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Hatshy [7]
3 years ago
14

When a 5.0-kilogram cart moving with a speed of 2.8 meters per second on a horizontal surface collides with a 2.0 kilogram cart

initially at rest, the carts locked together. What is the speed of the combined carts after the collision
Physics
1 answer:
mixer [17]3 years ago
8 0

Here in all such collision type question we can use momentum conservation as we can see that there is no external force on this system

m_1v_{1i} + m_2v_{2i} = m_1v_{1f} + m_2v_{2f}

as we know that

m_1 = 5 kg

m_2 = 2 kg

v_{1i} = 2.8 m/s

v_{2i} = 0 m/s

now from above equation we have

5(2.8) + 2(0) = m_1v+ m_2v

14 = (5+ 2) v

v = 2 m/s

so the speed of combined system is 2 m/s

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we see a magnet exerting a force on a current-carrying wire. Does a current-carrying wire exert a force on a magnet? Why or why
max2010maxim [7]

Answer:

Yes

Explanation: Electric and magnetic field are known to be inter-related, this implies that for any current carrying conductor there is a resulting magnetic field around the wire ( for example a current carrying conductor deflects a compass) and a magnetic field has been known to produce some amount current based on the<em> </em>principle of electromagnetic induction by Micheal Faraday.

The strength of magnetic field generated by a current carrying conductor is given by Bio-Savart law (purely mathematical) which is

B =\frac{u_{0}I }{2πr}

B= strength of magnetic field

I =current on conductor

r = distance on any point of the conductor relative to it center

If a current carrying could generate this magnitude of magnetic field, thus this magnetic field has the ability to interact (exert a force on any magnetic material) with any other magnetic material including a magnet.

Yes, a current carrying conductor can exert a force on a magnetic field

7 0
3 years ago
Read 2 more answers
Five lamp, each labbled "6V,3W" are operated at normal brightness. What is the total energy supplied to the lamps in five second
REY [17]

Answer:

E = 75 J

Explanation:

First, we will calculate the total power consumed by the five lamps:

Total\ Power = P = (5)(Power\ of\ one\ lamp)\\P = (5)(3\ W)\\P = 15\ W

Now, the energy supply can be calculated as follows:

E = Pt

where,

E = Energy = ?

t = time = 5 s

Therefore,

E = (15 W)(5 s)

<u>E = 75 J</u>

8 0
2 years ago
A circuit contains four light bulbs. One light bulb goes out but the other three stays it. This must be an)
Paul [167]

Answer:

It is a parallel connection

Explanation:

In parallel connection the

Cell is not easily used up because the cells share the total current generated together with all bulbs.

But a major problem is the bulbs must not be left together undisconnected to avoid exhaustion arising from short fall in the strength of one cell as this bounds to affect others

3 0
3 years ago
A 1.2 x10 3 kilogram automobile in motion strikes a 1.0 x 10 -4 kilogram insect as a result the insect is accelerated at a rate
Mariana [72]
According to Newton's second law, the force applied to an object is equal to the product between the mass of the object and its acceleration:
F=ma
where F is the magnitude of the force, m is the mass of the object and a its acceleration.

In this problem, the object is the insect, with mass m=1.0 \cdot 10^{-4} kg. The acceleration of the insect is a=1.0 \cdot 10^2 m/s^2, therefore we can calculate the force exerted by the car on the insect:
F=ma=(1.0 \cdot 10^{-4} kg)(1.0 \cdot 10^2 m/s^2)=0.01 N

How do we find the force exerted by the insect on the car?
According to Newton's third law (known as action-reaction law), when an object A exerts a force on an object B, object B also exerts a force equal and opposite on object A. Therefore, the force exerted by the insect on the car is equal to the force exerted by the car on the object, so it is 0.01 N.
6 0
3 years ago
A spacecraft at rest has moment of inertia of 100 kg-m^2 about an axis of interest. If a 1 newton thruster with a 1 meter moment
snow_lady [41]

Answer:

18 radians

Explanation:

The computation is shown below:

As we know that

Torque = Force × Moment arm

= 1N × 1M

= 1N-M

Torque = I\alpha

\alpha = \frac{torque}{I}\\\\= \frac{1}{100}\\\\= 0.01 rad/s^2

Now

\theta = w_ot + \frac{1}{2} \alpha t^2\\\\w_o = 0\\\\\theta = 0 \times 60 + \frac{1}{2} \times 0.01 \times 60^2\\\\= 18\ radians

Here t = 1 minutes = 60 seconds

3 0
3 years ago
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