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MatroZZZ [7]
2 years ago
15

A bicyclist covers the first leg of a journey that is d1meters in t1seconds at a speed of v1m/s and the second leg of d2meters i

n t2seconds at a speed of v2m/sIf his average speed is equal to the average of v1 and v2
then which of the following is true?
1. t1=t2
2. t1≠t2
3. d1=d2
4. d1≠d2
Physics
1 answer:
Murljashka [212]2 years ago
4 0

According to motion in straight line  t1≠t2


A biker travels d1 meters in t1 seconds at v1 m/s for the first leg and d2 meters in t2 seconds at v2 m/s for the second leg. It's possible that t1t2 if his average speed is equal to the average of v1 and v2.
An object is said to be in motion if its position in relation to its surroundings changes over time. It is a shift in an object's position over time. The only type of motion that exists is motion in a straight line.


A reference system is constantly used to describe a particle's motion. An arbitrary origin point is used to create a reference system, and a coordinate system is imagined to be connected to it. The reference system for a specific problem is the coordinate system that has been selected for it. For the majority of the problems, we typically select an earth-based coordinate system as the reference system.


To learn more about Motion in straight lines please visit -brainly.com/question/17675825
#SPJ1

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Use the circuit diagram to decide if the lightbulb will light. Justify your answer.
podryga [215]

Answer:

The lightbulb will NOT light.

Explanation:

You put me in a difficult position.  I can't help it, but the "sample answer" is by far the best way to explain this, briefly and correctly.  There's no other choice but to copy it.

This is a short circuit. The branch without the bulb has almost no resistance, so all the current will flow through that branch instead of flowing through the bulb.

<em>If</em> the lower switch were <u>opened</u>, THEN we would have a series circuit.  Current would no longer have any other choice but to flow through the bulb, and the bulb would light.

5 0
3 years ago
Read 2 more answers
A current of 4.00 mA flows through a copper wire. The wire has an initial diameter of 4.00 mm which gradually tapers to a diamet
lesya692 [45]

The change in mean drift velocity for electrons as they pass from one end of the wire to the other is 3.506 x 10⁻⁷ m/s and average acceleration of the electrons is 4.38 x 10⁻¹⁵ m/s².

The given parameters;

  • <em>Current flowing in the wire, I = 4.00 mA</em>
  • <em>Initial diameter of the wire, d₁ = 4 mm = 0.004 m</em>
  • <em>Final diameter of the wire, d₂ = 1 mm = 0.001 m</em>
  • <em>Length of wire, L = 2.00 m</em>
  • <em>Density of electron in the copper, n = 8.5 x 10²⁸ /m³</em>

<em />

The initial area of the copper wire;

A_1 = \frac{\pi d^2}{4} = \frac{\pi \times (0.004)^2}{4} =1.257\times 10^{-5} \ m^2

The final area of the copper wire;

A_2 = \frac{\pi d^2}{4} = \frac{\pi (0.001)^2}{4} = 7.86\times 10^{-7} \ m^2

The initial drift velocity of the electrons is calculated as;

v_d_1 = \frac{I}{nqA_1} \\\\v_d_1 = \frac{4\times 10^{-3} }{8.5\times 10^{28} \times 1.6\times 10^{-19} \times 1.257\times 10^{-5}} \\\\v_d_1 = 2.34 \times 10^{-8} \ m/s

The final drift velocity of the electrons is calculated as;

v_d_2 = \frac{I}{nqA_2} \\\\v_d_2 = \frac{4\times 10^{-3} }{8.5\times 10^{28} \times 1.6\times 10^{-19} \times 7.86\times 10^{-7}} \\\\v_d_2 = 3.74\times 10^{-7}  \ m/s

The change in the mean drift velocity is calculated as;

\Delta v = v_d_2 -v_d_1\\\\\Delta v = 3.74\times 10^{-7} \ m/s \ -\ 2.34 \times 10^{-8} \ m/s = 3.506\times 10^{-7} \ m/s

The time of motion of electrons for the initial wire diameter is calculated as;

t_1 = \frac{L}{v_d_1} \\\\t_1 = \frac{2}{2.34\times 10^{-8}} \\\\t_1 = 8.547\times 10^{7} \ s

The time of motion of electrons for the final wire diameter is calculated as;

t_2 = \frac{L}{v_d_1} \\\\t_2= \frac{2}{3.74 \times 10^{-7}} \\\\t_2 = 5.348 \times 10^{6} \ s

The average acceleration of the electrons is calculated as;

a = \frac{\Delta v}{\Delta t} \\\\a = \frac{3.506 \times 10^{-7} }{(8.547\times 10^7)- (5.348\times 10^6)} \\\\a = 4.38\times 10^{-15} \ m/s^2

Thus, the change in mean drift velocity for electrons as they pass from one end of the wire to the other is 3.506 x 10⁻⁷ m/s and average acceleration of the electrons is 4.38 x 10⁻¹⁵ m/s².

Learn more here: brainly.com/question/22406248

7 0
2 years ago
Mercury is in the 80th position in the periodic table. How many protons does it have?
aalyn [17]
Mercury has 80 protons.
4 0
3 years ago
Read 2 more answers
A flywheel with radius of 0.400 mm starts from rest and accelerates with a constant angular acceleration of 0.600 rad/s2rad/s2.
charle [14.2K]

Answer: 0.00024\ m/s^2

Explanation:

Given

Radius of flywheel is r=0.4\ mm

Angular acceleration \alpha=0.6\ rad/s^2

For no change in radius, tangential acceleration is  given as

\Rightarrow a_t=a\lpha \times r

Insert the values

\Rightarrow a_t=0.6\times 0.4\times 10^{-3}\ m/s^2\\\Rightarrow a_t=2.4\times 10^{-4}\ m/s^2\ \text{or}\ 0.00024\ m/s^2

5 0
2 years ago
Consider a cyclotron in which a beam of particles of positive charge q and mass m is moving along a circular path restricted by
Ulleksa [173]

A) v=\sqrt{\frac{2qV}{m}}

B) r=\frac{mv}{qB}

C) T=\frac{2\pi m}{qB}

D) \omega=\frac{qB}{m}

E) r=\frac{\sqrt{2mK}}{qB}

Explanation:

A)

When the particle is accelerated by a potential difference V, the change (decrease) in electric potential energy of the particle is given by:

\Delta U = qV

where

q is the charge of the particle (positive)

On the other hand, the change (increase) in the kinetic energy of the particle is (assuming it starts from rest):

\Delta K=\frac{1}{2}mv^2

where

m is the mass of the particle

v is its final speed

According to the law of conservation of energy, the change (decrease) in electric potential energy is equal to the increase in kinetic energy, so:

qV=\frac{1}{2}mv^2

And solving for v, we find the speed v at which the particle enters the cyclotron:

v=\sqrt{\frac{2qV}{m}}

B)

When the particle enters the region of magnetic field in the cyclotron, the magnetic force acting on the particle (acting perpendicular to the motion of the particle) is

F=qvB

where B is the strength of the magnetic field.

This force acts as centripetal force, so we can write:

F=m\frac{v^2}{r}

where r is the radius of the orbit.

Since the two forces are equal, we can equate them:

qvB=m\frac{v^2}{r}

And solving for r, we find the radius of the orbit:

r=\frac{mv}{qB} (1)

C)

The period of revolution of a particle in circular motion is the time taken by the particle to complete one revolution.

It can be calculated as the ratio between the length of the circumference (2\pi r) and the velocity of the particle (v):

T=\frac{2\pi r}{v} (2)

From eq.(1), we can rewrite the velocity of the particle as

v=\frac{qBr}{m}

Substituting into(2), we can rewrite the period of revolution of the particle as:

T=\frac{2\pi r}{(\frac{qBr}{m})}=\frac{2\pi m}{qB}

And we see that this period is indepedent on the velocity.

D)

The angular frequency of a particle in circular motion is related to the period by the formula

\omega=\frac{2\pi}{T} (3)

where T is the period.

The period has been found in part C:

T=\frac{2\pi m}{qB}

Therefore, substituting into (3), we find an expression for the angular frequency of motion:

\omega=\frac{2\pi}{(\frac{2\pi m}{qB})}=\frac{qB}{m}

And we see that also the angular frequency does not depend on the velocity.

E)

For this part, we use again the relationship found in part B:

v=\frac{qBr}{m}

which can be rewritten as

r=\frac{mv}{qB} (4)

The kinetic energy of the particle is written as

K=\frac{1}{2}mv^2

So, from this we can find another expression for the velocity:

v=\sqrt{\frac{2K}{m}}

And substitutin into (4), we find:

r=\frac{\sqrt{2mK}}{qB}

So, this is the radius of the cyclotron that we must have in order to accelerate the particles at a kinetic energy of K.

Note that for a cyclotron, the acceleration of the particles is achevied in the gap between the dees, where an electric field is applied (in fact, the magnetic field does zero work on the particle, so it does not provide acceleration).

6 0
3 years ago
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