Ask question

Ask question

All categories

3 years ago

12

A heavy ball with a weight of 100 N is hung from the ceiling of a lecture hall on a 4.3-m-long rope. The ball is pulled to one s

ide and released to swing as a pendulum, reaching a speed of 5.0 m/s as it passes through the lowest point. Part A What is the tension in the rope at that point?

1 answer:

Vlada [557]3 years ago

6 0

Answer: T= 159.3N

Explanation:

The total Tension on the rope at this point=

Sum of the weight of the ball + the centripetal force on the ball

T= W+ mv^2/r

Where T is tension, W is weight of the ball, m is mass of the ball, r is radius of rotation, v is the velocity of rotation. g= acceleration due to gravity

W= 100N

m= W/g= 100/9.8 = 10.2kg

v= 5m/s

r= 4.3m

Therefore,

T= 100 + 10.2×5^2/4.3

T= 100 + 59.3

T= 159.3N

You might be interested in

We can model a certain battery as a voltage source in series with a resistance. The open-circuit voltage of the battery is 10 V

Angelina_Jolie [31]

Answer:

51.4 Ohms

Explanation:

By applying voltage division rule

$V_f=v_i\times \frac {R_l}{R_l+R_m}$ where v is voltage, subscripts i and f represnt initial and final, R is resistance, m is internal and l is external.Substituting 7V for final voltage, 10V for initial voltage and the external resistance as 120 Ohms then

$7=10*\frac {120}{120+R_m}\\7R_m+840=1200\\R_m={1200-840}{7}=51.428571\approx 51.4 Ohms$

3 0

3 years ago

A long rod of 60-mm diameter and thermophysical properties rho= 8000 kg/m3, c= 500 J/kg·K, and k= 50 W/m·K is initially at a uni

Dvinal [7]

Answer:

Tc = = 424.85 K

Explanation:

Data given:

D = 60 mm = 0.06 m

$\rho = 8000 kg/m^3$

k = 50 w/m . k

c = 500 j/kg.k

$h_{\infty} = 1000 w/m^2$

$t_{\infity} = 750 k$

$t_w = 500 K$

$surface area = As = \pi dL$

$\frac{As}{L} = \pi D = \pi \timeS 0.06$

HEAT FLOW Q is

$Q = h_{\infty} As (T_[\infty} - Tw)$

$= 1000 \pi\times 0.06 (750-500)$

= 47123.88 w per unit length of rod

volumetric heat rate

$q = \frac{Q}{LAs}$

$= \frac{47123.88}{\frac{\pi}{4} D^2 \times 1}$

$q = 1.66\times 10^{7} w/m^3$

$Tc = \frac{- qR^2}{4K} + Tw$

$= \frac{ - 1.67\times 10^7 \times (\frac{0.06}{2})^2}{4\times 56} + 500$

= 424.85 K

7 0

3 years ago

iren [92.7K]

Answer:

7

Explanation:

A quotient is the answer to a division.For example,the quotient of 10 is 2 and 5 because 5÷10=2.

5 0

3 years ago

Did you know whales have hip leg and shin bones? well now you do

storchak [24]

Answer:

No, I did not. Thank you for educating me!

Explanation:

Have a great day!

6 0

2 years ago

As part of a heat treatment process, cylindrical, 304 stainless steel rods of 100-mm diameter are cooled from an initial tempera

saveliy_v [14]

Answer:

Explanation:

Given that:

diameter = 100 mm

initial temperature = 500 ° C

Conventional coefficient = 500 W/m^2 K

length = 1 m

We obtain the following data from the tables A-1;

For the stainless steel of the rod $\overline T = 548 \ K$

$\rho = 7900 \ kg/m^3$

$K = 19.0 \ W/mk \\ \\ C_p = 545 \ J/kg.K$

$\alpha = 4.40 \times 10^{-6} \ m^2/s \\ \\ B_i = \dfrac{h(\rho/4)}{K} \\ \\ =0.657$

Here, we can't apply the lumped capacitance method, since Bi > 0.1

$\theta_o = \dfrac{T_o-T_{\infty}}{T_i -T_\infty}} \\ \\ \theta_o = \dfrac{50-30}{500 -30}} \\ \\ \theta_o = 0.0426\\$

$0.0426 = c_1 \ exp (- E^2_1 F_o_)\\ \\ \\ 0.0426 = 1.1382 \ exp (-10.9287)^2 \ f_o \\ \\ = f_o = \dfrac{In(0.0374)}{0.863} \\ \\ f_o = 3.81$

$t_f = \dfrac{f_o r^2}{\alpha} \\ \\ t_f = \dfrac{3.81 \times (0.05)^2}{4.40 \times 10^{-6}} \\ \\ t_f= 2162.5 \\ \\ t_f = 36 mins$

However, on a single rod, the energy extracted is:

$\theta = pcv (T_i - T_{\infty} )(1 - \dfrac{2 \theta}{c} J_1 (\zeta) ) \\ \\ = 7900 \\times 546 \times 0.007854 \times (500 -300) (1 - \dfrac{2 \times 0.0426}{1.3643}) \\ \\ \theta = 1.54 \times 10^7 \ J$

Hence, for centerline temperature at 50 °C;

The surface temperature is:

$T(r_o,t) = T_{\infty} +(T_1 -T_{\infty}) \theta_o \ J_o(\zeta_1) \\ \\ = 30 + (500-30) \times 0.0426 \times 0.5386 \\ \\ \mathbf{T(r_o,t) = 41.69 ^0 \ C}$

5 0

3 years ago