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igor_vitrenko [27]
3 years ago
9

A 2.0-kg cart is rolling along a frictionless, horizontal track towards a 1.8-kg cart that is held initially at rest. The carts

are loaded with strong magnets that cause them to attract one another. Thus, the speed of each cart increases. At a certain instant before the carts collide, the first cart's velocity is 5.9 m/s, and the second cart's velocity is -2.7 m/s. (a) What is the total momentum of the system of the two carts at this instant
Physics
1 answer:
Reil [10]3 years ago
3 0

Answer:

the total momentum of the system before collision is 6.94 kgm/s

Explanation:

Given;

mass of the first cart, m₁ = 2.0 kg

mass of the second cart, m₂ = 1.8 kg

velocity of the first cart before collision, u₁ = 5.9 m/s

velocity of the second cart before collision, u₂ = -2.7 m/s

The total momentum of the system before collision is calculated as follows;

P_t = P_1 + P_2 \\\\P_t = m_1u_1 + m_2u_2\\\\P_t = (2\times 5.9) + (1.8 \times -2.7)\\\\P_t = 11.8 - 4.86\\\\P_t = 6.94 \ kgm/s

Therefore, the total momentum of the system before collision is 6.94 kgm/s

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If the velocity of a long jumper during take-offat an angle of 30 degrees is 42 f/s, how fast is he moving forward (Vx) and how
Arturiano [62]

Answer:

Vx=  11.0865(m/s)

Vy=  6.4008(m/s)

Explanation:

Taking into account that 1m is equal to 0.3048 ft, the takeoff speed in m / s will be:

V= 42(ft/s) × 0.3048(m/ft) = 12.8016(m/s)

The take-off angle is equal to 30 °, taking into account the Pythagorean theorem the velocity on the X axis will be:

Vx= 12.8016 (m/s) × cos(30°)= 11.0865(m/s)

And for the same theorem the speed on the Y axis will be:

Vy= 12.8016 (m/s) × sen(30°)= 6.4008(m/s)

5 0
3 years ago
A 15x10^-6c charge is placed at the origin and a 9x10^-6C charge is placed on the x-axis at x=1.00m. where, on the x-axis is the
Harlamova29_29 [7]

The Electric field is zero at a distance 2.492 cm from the origin.

Let A be point where the charge 15\times10^-6 C is placed which is the origin.

Let B be the point where the charge 9\times 10^-6 C is placed. Given that B is at a distance 1 cm from the origin.

Both the charges are positive. But charge at origin is greater than that of B. So we can conclude that the point on the x-axis where the electric field = 0 is after B on x - axis.

i.e., at distance 'x' from B.

Using Coulomb's law, \frac{kQ_A^2}{d_A^2} = \frac{kQ_B^2}{d_B^2},

Q_A = 15\times 10^-6 C

Q_B=9\times10^-6C

d_A = 1+x cm

d_B=x cm

k is the Coulomb's law constant.

On substituting the values into the above equation, we get,

\frac{(15\times10^-6)^2}{(1+x)^2} =\frac{(9\times10^-6)^2}{x^2}

Taking square roots on both sides and simplifying and solving for x, we get,

1.67x = 1+x

Therefore, x = 1.492 cm

Hence the electric field is zero at a distance 1+1.492 = 2.492 cm from the origin.

Learn more about Electric fields and Coulomb's Law at brainly.com/question/506926

#SPJ4

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1 year ago
In preparation for an electrophoresis procedure, enzymes are added to DNA in order to
lapo4ka [179]
It is 2.) Cut the DNA into fragments
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3 years ago
An object travels with a constant speed in a circular path. The net force on the object is
Pepsi [2]

Answer:

toward the center

Explanation:

Before answering, let's remind the first two Newton Laws:

1) An object at rest tends to stay at rest and an object moving at constant velocity tends to continue its motion at constant velocity, unless acted upon a net force

2) An object acted upon a net force F experiences an acceleration a according to the equation

F=ma

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In this problem, we have an object travelling at constant speed in a circular path. The fact that the trajectory of the object is circular means that the direction of motion of the object is constantly changing: this means that its velocity is changing, so it has an acceleration. And therefore, a net force is acting on it. The force that keeps the object travelling in the circular path is called centripetal force, and it is directed towards the center of the circle (because it prevents the object from continuing its motion straight away).

So, the correct answer is

toward the center

8 0
3 years ago
An ant crawls 50 cm to the right, stops, and then crawls 30 cm to the left. Calculate the distance that the ant travels, and cal
borishaifa [10]

Answer and Explanation:

If the ant was to crawl 50cm to the right, then come back 30 cm, then the total distance walked would be <u>80cm</u>.

- Combine 50cm and 30cm to get 80 cm.

For displacement, the answer is <u>20 cm.</u>

- When calculating displacement, you use the initial (starting) distance. and subtract that from the final distance, giving you the displacement, or the amount traveled from the starting point to the final point if you were to make a straight line from the starting point to end point. (0 to 50, then back 30 the same direction, so subtract 30 from 50 to get 20)

<em><u>#teamtrees #PAW (Plant And Water)</u></em>

<em><u>I hope this helps!</u></em>

8 0
2 years ago
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