Please help me answer this it is due by Sunday night.

A spectral line from a star is observed to have a wavelength of 656.5 nm. The rest wavelength of this line is 656.8 nm. (a) What

nignag [31]

Answer:

speed of star is 1.37 × $10^{5}$ m/s

it is approaching to earth because wavelength of star is decreasing than rest

if emit same wavelength it does not move anywhere

it will remain steady condition with respect earth

Explanation:

given data

wavelength = 656.5 nm

rest wavelength = 656.8 nm

to find out

the speed of the star , is it approaching

solution

we know here equation that is

wavelength shift / wavelength at rest = velocity of source / speed of light

so put all value and find

velocity of source = 3 × $10^{-9}$ × 3 × $10^{8}$ / 656.8 × $10^{-9}$

velocity of source star = 1.37 × $10^{5}$ m/s

and

it is approaching to earth because wavelength of star is decreasing than rest

and

if emit same wavelength it does not move anywhere

it will remain steady condition with respect earth

6 0

3 years ago

What is the momentum of an object that is moving in a straight line at a speed of 10 m/s and has a mass of 10 kilograms?

Aleksandr-060686 [28]

The momentum of an object is given by the product between its mass and its velocity:
$p=mv$
where m is the mass and v the velocity.

For the object in our problem, m=10 kg and v=10 m/s, therefore its momentum is
$p=mv=(10 kg)(10 m/s)=100 kg m/s$
So, the correct answer is B).

8 0

3 years ago

The lower the value of the coefficient of friction,the_the resistance to sliding

malfutka [58]

Answer:

lower

Explanation:

The lower the value of the coefficient of friction, the lower the resistance to sliding.

The coefficient of friction is the ratio of the frictional force and the normal force pressing two surfaces in contact together.

U = $\frac{F}{N}$

U is the coefficient of friction

F is the frictional force

N is the normal force

We see that coefficient of friction is directly proportional to frictional force.

7 0

3 years ago

Physics. Need help. Brainlieast answer for most/ all of the answers answered

Mumz [18]

<u>ALL of the following work assumes NO AIR RESISTANCE:</u>

1). an object moving under the influence of only gravity, and not in orbit; its horizontal velocity is constant, and its vertical motion is accelerated downward at 9.8 m/s²

2). a parabola

3). Horizontal: velocity is constant, acceleration is zero. . . . Vertical: acceleration is 9.8 m/s² downward, velocity depends on whether it was launched, thrown up, thrown down, dropped, etc.

4). a). the one that was thrown horizontally; b). both hit the ground at the same time; c). both hit the ground with the same vertical velocity

5). a). zero; b). zero; c). gravity ... 9.8 m/s² down; d). 3.06 seconds; e). 4.38 m/s; f). 30 m/s g). no; gravity has no effect on horizontal motion

6). a). 1.8 seconds; b). 13.1 meters; c). 17.6 m/s down; d). 7.3 m/s; gravity has no effect on horizontal motion

7). 45 m/s

8). without air resistance, the ball is traveling horizontally at 13 km/hr, and it lands back in your hand

9). a). 4.49 m/s; b). 29.7 m/s

10). 7.24 meters

11). 700 meters

12). A). 103.7 meters ( ! she's in big trouble ! ); B). 17.5 meters

3 0

3 years ago

Starting from rest, a basketball rolls from the top to the bottom of a hill, reaching a translational speed of 6.1 m/s. Ignore f

tatiyna

Answer:

a) h=3.16 m, b) v_{cm }^ = 6.43 m / s

Explanation:

a) For this exercise we can use the conservation of mechanical energy

Starting point. Highest on the hill

Em₀ = U = mg h

final point. Lowest point

$Em_{f}$ = K

Scientific energy has two parts, one of translation of center of mass (center of the sphere) and one of stationery, the sphere

K = ½ m $v_{cm }^{2}$ + ½ $I_{cm}$ w²

angular and linear speed are related

v = w r

w = v / r

K = ½ m v_{cm }^{2} + ½ I_{cm} v_{cm }^{2} / r²

Em_{f} = ½ v_{cm }^{2} (m + I_{cm} / r2)

as there are no friction losses, mechanical energy is conserved

Em₀ = Em_{f}

mg h = ½ v_{cm }^{2} (m + I_{cm} / r²) (1)

h = ½ v_{cm }^{2} / g (1 + I_{cm} / mr²)

for the moment of inertia of a basketball we can approximate it to a spherical shell

I_{cm} = ⅔ m r²

we substitute

h = ½ v_{cm }^{2} / g (1 + ⅔ mr² / mr²)

h = ½ v_{cm }^{2}/g 5/3

h = 5/6 v_{cm }^{2} / g

let's calculate

h = 5/6 6.1 2 / 9.8

h = 3.16 m

b) this part of the exercise we solve the speed of equation 1

v_{cm }^{2} = 2m gh / (1 + I_{cm} / r²)

in this case the object is a frozen juice container, which we can simulate a solid cylinder with moment of inertia

I_{cm} = ½ m r²

we substitute

v_{cm } = √ [2gh / (1 + ½)]

v_{cm } = √(4/3 gh)

let's calculate

v_{cm } = √ (4/3 9.8 3.16)

v_{cm }^ = 6.43 m / s

4 0

3 years ago