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Luden [163]
3 years ago
6

One light-year is the distance light travels in one year. This distance is equal to 9.461 1015 m. After the sun, the star neares

t to Earth is Alpha Centauri, which is about 4.35 light-years from Earth. Express this distance in a. megameters. b. picometers.
Physics
1 answer:
Oliga [24]3 years ago
4 0

Answer:

A.) 411.6 Mega metres

B.) 411.6 × 10^18 Picometers

Explanation:

Given that One light-year is the distance light travels in one year. This distance is equal to 9461 1015 m.

After the sun, the star nearest to Earth is Alpha Centauri, which is about 4.35 light-years from Earth.

The distance travelled will be

Distance = 9461 1015 × 4.35

Distance = 411557915.3 m

To Express this distance in

a. megameters, divide the value by 1000000

411557915.3 / 1000000

411.5579153 Mega metre

411.6 Mega metres approximately

b. picometers, multiply by 10^12

That is, 411557915.3 × 10^12

411.6 × 10^18 Picometers

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c hydrogen

Explanation:

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A 3.0-kg brick rests on a perfectly smooth ramp inclined at 34° above the horizontal. The brick is kept from sliding down the pl
Firdavs [7]

Answer:

d=0.137 m ⇒13.7 cm

Explanation:

Given data

m (Mass)=3.0 kg

α(incline) =34°

Spring Constant (force constant)=120 N/m

d (distance)=?

Solution

F=mg

F=(3.0)(9.8)

F=29.4 N

As we also know that

Force parallel to the incline=FSinα

F=29.4×Sin(34)

F=16.44 N

d(distance)=F/Spring Constant

d(distance)=16.44/120

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3 years ago
Which of the following represents a fully plastic collision? 1. Two billiard balls hit each other then move in opposite directio
Nutka1998 [239]

Answer:

Option (4)

Explanation:

There are two types of collision.

Perfectly elastic collision: the collision in which the momentum and kinetic energy is conserved. There is no loss of energy in other forms of energy.

Perfectly plastic collision: The collision in which the momentum is conserved and kinetic energy is not conserved. The two bodies stick after the collision.

Here, the bullet hits the block and then embedded in the block, it is the example of plastic collision.

3 0
3 years ago
An artillery shell is fired with an initial velocity of 300 m/s at 52.0° above the horizontal. To clear an avalanche, it explode
sasho [114]

The x- and y-coordinates are 9142.57 m and -304.425 m

<u>Explanation:</u>

As the motion of the shell is in a plane (two dimensional space) and the acceleration is that due to gravity which is vertically downward, we resolve initial velocity of the shell v_{0} in horizontal and vertical directions. If the initial velocity of the shell is making angle with the horizontal, the horizontal component of initial velocity will be

                v_{x}=v_{0} \times \cos \theta

As the acceleration of the shell is vertical having no horizontal component, the shell may be considered to move horizontally with constant velocity of v_{x} and hence the horizontal distance covered (or the x coordinate of the shell with point of projection as origin) is given by

           v_{x}=v_{o} \times \cos \theta=300 \times \cos \left(52^{\circ}\right)=184.69 \mathrm{m} / \mathrm{s}

           v_{y}=v_{o} \times \sin \theta==300 \times \sin \left(52^{\circ}\right)=236.4 \mathrm{m} / \mathrm{s}

For motion with constant acceleration, we know

            s=s_{0}+v_{0} t+\left(\frac{(1)}{2}\right) a t^{2}

Along the horizontal, x-axis, we might write this as

            x=x_{0}+v_{x 0} t+\left(\frac{1}{2}\right) a_{x} t^{2}

Measuring distances relative to the firing point means

               x_{0}=0

we know that,

              a_{x}=0

or,

             v_{x}=v_{x 0}=\text { constant }

By applying the values, we get,

           x=0+(184.69 \times 49.5)+\left(\left(\frac{1}{2}\right) \times 0 \times(49.5)^{2}\right)=9142.57 \mathrm{m}

The acceleration of gravity is vertically downward and is g=-9.8 \mathrm{m} / \mathrm{s}^{2} , hence the vertical distance covered (or y coordinate of the shell) is given by the second equation of motion

           y=y_{0}+v_{y 0} t+\left(\frac{1}{2}\right) a_{y} t^{2}

we know, y_{0}=0 and a_{y}=-9.8 \mathrm{m} / \mathrm{s}^{2}, so,

          y=0+(236.4 \times 49.5)+\left(\left(\frac{1}{2}\right) \times(-9.8) \times(49.5)^{2}\right)

                 y = 11701.8 - 4.9(2450.25)= 11701.8 - 12006.225 = - 304.425 m

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