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4 years ago

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Consider a 7 m stretched string that is clamped at both ends. What is the longest wavelength standing wave that it can support (

in m)?

1 answer:

user100 [1]4 years ago

7 0

A vibrating stretched string has nodes or fixed points at each end. The string will vibrate in its fundamental frequency with just one anti node in the middle - this gives half a wave.

$l=\frac{\lambda }{2}$

Rearranging for the wavelength

$\lambda=2l$

$\lambda =2(7)$

$\lambda = 14m$

Therefore the longest wavelength standing wave that it can support is 14m

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Suppose you have two small pith balls that are 5.5 cm apart and have equal charges of -29 nc?

zysi [14]

The question is missing, however, I guess the problem is asking for the value of the force acting between the two balls.

The Coulomb force between the two balls is:
$F= k_e \frac{ q_1 q_2}{r^2}$
where $k_e=8.99\cdot10^9~N m^2 C^{-2}$ is the Coulomb's constant, $q_1=q_2=29~nC=29\cdot 10^{-9}~C$ is the intensity of the two charges, and $r=5.5~cm=0.055~m$ is the distance between them.

Substituting these numbers into the equation, we get
$F=2.5~10^{-3}~N$

The force is repulsive, because the charges have same sign and so they repel each other.

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3 years ago

A jumbo jet must reach a speed of 360 km/h on the runway for takeoff. What is the lowest constant acceleration needed for takeof

weeeeeb [17]

The lowest constant acceleration needed for takeoff from a 1.80 km runway is 2.8 m/s².

To find the answer, we need to know about the Newton's equation of motion.

<h3>What's the Newton's equation of motion to find the acceleration in term of initial velocity, final velocity and distance?</h3>

The Newton's equation of motion that connects velocity, distance and acceleration is V² - U²= 2aS
V= final velocity, U= initial velocity, S= distance and a= acceleration

<h3>What's the acceleration, if the initial velocity, final velocity and distance are 0 m/s, 360km/h and 1.8 km respectively?</h3>

Here, S= 1.8 km or 1800 m, V= 360km/h or 100m/s , U= 0 m/s
So, 100²-0= 2×a×1800

=> 10000= 3600a

=> a= 10000/3600 = 2.8 m/s²

Thus, we can conclude that the lowest constant acceleration needed for takeoff from a 1.80 km runway is 2.8 m/s².

Learn more about the Newton's equation of motion here:

brainly.com/question/8898885

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2 years ago

A motor does 1000J of work in 18 seconds. What is the power of the<br> motor?

solong [7]

Answer:

55.56 watt

Explanation:

Work (W) = 1000 J

time (T) = 18 sec

power (P) = ?

We know power is the rate of doing work so

p = w / t

= 1000 / 18

= 55. 56 watt

7 0

3 years ago

Very far from earth (at R- oo), a spacecraft has run out of fuel and its kinetic energy is zero. If only the gravitational force

Margaret [11]

Answer:

Speed of the spacecraft right before the collision: $\displaystyle \sqrt{\frac{2\, G\cdot M_\text{e}}{R\text{e}}}$ .

Assumption: the earth is exactly spherical with a uniform density.

Explanation:

This question could be solved using the conservation of energy.

The mechanical energy of this spacecraft is the sum of:

the kinetic energy of this spacecraft, and
the (gravitational) potential energy of this spacecraft.

Let $m$ denote the mass of this spacecraft. At a distance of $R$ from the center of the earth (with mass $M_\text{e}$ ), the gravitational potential energy ( $\mathrm{GPE}$ ) of this spacecraft would be:

$\displaystyle \text{GPE} = -\frac{G \cdot M_\text{e}\cdot m}{R}$ .

Initially, $R$ (the denominator of this fraction) is infinitely large. Therefore, the initial value of $\mathrm{GPE}$ will be infinitely close to zero.

On the other hand, the question states that the initial kinetic energy ( $\rm KE$ ) of this spacecraft is also zero. Therefore, the initial mechanical energy of this spacecraft would be zero.

Right before the collision, the spacecraft would be very close to the surface of the earth. The distance $R$ between the spacecraft and the center of the earth would be approximately equal to $R_\text{e}$ , the radius of the earth.

The $\mathrm{GPE}$ of the spacecraft at that moment would be:

$\displaystyle \text{GPE} = -\frac{G \cdot M_\text{e}\cdot m}{R_\text{e}}$ .

Subtract this value from zero to find the loss in the $\rm GPE$ of this spacecraft:

$\begin{aligned}\text{GPE change} &= \text{Initial GPE} - \text{Final GPE} \\ &= 0 - \left(-\frac{G \cdot M_\text{e}\cdot m}{R_\text{e}}\right) = \frac{G \cdot M_\text{e}\cdot m}{R_\text{e}} \end{aligned}$

Assume that gravitational pull is the only force on the spacecraft. The size of the loss in the $\rm GPE$ of this spacecraft would be equal to the size of the gain in its $\rm KE$ .

Therefore, right before collision, the $\rm KE$ of this spacecraft would be:

$\begin{aligned}& \text{Initial KE} + \text{KE change} \\ &= \text{Initial KE} + (-\text{GPE change}) \\ &= 0 + \frac{G \cdot M_\text{e}\cdot m}{R_\text{e}} \\ &= \frac{G \cdot M_\text{e}\cdot m}{R_\text{e}}\end{aligned}$ .

On the other hand, let $v$ denote the speed of this spacecraft. The following equation that relates $v\!$ and $m$ to $\rm KE$ :

$\displaystyle \text{KE} = \frac{1}{2}\, m \cdot v^2$ .

Rearrange this equation to find an equation for $v$ :

$\displaystyle v = \sqrt{\frac{2\, \text{KE}}{m}}$ .

It is already found that right before the collision, $\displaystyle \text{KE} = \frac{G \cdot M_\text{e}\cdot m}{R_\text{e}}$ . Make use of this equation to find $v$ at that moment:

$\begin{aligned}v &= \sqrt{\frac{2\, \text{KE}}{m}} \\ &= \sqrt{\frac{2\, G\cdot M_\text{e} \cdot m}{R_\text{e}\cdot m}} = \sqrt{\frac{2\, G\cdot M_\text{e}}{R_\text{e}}}\end{aligned}$ .

6 0

3 years ago

riders on the tower of doom, an amusement park ride, experience 2.0 s of free fall, after which they are slowed to a stop in 0.5

Rainbow [258]

By Newton's second law, the apparent weight of rider coming at rest is <u>3548 N</u>.

According to formula, v = u + at

a = 39.2 m/ $s^{2}$

Force, F = ma

F = m(a+g)

F = apparent weight

m = mass, given = 65 kg

a = acceleration, given = 39.2 m/ $s^{2}$

g = 9.8 m/ $s^{2}$

Put these values in formula, F = m(a+g)

F = 65(39.2+9.8)

F = 3548 N

Therefore, the apparent weight of rider coming at rest is <u>3548 N</u>.

Learn more about apparent weight here:- brainly.com/question/24897276

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6 0

1 year ago