All categories

1 year ago

A A load of 1000 N can be lifted by applying

Physics

2 answers:

o-na [289]1 year ago

6 0

Answer:

Explanation:

$\text{Since the fulcrum lies in between the load and effort, it is class I lever.}$

$\text{Mechanical advantage (M.A.) of levers is given by,}$

$\text{M.A.=effort armload arm=50cm,10cm}$

$\text{Thus, mechanical advantage , M.A. = 5.}$

$\text{Given, effort (E)=10N, Load(L)=?}$

∴ $L=E \times M.A.$

$=10\times5=50N$

drek231 [11]1 year ago

6 0

Answer:

$\fbox {100 cm}$

Explanation:

Finding Mechanical Advantage :

MA = Load / Effort

MA = 1000/250

MA = 4

Finding length of effort arm :

Load arm = Effort arm / 4

Effort arm = 25 cm x 4

Effort arm = 100 cm

You might be interested in

An 20-cm-long Bicycle Crank Arm. With A Pedal At One End. Is Attached To A 25-cm-diameter Sprocket, The Toothed Disk Around Whic

malfutka [58]

To solve the problem, it is necessary to apply the concepts related to the kinematic equations of the description of angular movement.

The angular velocity can be described as

$\omega_f = \omega_0 + \alpha t$

Where,

$\omega_f =$ Final Angular Velocity

$\omega_0 =$ Initial Angular velocity

$\alpha =$ Angular acceleration

t = time

The relation between the tangential acceleration is given as,

$a = \alpha r$

where,

r = radius.

PART A ) Using our values and replacing at the previous equation we have that

$\omega_f = (94rpm)(\frac{2\pi rad}{60s})= 9.8436rad/s$

$\omega_0 = 63rpm(\frac{2\pi rad}{60s})= 6.5973rad/s$

$t = 11s$

Replacing the previous equation with our values we have,

$\omega_f = \omega_0 + \alpha t$

$9.8436 = 6.5973 + \alpha (11)$

$\alpha = \frac{9.8436- 6.5973}{11}$

$\alpha = 0.295rad/s^2$

The tangential velocity then would be,

$a = \alpha r$

$a = (0.295)(0.2)$

$a = 0.059m/s^2$

Part B) To find the displacement as a function of angular velocity and angular acceleration regardless of time, we would use the equation

$\omega_f^2=\omega_0^2+2\alpha\theta$

Replacing with our values and re-arrange to find $\theta,$

$\theta = \frac{\omega_f^2-\omega_0^2}{2\alpha}$

$\theta = \frac{9.8436^2-6.5973^2}{2*0.295}$

$\theta = 90.461rad$

That is equal in revolution to

$\theta = 90.461rad(\frac{1rev}{2\pi rad}) = 14.397rev$

The linear displacement of the system is,

$x = \theta*(2\pi*r)$

$x = 14.397*(2\pi*\frac{0.25}{2})$

$x = 11.3m$

5 0

3 years ago

falling objects drop with an average acceleration of 9.8m/sec/sec. if an object falls from a tall building how long will it take

sveta [45]

Answer:

5 seconds

Explanation:

Acceleration = (final velocity - initial velocity) ÷ time

$a = \frac{v - u}{t}$

$9.8 = \frac{49 - 0}{t}$

$9.8 = \frac{49}{t}$

$9.8t = 49$

$t = \frac{49}{9.8}$

$t = 5$

8 0

2 years ago

Gravity does not actually "pull" objects at

Radda [10]

I believe it’s (D. Any object)

5 0

2 years ago

A 50 n force is acting on a lever 1.5 m from the fulcrum balances an object 1m from the fulcrum on the other arm. what is the we

Angelina_Jolie [31]

Answer:

A 100 N force acting on a lever 2 m from the fulcrum balances an object 0.5 m from the fulcrum on. ... What is the weight of the object(in newtons)? What is its mass (in kg)? ... mass at the one end and effort arm is the distance between pivot and effort applied at the other end.

Explanation:

hpoe this helps you.

4 0

2 years ago

Adam is teeing off on hole number two. The hole is 390 yards away. It is a par four hole. What club should he use to tee off? Ex

kkurt [141]

Answer:

A driver.

Explanation:

Using a driver while at least 350 yds away is better than using a iron, because it will be a waste of the par 4 as it is not as powerful as the driver.

3 0

3 years ago