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NISA [10]
3 years ago
10

The diagram below shows the structural formula and 3D model of the same substance. Which substance is represented by the diagram

above?
A. H2O
B. OH
C. H2O2
D. OH2

PLZ HELP ASAP
Physics
1 answer:
Agata [3.3K]3 years ago
4 0
I feel like its C but mmmm idk
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A cruise ship makes its way from one island to another. The ship is in motion compared with which reference point?
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A reference point would be something not on the ship which could be used to calculate distance traveled.

Answer: C.) A lighthouse on a nearby Island

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Is Algae Biotic or Abiotic?
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It’s biotic because it’s a living thing
Abiotic would be like a rock non living things
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Please solve for 15 points. Please don’t input a link.
melomori [17]

Answer:

a). Single replacement.

Explanation:

Because one element replaces another element in a compound

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3 years ago
An object, when pushed with a net force F, has an acceleration of 4 m/s . Now twice the force is applied to an object that has f
hjlf

Answer:

B. 2 m/s

B. Acceleration = 4.05 m/s² and Tension = 297.5 N.

Explanation:

A force is applied on a mass m whose acceleration is 4 m/s

Force = mass × acceleration

a = F/m = 4 m/s

4 m/s = F/m

F = 4 m/s (m)

If  Force of 2F is applied on a mass of 4m ; it acceleration is as follows:

2F/4 m = F/ 2m

4m/s (m) / 2m = 2 m/s

a = 2 m/s

2.

Given that

mass m_1 = 30 kg

mass m_2 = 50 kg

\mu = 0.1

From the question; we can arrive at two cases;

That :

m_{2} a _ \ {net} }= m_2g - T   ----- equation (1)

m_{1} a _ \ {net} }=  T - mg sin \theta  - F ---- equation (2)

50 a = 50 g - T

30 a = T - 30 g sin 30 - 4 × 30 g cos 30

By summation

80 a =[ 50  - 30 * \frac{1}{2} - 0.1 *30* \frac{\sqrt{3}}{2}]g

80 a = 32. 4 × 10 m/s ²  (using g as 10m/s²)

80 a = 324 m/s ²

a = 324/80

a = 4.05 m/s²

From equation , replace a with 4.05

50 × 4.05 = 50 × 10 - T

T = 500 -202.5

T =297.5 N

8 0
3 years ago
A cart for hauling ore out of a gold mine has a mass of 413 kg, including its load. the cart runs along a straight stretch of tr
Andrei [34K]
Given from the problem :mass m = 413 kg;coefficient of friction u = 0.0163;acceleration due to gravity g = 9.8 m/s2;inclined angle @1 = 14.3;inclined angle  @2 = 4.69;distance travelled d = 175 m;applied fore F = 410 N;  the component of the force from the donkey in the direction of motion isF2 = F1 [email protected]= 397.2964498768165 N
Fy = N - mg [email protected] = 0N = mg [email protected] = 4037.964151113007 NFx = F2 - mg [email protected] - f = mahere f = u N=65.8188156631420141

F2 - mg [email protected] - f = maa = F2 - mg [email protected] - f/ m=0.31923412183075155 m/s^2
work done by donkeyW = F2 d=69526.8787284428875 J
8 0
3 years ago
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