000\cdot2}{111^2} \\ m=3408.81\text{ kg} \end{gathered}$

The mass of the helicopter is 3408.81 kilograms.

7 0

1 year ago

HELP ME PLEASEEEEEEEEEEEEEE

ozzi

Answer: The correct statements are:

The atoms are very attracted to one another.
The atoms are held tightly together.

Explanation:

Solid state: In this state, the molecules are closely packed and cannot move freely from one place to another that means no space between them and the intermolecular force of attraction between the molecules are strong.

In solid substance, the particles are very close to each other due to this the intermolecular forces of attraction are strongest.

The key point about solid are:

The atoms are very attracted to one another.
The atoms are not moving freely.
It will not spread out evenly to fill any container.
The atoms are held tightly together.
The forces of attraction are strong to bring molecules together.
The atoms are close and in fixed positions.

7 0

3 years ago