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3 years ago

What are good colleges to attend to become a botanist ??? HELP!!!!!!!!!!!!!

Physics

1 answer:

umka21 [38]3 years ago

3 0

Answer:Strayer university/michigan state university/kent state university

Explanation:

my uncle went there to michigan state university he said it was good to become a botanist

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Balancing carefully, three boys inch out onto a horizontal tree branch above a pond, each planning to dive in separately. The th

VikaD [51]

Answer:

what is this a riddle lol it breaks when he either jumps or lands

Explanation:

3 0

2 years ago

In a two-source circuit, one source acting alone produces 10 ma through a given branch. the other source acting alone produces 8

pashok25 [27]

Refer to the figure below.
R = resistance.

Case 1:
The voltage source is V₁ and the current is 10 mA. Therefore
V₁ = (10 mA)R

Case 2:
The voltage source is V₂ and the current is 8 mA. Therefore
V₂ = (8 mA)R

Case 3:
The voltage across the resistance is V₁ - V₂. Therefore the current I is given by
V₁ - V₂ = IR
10R - 8R = (I mA)R
2 = I
The current is 2 mA.

Answer: 2 mA

6 0

2 years ago

Explain why you cannot use mass or volume alone to identify substances.

Alex17521 [72]

Objects can have the same mass (but different <span>compositions). Only mass or volume cannot tell you if the object is solid or vo</span>lumes) or same volume (but different masses)

4 0

3 years ago

Honeybees acquire a charge while flying due to friction with the air. A 100 mg bee with a charge of +23 pC experiences an electr

Sedbober [7]

Answer:

(A) ratio of electric force to weight will be $23.469\times 10^{-10}$

(b) Electric field will be $E=4.26\times 10^{10}N/C$

Explanation:

We have given mass of bee = 100 mg = $m=100\times 10^{-3}=0.1kg$

Charge on bee $q=23pC=23\times 10^{-12}C$

Electric field E = 100 N/C

Weight of the bee $W=mg=0.1\times 9.8=0.98N$

Electric force on the bee $F=qE=23\times 10^{-12}\times 100=23\times 10^{-10}N$

So the ratio of electric force on the bee and weight is $=\frac{F}{W}=\frac{23\times 10^{-10}}{0.98}=23.469\times 10^{-10}$

(B) To hold the bee in air electric force must be equal to weight of bee

So $mg=qE$

$0.1\times 9.8=23\times 10^{-12}E$

$E=4.26\times 10^{10}N/C$

4 0

2 years ago

Desperate for help! please!!!!!!!!

lys-0071 [83]

$0.4029 \mathrm{m} / \mathrm{s}^{2}$ is the acceleration of the box.

<u>Explanation:</u>

Given data:

Mass of the box = 3.74 kg

Flat friction-less ground is pulled forward by a 4.20 N force at a 50.0 degree angle and pulled back by a 2.25 N force at a 122 degree angle.

First, we need to find the net horizontal force acting on the box. With the given data, the equation can be formed as below. Net horizontal force acting on the box (F) is given by

$F=\left(4.20 \times \cos 50^{\circ}\right)+\left(2.25 \times \cos 122^{\circ}\right)$

$F=(4.20 \times 0.64278)+(2.25 \times-0.5299)$

F = 2.699676 – 1.192275 = 1.507 N

Next, find acceleration of the box using Newton's second law of motion. This states that the link between mass (m) of an objects and the force (F) required to accelerate it. The equation can be given as

$F=m \times acceleration\ (a)$

$\text {acceleration }(a)=\frac{F}{m}=\frac{1.507}{3.74}=0.4029 \mathrm{m} / \mathrm{s}^{2}$

8 0

2 years ago