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anyanavicka [17]
2 years ago
14

Cart 1 has a mass of 1 kg. Cart 2 has a mass of 2 kg. They are pushed apart by a spring. The spring exerts 2N of force into cart

1. How much force is exerted into cart 2? Explain.
Physics
1 answer:
almond37 [142]2 years ago
5 0

The force exerted on the cart 2 during the collision is 2 N.

The given parameters:

  • <em>Mass of cart 1 = m1 = 1 kg</em>
  • <em>Mass of cart 2 = m2 = 2kg</em>
  • <em>Force applied on cart 1 = 2 N</em>

According to Newton's third law of motion, action and reaction are equal and opposite. The force exerted on cart 1 is equal in magnitude to the force exerted on cart 2 but in opposite direction.

F_1 = -F_2

F_2 = -F_1\\\\F_2 = -2 \ N\\\\|F_2| = 2 \ N

Thus, the force exerted on the cart 2 during the collision is 2 N.

Learn more about Newton's third law of motion here: brainly.com/question/13874955

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If the primary of a transformer were connected to a dc power source,
QveST [7]

Answer:

D. only briefly while being connected or disconnected.

Explanation:

As we know that transformer works on the principle of mutual inductance

here we know that as per the principle of mutual inductance when flux linked with the primary coil charges then it will induce EMF in secondary coil

So here when AC source is connected with primary coil then it will give output across secondary coil because AC source will have change in flux with time.

Now when we connect DC source across primary coil then it will not induce any EMF across secondary coil because DC source is a constant voltage source in which flux will remain constant always

So here in DC source the EMF will only induce at the time of connection or disconnection when flux will change in it while rest of the time it will give ZERO output

so correct answer will be

D. only briefly while being connected or disconnected.

8 0
3 years ago
Kyle is flying a helicopter at 125 m/s on a heading of 325 o . If a wind is blowing at 25 m/s toward a direction of 240.0 o , wh
frosja888 [35]

Answer:

The resultant velocity of the helicopter is \vec v_{H} = \left(89.894\,\frac{m}{s}, -93.348\,\frac{m}{s}\right).

Explanation:

Physically speaking, the resulting velocity of the helicopter (\vec v_{H}), measured in meters per second, is equal to the absolute velocity of the wind (\vec v_{W}), measured in meters per second, plus the velocity of the helicopter relative to wind (\vec v_{H/W}), also call velocity at still air, measured in meters per second. That is:

\vec v_{H} = \vec v_{W}+\vec v_{H/W} (1)

In addition, vectors in rectangular form are defined by the following expression:

\vec v = \|\vec v\| \cdot (\cos \alpha, \sin \alpha) (2)

Where:

\|\vec v\| - Magnitude, measured in meters per second.

\alpha - Direction angle, measured in sexagesimal degrees.

Then, (1) is expanded by applying (2):

\vec v_{H} = \|\vec v_{W}\| \cdot (\cos \alpha_{W},\sin \alpha_{W}) +\|\vec v_{H/W}\| \cdot (\cos \alpha_{H/W},\sin \alpha_{H/W}) (3)

\vec v_{H} = \left(\|\vec v_{W}\|\cdot \cos \alpha_{W}+\|\vec v_{H/W}\|\cdot \cos \alpha_{H/W}, \|\vec v_{W}\|\cdot \sin \alpha_{W}+\|\vec v_{H/W}\|\cdot \sin \alpha_{H/W} \right)

If we know that \|\vec v_{W}\| = 25\,\frac{m}{s}, \|\vec v_{H/W}\| = 125\,\frac{m}{s}, \alpha_{W} = 240^{\circ} and \alpha_{H/W} = 325^{\circ}, then the resulting velocity of the helicopter is:

\vec v_{H} = \left(\left(25\,\frac{m}{s} \right)\cdot \cos 240^{\circ}+\left(125\,\frac{m}{s} \right)\cdot \cos 325^{\circ}, \left(25\,\frac{m}{s} \right)\cdot \sin 240^{\circ}+\left(125\,\frac{m}{s} \right)\cdot \sin 325^{\circ}\right)\vec v_{H} = \left(89.894\,\frac{m}{s}, -93.348\,\frac{m}{s}\right)

The resultant velocity of the helicopter is \vec v_{H} = \left(89.894\,\frac{m}{s}, -93.348\,\frac{m}{s}\right).

8 0
3 years ago
An object is 50 cm from a converging lens with a focal length of 40 cm . A real image is formed on the other side of the lens, 2
Leno4ka [110]

Answer:

d) -4.0

Explanation:

The magnification of a lens is given by

M=-\frac{q}{p}

where

M is the magnification

q is the distance of the image from the lens

p is the distance of the object from the lens

In this problem, we have

p = 50 cm is the distance of the object from the lens

q = 250 cm - 50 cm is the distance of the image from the lens (because the image is 250 cm from the obejct

Also, q is positive since the image is real

So, the magnification is

M=-\frac{200 cm}{50 cm}=-4.0

7 0
3 years ago
Please Someone Help Me...
goldenfox [79]

Answer:

Hi myself Shrushtee.

Explanation:

The fuse is connected to the live wire so that the appliance will not become charged (have a potential difference of 230 V) after the fuse has melted due to excessive current. Fuses must be fitted onto the live wire so that when it blows, it will disconnect (isolate) the appliance from the high voltage live wire.

6 0
2 years ago
Consider the following FBDs.FBDsAn elevator is suspended by a cable and moves uniformly upward. Which of the above free body dia
Drupady [299]

The force of gravity F_g will act downwards.

Normal force F_N will act upwards equal to the force of gravity.

A force due to uniform acceleration F_a will act upwards to move the elevator upwards.

Thus, figure E is the correct answer.

6 0
1 year ago
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