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DochEvi [55]
3 years ago
10

8. A car travels at a constant velocity of 70 mph for one hour. By the end of the second hour, the car’s velocity was 60 mph. At

the end of the third hour, the car’s velocity was 80 mph. During which hour was their acceleration positive? During which hour was their acceleration negative?
Physics
1 answer:
Mrac [35]3 years ago
8 0

<u>Answer:</u>

  Positive acceleration is in third hour and negative acceleration is in second hour.

<u>Explanation:</u>

  Velocity of car in first hour =  70 mph

  Velocity of car in second hour = 60 mph

  Velocity of car in third hour = 80 mph

   Acceleration = Change in velocity / Time

   Acceleration in second hour = (60 - 70)/1 = -10 mph²

   Acceleration in third hour = (80 - 60)/1 = 20 mph²

   So positive acceleration is in third hour and negative acceleration is in second hour.

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Give a calculate answer to show that the two values (English system and metric system) for the Planck Constant are equivalent.
Nataly_w [17]

Answer:

Given values of Planck Constant are equivalent in English system and metric system.

Explanation:

Value of Planck's constant is given in English system as 4.14 x 10⁻¹⁵eV s.

Converting this in to metric system .

We have 1 eV = 1.6 x 10⁻¹⁹ J

Converting

     4.14 x 10⁻¹⁵eV s = 4.14 x 10⁻¹⁵x 1.6 x 10⁻¹⁹ = 6.63 x 10⁻³⁴ Joule s

So Given values of Planck Constant are equivalent in English system and metric system.

7 0
3 years ago
To practice Problem-Solving Strategy 23.2 for continuous charge distribution problems. A straight wire of length L has a positiv
Lesechka [4]

Answer:

             E = k Q / [d(d+L)]

Explanation:

As the charge distribution is continuous we must use integrals to solve the problem, using the equation of the elective field

       E = k ∫ dq/ r² r^

"k" is the Coulomb constant 8.9875 10 9 N / m2 C2, "r" is the distance from the load to the calculation point, "dq" is the charge element  and "r^" is a unit ventor from the load element to the point.

Suppose the rod is along the x-axis, let's look for the charge density per unit length, which is constant

         λ = Q / L

If we derive from the length we have

        λ = dq/dx       ⇒    dq = L dx

We have the variation of the cgarge per unit length, now let's calculate the magnitude of the electric field produced by this small segment of charge

        dE = k dq / x²2

        dE = k λ dx / x²

Let us write the integral limits, the lower is the distance from the point to the nearest end of the rod "d" and the upper is this value plus the length of the rod "del" since with these limits we have all the chosen charge consider

        E = k \int\limits^{d+L}_d {\lambda/x^{2}} \, dx

We take out the constant magnitudes and perform the integral

        E = k λ (-1/x){(-1/x)}^{d+L} _{d}

   

Evaluating

        E = k λ [ 1/d  - 1/ (d+L)]

Using   λ = Q/L

        E = k Q/L [ 1/d  - 1/ (d+L)]

 

let's use a bit of arithmetic to simplify the expression

     [ 1/d  - 1/ (d+L)]   = L /[d(d+L)]

The final result is

     E = k Q / [d(d+L)]

3 0
3 years ago
1 point
Yuliya22 [10]
PE= 3kg x 10N/kg x 10m
= 300J
8 0
3 years ago
What conditions made it possible for earth's interior to separate into layer?
mamaluj [8]

Answer:

Separation of the Earth into layers (crust, mantle, inner core, and outer core) was largely caused by gravitational differentiation (separating different constituents at temperature where materials are liquid or plastic, owing to differences in density) early in Earth's history.

Explanation:

hoped it helped!!

7 0
3 years ago
A vehicle is moving with 20m/s towards the east and another is moving 15m/s towards the west​
just olya [345]

Answer:

5 m/s

Explanation:

Given that,

A vehicle is moving with 20m/s towards the east and another is moving 15m/s towards the west​.

It is assumed to find the resultant velocity of the vehicle. Let east side is positive and west is negative. So,

v=v_1+v_2\\\\=20+(-15)\\\\=5\ m/s

Hence, the resultant velocity of the vehicle is equal to 5 m/s.

6 0
3 years ago
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