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kirill [66]
3 years ago
10

A positive charge of 3.2 x 10 -5 C is located 0.85 m away from another positive charge of 7.4 x 10 -6 C. What is the electric fo

rce between the two charges?
Physics
1 answer:
zmey [24]3 years ago
7 0

by using Coulumbs Law its 2.95N.

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Use the drop-down menus to complete the sentences.
Grace [21]

Answer:

1) thinner, diverge

2) thicker, converge

Explanation:

i got it right

5 0
3 years ago
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One end of a 34-m unstretchable rope is tied to a tree; the other end is tied to a car stuck in the mud. The motorist pulls side
anyanavicka [17]

Answer:

Fc = 89.67N

Explanation:

Since the rope is unstretchable, the total length will always be 34m.

From the attached diagram, you can see that we can calculate the new separation distance from the tree and the stucked car H as follows:

L1+L2=34m

L1^2=L2^2=L^2=2^2+(H/2)^2  Replacing this value in the previous equation:

\sqrt{2^2+H^2/4}+ \sqrt{2^2+H^2/4}=34  Solving for H:

H=\sqrt{52}

We can now, calculate the angle between L1 and the 2m segment:

\alpha = atan(\frac{H/2}{2})=60.98°

If we make a sum of forces in the midpoint of the rope we get:

-2*T*cos(\alpha ) + F = 0  where T is the tension on the rope and F is the exerted force of 87N.

Solving for T, we get the tension on the rope which is equal to the force exerted on the car:

T=Fc=\frac{F}{2*cos(\alpha) } = 89.67N

7 0
3 years ago
In an intergalactic competition, spaceship pilots compete to see who can cover the distance between two asteroids in the short-
pogonyaev

Answer:

a)  truc is C,  b) correct result is the B

Explanation:

As the speed of the competition is very high, for the judges the speed is

           v = d / t

           v = 3 109 m / 20

           v = 1.5 108 m / s

This is half the speed of light. For these high speeds we must use the relations of special relativity.

For the time          t = to γ

For distance         L = Lo / γ

                            γ = √ (1-v2 / c2)

Own time and distance (to and Lo) corresponds to the observer who is not moving the judges in this case

Let's look for the range value

                     γ = 1 / √ (1 - (1.5 / 3) 2) = 1 / 0.866 = 1.15

The time              t = 20 1.15 = 23 s

The distance       L = 3 10 9 /1.15 = 2.60 109 m

From these results we see that time increases and the distance is shorter.

Let's review the claims

A) False. It's the opposite

B) False

C) True. It is according to the result found

D) False.

In the nuclear fusion process, we will also use the special relativity that has a relationship between energy and mass

         ΔE = c² Δm

As in the process energy is released, for the law of conservation of the mass of energy to be fulfilled, the total mass of the products, He atom, must be reduced.

Therefore the correct result is the B

4 0
4 years ago
Given that the concentration of bovine carbonic anhydrase is 3.3 pmol ⋅ L − 1 and R max ( V max ) = 222 μmol ⋅ L − 1 ⋅ s − 1 , d
LuckyWell [14K]

Answer:

The turnover number of the enzyme molecule bovine carbonic anhydrase = 67,272,727.27 s^–1.

Explanation:

Given:

The concentration of bovine carbonic anhydrase = total enzyme concentration = Et = 3.3 pmol⋅L^–1 = 3.3 × 10^–12 mol.L^–1

The maximum rate of reaction = Rmax (Vmax) = 222 μmol⋅L^–1⋅s^–1 = 222 × 10^–6 mol.L^–1⋅s^–1

The formula for the turnover number of an enzyme (kcat, or catalytic rate constant) = Rmax ÷ Et = 222 × 10^–6 mol.L^–1⋅s^–1 ÷ 3.3 × 10^–12 mol.L^–1 = 67,272,727.27 s^–1

Therefore, the turnover number of the enzyme molecule bovine carbonic anhydrase = 67,272,727.27 s^–1

3 0
3 years ago
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timama [110]
This answer should be gas
4 0
3 years ago
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