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kherson [118]
3 years ago
15

How do most primary producers make their own food?​

Physics
2 answers:
Murljashka [212]3 years ago
5 0
They use the sun to make their own food
MrRissso [65]3 years ago
3 0

Answer:

in a kitchen

Explanation:

only if they're producing

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You are standing outside during the daytime. How can you see a blade of grass?
pashok25 [27]

Answer:

C: Light travels from the Sun to the grass and is then reflected to your eyes.

Explanation:

Our eyes don't produce light but detect light (so answer A is not correct)

The grass doesn't produce light unless it is burning (discard answer B)

Answer C is the correct one.

The light we detect from the grass is not light bounced off the sky, it comes directly from the grass to your eyes (you are not looking at the sky when you see the grass) Discard answer D

8 0
3 years ago
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10 Select the correct answer. What property of a wave remains unchanged when a wave enters a different medium? O A amplitude OB.
Mama L [17]

a and b ik thats the awser becuse imjust did it

8 0
3 years ago
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A car possesses 20,000 units of momentum. what would be the car's new momentum if ... its velocity was doubled?
pochemuha
10,000 units of momentum.
p=mv
20,000=m(2v)
10,000=mv
7 0
4 years ago
A capacitor is formed from two concentric spherical conducting shells separated by vacuum. The inner sphere has radius 11.0 cm ,
viktelen [127]
Part A)
First of all, let's convert the radii of the inner and the outer sphere:
r_A = 11.0 cm = 0.110 m
r_B = 16.5 cm=0.165 m
The capacitance of a spherical capacitor which consist of two shells with radius rA and rB is
C=4 \pi \epsilon _0  \frac{r_A r_B}{r_B- r_A}=4\pi(8.85 \cdot 10^{-12}C^2m^{-2}N^{-1}) \frac{(0.110m)(0.165m)}{0.165m-0.110m}=
=3.67\cdot 10^{-11}F

Then, from the usual relationship between capacitance and voltage, we can find the charge Q on each sphere of the capacitor:
Q=CV=(3.67\cdot 10^{-11}F)(100 V)=3.67\cdot 10^{-9}C

Now, we can find the electric field at any point r located between the two spheres, by using Gauss theorem:
E\cdot (4 \pi r^2) =  \frac{Q}{\epsilon _0}
from which
E(r) =  \frac{Q}{4 \pi \epsilon_0 r^2}
In part A of the problem, we want to find the electric field at r=11.1 cm=0.111 m. Substituting this number into the previous formula, we get
E(0.111m)=2680 N/C

And so, the energy density at r=0.111 m is
U= \frac{1}{2} \epsilon _0 E^2 =  \frac{1}{2} (8.85\cdot 10^{-12}C^2m^{-2}N^{-1})(2680 N/C)^2=3.17 \cdot 10^{-5}J/m^3

Part B) The solution of this part is the same as part A), since we already know the charge of the capacitor: Q=3.67 \cdot 10^{-9}C. We just need to calculate the electric field E at a different value of r: r=16.4 cm=0.164 m, so
E(0.164 m)= \frac{Q}{4 \pi \epsilon_0 r^2}=1228 N/C

And therefore, the energy density at this distance from the center is
U= \frac{1}{2}\epsilon_0 E^2 =  \frac{1}{2} (8.85\cdot 10^{-12}C^2m^{-2}N^{-1})(1228 N/C)^2=6.68 \cdot 10^{-6}J/m^3
8 0
3 years ago
The total resistance of the circuit <br>​
ZanzabumX [31]

The total resistance of a series circuit is equal to the sum of individual resistances. Voltage applied to a series circuit is equal to the sum of the individual voltage drops. The voltage drop across a resistor in a series circuit is directly proportional to the size of the resistor.

If you know the total current and the voltage across the whole circuit, you can find the total resistance using Ohm's Law: R = V / I. For example, a parallel circuit has a voltage of 9 volts and total current of 3 amps. The total resistance RT = 9 volts / 3 amps = 3 Ω

Current: The total circuit current is equal to the sum of the individual branch currents. Resistance: Individual resistances diminish to equal a smaller total resistance rather than add to make the total.

4 0
3 years ago
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