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3 years ago

In the illustration, which two simple machines are being used to enable the student to reach the door?

Physics

2 answers:

lisov135 [29]3 years ago

4 0

The answer will be D the wheel and axle her wheel chair and the ramp is the inclined plane.

vodka [1.7K]3 years ago

3 0

D. wheel-and-axle AND an incline plane

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An engineer wishes to design a roller coaster so that

nataly862011 [7]

To prevent cars from falling, the radius at the top of the circle should be

small such cars inverted at the top remain attached during motion.

Correct response;

The radius of the coaster can be C) 20 m

<h3>Method by which the above option is selected</h3>

Mass of roller coaster car, m = 500 kg

Speed at the top of the circle, v = 20 m/s

Required:

The maximum radius of the circular path the roller coaster car.

Solution:

$\displaystyle Centrifugal \ force, \, F_{c} = \mathbf{\frac{m \cdot v^2}{r}}$

Where;

r = The radius of the circular path.

Weight of the roller coaster car = m·g = The centripetal force

Where;

g = Acceleration due to gravity = 9.81 m/s²

At equilibrium, we have;

Centrifugal force = Centripetal force

$\displaystyle \frac{m \cdot v^2}{r} = \mathbf{ m \cdot g}$

Therefore;

$\displaystyle r = \mathbf{ \frac{v^2}{g}}$

Which gives;

$\displaystyle r = \frac{20^2}{9.81} \approx 40.77$

The maximum radius for safety of a roller coaster, r ≈ 40.77 meters

$\displaystyle Range \ of \ radius \ of \ the \ circle = \frac{40.77}{4} \leq Radius \ of \ circle \leq 40.77$

Which gives;

Range of the radius of the circle = 10.2 ≤ Radius of circle ≤ 40.77

The correct option for safety considerations is therefore;

C) 20 m

The possible question options are;

A) 5 m B) 10 m C) 20 m D) 40 m E) 80 m

Learn more about centripetal force here:

brainly.com/question/12674230

7 0

2 years ago

The cylindrical tub of a dryer in a laundromat rotates counterclockwise about a horizontal axis at 41.5 rev/min as it dries the

frozen [14]

Answer:

$\theta = 49.81^0$

Explanation:

Given that:

$\omega = 41.5 \ rev/min\\\\\omega = 41.5 *\frac{1}{60}* 2 \pi\\\\\omega = 4.45 \ rad/s\\\\\\diameter = 0.748 m$

If we let the piece of the close lose contact at ∠θ;

Then ; from force balance;

we have:

$\\\\mg sin \theta = \frac{mv^2}{r}\\\\sin \theta = \frac{2v^2}{dg}\\\\\theta = sin^{-1} (\frac{2v^2}{dg})$

where;

$v = \frac{\omega d}{2}\\\\v = \frac{4.45 *0.748}{2}\\\\v = 1.6643\\\\v^2 = 2.77$

Again:

$\theta = sin^{-1}(\frac{2v^2}{dg})\\\\\theta = sin^{-1}( \frac{2*2.77}{0.74*9.8})\\\\\theta = 49.81^0$

6 0

3 years ago

What stress causes this type of fault to form?

worty [1.4K]

Answer:

gravity

Explanation:

3 0

3 years ago

Read 2 more answers

A bin is given a push across a horizontal surface. The bin has a mass m, the push gives it an initial speed of 1.60 m/s, and the

emmasim [6.3K]

Answer:

The bin moves 0.87 m before it stops.

Explanation:

If we analyze the situation and apply the law of conservation of energy to this case, we get:

Energy Dissipated through Friction = Change in Kinetic Energy of Bin (Loss)

F d = (0.5)(m)(Vi² - Vf²)

where,

F = Frictional Force = μR

but, R = Normal Reaction = Weight of Bin = mg

Therefore, F = μmg

Hence, the equation becomes:

μmg d = (0.5)(m)(Vi² - Vf²)

μg d = (0.5)(Vi² - Vf²)

d = (0.5)(Vi² - Vf²)/μg

where,

Vf = Final Velocity = 0 m/s (Since, bin finally stops)

Vi = Initial Velocity = 1.6 m/s

μ = coefficient of kinetic friction = 0.15

g = 9.8 m/s²

d = distance moved by bin before coming to stop = ?

Therefore,

d = (0.5)[(1.6 m/s)² - (0 m/s)²]/(0.15)(9.8 m/s²)

d = 0.87 m

5 0

3 years ago

Northrop aircraft developed and built a deceleration sled to test the effects of the extreme forces on humans and equipment. In

dolphi86 [110]

20 hours i think that what they do

4 0

3 years ago