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Roman55 [17]
3 years ago
8

How to solve this.

Physics
1 answer:
viva [34]3 years ago
5 0

Answer:

8.1 s

Explanation:

Draw a free body diagram of the small block.  There are four forces acting on the block:

Applied force F pushing to the right,

Weight force mg pulling down,

Normal force N pushing up,

and friction force Nμ pushing left.

Sum of forces in the y direction:

∑F = ma

N − mg = 0

N = mg

Sum of forces in the x direction:

∑F = ma

F − Nμ = ma

F − mgμ = ma

a = (F − mgμ) / m

Plug in values:

a = (12.2 N − (3.0 kg) (9.8 m/s²) (0.320)) / (3.0 kg)

a = 0.931 m/s²

Next, draw a free body diagram of the larger block.  There are four forces:

Normal force N pushing down (equal and opposite),

Friction force Nμ pushing right (equal and opposite),

Weight force Mg pulling down,

and normal force N₂ pushing up.

Sum of forces in the x direction:

∑F = ma

Nμ = Ma

mgμ = Ma

a = mgμ / M

Plug in values:

a = (3.0 kg) (9.8 m/s²) (0.320) / (11.0 kg)

a = 0.855 m/s²

So the acceleration of the smaller block relative to the larger block is 0.931 m/s² − 0.855 m/s² = 0.0754 m/s².

Given:

Δx = 2.5 m

v₀ = 0 m/s

a = 0.0754 m/s²

Find: t

Δx = v₀ t + ½ at²

2.5 m = (0 m/s) t + ½ (0.0754 m/s²) t²

t = 8.14 s

Rounded to 2 significant figures, it takes 8.1 seconds.

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liq [111]

Answer:

cost of running the furnace during January is $5619.62

Explanation:

given data

runs a day = 13 hours

January days = 31 days

resistance = 7.2 ohm

current = 16.7 A

cost of electricity = $0.10/kWh

to find out

cost of running the furnace during January

solution

first we get her power consumed by furnace that is

Power consumed = \frac{I^2}{R}  ........1

put here value we get

Power consumed = \frac{16.7^2}{7.2}

Power consumed = 38.7347 W

and

Power consumed by furnace in one hour is

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Power consumed by furnace in one hour is = 38.7347 × 3600  

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and

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Power consumed by furnace in the month of January = 139.445kWh × 13 hours × 31 days

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cost of running the furnace during January is = $0.10/kWh × 56196.335 kWh

cost of running the furnace during January is $5619.62

4 0
3 years ago
Read 2 more answers
how do you find work when only given the angle a sled is pulled, the mass, the coefficent of kinetic friction and distance
Sergio039 [100]

Answer:

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s = distance traveled

[μ M g (1 - sin θ)] cos θ * s = work done in moving sled

Note that F = μ M g    if applied force is in the horizontal direction

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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
igomit [66]
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.

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

F_{e} = ?

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Plug-in the values in the above equation:
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