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3 years ago

A particle having a speed of 0.87c has a momentum of 10-16 kg·m/s. what is its mass?

1 answer:

8 0

The momentum p of a moving particle is the product between its mass, m, and tis velocity, v:
$p=mv$
In our problem, we know $p=10^{-16}~Kg \cdot m/s$ and $v=0.87c=0.87\cdot 3\cdot10^8~m/s=2.61\cdot 10^8~m/s$ , and using the relationship mentioned above, we can find the mass m of the particle:
$m= \frac{p}{v} =3.8\cdot10^{-25}~Kg$

You might be interested in

List down all the jovian planets in order of increasing distance from the sun

Olenka [21]

Answer:

There are total eight planets in the solar system and the average distance from the sun to each planet in increasing order is given below.

Explanation:

The average distance from the sun is listed below in increasing order.

1. Mercury - It is the most closet planet to Sun, 57 million km

2. Venus - 108 million km

3. Earth - 150 million km

4. Mars - 228 million km

5. Jupiter - 779 million km

6. Saturn - 1.43 billion km

7. Uranus - 2.88 billion km

8. Neptun - It is the most farthest from the Sun, 4.50 billion km

8 0

3 years ago

A seagull flies at a velocity of 9.00 m/s straight into the wind.

RideAnS [48]

a)If it takes the bird 18.0 minutes to fly 6 km away from the earth, the wind's speed will be 4 m/s.

b) The bird would need 7 minutes and 42 seconds to fly back 6 kilometers if he turned around and flew with the wind.

c)Compared to the 133.33 seconds it would take without the wind, the overall round-trip time is affected by the wind.

<h3>What is velocity?</h3>

The change of distance with respect to time is defined as speed. Speed is a scalar quantity. It is a time-based component. Its unit is m/sec.

The given data in the problem is

A seagull flies at a velocity, $\rm v_{SA} = 9 \ m/sec$

The time the bird takes,t=18.0 min

The distance traveled relative to the earth = 6.00 km

The seagull's relative velocity with reference to the ground as;

$\rm v_{sg} = \frac{6.00 \times 10^3 \ m }{(20 min) \times \frac{60 s }{1 \ min}} \\\\ v_{sg}= 5.00 \ m/sec$

Air velocity with reference to the ground is;

$\rm v_{AG}= v_{SG}-v_{SA} \\\\ v_{AG} = 5.00 \ m/sec - 9.00 \ m/sec \\\\ v_{AG} = -4.00 \ m/sec$

If the bird turns around and flies with the wind, The time will he take to return 6.00 km is;

$\rm v_{SG}=v_{SA}+v_{AG} \\\\ v_{SG}=-900 \ m/sec +(-4.00 \ m/sec) \\\\ v_{SG}= -13.00 \ m/sec$

The time the bird takes;

$\rm t = \frac{x_{SG}}{v_{SG}} \\\\ t = \frac{6.00 \times 10^3 \ m }{13.00 \ m/sec } \\\\ t = 462 m/sec \\\\ t = 7 \ min \ and \ 42 \ sec$

c)\

The total round-trip time compared to what it would be with no wind. is;

$\rm t = 20 \ min( \frac{60 \ sec }{1 \ min} )+ 462 \ sec \\\\ t = 1200 \ sec +6 462 \ ec \\\\ t= 1662 \ sec$

The time for the round trip is;

$\rm t = \frac{12 \times 10^ 3 }{ 9 \ m/sec } \\\\ t = 1333.33 \ sec$

Hence the wind's speed, the time bird would need to fly back the total round-trip time will be 4 m/s, 7 minutes and 42 seconds and 1333.33 sec.

To learn more about the velocity, refer to the link: brainly.com/question/862972.

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4 0

2 years ago

A projectile is fired directly upward from the ground with an initial velocity of 112 f/sec its distance s above the ground afte

Travka [436]

Answer:

3.48 seconds

Explanation:

At maximum height Vf=0 m/s

Vf= Vi - g*t

⇒g*t= Vi