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Liono4ka [1.6K]
3 years ago
11

You're in the middle of moving and need to bring furniture and into your new house. You want to prop the door open, but your dog

ate the doorstopper (and your homework). All you can find is a large brick. If you want to impart the largest torque on the door to keep it open, where should you place the brick?
A) Put the brick as close to the hinges as possible.
B) Put the brick in the middle of the door.
C) Put the brick as far from the hinges as possible.
Physics
1 answer:
Mamont248 [21]3 years ago
3 0

Answer:

C) Put the brick as far from the hinges as possible

Explanation:

As torque is the product of the force around the rotation point and the distance to the pivot point, and the mass (force) of the brick stays constant, what we can do to maximize the torque is maximize the distance to the pivot point, aka the hinge. So we should put the brick as far from the hinges as possible.

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A 175-kg roller coaster car starts from rest at the top of an 18.0-m hill and rolls down the hill, then up a second hill that ha
Anni [7]

Answer:

The work done by non-conservative forces on the car from the top of the first hill to the top of the second hill is 6574.75 joules.

Explanation:

By Principle of Energy Conservation and Work-Energy Theorem we present the equations that describe the situation of the roller coaster car on each top of the hill. Let consider that bottom has a height of zero meters.

From top of the first hill to the bottom

m\cdot g \cdot h_{1} = \frac{1}{2}\cdot m\cdot v_{1}^{2} +W_{1, loss} (1)

From the bottom to the top of the second hill

\frac{1}{2}\cdot m\cdot v_{1}^{2} = m\cdot g \cdot h_{2} + \frac{1}{2}\cdot m \cdot v_{2}^{2}+W_{2,loss} (2)

Where:

m - Mass of the roller coaster car, in kilograms.

v_{1} - Speed of the roller coaster car at the bottom between the two hills, in meters per second.

g - Gravitational acceleration, in meters per square second.

h_{1} - Height of the first top of the hill with respect to the bottom, in meters.

W_{1, loss} - Work done by non-conservative forces on the car between the top of the first hill and the bottom, in joules.

v_{2} - Speed of the roller coaster car at the top of the second hill, in meters per seconds.

h_{2} - Height of the second top of the hill with respect to the bottom, in meters.

W_{2, loss} - Work done by non-conservative forces on the car bewteen the bottom between the two hills and the top of the second hill, in joules.

By using (1) and (2), we reduce the system of equation into a sole expression:

m\cdot g\cdot h_{1} = m\cdot g\cdot h_{2} + \frac{1}{2}\cdot m \cdot v_{2}^{2} + W_{loss} (3)

Where W_{loss} is the work done by non-conservative forces on the car from the top of the first hill to the top of the second hill, in joules.

If we know that m = 175\,kg, g = 9.807\,\frac{m}{s^{2}}, h_{1} = 18\,m, h_{2} = 8\,m and v_{2} = 11\,\frac{m}{s}, then the work done by non-conservative force is:

W_{loss} = m\cdot\left[ g\cdot \left(h_{1}-h_{2}\right)-\frac{1}{2}\cdot v_{2}^{2} \right]

W_{loss} = 6574.75\,J

The work done by non-conservative forces on the car from the top of the first hill to the top of the second hill is 6574.75 joules.

8 0
2 years ago
Acceleration usually has the symbol a. It is a vector. What is the correct way
Ghella [55]

Explanation:

a straight line under the letter

3 0
2 years ago
Two fishing boats depart a harbor at the same time, one traveling east, the other south. the eastbound boat travels at a speed 1
Novosadov [1.4K]
<span>they are travelling at right angles to each other.
 At any given instant they form a right triangle with their starting point
 </span>South bound <span>= x  [mi/h]
</span> East bound <span> = x+1 [mi/h]
 after five hours they will be
 d=5x
 and
 d=5(x+1)
 miles away from the starting point 
 (5x)^2+(5(x+1))^2=625
 25x^2+(5x+5)^2=625
 25x^2+25x^2+50x+25=625
 50</span>x^2+50x-600=0
<span> x^2+ x - 12=0
 (x+4)(x-3)=0
 take the postive value
 x= 3 mph the speed of south bound
 4mph east bound </span>
6 0
3 years ago
A net force of 125 n is applied to a certain object. as a result, the object accelerates with an acceleration of 24.0 m/s2. the
pantera1 [17]
Newton's second law states that Fnet = ma, where Fnet is the net force applied, m is the mass of the object, and a is the object's acceleration. You have the values for Fnet and a, so you simply use this equation to solve for m, mass.
8 0
3 years ago
Read 2 more answers
You are moving a wagon with a friend's help you push on the left side of the wagon with 25 of force while your friend pulls from
Pavel [41]

Answer:

10N to the left side towards you

Explanation:

The net force is the resultant force that acts on a body.

Force is a push or pull on a body.    

 Push to left side  = 25N

 Pull to the right  = 15N

Net force  = Push to left side   -  Pull to the right  = 25N  - 15N

 Net force  = 10N to the left side towards you

The net force is therefore 10N to the left side towards you

5 0
3 years ago
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