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Lina20 [59]
3 years ago
6

a car initially at rest move with the constant accerates along straght line read after it's spread increase and finally related

uniformly. the time interval for the three parts of the jounry are in the ratio 1:3:1 find average velocity ?​
Physics
1 answer:
nasty-shy [4]3 years ago
8 0

Answer:

32km per hour

Explanation:

Explanation:

In first case v = a t

==> a t = 40 km p h

Now distance covered S1 + S2 + S3

S1 = 1/2 a t^2 and S3 = 1/2 a t^2

But S2 = 3t * 40 = 120 t km

Hence total distance = at^2 + 120 t

Time taken (total) = t + 3t + t = 5 t

Hence average speed = at^2 + 120 t / 5 t

Cancelling t we have at + 120 / 5 = 40 + 120 / 5 = 160/5 = 32 km per hour

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I am planning to make two dives. The first dive is to 60 feet for 45 minutes, and the second dive is to 60 feet for 35 minutes.
Tpy6a [65]

Answer:

1:04-1:10 hours

Explanation:

You'll need a <em>Recreational dive planner</em> table, I annexed a copy, now you'll follow the next steps:

  1. In the first part of your table, you'll look for the distance row (in feet) of your first dive, for this specific exercise you'll find 60, once you locate it you'll go down that column until you reach the time you'll dive, in this case, 45 (minutes) or the closest value (47).
  2. You'll check and keep the letter in that 47 row (S) for future use.
  3. Now you have to go to the second part of your table and look for the distance column, in feet, of your second dive. We find 60 and then going right in the blue row, we'll look for the time (35) or its closest value (36).
  4. Finally, we have to check the letter for 36 minutes (F) and we'll make it met with the letter S in the first portion of your tables. This will give us an interval of time, 1:04-1:10 in this case.

I hope you find this information useful and interesting! Good luck!

4 0
3 years ago
If the acceleration of an object is zero at some instant in time, what can be said about its velocity at that time? 1. It is neg
Mkey [24]

3. It is not changing at that time

Explanation:

If the acceleration of a body is zero at some instant in time, it implies that the velocity is not changing at that point in time. Velocity is the rate of change of displacement with time.

✓Acceleration and velocity shares a very close relationship.

✓ For a body to accelerate, the velocity must change. Acceleration is defined as the rate of change of velocity with time.

✓If at any point, a body moves with constant velocity i.e the velocity does not change with time, the acceleration becomes zero.

✓ For acceleration to occur, a body must change velocity.

Learn more:

Acceleration brainly.com/question/6323625

#learnwithBrainly

5 0
3 years ago
The standard measure used to compare sound intensity is the_____
BartSMP [9]
Frequency? Possibly I’m not 100% sure
5 0
3 years ago
Why is the weight of an object on the moon 1/6 th its weight on the earth?
ehidna [41]

The mass of moon is 1/100 times and its radius 1/4 times that of earth as a result the gravitational attraction on the moon is about one sixth when compared to earth.

7 0
3 years ago
What is the energy of a photon that has the same wavelength as an electron having a kinetic energy of 15 ev?
serg [7]

Answer: 6.268(10)^{-16}J

Explanation:

The kinetic energy of an electron K_{e} is given by the following equation:

K_{e}=\frac{(p_{e})^{2} }{2m_{e}}   (1)

Where:

K_{e}=15eV=2.403^{-18}J=2.403^{-18}\frac{kgm^{2}}{s^{2}}

p_{e} is the momentum of the electron

m_{e}=9.11(10)^{-31}kg  is the mass of the electron

From (1) we can find p_{e}:

p_{e}=\sqrt{2K_{e}m_{e}}    (2)

p_{e}=\sqrt{2(2.403^{-18}J)(9.11(10)^{-31}kg)}  

p_{e}=2.091(10)^{-24}\frac{kgm}{s}   (3)

Now, in order to find the wavelength of the electron \lambda_{e}   with this given kinetic energy (hence momentum), we will use the De Broglie wavelength equation:

\lambda_{e}=\frac{h}{p_{e}}    (4)

Where:

h=6.626(10)^{-34}J.s=6.626(10)^{-34}\frac{m^{2}kg}{s} is the Planck constant

So, we will use the value of p_{e} found in (3) for equation (4):

\lambda_{e}=\frac{6.626(10)^{-34}J.s}{2.091(10)^{-24}\frac{kgm}{s}}    

\lambda_{e}=3.168(10)^{-10}m    (5)

We are told the wavelength of the photon  \lambda_{p} is the same as the wavelength of the electron:

\lambda_{e}=\lambda_{p}=3.168(10)^{-10}m    (6)

Therefore we will use this wavelength to find the energy of the photon E_{p} using the following equation:

E_{p}=\frac{hc}{lambda_{p}}    (7)

Where c=3(10)^{8}m/s  is the spped of light in vacuum

E_{p}=\frac{(6.626(10)^{-34}J.s)(3(10)^{8}m/s)}{3.168(10)^{-10}m}  

Finally:

E_{p}=6.268(10)^{-16}J    

4 0
4 years ago
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