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2 years ago

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Physics

2 answers:

boyakko [2]2 years ago

6 0

Answer:

yessir

Explanation:

REY [17]2 years ago

4 0

Answer:

Convex lens are in both a magnifying glass and a microscope

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How many turns should a 10-cm long ideal solenoid have if it is to generate a 1.5-mT magnetic field when 1.0 A of current runs t

Ira Lisetskai [31]

Answer:

N=119.34 turns

Explanation:

The magnetic field of a solenoid is calculated using the formula:

B= µo* $\frac{I*N}{L}$ Equation 1

Where:

B: magnetic field in Teslas (T)

µo: free space permeability in T*m/A

I= Intensity of the current flowing through the conductor in ampere (A)

N= number of turns

L= solenoid length in meters (m)

Data of the problem:

L=10cm= $10*10^{-2}$ , B= 1.5mT= $1.5*10^{-3} T$ ,I=1A

µo= $4\pi *10^{-7} \frac{T*m}{A}$

We cleared N of the equation (1):

N=B*L/ µo*I

N= $(1.5*10^{-3} *10*10^{-2} )/(4\pi *10^{-7} *1$

$N=0.1193*10^{3}$

Answer

N=119.34 turns

3 0

2 years ago

A spring stretches by 15cm when a mass of 300g hangs down from it,if the spring is then stretched an additional 10cm and release

bearhunter [10]

Answer:

0.1 m

Explanation:

It is given that,

Mass of the object, m = 350 g = 0.35 kg

Spring constant of the spring, k = 5.2 N/m

Amplitude of the oscillation, A = 10 cm = 0.1 m

Frequency of a spring mass system is given by :

Time period:

4 0

2 years ago

A standing wave of the third overtone is induced in a stopped pipe, 2.5 m long. The speed of sound is The frequency of the sound

NemiM [27]

Answer:

f3 = 102 Hz

Explanation:

To find the frequency of the sound produced by the pipe you use the following formula:

$f_n=\frac{nv_s}{4L}$

n: number of the harmonic = 3

vs: speed of sound = 340 m/s

L: length of the pipe = 2.5 m

You replace the values of n, L and vs in order to calculate the frequency:

$f_{3}=\frac{(3)(340m/s)}{4(2.5m)}=102\ Hz$

hence, the frequency of the third overtone is 102 Hz

8 0

2 years ago

Official (Closed) - Non Sensitive

Pavlova-9 [17]

Answer:

The minimum running time is 319.47 s.

Explanation:

First we find the distance covered and time taken by the train to reach its maximum speed:

We have:

Initial Speed = Vi = 0 m/s (Since, train is initially at rest)

Final Speed = Vf = 29.17 m/s

Acceleration = a = 0.25 m/s²

Distance Covered to reach maximum speed = s₁

Time taken to reach maximum speed = t₁

Using 1st equation of motion:

Vf = Vi + at₁

t₁ = (Vf - Vi)/a

t₁ = (29.17 m/s - 0 m/s)/(0.25 m/s²)

t₁ = 116.68 s

Using 2nd equation of motion:

s₁ = (Vi)(t₁) + (0.5)(a)(t₁)²

s₁ = (0 m/s)(116.68 s) + (0.5)(0.25 m/s²)(116.68 s)²

s₁ = 1701.78 m = 1.7 km

Now, we shall calculate the end time and distance covered by train, when it comes to rest on next station.

We have:

Final Speed = Vf = 0 m/s (Since, train is finally stops)

Initial Speed = Vi = 29.17 m/s (The train must maintain max. speed for min time)

Deceleration = a = - 0.7 m/s²

Distance Covered to stop = s₂

Time taken to stop = t₂

Using 1st equation of motion:

Vf = Vi + at₂

t₂ = (Vf - Vi)/a

t₂ = (0 m/s - 29.17 m/s)/(- 0.7 m/s²)

t₂ = 41.67 s

Using 2nd equation of motion:

s₂ = (Vi)(t₂) + (0.5)(a)(t₂)²

s₂ = (29.17 m/s)(41.67 s) + (0.5)(- 0.7 m/s²)(41.67 s)²

s₂ = 607.78 m = 0.6 km

Since, we know that the rest of 7 km, the train must maintain the maximum speed to get to the next station in minimum time.

The remaining distance is:

s₃ = 7 km - s₂ - s₁

s₃ = 7 km - 0.6 km - 1.7 km

s₃ = 4.7 km

Now, for uniform speed we use the relation:

s₃ = vt₃

t₃ = s₃/v

t₃ = (4700 m)/(29.17 m/s)

t₃ = 161.12 s

So, the minimum running time will be:

t = t₁ + t₂ + t₃

t = 116.68 s + 41.67 s + 161.12 s

5 0

2 years ago

You have been abducted by aliens and find yourself on a strange planet. Fortunately, you have a meter stick with you. You observ

defon

Answer:

$36.4276\ m/s^2$

Explanation:

m = Mass of stick

L = Length of stick = 1 m

h = Center of mass of stick = $\dfrac{1}{2}=0.5\ m$

g = Acceleration due to gravity

T = Time period = 0.85 s

Time period is given by

$T=2\pi\sqrt{\dfrac{I}{mgh}}$

Moment of inertia is given by

$I=m\dfrac{L^2}{3}$

$T=2\pi\sqrt{\dfrac{m\dfrac{L^2}{3}}{mgh}}\\\Rightarrow T=2\pi\sqrt{\dfrac{L^2}{3gh}}\\\Rightarrow g=\dfrac{4\pi^2L^2}{3hT^2}\\\Rightarrow g=\dfrac{4\pi^2\times 1^2}{3\times 0.5\times 0.85^2}\\\Rightarrow g=36.4276\ m/s^2$

The acceleration of gravity on this planet is $36.4276\ m/s^2$

0 0

2 years ago

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