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bonufazy [111]
3 years ago
15

Determine whether the following statements are true or false and give an explanation or counter example. a. If the acceleration

of an object remains​ constant, then its velocity is constant. b. If the acceleration of an object moving along a line is always​ 0, then its velocity is constant. c. It is impossible for the instantaneous velocity at all times atb to equal the average velocity over the interval atb. d. A moving object can have negative acceleration and increasing speed.\
Physics
1 answer:
stiks02 [169]3 years ago
8 0

Answer:

A.) False

B.) True

C.) True

D.) False

Explanation:

a. If the acceleration of an object remains​ constant, then its velocity is constant. False because if velocity is constant, acceleration will be equal to zero.

b. If the acceleration of an object moving along a line is always​ 0, then its velocity is constant. Yes. According to definition of acceleration, saying it is the rate of change of velocity. So if velocity is not changing, acceleration = 0

c. It is impossible for the instantaneous velocity at all times at b to equal the average velocity over the interval at b. Yes. Because of constant velocity

d. A moving object can have negative acceleration and increasing speed. False. Because when an object is coming to rest velocity is always reducing.

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Answer:

Polished surfaces reduce heat loss

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2 years ago
A beam of unstable K mesons, traveling at a speed of 32c, passes through two counters 9.00 m apart. The particles suffer a negli
Anna71 [15]

Answer:

8.66 nano seconds

Explanation:

speed of k mesons ( v ) = c \sqrt{3} /2  m/s

Distance between counters ( d ) = 9.00 m

number of countable electrical pulses = 1000 counts in first counter and 250 in second counter

time of travel = d / v  = 18 / \sqrt{3} C  secs

Next write the decay of particles in lab frame

finally calculate the half life of Meson in its own frame

( t 1/2) of meson in its own frame = 8.66 n-secs

attached below is a detailed solution

5 0
2 years ago
Find the value of x.<br> 6x-1<br> 42<br> 10x+5<br> a<br> 78<br> x = [?]
Arada [10]

Answer: x=6

Explanation:

6 0
2 years ago
Water flows over a section of Niagara Falls at the rate of 1.1 × 106 kg/s and falls 50.0 m. How much power is generated by the f
Kryger [21]
<h3>Answer:</h3>

5.395 × 10^8 Watts

<h3>Explanation:</h3>

<u>We are given;</u>

  • Rate of flow is 1.1 × 10^6 kg/s
  • Distance is 50.0 m
  • Gravitational acceleration is 9.8 m/s²

We are required to calculate the power that is generated by the falling water

  • Power is the rate of work done
  • It is given by dividing the energy or work done by time
  • Power = Work done ÷ time

But; work done = Force × distance

Therefore;

Power = (F × d) ÷ time

The rate is 1.1 × 10^ 6 Kg/s

But, 1 kg = 9.81 N

Therefore, the rate is equivalent to 1.079 × 10^7 N/s

Thus,

Power = Rate (N/s) × distance

           = 1.079 × 10^7 N/s × 50.0 m

           = 5.395 × 10^8 Watts

The power generated from the falling water is 5.395 × 10^8 Watts

7 0
3 years ago
An object has a position given by the radius vector r = [2.0 m + (3.00 m/s)t](i)+ [3.0 m - (2.00 m/s^2)t^2](j). Here (i) and (j)
jenyasd209 [6]

Answer:

The speed of the object is (3i - 4.00tj)m/s

The magnitude of the acceleration is 4.00m/s²

Explanation:

Given - position vector;

r = (2.0 + 3.00t)i + (3.0 - 2.00t²)j       -------------------(i)

To get the speed vector (v), take the first derivative of equation (i) with respect to time t as follows;

v = \frac{dr}{dt}

 v = \frac{d[(2.0 + 3.00t)i + (3.0 - 2.00t^2)j]  }{dt}  

v  = 3i - 4.00tj      ------------------------(ii)

To get the acceleration vector (a), take the first derivative of the speed vector in equation(ii) as follows;

a = \frac{dv}{dt}

a = \frac{d(3i - 4.00tj)}{dt}

a = -4.00j

The magnitude of the acceleration |a| is therefore given by

|a| = |-4.00|

|a| = 4.00 m/s²

In conclusion;

the speed of the object is (3i - 4.00tj)m/s

the magnitude of the acceleration is 4.00m/s²

3 0
3 years ago
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