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3 years ago

Below is a physics question

Physics

1 answer:

Allisa [31]3 years ago

8 0

Explanation:

i1=1ampere

i1+i2=I

VA−VB=i1×3Ω=i2×6

⇒1×3=i2×6

2i=21ampere

Hence I=i1+i2=1+0.5=1.5ampere

Req=2+3+63×6=2+93×6=4Ω

equivalent circuit

Refer image .2

From KVL

1req=v

v=1.5×4

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What distance is required for a train to stop if its initial velocity is 23 m/s and its

Varvara68 [4.7K]

-- The train starts at 23 m/s and slows down by 0.25 m/s every second.

So it'll take (23/0.25) = 92 seconds to stop.

-- Its average speed during that time will be (1/2)(23+0) = 11.5 m/s

-- Moving at an average speed of 11.5 m/s for 92 sec, the train will cover

(11.5 m/s) x (92 sec) = <em>1,058 meters</em> .

7 0

3 years ago

A 0.6 kg block attached to a spring of force constant 13.6 N/m oscillates with an amplitude of 9 cm. Find the maximum speed of t

mash [69]

Answer:

1) 0.43 meters per second

2) 0.21 meters per second

3) 1.02 $\frac{m}{s^{2}}$

4) 0.66 seconds

Explanation:

part 1

By conservation of energy, the maximum kinetic energy (K) of the block is at equilibrium point where the potential energy is zero. So, at the equilibrium kinetic energy is equal to maximum potential energy (U):

$K=U$

$\frac{mv^2}{2}=\frac{kx_{max}^2}{2}$

With m the mass, v the speed, k the spring constant and xmax the maximum position respect equilibrium position. Solving for v

$v=\sqrt{\frac{kx_{max}^2}{m}}=\sqrt{\frac{(13.6)(0.09m)^2}{0.6}}=0.43\frac{m}{s}$

part 2

Again by conservation of energy we have kinetic energy equal potential energy:

$\frac{mv^2}{2}=\frac{kx_{max}^2}{2}=$

$v=\sqrt{\frac{kx_{max}^2}{m}}=\sqrt{\frac{(13.6)(0.045m)^2}{0.6}}=0.21\frac{m}{s}$

part 3

Acceleration can be find using Newton's second law:

$F=ma$

with F the force, m the mass and a the acceleration, but elastic force is -kx, so:

$-kx=ma$

$a= -\frac{kx}{m}=-\frac{(13.6)(0.045)}{0.6}=-1.02\frac{m}{s^{2}}$

part 4

The period of an oscillator is the time it takes going from one extreme to the other one, that is going form 4.5 cm to -4.5 cm respect the equilibrium position. That period is:

$T=2\pi\sqrt{\frac{m}{k}}=T=2\pi\sqrt{\frac{0.6}{13.6}}=1.32s$

So between 0 and 4.5 cm we have half a period:

$t=\frac{T}{2}=0.66s$

7 0

3 years ago

Let's examine over the next few questions an NFL kick as described in the 3rd Law video. The announcer claimed that the ball is

hichkok12 [17]

Answer:

8451.62109367671 Newtons

Explanation:

$1\ kg=1\times 9.8066500286389=9.8066500286389\ N$

$1 \lb=2.2046226218488\ kg$

It is known that

$1\ lbs=\dfrac{9.8066500286389}{2.2046226218488}=4.4482216282508\ N$

So,

$1900\ lbs=1900\times 4.4482216282508\\\Rightarrow 1900\ lbs=8451.6210936765\ N$

The force in Newtons is 8451.6210936765 Newtons

4 0

3 years ago

The atomic number of magnesium is 12. This means that its nucleus must contain

nasty-shy [4]

12 protons in the nucleus

5 0

3 years ago

An object moves uniformly around a circular path of radius 19.0 cm, making one complete revolution every 2.40 s.

klio [65]

Answer:

v = 0.5 m/s

f = 0.42 Hz

ω = 2.6 rad/sec

Explanation:

By definition, the translational speed is the rate of change of the position with respect to time.
The change in position along a complete revolution is just the following:
Δs = 2*π*r = 2*π*0.19 m = 1.19 m
The time needed to complete a revolution is 2.4 s, so the translational speed can be written as follows:

$v =\frac{\Delta s}{\Delta t} = \frac{1.19m}{2.4s} = 0.5 m/s (1)$

The frequency in Hz is just the inverse of the time needed to complete a revolution (known as the period T), as follows:
f = 1/T = 1/2.4s = 0.42 Hz (2)
Finally, the angular speed is the rate of change of the angle rotated with respect to time, as follows:

$\omega = \frac{\Delta\theta}{\Delta t} = \frac{2*\pi}{2.4s} = 2.6 rad/sec (3)$

5 0

3 years ago