Answer:
1800/300 = 6ropes
Explanation:
The engine weighs 1800N and the person exerts a force of 300N, so for him to lift the engine and exerting a force of 300N all through we divide the weight of the engine by the force exerted to know how many ropes are used. Which makes it 6 thereby each rope uses 300N to lift the engine.
Answer:
option A is correct
Explanation:
Given:
The length to be painted = m miles
The width to be painted = t inches
Area painted in 1 gallon = p square feet
Converting the every given dimension in feet, we have
length to be painted = m × 5280 feet
width to be painted = t/12 feet
area to be painted = (m × 5280 feet) × t/12 feet
now, applying the unitary method, we have
p square feet is painted ⇒ 1 gallon
1 square feet is painted ⇒ 1/p gallon
(m × 5280 feet) × t/12 feet square feet is painted ⇒ [(m × 5280 feet) × t/12 feet ] × 1/p gallon
thus, we get the gallons of paint required as 5820 mt/12p
hence option A is correct
In the Missouri Compromise, the slavery line for future US states ran along the southern border of Missouri at 36 degrees north 30 minutes
Answer:
(a) 
(b) 
Explanation:
Hello.
(a) In this case since the car is moving at an initial velocity of 18 m/s due north, the final velocity is computed considering the acceleration as positive since it is due north as well:
(b) In this case, since the car is moving due north by the acceleration is due south it is undergoing a slowing down process, thereby the acceleration is negative therefore the final velocity turns out:
Best regards.
Let <em>F</em> be the magnitude of the force applied to the cart, <em>m</em> the mass of the cart, and <em>a</em> the acceleration it undergoes. After time <em>t</em>, the cart accelerates from rest <em>v</em>₀ = 0 to a final velocity <em>v</em>. By Newton's second law, the first push applies an acceleration of
<em>F</em> = <em>m a</em> → <em>a</em> = <em>F </em>/ <em>m</em>
so that the cart's final speed is
<em>v</em> = <em>v</em>₀ + <em>a</em> <em>t</em>
<em>v</em> = (<em>F</em> / <em>m</em>) <em>t</em>
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If we force is halved, so is the accleration:
<em>a</em> = <em>F</em> / <em>m</em> → <em>a</em>/2 = <em>F</em> / (2<em>m</em>)
So, in order to get the cart up to the same speed <em>v</em> as before, you need to double the time interval <em>t</em> to 2<em>t</em>, since that would give
(<em>F</em> / (2<em>m</em>)) (2<em>t</em>) = (<em>F</em> / <em>m</em>) <em>t</em> = <em>v</em>
