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Marta_Voda [28]

3 years ago

8

A member of the marching band tunes their trombone inside their school. When they walk out to the field the instrument goes out

of tune. What is the best explanation for this?

2 answers:

Alona [7]3 years ago

4 0

The best explanation is the <em>difference</em> between the inside <em>temperature</em> and the outside temperature.

If the player doesn't change his emboucher (muscles and position of his lips), then the pitch produced by the instrument depends only on the physical dimensions of its plumbing, and the speed of sound in the tube.

BOTH of those things change slightly when the temperature changes.

NNADVOKAT [17]3 years ago

4 0

Answer:

the answer is d

Explanation:

just being smart

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When light hits the water droplets in a cloud, the cloud appears white. The light waves are being:

Tamiku [17]

Answer:

scattered

Explanation:

7 0

3 years ago

What is acceleration of a body moving with uniform velocity? (convert 50km/hr into m/s)

gregori [183]

Answer:

answer will be 0

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5 0

2 years ago

The horizontal beam in Fig. E11.14 weighs 190 N, and its center of gravity is at its center. Find (a) the tension in the cable a

grandymaker [24]

Answer:

(a). The tension in the cable is 658.33 N.

(b). The horizontal components of the force exerted on the beam at the wall is 526.66 N.

(c). The vertical components of the force exerted on the beam at the wall is 95.002 N.

Explanation:

Given that,

Weight of beam= 190 N

Here, The center of gravity is at its center

According to figure,

The angle is

$\sin\theta=\dfrac{3}{5}$

The horizontal component is

$T_{x}=T\cos\theta$

The vertical component is

$T_{y}=T\sin\theta$

(a). We need to calculate the tension in the cable

Using formula of net torque acting on the pivot

$\sum\tau=F_{b}\times r+F_{w}\times r'-T\sin \theta\times r'$

Put the value into the formula

$0=190\times2+300\times 4-T\sin\theta\times 4$

$T\sin\theta\times 4=380+1200$

$T=\dfrac{1580\times5}{3\times 4}$

$T=658.33\ N$

(b). We need to calculate the horizontal components of the force exerted on the beam at the wall

Using formula of horizontal component

$F_{x}=T\cos\theta$

Put the value into the formula

$F_{x}=658.33\times\dfrac{4}{5}$

$F_{x}=526.66\ N$

(c). We need to calculate the vertical components of the force exerted on the beam at the wall

Using formula of vertical component

$F_{y}=F_{b}+F_{w}-T\sin\theta$

Put the value into the formula

$F_{y}=190+300-658.33\times\dfrac{3}{5}$

$F_{y}=95.002\ N$

Hence, (a). The tension in the cable is 658.33 N.

(b). The horizontal components of the force exerted on the beam at the wall is 526.66 N.

(c). The vertical components of the force exerted on the beam at the wall is 95.002 N.

3 0

3 years ago

Which statement best describes the energy of activation?

Varvara68 [4.7K]

hi <3

i believe the answer would be D, as when the temperature increases the particles have more energy and can overcome the activation energy more rapidly.

hope this helps :)

4 0

2 years ago

A jet liner must reach a speed of 82 m/s for takeoff. If the

SIZIF [17.4K]

Answer:

The acceleration that the jet liner that must have is 2.241 meters per square second.

Explanation:

Let suppose that the jet liner accelerates uniformly. From statement we know the initial ( $v_{o}$ ) and final speeds ( $v_{f}$ ), measured in meters per second, of the aircraft and likewise the runway length ( $d$ ), measured in meters. The following kinematic equation is used to calculate the minimum acceleration needed ( $a$ ), measured in meters per square second:

$a = \frac{v_{f}^{2}-v_{o}^{2}}{2\cdot d}$

If we know that $v_{o} = 0\,\frac{m}{s}$ , $v_{f} = 82\,\frac{m}{s}$ and $d = 1500\,m$ , then the acceleration that the jet must have is:

$a = \frac{\left(82\,\frac{m}{s} \right)^{2}-\left(0\,\frac{m}{s} \right)^{2}}{2\cdot (1500\,m)}$

$a = 2.241\,\frac{m}{s^{2}}$

The acceleration that the jet liner that must have is 2.241 meters per square second.

3 0

2 years ago