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True [87]
3 years ago
10

In case 1, a force f is pushing perpendicular on an object a distance l/2 from the rotation axis. in case 2 the same force is pu

shing at an angle of 30 degrees a distance l from the axis. 1 in which case is the torque due to the force about the rotation axis biggest?
Physics
1 answer:
Serggg [28]3 years ago
6 0
In case 1, the torque is given by the product between the force and the arm:
\tau_1 = F \cdot  \frac{L}{2}

In case 2, the torque is given by the product between the component of the force perpendicular to the arm and the arm itself, so we have:
\tau_2 = F \cos 30^{\circ} L=F  \frac{ \sqrt{3} }{2} L =  \sqrt{3} F  \frac{L}{2}= \sqrt{3} \tau_1

and since \sqrt{3} is larger than 1, than the torque in case 2 is larger than the torque in case 1.
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Satellite A has an orbital radius 3.00 times greater than that of satellite B. Satellite B's orbital period around Earth is 120
igomit [66]

Answer:

To find the circumference (orbit) of an object, you use Pi x Diameter. 

As you have the circumference of B, you divide it by Pi to get the Diameter. 

So 120 divided by 3.141592654 = 38.2 minutes for the Diameter. 

As' radius and Diameter will be 3x greater than B. 

38.2 x 3 = 114.6 

To get back to the orbital period, times 114.6 by Pi, and you will get 360 minutes

HOPE THIS HELPS AND PLS MARK AS BRAINLIEST

THNXX :)

7 0
3 years ago
An aluminum bar 600mm long, with diameter 40mm, has a hole drilled in the center of the bar. The hole is 40mm in diameter and 10
s2008m [1.1K]

Answer:

<em>1.228 x </em>10^{-6}<em> mm </em>

<em></em>

Explanation:

diameter of aluminium bar D = 40 mm  

diameter of hole d = 30 mm

compressive Load F = 180 kN = 180 x 10^{3} N

modulus of elasticity E = 85 GN/m^2  = 85 x 10^{9} Pa

length of bar L = 600 mm

length of hole = 100 mm

true length of bar = 600 - 100 = 500 mm

area of the bar A = \frac{\pi D^{2} }{4} =  \frac{3.142* 40^{2} }{4} = 1256.8 mm^2

area of hole a = \frac{\pi(D^{2} - d^{2}) }{4} = \frac{3.142*(40^{2} - 30^{2})}{4} = 549.85 mm^2

Total contraction of the bar = \frac{F*L}{AE} + \frac{Fl}{aE}

total contraction = \frac{F}{E} * (\frac{L}{A} +\frac{l}{a})

==> \frac{180*10^{3}}{85*10^{9}} *( \frac{500}{1256.8} + \frac{100}{549.85}) = <em>1.228 x </em>10^{-6}<em> mm </em>

6 0
3 years ago
A projectile is launched into the air with the initial speed of vi = 40 m/s at a launch angle of 20 degrees above the horizontal
Sphinxa [80]

The range of the projectile is 188 m

Explanation:

The motion of the arrow in this problem is a projectile motion, so it follows a parabolic path which consists of two independent motions:  

- A uniform motion (constant velocity) along the horizontal direction  

- An accelerated motion with constant acceleration (acceleration of gravity) in the vertical direction  

The path of a projectile is the combination of these two motions: see figure in attachment.

In order to find the horizontal range of the projectile, we just need to calculate the horizontal distance travelled.

We have:

t = 5.0 s (time of fligth of the projectile)

and the horizontal velocity is constant, and it is given by

v_x = v_i cos \theta

where

v_i = 40 m/s is the initial velocity

\theta=20^{\circ} is the angle of projection

Substituting,

v_x = (40)(cos 20^{\circ})=37.6 m/s

And therefore, the range of the projectile is:

d=v_x t = (37.6)(5.0)=188 m

Learn more about projectile motion:

brainly.com/question/8751410

#LearnwithBrainly

5 0
3 years ago
What is the purpose of humping
stiv31 [10]

Explanation:

I'm not sure to be honest lol

4 0
3 years ago
Read 2 more answers
The area of the piston to the master cylinder in a hydraulic braking system of a car is 0.4 square inches. If a force of 6.4 lb
Anit [1.1K]

Answer:

The force applied on one wheel during braking = 6.8 lb

Explanation:

Area of the piston (A) = 0.4 in^{2}

Force applied on the piston(F) = 6.4 lb

Pressure on the piston (P) = \frac{F}{A}

⇒ P = \frac{6.4}{0.4}

⇒ P = 16 \frac{lb}{in^{2} }

This is the pressure inside the cylinder.

Let force applied on the brake pad = F_{1}

Area of the brake pad (A_{1})= 1.7 in^{2}

Thus the pressure on the brake pad (P_{1}) =  \frac{F_{1} }{A_{1} }

When brake is applied on the vehicle the pressure on the piston is equal to pressure on the brake pad.

⇒ P = P_{1}

⇒ 16 = \frac{F_{1} }{A_{1} }

⇒ F_{1} = 16 × A_{1}

Put the value of A_{1} we get

⇒ F_{1} = 16 × 1.7

⇒ F_{1} = 27.2 lb

This the total force applied during braking.

The force applied on one wheel = \frac{F_{1} }{4} = \frac{27.2}{4} = 6.8 lb

⇒ The force applied on one wheel during braking.

7 0
3 years ago
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