What would happen to the wavelength if the frequency was doubled?

Harry and ron set up this experiment with a glider, whose mass they have measured to be 565 g, and seven washers hanging from th

svetlana [45]

Let's call $m=565~g=0.565~kg$ the mass of the glider and $m_w=7\cdot12~g =84~g=0.084~kg$ the total mass of the seven washers hanging from the string.
The net force on the system is given by the weight of the hanging washers:
$F_{net} = m_w g$
For Newton's second law, this net force is equal to the product between the total mass of the system (which is $m+m_w$ ) and the acceleration a:
$F_{net}=(m+m_w)a$
So, if we equalize the two equations, we get
$m_w g = (m+m_w)a$
and from this we can find the acceleration:
$a= \frac{m_w g}{(m+m_w)} = \frac{0.084~kg \cdot 9.81~m/s^2}{(0.565~kg+0.084~kg)}=1.27~m/s^2$

5 0

3 years ago

un avión aterriza en la superficie de un portaaviones a 50 m/s y se detiene por completo en 120 metros, ¿cuál es la aceleración

sergejj [24]

Answer:

La aceleración necesaria para detener el avión es - 10.42 m/s².

Explanation:

Un movimiento uniformemente acelerado (M.U.A) es aquél cuya aceleración es constante y la velocidad de un objeto cambia a medida que el movimiento evoluciona.

Siendo la aceleración "a" el cambio de velocidad al tiempo transcurrido en un punto A a B, la velocidad inicial la velocidad que tiene un cuerpo al iniciar su movimiento en un período de tiempo y la velocidad final la velocidad que tiene un cuerpo al finalizar su movimiento en un período de tiempo, entonces en M.U.A se cumple:

Vf² - Vo² = 2*a*d

donde:

Vf: Velocidad final
Vo: Velocidad inicial
a: Aceleración
d: Distancia recorrida

En este caso:

Vf: 0 m/s, porque el avión se detiene
Vo: 50 m/s
a: ?
d: 120 m

Reemplazando:

(0 m/s)² - (50 m/s)² = 2*a*120 m

Resolviendo:

$a=\frac{(0 m/s)^{2} -(50 m/s)^{2} }{2*120 m}$

a= - 10.42 m/s²

La aceleración necesaria para detener el avión es - 10.42 m/s².

5 0

3 years ago

A heat engine does 200 j of work per cycle while exhausting 600 j of heat to the cold reservoir. what is the engine's thermal ef

AveGali [126]

The thermal efficiency of an engine is
$\eta= \frac{W}{Q}$
where
W is the work done by the engine
Q is the heat absorbed by the engine to do the work

In this problem, the work done by the engine is W=200 J, while the heat exhausted is Q=600 J, so the efficiency of the machine is
$\eta= \frac{W}{Q}= \frac{200 J}{600 J}=0.33 = 33 \%$

8 0

3 years ago

How does your power output in climbing the stairs compare to the power output of a 100-watt light bulb? if your power could have

cricket20 [7]

1) Assuming an adult person has an average mass of m=80 kg, and assuming it takes about 30 seconds to climb 5 meters of stairs, the energy used by the person is
$E=mgh=(80 kg)(9.81 m/s^2)(5 m)=3924 J$
So the power output is
$P= \frac{E}{t}= \frac{3924 J}{30 s} \sim 130 W$

And since the estimate we made is very rough, we can say that the power output of the person is comparable to the power output of the light bulb of 100 W.

2) Based on the results we found in the previous part of the exercise, since the power output of the person is comparable to the power output of 1 light bulb of 100 W, we can say that the person could have kept burning only one 100-W light bulb during the climb.

6 0

2 years ago

Read 2 more answers

A fairground ride consists of a large vertical drum that spins so

expeople1 [14]

Answer:

v = 3.84 m/s

Explanation:

In order for the riders to stay pinned against the inside of the drum the frictional force on them must be equal to the centripetal force:

$Centripetal\ Force = Frictional\ Force\\\\\frac{mv^2}{r} = \mu R = \mu W\\\\\frac{mv^2}{r} = \mu mg\\\\\frac{v^2}{r} = \mu g\\\\v = \sqrt{\mu gr}$

where,

v = minimum speed = ?

g = acceleration due to gravity = 9.81 m/s²

r = radius = 10 m

μ = coefficient of friction = 0.15

Therefore,

$v=\sqrt{(0.15)(9.81\ m/s^2)(10\ m)}$

v = 3.84 m/s

6 0

3 years ago