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3 years ago

Which of the following BEST describes all planets in the universe?

Physics

1 answer:

Harman [31]3 years ago

7 0

Answer:

c large, spherical body that orbits in a clear path around a star

Explanation:

you can not say b because the sun is a star and you cant say a and d because all planets are not made of rock and all planets are not made of gas

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~ is the following statement true or false? Explain your answer.

USPshnik [31]

Answer: False

Explanation: The sun is one of earths primary energy sources. Without the sun, all animals, plants, humans would die. The sun's energy provides warmth for humans and plants and animals cannot grow without the sun.

7 0

3 years ago

Question 4

mamaluj [8]

Answer:

What is an electromagnetic wave

Explanation:

Electromagnetic radiation is also more commonly known as light, light travels in waves- and light travels at the same speed as electromagnetic waves. :)

8 0

3 years ago

Read 2 more answers

Determine the average acceleration for x(t)=19t^2+7t^3 for a time interval between 3 and 9 seconds

AleksandrR [38]

Answer:

290

Explanation:

Average acceleration is the change in velocity over change in time.

First, find the velocity by taking the derivative of position.

v(t) = dx/dt

v(t) = 38t + 21t²

At t = 3 and t = 9:

v(3) = 303

v(9) = 2043

So the average acceleration is:

a = Δv / Δt

a = (2043 − 303) / (9 − 3)

a = 290

Use appropriate units.

7 0

3 years ago

Hockey puck B rests on a smooth ice surface and is struck by a second puck A, which has the same mass. Puck A is initially trave

Marina86 [1]

Answer:

$a.v_{b2}=6.8544 \ m/s\\\\b. v_{a2}=12.891 \ m/s\\\\c. \theta _b=62\textdegree$

Explanation:

Puck A's initial speed is $v_a_1=14.6m/s$ and move in a direction $\theta_b=28.0\textdegree$ after the collision.

# $P_1=P_2$ since there's no external force on the system( $P=mv).$

#The collision equation can be written as;

$m_av_a_1+m_bv_b_1=m_av_a_2+m_bv_b_2$

The kinetic energies before and after the collision are expressed as:

$K_a_1+K_{b1}=K_a_2+K_{b2}, \ K=0.5mv^2$

$0.5m_av_a_1+0.5m_b(0)=0.5m_av_a_2+0.5m_bv_b_2\\\\14.6^2=v_{a2}^2+v_{b2}^2\\\\v_{b2}^2=213.16-v_{a2}^2$

Let +x along A's initial direction and +y along A's final direction makes the angle $62\textdegree$

$\dot v_{a1}=14.6i\\\\\dot v_{a2}=(v_{a2} \ cos \ 28\textdegree)i+(v_a_2\ sin \ 28\textdegree)j\\\\\dot v_{b2}=v_{b2x}i+v_{b2y}j$

#Substitute in $v_{b2}^2=213.16-v_{a2}^2$ :

$\dot v_{b2}=(14.6i)-\dot v{a2}\\\\\dot v_{b2}.\dot v_{b2}=v_{b2}^2\\\\\#Right \ side\\\\(14.6i-\dot v{a2}).(14.6i-\dot v{a2})=(14.6i)^2+\dot v_{a2}^2.\dot v_{a2}^2-2(14.6i).\dot v_{a2}\\\\=213.16+v_{a2}^2-2(14.6i).(v_{a2}\ cos 28\ \textdegree i+v_{a2} \ sin \ 28\textdegree j)\\\\v_{b2}^2=213.16+v_{a2}^2-29.2\ cos \ 28\textdegree v_{a2}\ \ v_{b2}^2=213.16-v_{a2}^2\\\\213.16-v_{a2}^2=213.16+v_{a2}^2-29.2\ cos \ 28\textdegree v_{a2}\\\\2v_a2}^2=29.2\ cos\ 28 \textdegree v_{a2}\\\\$

$v_{a2}=12.891\ m/s$

Hence, the the speed of puck A after the collision is 12.891 m/s

#. The velocity of A after the collision is;

$\dot v_{a2}=12.891 \ cos \ 28 \textdegree i+12.891 \ sin \ 28\textdegree j\\\\=11.382i+6.052j$

Substitute $\dot v_{a2}$ into $\dot v_{b2}=(14.6i)-\dot v{a2}$ :

$\dot v_{b2}=14.6i-(11.382i+6.052j)\\\\=3.218i-6.052j$

This is the velocity of puck B after the collision, it's speed is:

$v_{b2}=\sqrt{v_{b2x}^2+v_{b2y}^2}\\\\=\sqrt{3.218^2+(-6.052)^2}\\\\v_{b2}=6.8544 \ m/s$

The velocity of puck B after the collision is 6.8544 m/s

c. The direction of puck B after the collision is:

$\theta _b=tan^{-1}\frac{v_{b2y}}{v_{b2x}}\\\\=tan^{-1} \frac{-6.052}{3.218}\\\\\approx 62\textdegree$

Hence, the direction of B's velocity after the collision is 62°

4 0

4 years ago

Suppose there are only two charged particles in a particular region. Particle 1 carries a charge of +q and is located on the pos

liubo4ka [24]

Answer:

x₂ = 0.1715 d

1) false

2) True

3) True

4) false

5) True

Explanation:

The field electrifies a vector quantity, so we can add the creative field by these two charges

E₂-E₁ = 0

k q₂ / r₂² - k q₁ / r1₁²= 0

q₂ / r₂² = q₁ / r₁²

suppose the sum of the fields is zero at a place x to the right of zero

r₂ = d + x

r₁ = d -x

we substitute

q₂ / (d + x)² = q₁ / (d-x)²

we solve the equation

q₂ / q₁ (d-x)² = (d + x) ²

let's replace the value of the charges

q₂ / q₁ = + 2q / + q = 2

2 (d²- 2xd + x²) = d² + 2xd + x²

x² -6xd + d² = 0

we solve the quadratic equation

x = [6d ± √ (36d² - 4 d²)] / 2

x = [6d ± 5,657 d] / 2

x₁ = 5.8285 d

x₂ = 0.1715 d

with the total field value zero it is between the two loads the correct solution is x₂ = 0.1715 d

this value remains on the positive part of the x axis, that is, near charge 1

now let's examine the different proposed outcomes

1) false

2) True

3) True

4) false

5) True

6 0

4 years ago