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Sedaia [141]
3 years ago
13

A boy of mass 46.2 kg is initially on a skateboard of mass 2.00 kg, moving at a speed of 10.2 m/s. The boy falls off the skatebo

ard, and his center of mass moves forward at a speed of 10.9 m/s. Find the final velocity of the skateboard.
Physics
1 answer:
gladu [14]3 years ago
5 0

Answer:

v=-5.97\,\frac{m}{s}

Explanation:

The system boy-skateboard can be modelled by means of the Principle of Momentum Conservation:

(48.2\,kg)\cdot (10.2\,\frac{m}{s} ) = (46.2\,kg)\cdot (10.9\,\frac{m}{s} )+(2\,kg)\cdot v

The final velocity of the skateboard:

v=-5.97\,\frac{m}{s}

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iragen [17]

If the first thing that the goslings saw was a dog, they would have followed the dog as a mother.

Imprinting refers to the process of training an animal to bond with anything it sees after birth even if it is not its real mother. Lorenz first achieved imprinting in 1935 using geese which followed him as their mother shortly after they were born.

If the geese were exposed to a dog, they could also have seen the dog as their mother and followed it accordingly shortly after birth.

Learn more: brainly.com/question/11401513

7 0
2 years ago
a train engineer started the train from a standstill and sped up to 5 meters per second, she then rounded a corner at a constant
olasank [31]

for acceleration we can define that rate of change in velocity is know as acceleration

So whenever velocity of train is changing with time we can say train is accelerating

Now here if initially train is standstill then after some time its speed is 5 m/s

so here the train is accelerated first time

Then on straight path its speed changed from 5 m/s to 10 m/s so here train gets accelerated second time

After this train chugged around a curve with same speed 10 m/s

SO here since train is moving in curve so here its direction of velocity is continuously changing and this type of acceleration is known as centripetal acceleration

SO this is accelerated Third time

Then its speed decreases and it comes to speed of 5 m/s from 10 m/s

So here it is acceleration of train for Fourth time

Then finally train comes to stop so again its speed changed from 5 m/s to 0

so this is acceleration of train Fifth time

So total train will accelerate 5 times in whole path

3 0
3 years ago
For this problem, we assume that we are on planet-i. the radius of this planet is r =4200 km, the gravitational acceleration at
Minchanka [31]
The expression commonly used for potential gravitational energy is just simplification. It is actually just the first term in Taylor expansion of the real expression. 
In general, the potential energy of gravitational field is defined as:
U=-G \frac{mM}{r}
Where G is universal gravitational constant, and r is the distance between the objects centers of mass. Negative sign represents the bound state.
Since we are not given the mass of the planet we have to calculate it.
F_g=G\frac{mM}{r_p^2}\\ mg=G\frac{mM}{r_p^2}\\ g=G\frac{M}{r_p^2}
This formula can be used for any planet. It gives you the gravitational acceleration on the planet's surface. We can use it to calculate the planet's mass:
g=G\frac{M}{r_p^2}\\ M=\frac{gr_p^2}{G}=2.41\cdot 10^{24}kg
Now we can calculate the potential energy of that cannonball when it reaches its maximum height.
U=-G \frac{mM}{r}\\ U=-G \frac{mM}{r_p+h}
When we plug in the numbers we get:
U=-4.99\cdot 10^{10} J
The potential energy has to be equal to the kinetic energy.
E_k=4.99\cdot 10^{10} J

3 0
3 years ago
In the diagram below, point A is southwest of point B. Jaleh walks for 400 seconds along path 1, going from point A to point B.
timama [110]

Answer:

C

Explanation:

velocity = displacement (m) / change of time (s)

velocity = (400 + 300) / (100 + 400)

velocity = 1.4 m/s

4 0
2 years ago
Read 2 more answers
How are distance and displacement the same?
ratelena [41]
In the case of straight line, when the object is not repeating it's path then distance and displacement would be same.
3 0
3 years ago
Read 2 more answers
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