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alina1380 [7]
2 years ago
5

A stationary skateboarder I with a mass of 50 kg pushes a stationary skateboarder II with a mass of 75 kg. After the push the sk

ateboarder II moves with a velocity of 2 m/s to the right. What is the velocity of the skateboarder I?
Physics
1 answer:
Mice21 [21]2 years ago
8 0

The final velocity of the skateboarder I after the collision is 3 m/s to the left.

<h3>Velocity of skateboarder I</h3>

The velocity of skateboarder I is determined from the principle of conservation of linear momentum.

m1u1 + m2u2 = m1v1 + m2v2

50(0) + 75(0) = 50v1 + 75(2)

0 = 50v1 + 150

v1 = (-150)/50

v1 = - 3

v1 = 3 m/s to the left

Thus, the final velocity of the skateboarder I after the collision is 3 m/s to the left.

Learn more about linear momentum here: brainly.com/question/7538238

#SPJ1

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A student sits at rest on a piano stool that can rotate without friction. The moment of inertia of the student-stool system is 4
irina [24]

Here We can use principle of angular momentum conservation

Here as we know boy + projected mass system has no external torque

Since there is no torque so we can say the angular momentum is conserved

mvL = (I + mL^2)\omega

now we know that

m = 2 kg

v = 2.5 m/s

L = 0.35 m

I = 4.5 kg-m^2

now plug in all values in above equation

2\times 2.5 \times 0.35 = (4.5 + (2\times 0.35^2))\omega

1.75 = [4.5 + 0.245]\omega

1.75 = 4.745\omega

\omega = 0.37 rad/s

so the final angular speed will be 0.37 rad/s

4 0
3 years ago
Which of the following diagrams correctly shows the electron configuration of Sulfur, with atomic number 16?
nika2105 [10]

Answer:

Diagram C

Explanation:

We are given  that Sulfur with atomic number 16.

We have to find that which diagram shows the electronic configuration of sulfur.

S=16

Its Diagram C

6 0
3 years ago
Read 2 more answers
Energy is conventionally measured in Calories as well as in joules. One Calorie in nutrition is one kilocalorie, defined as 1 kc
Sergeeva-Olga [200]

Answer:

a) The student must run flight of stairs to lose 1.00 kg of fat 709.5 times.

b) Average power

P(w)= 1062.07 [w]

P(hp)=1.42 [hp]

c) This activity is highly unpractical, because the high amount of repetitions he has to due in order to lose, just 1 Kg of fat.

Explanation:

First, lets consider the required amount of work to move the mass of the student. (considering running stairs just as a vertical movement)

Work:

W= F*d= m*g*d

Where m is the mass of the student, g is gravity (9.8 m/s) and d is the total distance going up the stairs (0.15m *85steps= 12.75m )

W= F*d= m*g*d=85* 9.8*12.75=10620.75 [J]

Converting from Joules to Kcals:

\frac{10620.75}{4186} =2.537 Kcal

Now lets take into account the efficiency of the human body (20%)

2.537 ---> 20%

 x       ---> 100%

x=\frac{2.537*100}{20} =12.685

So the student is consuming 12.685 KCals each time he runs up the stairs.

Now,

1 g --> 9 Kcals

1000 g --> 9000KCals

Burning 1 g of fat, requieres 9 KCals, 1000g burns 9000KCals. So in order to burn a 1Kg of fat:

\frac{9000Kcals}{12.685Kcals} =709.5 times

He must run up the stairs 709.5 times, to burn 1 Kg of fat.

********************

For b) just converting units, taking into account the time lapse. (53103.75 is the 100% of the energy in joules, from converting 12.685Kcals to joules)

Power=\frac{Joules}{Seconds} =\frac{53103.75}{50} =1062.075 [W]\\

P(hp)=\frac{P(w)}{745.7} =\frac{1062.075}{745.7} =1.42[hp]

*****

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Calculate the speed of a train that traveled 810 kilometers in 9 hours.
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Answer:

90 kilometers an hour

let me know if I'm wrong

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What is the speed of light in meters per second?
MAXImum [283]
The answer is approximately 2.998e+8
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