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ankoles [38]
3 years ago
9

Suppose you lived in a pre-industrial society and needed to lift a heavy (20 kg) block a height of 5 m and had two choices for h

ow to accomplish your task: You could use a lever with a short arm of 1 m and a long arm of 10 m. You'd place the block on the end of the short arm and apply force at the very end of the long arm. You could use a ramp with a length of 8 m and place the block on a cart with wheels. Which choice would you make? Be sure to discuss differences between the effort force of the two options in your answer.
Physics
1 answer:
igomit [66]3 years ago
4 0
Let's break the question into two parts:

1) The force needed in Ramp scenario.
2) The effort force needed in the lever scenario.

1. Ramp Scenario: 
In an incline, the only component of cart's weight(mg) that is in the direction of motion is mgsin \alpha. Therefore the effort force in this case must be equal or greater than mgsin \alpha.

Now we need to find \alpha. \alpha is the angle between the incline of the ramp and the ground. 

Since the height is 5m and the length of the ramp is 8m, sin \alpha would be 5/8 or 0.625. Now that you have sin \alpha, mutiple it with mg.

=> m*g*sin \alpha  = 20 * 10 * 5 / 8. (Taking g = 10 m/s² for simplicity) = 125N
Therefore, the minimum Effort force you would require in this case is 125N.

2. Lever Scenario:
Just apply "moment action" in this case, which is:
F_{e}  d_{e}  = F_{r}  d_{r}

F_{e} = ?

F_{r} = mg = 20 * 10 = 200N
d_{e} = 10m
d_{r} = 1m


Plug-in the values in the above equation:
F_{e} = 200/10= 20N


As 20N << 125N, the best choice is to use lever.

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The total electric field consists of the vector sum of two parts. One part has a magnitude of E1 = 1300 N/C and points at an ang
V125BC [204]

Answer:

2954.6 N/C, 46.36 degree from positive  axis

Explanation:

E1 = 1300 N/C, θ1 = 35 degree

E2 = 1700 N/C, θ2 = 55 degree

Now write the electric fields in vector form

E1 = 1300 ( Cos 35 i + Sin 35 j) = 1064.9 i + 745.6 j

E2 = 1700 ( Cos 55 i + Sin 55 j) = 975.08 i + 1392.6 j

Resultant electric field

E = E1 + E2

E =  1064.9 i + 745.6 j + 975.08 i + 1392.6 j

E = 2039.08 i + 2138.2 j

Magnitude of E

E = sqrt (2039.08^2 + 2138.2^2)

E = 2954.6 N/C

Let it makes an angle Φ from X axis

tan Φ = 2138.2 / 2039.08 = 1.049

Φ = 46.36 degree from positive X axis.

5 0
3 years ago
How does approaching a temperature of absolute zero affect kinetic energy.?
Alexus [3.1K]

First of all the kinetic energy is when the particles move in continuous random motion.

If the temperature is high the colliding particles will collide more. and if the temperature is low the colliding particles will collide less.

Low temperature result in low kinetic energy 
High temperature result in high kinetic energy

Absolute zero is the point where where all molecules have no kinetic energy. It is a theoretical value (it has never been reached).

The Kelvin temperature scale is based on absolute zero being the lowest possible temperature that could theoretically be reached. That is why there is no such thing as a negative Kelvin temperature value.

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Calculate the momentum of a 10kg bowling ball rolling at 2 m/s.
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Answer:

20 kg. m/s

Explanation:

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p = 10 × 2

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What force is required to accelerate a 1840 kg car from 4.77 m/s to 23.5 m/s,
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A :-) for this question , we should apply
a = v - u by t
Given - u = 4.77 m/s
v = 23.5 m/s
t = 5.18 m/s
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3 years ago
A dart is inserted into a spring-loaded dart gun by pushing the spring in by a distance x. For the next loading, the spring is c
bezimeni [28]

Answer:

The second dart leaves the gun two times as faster than the first one.

Explanation:

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