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Varvara68 [4.7K]
3 years ago
12

A 3.70 kg block moving 2.09 m/s right hits a 2.42 kg block moving 3.92 m/s left. Find the total momentum of the system.

Physics
1 answer:
Novosadov [1.4K]3 years ago
5 0

Answer:

-1.7534 Kg.m/s

Explanation:

Momentum, p is a product of mass and velocity, expressed as p= mv where m is the mass and v is the velocity which is dependent on the direction. Taking left as negative and right as positive then the initial momentum will be

p=-2.42*3.92+(3.7*2.09)=-1.7534 Kg.m/s

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EX
Serjik [45]

Given the distance traveled and time elapsed, the average speed of the train is approximately 26.944m/s.

<h3>What is the average speed of the train?</h3>

Speed is simply referred to as distance traveled per unit time.

Mathematically, Speed = Distance ÷ time.

Given the data in the question;

  • Distance traveled = 221miles
  • Elapsed time = 3 hours and 40 minutes

First we convert miles to meters and Hours minutes to seconds.

221 miles = ( 221 × 1609.344 )m = 355665.024 meters

3 hours and 40 minutes = ( 3×60×60)s + ( 40×60)s

= 10800s + 2400s

= 13200s

Now, determine the average speed.

Speed = Distance ÷ time

Speed = 355665.024m / 13200s

Speed = 26.944m/s

Given the distance traveled and time elapsed, the average speed of the train is approximately 26.944m/s.

Learn more about speed here: brainly.com/question/7359669

#SPJ1

4 0
2 years ago
Certain neutron stars (extremely dense stars) are believed to be rotating at about 500 rev/s. If such a star has a radius of 17
Alexus [3.1K]

Answer:

7.22 × 10²⁹ kg

Explanation:

For the material to be in place, the gravitational force on the material must equal the centripetal force on the material.

So, F = gravitational force = GMm/R² where M = mass of neutron star, m = mass of object and R = radius of neutron star = 17 km

The centripetal force F' = mRω² where R = radius of neutron star and ω  = angular speed of neutron star

So, since F = F'

GMm/R² = mRω²

GM = R³ω²

M = R³ω²/G

Since ω = 500 rev/s = 500 × 2π rad/s = 1000π rad/s = 3141.6 rad/s = 3.142 × 10³ rad/s and r = 17 km = 17 × 10³ m and G = universal gravitational constant = 6.67 × 10⁻¹¹ Nm²/kg²

Substituting the values of the variables into M, we have

M = R³ω²/G

M = (17 × 10³ m)³(3.142 × 10³ rad/s)²/6.67 × 10⁻¹¹ Nm²/kg²

M = 4913 × 10⁹ m³ × 9.872 × 10⁶ rad²/s²/6.67 × 10⁻¹¹ Nm²/kg²

M = 48,501.942 × 10¹⁵ m³rad²/s² ÷ 6.67 × 10⁻¹¹ Nm²/kg²

M = 7217.66 × 10²⁶ kg

M = 7.21766 × 10²⁹ kg

M ≅ 7.22 × 10²⁹ kg

8 0
3 years ago
Think Critically Suppose your teacher gave you a slide of an unknown cell. How would you tell whether the cell was from an anima
topjm [15]

Answer:

You will be able to tell whether the unknown cell is from an animal or from a plant through the knowledge of difference between plants and animal cells.

Explanation:

A cell can be defined as the simplest bit of living matter which exhibits a variety of shapes and sizes and that can exist independently.

When a slide of an unknown cell is studied under a microscope, different cell structures would be observed which would be used to conclude if the cell on the slide is a plant or animal cell.

The following features are observed in a plant cell:

--> cellulose cell wall is present

--> Chloroplast is present

--> A large central vacuole is present

--> Centriole is absent

The following features are observed in animal cell:

--> Cellulose cell wall is absent

--> Chloroplast is absent

--> Small vacuoles is present

--> Centriole is present

The difference between a plant and an animal cell can be seen from the features above and a clear knowledge of this will help the student tell whether the unknown cell is from an animal or from a plant.

4 0
3 years ago
A 2000 kg truck traveling north at 34 km/h turns east and accelerates to 58 km/h. (a) What is the change in the truck's kinetic
barxatty [35]

Explanation:

It is given that,

Mass of the truck, m = 2000 kg

Initial velocity of the truck, u = 34 km/h = 9.44 m/s

Final velocity of the truck, v = 58 km/h = 16.11 m/s

(a) Change in truck's kinetic energy, \Delta E=\dfrac{1}{2}m(v^2-u^2)

\Delta E=\dfrac{1}{2}\times 2000\ kg\times (16.11^2-9.44^2)

\Delta E=170418.5\ J

\Delta E=1.7\times 10^5\ J

(b) Change in momentum of the truck, \Delta p=m(v-u)

\Delta p=2000\ kg\times (16.11-9.44)

\Delta p=13340\ kg-m/s

Hence, this is the required solution.

6 0
4 years ago
a hunter on level ground fires a bullet perfectly level with the ground. at the same time the bullet leaves the gun, he drops an
padilas [110]
Mark brainliest and I will help you
5 0
3 years ago
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