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2 years ago

When a 1.0-kilogram cart moving with a speed of

2 answers:

5 0

Hello there.

<span>When a 1.0-kilogram cart moving with a speed of
0.50 meter per second on a horizontal surface
collides with a second 1.0-kilogram cart initially
at rest, the carts lock together. What is the speed
of the combined carts after the collision?

</span><span>(3) 0.25 m/s</span>

neonofarm [45]2 years ago

4 0

The type of collision in this system is called inelastic collision where after collision the two objects stick together. We can solve for the final velocity after collision by the expression:

v = m1v1 + m2V2 / (m1+m2)

v = 1(0.50) + (1)(0) / (1+1) = 0.25 m/s

Therefore, option 3 is the correct answer.

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In the diagram shown, satellite S moves in a circular orbit around the Earth (E) in a counterclockwise direction.

zubka84 [21]

The characteristics of the velocity vector used to find the results for the direction of acceleration and velocity are:

Acceleration is towards the center of the circle

The velocity is tangent to the circle counterclockwise.

Newton's Second Law establishes a relationship between force, mass and acceleration of bodies.

<h3>Centripetal acceleration. </h3>

In the case of circular motion there is a constant change in the direction of the velocity vector, even when its module remains constant, this change in direction points towards the center of the circle, so that the module is constant.

They indicate that the satellite is moving counterclockwise, therefore the speed must go to the left (counterclockwise) tangential to the circle.

In conclusion using the characteristics of the velocity vector we can find the results for the direction of acceleration and velocity are:

Acceleration is towards the center of the circle

The velocity is tangent to the circle counterclockwise.

Learn more about centripetal acceleration here: brainly.com/question/25243603

6 0

2 years ago

An electron enters a region between two large parallel plates made of aluminum separated by a distance of 2.0 cm and kept at a p

kotykmax [81]

Answer:

a) $v=1.77\times 10^6\ m/s$

b) $v=3.872\times 10^6\ m/s$

c) $v=5.5\times 10^6\ m/s$

d) $v=6.7\times 10^6\ m/s$

e) $v=7.7\times 10^6\ m/s$

Explanation:

Given that

d = 2 cm

V = 200 V

$u=4\times 10^5\ m/s$

We know that

F = E q

F = m a

E = V/d

m a = q .V/d b

$a=\dfrac{q.V}{m.d}$ ---------1

The mass of electron

$m=9.1\times 10^{-31}\ kg$

The charge on electron

$q=1.6\times 10^{-19}\ C$

Now by putting the all values in equation 1

$a=\dfrac{1.6\times 10^{-19}\times 200}{9.1\times 10^{-31}\times 0.02}\ m/s^2$

$a=1.5\times 10^{15}\ m/s^2$

We know that

$v^2=u^2+2as$

s = 0.1 cm

$v^2=(4\times 10^5)^2+2\times 1.5\times 10^{15}\times 0.1\times 10^{-2}$

$v=\sqrt{3.16\times 10^{12}}\ m/s$

$v=1.77\times 10^6\ m/s$

s = 0.5 cm

$v^2=(4\times 10^5)^2+2\times 1.5\times 10^{15}\times 0.5\times 10^{-2}$

$v=\sqrt{1.5\times 10^{13}}\ m/s$

$v=3.872\times 10^6\ m/s$

s = 1 cm

$v^2=(4\times 10^5)^2+2\times 1.5\times 10^{15}\times 1\times 10^{-2}$

$v=\sqrt{3.06\times 10^{13}}\ m/s$

$v=5.5\times 10^6\ m/s$

s = 1.5 cm

$v^2=(4\times 10^5)^2+2\times 1.5\times 10^{15}\times 1.5\times 10^{-2}$

$v=\sqrt{4.5\times 10^{13}}\ m/s$

$v=6.7\times 10^6\ m/s$

s = 2 cm

$v^2=(4\times 10^5)^2+2\times 1.5\times 10^{15}\times 2\times 10^{-2}$

$v=\sqrt{6.06\times 10^{13}}\ m/s$

$v=7.7\times 10^6\ m/s$

5 0

3 years ago

Read 2 more answers

In the image which element is the most massive ?

ehidna [41]

Answer:

that one

Explanation:

actually if u look very closely u can notice that the taco has some dog tird cover up with the rat tail and topped with some soymilk

3 0

2 years ago

If your goal is to improve your personal relationships with your friends at school, which category would it fall into?

pogonyaev

it would fall into the social category

4 0

2 years ago

A toy car attached to a rope is traveling around in a circle. It makes 4 laps in 12s.What is the angular velocity of the toy car

umka21 [38]

Answer:

2.09 rad/s

Explanation:

Given that,

A toy car makes 4 laps in 12 s.

We need to find the angular velocity of the toy car. We need to find the angular velocity of the toy car.

4 laps = 25.13 rad

So,

$v=\dfrac{25.13\ rad}{12\ s}\\\\v=2.09\ rad/s$

So, the angular velocity of the toy car is 2.09 rad/s.

3 0

2 years ago