From the frame of reference of car 2, what is the velocity of car

A magician quickly pulls a silk tablecloth out from under a set table without disturbing the dishes and silverware placed on it.

expeople1 [14]

<h3>Answer:</h3>

the dishes and silverware are in inertia of rest

<h3>Explanation:</h3>

Newton's first law of motion is that things at rest tend to stay at rest unless acted upon by an external force.

A silk tablecloth will have a fairly slippery surface, so there is not much friction to cause tableware to be accelerated.

8 0

3 years ago

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Which of the following statements is consistent with Thomson’s and Millikan’s work with cathode rays and electrons? (A) Cathode

victus00 [196]

Answer:

Dalton's ideas proved foundational to modern atomic theory. However, one of his underlying assumptions was later shown to be incorrect. Dalton thought that atoms were the smallest units of matter-−minustiny, hard spheres that could not be broken down any further. This assumption persisted until experiments in physics showed that the atom was composed of even smaller particles. In this article, we will discuss some of the key experiments that led to the discovery of the electron and the nucleus.

J.J. Thomson and the discovery of the electron

In the late 19^{\text{th}}19

th

19, start superscript, start text, t, h, end text, end superscript century, physicist J.J. Thomson began experimenting with cathode ray tubes. Cathode ray tubes are sealed glass tubes from which most of the air has been evacuated. A high voltage is applied across two electrodes at one end of the tube, which causes a beam of particles to flow from the cathode (the negatively-charged electrode) to the anode (the positively-charged electrode). The tubes are called cathode ray tubes because the particle beam or "cathode ray" originates at the cathode. The ray can be detected by painting a material known as phosphors onto the far end of the tube beyond the anode. The phosphors spark, or emit light, when impacted by the cathode ray.

A diagram of a cathode ray tube.

A diagram of J.J. Thomson's cathode ray tube. The ray originates at the cathode and passes through a slit in the anode. The cathode ray is deflected away from the negatively-charged electric plate, and towards the positively-charged electric plate. The amount by which the ray was deflected by a magnetic field helped Thomson determine the mass-to-charge ratio of the particles. Image from Openstax, CC BY 4.0.

To test the properties of the particles, Thomson placed two oppositely-charged electric plates around the cathode ray. The cathode ray was deflected away from the negatively-charged electric plate and towards the positively-charged plate. This indicated that the cathode ray was composed of negatively-charged particles.

Thomson also placed two magnets on either side of the tube, and obser

Explanation:

5 0

3 years ago

At one instant a bicyclist is 21.0 m due east of a park's flagpole, going due south with a speed of 13.0 m/s. Then 21.0 s later,

andre [41]

Answer:

Part a)

$d = 21\sqrt2 = 29.7 m$

Part b)

Direction is 45 degree North of West

Part c)

$v_{avg} = 1.41 m/s$

Part d)

direction of velocity will be 45 Degree North of West

Part e)

$a = 0.875 m/s^2$

Part f)

$\theta = 45 degree$ North of East

Explanation:

Initial position of the cyclist is given as

$r_1 = 21.0 m$ due East

final position of the cyclist after t = 21.0 s

$r_2 = 21.0 m$ due North

Part a)

for displacement we can find the change in the position of the cyclist

so we have

$d = r_2 - r_1$

$d = 21\hat j - 21\hat i$

so magnitude of the displacement is given as

$d = 21\sqrt2 = 29.7 m$

Part b)

direction of the displacement is given as

$\theta = tan^{-1}\frac{y}{x}$

$\theta = tan^{-1}\frac{21}{-21}$

so it is 45 degree North of West

Part c)

For average velocity we know that it is defined as the ratio of displacement and time

so here the magnitude of average velocity is defined as

$v_{avg} = \frac{\Delta x}{t}$

$v_{avg} = \frac{29.7}{21}$

$v_{avg} = 1.41 m/s$

Part d)

As we know that average velocity direction is always same as that of average displacement direction

so here direction of displacement will be 45 Degree North of West

Part e)

Here we also know that initial velocity of the cyclist is 13 m/s due South while after t = 21 s its velocity is 13 m/s due East

So we have

change in velocity of the cyclist is given as

$\Delta v = v_f - v_i$

$\Delta v = 13\hat i - (-13\hat j)$

now average acceleration is given as

$a = \frac{\Delta v}{\Delta t}$

$a = \frac{13\hat i + 13\hat j}{21}$

so the magnitude of acceleration is given as

$a = \frac{13\sqrt2}{21} = 0.875 m/s^2$

Part f)

direction of acceleration is given as

$\theta = tan^{-1}\frac{y}{x}$

$\theta = tan^{-1}\frac{13}{13}$

$\theta = 45 degree$ North of East

4 0

3 years ago

A railroad cart with a mass of m1 = 11.6 t is at rest at the top of an h = 10.9 m high hump yard hill.

leonid [27]

The final common speed of the two carts will be 69.3 m/sec.The momentum conservation principle is applied.

<h3>What is the law of conservation of momentum?</h3>

According to the law of conservation of momentum, the momentum of the body before the collision is always equal to the momentum of the body after the collision.

Unit conversion:

1 metric tons = 1000 kg

Given data;

m₁= 11.6 metric ton =11600 kg

m₂ = 23.2 metric ton = 23200 kg

Let v represent the combined velocity of the two carts once they are connected, and let u represent the starting velocity of cart 1 when it reaches the bottom.

Considering energy conservation;

$\rm m_1 g h = \frac{1}{2} m_1 \times u^2 \\\\ u^2 = 2gh\\\\ u^2 = 2 \times 9.8 \times 10.6 \\\\ u = 207.972 \ m/s$