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Morgarella [4.7K]
3 years ago
11

Which product is typically made using soft wood

Physics
2 answers:
kolezko [41]3 years ago
8 0

Answer:

i think it is c

Explanation:

8090 [49]3 years ago
3 0

Answer:its plywood

Explanation:i just took the test

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An airplane starts from rest and accelerates at a constant rate of 3.00 m/s2 for 30.0 s before leaving the ground. a. How far di
rosijanka [135]

The formula we can use in this case is:

d = v0t + 0.5 at^2

v = at + v0

where,

d = distance travelled

v0 = initial velocity = 0 since at rest

t = time travelled

a = acceleration

v = final velocity when it took off

 

a. d = 0 + 0.5 * 3 * 30^2

d = 1350 m

 

b. v = 3 * 30 + 0

<span>v = 90 m/s</span>

8 0
3 years ago
Planet 1 orbits Star 1 and Planet 2 orbits Star 2 in circular orbits of the same radius. However, the orbital period of Planet 1
hichkok12 [17]

Answer:

The mass of Star 2 is Greater than the mass of Start 1. (This, if we suppose the masses of the planets are much smaller than the masses of the stars)

Explanation:

First of all, let's draw a free body diagram of a planet orbiting a star. (See attached picture).

From the free body diagram we can build an equation with the sum of forces between the start and the planet.

\sum F=ma

We know that the force between two bodies due to gravity is given by the following equation:

F_{g} = G\frac{m_{1}m_{2}}{r^{2}}

in this case we will call:

M= mass of the star

m= mass of the planet

r = distance between the star and the planet

G= constant of gravitation.

so:

F_{g} =G\frac{Mm}{r^{2}}

Also, if the planet describes a circular orbit, the centripetal force is given by the following equation:

F_{c}=ma_{c}

where the centripetal acceleration is given by:

a_{c}=\omega ^{2}r

where

\omega = \frac{2\pi}{T}

Where T is the period, and \omega is the angular speed of the planet, so:

a_{c} = ( \frac{2\pi}{T})^{2}r

or:

a_{c}=\frac{4\pi^{2}r}{T^{2}}

so:

F_{c}=m(\frac{4\pi^{2}r}{T^{2}})

so now we can do the sum of forces:

\sum F=ma

F_{g}=ma_{c}

G\frac{Mm}{r^{2}}=m(\frac{4\pi^{2}r}{T^{2}})

in this case we can get rid of the mass of the planet, so we get:

G\frac{M}{r^{2}}=(\frac{4\pi^{2}r}{T^{2}})

we can now solve this for T^{2} so we get:

T^{2} = \frac{4\pi ^{2}r^{3}}{GM}

We could take the square root to both sides of the equation but that would not be necessary. Now, the problem tells us that the period of planet 1 is longer than the period of planet 2, so we can build the following inequality:

T_{1}^{2}>T_{2}^{2}

So let's see what's going on there, we'll call:

M_{1}= mass of Star 1

M_{2}= mass of Star 2

So:

\frac{4\pi^{2}r^{3}}{GM_{1}}>\frac{4\pi^{2}r^{3}}{GM_{2}}

we can get rid of all the constants so we end up with:

\frac{1}{M_{1}}>\frac{1}{M_{2}}

and let's flip the inequality, so we get:

M_{2}>M_{1}

This means that for the period of planet 1 to be longer than the period of planet 2, we need the mass of star 2 to be greater than the mass of star 1. This makes sense because the greater the mass of the star is, the greater the force it applies on the planet is. The greater the force, the faster the planet should go so it stays in orbit. The faster the planet moves, the smaller the period is. In this case, planet 2 is moving faster, therefore it's period is shorter.

6 0
3 years ago
If you push a 4-kg mass...
Lena [83]

Answer:

B

Explanation:

F = ma , a = F/m

a1 = F/10 and a2 = F/4

Since Force is constant, a2 will we greater than a1

4 0
3 years ago
Gwen is researching the effects of a small asteroid impact in an area that has started to regrow after being hit ten years ago.
Bumek [7]

Answer:

Gwen’s assumption of asteroid hit as long term change is incorrect. Asteroid hit is not a long term change, instead, it is a short term change.  

Explanation:

Examples of short term changes are drought, flood, volcanic eruption, etc. A short term change occurs quickly and can immediately affect organisms but it doesn’t become a reason for species extinction. The effects of a short term change don’t prevail over a long span of time.

Examples of long term changes are ice age, global warming, deforestation, etc. Unlike a short term change, it takes time but the consequences are far-reaching. It can lead to species extinction.

In this question, asteroid hit is a quick and unexpected hazard, unlike the slow long term environmental changes.

3 0
3 years ago
Read 2 more answers
A shot-putter projects the shot at 42.00˚ to the horizontal from a height of 2.100 m. It lands 17.00 m away horizontally. Next,
Tamiku [17]

Answer:

Explanation:

Let 100 m/s  be the velocity of projection.

So horizontal component

= 100 cos42

= 74.31 m /s

Vertical component = - 100 sin 42 . in upward direction

66.91 m/s

Net displacement = 2.1 downwards ( + ve )

Using s = ut + 1/2 gt²

2.1 = - 66.91 t  + .5 x 9.8 x t²

4.9 t² -  66.91 t - 2.1 = 0

t = 13.685 s

Horizontal distance covered

= 13.685 x 74.31

= 1016.93 m

If angle of projction is 40°

So horizontal component

= 100 cos40

= 76.60 m /s

Vertical component = - 100 sin 42 . in upward direction

64.27 m/s

Net displacement = 2.1 downwards ( + ve )

Using s = ut + 1/2 gt²

2.1 = -76.60 t  + .5 x 9.8 x t²

4.9 t² -  76.60 t - 2.1 = 0

t = 15.659  s

Horizontal distance covered

= 15.659 x 76.60

= 1199.49  m

So horizontal range is increased , if angle of projection is increased .

8 0
3 years ago
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