Select the correct answer.

A merry-go-round is spinning at a rate of 4.04.0 revolutions per minute. Cora is sitting 0.50.5 m from the center of the merry-g

dsp73

Answer:

angular speed of both the children will be same

Explanation:

Rate of revolution of the merry go round is given as

f = 4.04 rev/min

so here we have

$f = \frac{4.04}{60} =0.067 rev/s$

here we know that angular frequency is given as

$\omega = 2\pi f$

$\omega = 2\pi(0.067)$

$\omega = 0.42 rad/s$

now this is the angular speed of the disc and this speed will remain same for all points lying on the disc

Angular speed do not depends on the distance from the center but it will be same for all positions of the disc

7 0

3 years ago

A worker drives a 0.562 kg spike into a rail tie with a 2.26 kg sledgehammer. The hammer hits the spike with a speed of 64.4 m/s

zubka84 [21]

Answer:

Explanation:

Given that,

Mass of sledge hammer;

Mh =2.26 kg

Hammer speed;

Vh = 64.4 m/s

The expression fot the kinetic energy of the hammer is,

K.E(hammer) = ½Mh•Vh²

K.E(hammer) = ½ × 2.26 × 64.4²

K.E ( hammer) = 4686.52 J

If one forth of the kinetic energy is converted into internal energy, then

ΔU = ¼ × K.E(hammer)

∆U = ¼ × 4686.52

∆U = 1171.63 J

Thus, the increase in total internal energy will be 1171.63 J.

4 0

4 years ago

Please help answer question

nika2105 [10]

Answer:

C = 1.01

Explanation:

Given that,

Mass, m = 75 kg

The terminal velocity of the mass, $v_t=60\ m/s$

Area of cross section, $A=0.33\ m^2$

We need to find the drag coefficient. At terminal velocity, the weight is balanced by the drag on the object. So,

R = W

or

$\dfrac{1}{2}\rho CAv_t^2=mg$

Where

$\rho$ is the density of air = 1.225 kg/m³

C is drag coefficient

So,

$C=\dfrac{2mg}{\rho Av_t^2}\\\\C=\dfrac{2\times 75\times 9.8}{1.225\times 0.33\times (60)^2}\\\\C=1.01$

So, the drag coefficient is 1.01.

4 0

3 years ago

Please help! i'm horrible at this

frez [133]

Answer:

a = Δv/t = (vf - vi)/t = (0 - 5)/4 = -1.25 m/s²

Explanation:

You may or may not need the negative sign, depending on how the question designer was thinking about the problem.

4 0

3 years ago

Un camion de envios se encuentra detenido en una señal de pare, permitiendo que pase una ambulancia. Inicia su recorrido y al ca

Nesterboy [21]

Answer:

$0.741\ \text{m/s}^2$

Explanation:

v = Velocidad final = $40\ \text{km/h}=\dfrac{40}{3.6}\ \text{m/s}$

u = Velocidad inicial = 0

t = Tiempo empleado = 15 s

a = Aceleración

De las ecuaciones cinemáticas tenemos

$v=u+at\\\Rightarrow a=\dfrac{v-u}{t}\\\Rightarrow a=\dfrac{\dfrac{40}{3.6}-0}{15}\\\Rightarrow a=0.741\ \text{m/s}^2$

La aceleración del camión en el primer intervalo de tiempo es $0.741\ \text{m/s}^2$ .

4 0

3 years ago