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ANTONII [103]
2 years ago
12

Are lions stronger then tigers?

Physics
1 answer:
Oliga [24]2 years ago
6 0

Answer:

I would think no, because a tiger has a more difficult truain to live in while the lion lives in a flat plain, also lions are a pack animal, only the female lions do the hunting, thus making the lions not as active all of the time, where as the tiger is a solitary animal and does all of the hunting by it's self.

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An organ pipe open at both ends is 1.5 m long. A second organ pipe that is closed at one end and open at the other is 0.75 m lon
telo118 [61]

Answer:

A. 110Hz,220Hz, 330 Hz

Explanation:

for organ open at open both ends;

the length of the organ for the fundamental frequency, L = A---->N + N----->A

A---->N  = λ /4 and N----->A = λ /4

L = λ /4 + λ /4 = λ /2

L = \frac{\lambda}{2} \\\\\lambda = 2L

λ  = 2 x 1.5m = 3.0 m

Wave equation is given by;

V = Fλ

Where;

V is the speed of sound

F is the frequency of the wave

F = V/ λ

F₀ = V / 2L

Where;

F₀  is the fundamental frequency

F₀ = 330 / 2(1.5)

F₀ = 330 / 3

F₀ = 110 Hz

the length of the organ for the first overtone, L = A---->N + N----->A + A----->N +  N----->A

L = 4λ /4

L = λ

λ = 1.5 m

F₁ = 330 / 1.5

F₁ = 220 Hz

Thus, F₁ = 2F₀

For open organ at one end

the length of the organ for the fundamental frequency, L = N------A

L = λ /4

λ = 4L

F₀ = V/4L

F₀ = 330 / (4 x 0.75)

F₀ = 110 Hz

the length of the organ for the first overtone, L = N-----N + N-----A

L = λ/2 + λ / 4

L = 3λ /4

F₁ = 3F₀

F₁ = 3 x 110

F₁ = 330 Hz

Thus the fundamental frequency for both organs is 110 Hz,

The first overtone for the organ open at both ends is 220 Hz

The first overtone for the organ open at one end is 330 Hz

The correct option is "A. 110Hz,220Hz, 330 Hz"

6 0
2 years ago
1.- An elevator is being lowered at a constant speed by a steel cable attached to an electric motor. Which statement is correct?
Nady [450]

Answer:

the correct one is C

Explanation:

For this exercise we must use the work definition

    W = F. s

Where the bold characters indicate vectors and the point is the scalar producer

    W = F s cos θ

Where θ is the angles between force and displacement.

Let us support this in our case. The cable creates an upward tension and with the elevator going down the angle between them is 180º so the work of the cable on the elevator is negative.

The evade has a downward force, its weight so the force goes down and the displacement goes down, as both are in the same direction the work is positive

When examining the statements the correct one is C

4 0
3 years ago
NaBr + CaF2 - NaF + CaBrz<br> es<br> What coefficients are needed to balance the chemical equation?
Dovator [93]

Answer:

4NaBr + 2CaF2 ---------- 4 NaF + 2CaBrz

Explanation:

8 0
3 years ago
In which type of circuit does charge move in only one direction?
mihalych1998 [28]

Answer:

<h2>the ans...is dc circuit...</h2>
6 0
3 years ago
Read 2 more answers
A 1870 kg car traveling at 13.5 m/s collides with a 2970 kg car that is initally at rest at a stoplight. The cars stick together
Inga [223]

Answer:

The value  is \mu  = 0.72

Explanation:

From the question we are told that

   The mass of the first car is  m_1  =  1870\ kg

    the initial  speed of the car  is  u  =  13.5 \  m/s

    The  mass of the second car is  m_2 =  2970\  kg

    The distance move by both cars is  s =  1.93  m

Generally from the law of momentum conservation

    m_1 * u_1 + m_2 *  u_2  =  (m_1 + m_2 ) *  v_f

Here u_2  =  0 because the second car is at rest

and  v_f is the final  velocity of the the two car

So

    1870*  13.5+ 0=  ( 1870 + 2970 ) *  v_f      

=> v_f  =  5.22\  m/s

Generally from kinematic equation

    v_f^2 = u_2^2  +  2as

here a is the deceleration

So

    5.22^2 = 0  +  2 *a  *  1.93

=> a =  7.06 \  m/s^2

Generally the frictional  force is equal to the force propelling the car , this can be mathematically represented as

   F_f  =  F

Here  F is mathematically represented as

F =  (m_1 + m_2) *  a

F =  (1870 + 2970) *  7.06    

F =34170.4 \ N  

and

F_f  =  \mu *  (m_1 + m_2 ) *  g

F_f  =  47432 * \mu

So

47432 * \mu   = 34170.4

=> 47432 * \mu   = 34170.4

=> \mu  = 0.72

6 0
2 years ago
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