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3 years ago

In a tropic level, energy flows from the producer to the primary consumer and to a secondary consumer. True False

Physics

1 answer:

Ganezh [65]3 years ago

4 0

Answer:

True

Explanation:

This is true because in an energy flow, the primary producers such as plants produce their own foods by using energy from the sun. These primary producers are consumed by primary consumers such as rabbits for food, while secondary consumers such as snakes would consume the primary consumers.

By so doing, energy flows from one trophic level to the other.

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2 years ago

Determine the angular speed, in rad/s of:

ryzh [129]

Answer:

Explanation:

A. The earth about its axis:

The earth makes one revolution in 24 hours. to know the number of revolutions per second it makes, we need to convert hours to seconds and the revolution to rad.

$\frac{1 rev}{24hours}\times \frac{1 hr}{3600s}\times \frac{2\pi}{1}= 7.27 \times10^-5 rad/s$

B. The minute hand of the clock makes one revolution in 60 minutes

To convert this to rad per second, we have

$\frac{1 rev}{60mins}\times \frac{1 min}{60s}\times \frac{2\pi}{1}= 1.745 \times10^-3rad/s$

C. The hour hand of a clock completes one revolution in 12 hours

$\frac{1 rev}{12hours}\times \frac{1 hr}{3600s}\times \frac{2\pi}{1}= 1.454 \times10^-4 rad/s$

D. an egg beater turning at 300 rpm.

$\frac{300 rev}{1minute}\times \frac{1 min}{60s}\times \frac{2\pi}{1}= 7.27 \times10^-5 rad/s=0.0218rad/s$

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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

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

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

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 $N = 50.9\:\text{N}$

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2 years ago