29. The x-axis of a velocity-time graph represents

A 42.6kg lamp is hanging from wires as shown in figure.The ring has negligible mass. Find tensionsT1, T2,T3 if the object is in

IRISSAK [1]

Answer:

T1 = 417.48N

T2 = 361.54N

T3 = 208.74N

Explanation:

Using the sin rule to fine the tension in the strings;

Given

amass = 42.6kg

Weight = 42.6 * 9.8 = 417.48N

The third angle will be 180-(60+30)= 90 degrees

Using the sine rule

W/Sin 90 = T3/sin 30 = T2/sin 60

Get T3;

W/Sin 90 = T3/sin 30

417.48/1 = T3/sin30

T3 = 417.48sin30

T3 = 417.48(0.5)

T3 = 208.74N

Also;

W/sin90 = T2/sin 60

417.48/1 = T2/sin60

T2 = 417.48sin60

T2 = 417.48(0.8660)

T2 = 361.54N

The Tension T1 = Weight of the object = 417.48N

8 0

3 years ago

Which diseases are most likely to be treated with antibiotics?

Vitek1552 [10]

Answer:

c) those caused by parasites

4 0

3 years ago

What are the benefits of the cool-down period following exercise?

kvv77 [185]

The benefits of the cool down period are quite important, it allows your body to slow your heart rate at a nice healthy safe pace, if you stop right away it can cause breathing, heart, and muscle problems.

5 0

3 years ago

A train has a length of 81.1 m and starts from rest with a constant acceleration at time t = 0 s. At this instant, a car just re

arlik [135]

Answer: a) vcar= 7 m/s ; b) a train= 0.65 m/s^2

Explanation: By using the kinematic equation for the car and the train we can determine the above values of the car velocity and the acceletarion of the train, respectively.

We have for the car

distance = v car* t, considering the length of train (81.1 m) travel by the car during the first 11.6 s

the v car = distance/time= 81.1 m/11.6s= 7 m/s

In order to calculate the acceleration we have to use the kinematic equation for the train from the rest

distance train = (a* t^2)/2

distance train : distance travel by the car at constant speed

so distance train= (vcar*36.35)m=421 m

the a traiin= (2* 421 m)/(36s)^2=0.65 m/s^2

4 0

3 years ago

I would like help with this physics problem

Darina [25.2K]

(a) This is a freefall problem in disguise - when the ball returns to its original position, it will be going at the same speed but in the opposite direction. So the ball's final velocity is the negative of its initial velocity.

Recall that

$v_f=v_i+at$

We have $v_f=-v_i$ , so that

$-2v_i=at\implies-2\left(8\,\dfrac{\mathrm m}{\mathrm s}\right)=\left(-2\,\dfrac{\mathrm m}{\mathrm s^2}\right)t\implies t=8\,\mathrm s$

(b) The speed of the ball at the start and at the end of the roll are the same 8 m/s, so the average speed is also 8 m/s.

(c) The ball's average velocity is 0. Average velocity is given by $\dfrac{v_i+v_f}2$ , and we know that $v_f=-v_i$ .

(d) The position of the ball $x_f$ at time $t$ is given by

$x_f=x_i+v_it+\dfrac12at^2$

Take the starting position to be the origin, $x_i=0$ . Then after 6 seconds,

$x_f=\left(8\,\dfrac{\mathrm m}{\mathrm s}\right)(6\,\mathrm s)+\dfrac12\left(-2\,\dfrac{\mathrm m}{\mathrm s^2}\right)(6\,\mathrm s)^2=42\,\mathrm m$

so the ball is 42 m away from where it started.

We're not asked to say in which direction it's moving at this point, but just out of curiosity we can determine that too:

$x_f-x_i=\dfrac{v_i+v_f}2t\implies42\,\mathrm m=\dfrac{8\,\frac{\mathrm m}{\mathrm s}+v_f}2(6\,\mathrm s)\implies v_f=6\,\dfrac{\mathrm m}{\mathrm s}$

Since the velocity is positive, the ball is still moving up the incline.

8 0

3 years ago