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pishuonlain [190]
3 years ago
7

You drop a rock off a bridge. The rock's height, h (in feet above the water), after t seconds is modeled by h = – 16 t2 + 541. W

hat is the height of the rock after 2 seconds?
Physics
1 answer:
marin [14]3 years ago
5 0
<span>h ( t) = h(1 sec) = -16t^2 + 541

so h (2 sec) = -16*(2)^2 + 541 = -64 + 541 = <span>477 ft

Therefore, </span></span>the height of the rock after 2 seconds is 477 feet.

I hope my answer has come to your help. Thank you for posting your question here in Brainly.
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8. two +1 C charges are separated by 3000m. What is the magnitude of the electric force between them?
Sidana [21]

Answer:

1000 N

Explanation:

The magnitude of the electrostatic force between two charged object is given by

F=k\frac{q_1 q_2}{r^2}

where

k is the Coulomb constant

q1, q2 is the magnitude of the two charges

r is the distance between the two objects

Moreover, the force is:

- Attractive if the two forces have opposite sign

- Repulsive if the two forces have same sign

In this problem:

q_1=q_2=+1C are the two charges

r = 3000 m is their separation

Therefore, the electric force between the charges is:

F=(9\cdot 10^9)\frac{(1)(1)}{3000^2}=1000 N

8 0
3 years ago
In a compoumd are atoms physically or chemically combined
Andrei [34K]

Answer:

They are...if I'm correct Chemically combined, sorry if I'm wrong.

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2 years ago
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joja [24]
Well i think the answer is impossible to find because there is no picture
4 0
3 years ago
How can you determine the rate of a chemical reaction?
DanielleElmas [232]
Take a look at a simple reaction like the one below:

In this reaction some reactant A is turned into some product B. The rate of reaction can be represented by a decrease in concentration of A over time or as the increase of B over time. This is written:

6 0
3 years ago
Chapter 21, Problem 009 Two identical conducting spheres, fixed in place, attract each other with an electrostatic force of 0.12
PilotLPTM [1.2K]

Answer:

a) -1.325 μC

b) 4.17μC

Explanation:

First, you need to know that charge is conserved. So, the adition of the charges, as there is no lost in charge, should always be the same. Also, after the wire is removed, both spheres will have the same charge, trying to find equilibrium. In summary:

q_1 + q_2 = constant\\q_1_f = q_2_f |Then\\q_1_f + q_2_f = 2q_1_f = q_1_o+q_2_o\\q_1_f = q_2_f = \frac{q_1_o+q_2_o}{2}

We know both q1f and q2f must be positive, because the negative charge at the beginning was the the smaller.

The electrostatic force is equal to:

F_e = k\frac{q_1q_2}{r^2}

K is the Coulomb constant, equal to 9*10^9 Nm^2/C^2

Now, we are told that the electrostatic force after the wire is equal to 0.0443 N:

F_e_f = k \frac{q_1_fq_2_f}{r^2} = k\frac{\frac{q_1_o+q_2_o}{2}\frac{q_1_o+q_2_o}{2}}{r^2} = k\frac{(q_1_o+q_2_o)^2}{4r^2}  \\(q_1_o+q_2_o) = \sqrt{\frac{F_e_f*4r^2}{k}} = \sqrt{\frac{0.0443N *4(0.641m)^2}{9*10^9Nm^2/C^2} } = 2.844 *10^{-6}C \\ q_1_o = 2.844*10^{-6}C - q_2_o

Originally, the force is negative because it was an attraction force, therefore, its direction was opposite to the direction of the repulsive force after the wire:

F_e_o = k\frac{q_1_oq_2_o}{r^2}\\ q_1_oq_2_o = \frac{F_e_o*r^2}{k} = \frac{-0.121N(0.641m)^2}{9*10^9Nm^2/C^2} = -5.524*10^{-12}

(2.844*10^{-6}C - q_2_o)q_2_o = -5.524*10^{-12}\\0 = q_2_o^2 - 2.844*10^{-6}q_2_o - 5.524*10^{-12}

Solving the quadratic equation:

q_2_o = 4.17*10^{-6}C | -1.325 * 10^{-6}C

for this values q_1 wil be:

q_1_o =  -1.325 *10^{-6}C | 4.17*10^{-6}C

So as you can see, the negative charge will always be -1.325 μC and the positive 4.17μC

5 0
3 years ago
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