AM Radio Waves

Does the Earth stay in one place throughout the year? If it doesn't, describe its motion and where it's located in the solar syt

stiv31 [10]

the earth moves throughout the year such as rotate around the sun, so yes the it does move and it sits roughly at 93.048 million miles away from the sun. I hope this helps you out! :)

6 0

3 years ago

Arbeitsauftrag 2

kramer

Explanation:

The height of the pendulum is measured from the lowest point it reaches (point 3).

At 1, the kinetic energy of the pendulum is zero (because it is not moving), and it has maximum potential energy.

At 2, the pendulum has both kinetic and potential energy, and how much of each it has depends on its height—smaller the height greater the kinetic energy and lower the potential energy.

At 3, the height is zero; therefore, the pendulum has no potential energy, and has maximum kinetic energy.

At 4, the pendulum again gains potential energy as it climbs back up, Again how much of each forms of energy it has depends on its height.

At 5, the maximum height is reached again; therefore, the pendulum has maximum potential energy and no kinetic energy.

Hope this helps :)

8 0

2 years ago

~~~NEED HELP ASAP~~~ Please solve each section and show all work for each section.

anastassius [24]

Explanation:

Forces on Block A:

Let the x-axis be (+) towards the right and y-axis be (+) in the upward direction. We can write the net forces on mass $m_A$ as

$x:\:\:(F_{net})_x = f_N - T = -m_Aa\:\:\:\:\:\:\:(1)$

$y:\:\:(F_{net})_y = N - m_Ag = 0 \:\:\:\:\:\:\:\:\:(2)$

Substituting (2) into (1), we get

$\mu_km_Ag - T = -m_Aa \:\:\:\:\:\:\:\:\:(3)$

where $f_N= \mu_kN$ , the frictional force on $m_A.$ Set this aside for now and let's look at the forces on $m_B$

Forces on Block B:

Let the x-axis be (+) up along the inclined plane. We can write the forces on $m_B$ as

$x:\:\:(F_{net})_x = T - m_B\sin30= -m_Ba\:\:\:\:\:\:\:(4)$

$y:\:\:(F_{net})_y = N - m_Bg\cos30 = 0 \:\:\:\:\:\:\:\:\:(5)$

From (5), we can solve for N as

$N = m_B\cos30 \:\:\:\:\:\:\:\:\:(6)$

Set (6) aside for now. We will use this expression later. From (3), we can see that the tension T is given by

$T = m_A( \mu_kg + a)\:\:\:\:\:\:\:\:\:(7)$

Substituting (7) into (4) we get

$m_A(\mu_kg + a) - m_Bg\sin 30 = -m_Ba$

Collecting similar terms together, we get

$(m_A + m_B)a = m_Bg\sin30 - \mu_km_Ag$

or

$a = \left[ \dfrac{m_B\sin30 - \mu_km_A}{(m_A + m_B)} \right]g\:\:\:\:\:\:\:\:\:(8)$

Putting in the numbers, we find that $a = 1.4\:\text{m/s}$ . To find the tension T, put the value for the acceleration into (7) and we'll get $T = 21.3\:\text{N}$ . To find the force exerted by the inclined plane on block B, put the numbers into (6) and you'll get $N = 50.9\:\text{N}$

8 0

3 years ago

Ô tô chuyển động trên đường cong có bán kính R = 300(m). Nếu vận tốc ôtô tăng đều từ v1 =

kow [346]

Answer:

Mc ,12,4

Explanation:

7 0

2 years ago

A way of showing how energy moves through an ecosystem _________

Nataly_w [17]

Explanation:

Plants, as a autotrophs have chlorophyll to capture light energy from sun to make starch and sugar. Then, consumers eat plants, and the sugar is transferred to higher trophic level in a form of organic food. Nevertheless, energy is lost by uneaten food, indigestible food, unabsorbed food, excretory waste (eg co2) and heat loss by respiration.

7 0

3 years ago

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