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Sophie [7]
3 years ago
9

Calculate net force 532N 215N

Physics
1 answer:
lorasvet [3.4K]3 years ago
8 0

Answer:

FN is the forces acting on a body. When the body is at rest, the net force formula is given by, FNet = Fa + Fg.

Im in 7th and thats all I know so I hope it's enough

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Explain two reason why SI is easier than the English System.
kifflom [539]

Answer:

There are several reasons why SI is preferred to the old English system of measurement: ... SI uses base 10, just like our number system, so it is much easier to learn, remember and convert between units. The prefixes used in SI are from Latin and Greek, and they refer to the numbers that the terms represent.

Explanation:

5 0
3 years ago
Given 1 inch ≡ 2.54 cm and 1 foot ≡
d1i1m1o1n [39]

Answer:

2991.47 [cm^2]

Explanation:

To solve this problem we must perform a dimensional analysis and use the corresponding conversion values:

3.22[ft^{2}]*\frac{12^{2}in^{2} }{1^{2}ft^{2}} *\frac{2.54^{2}cm^{2}  }{1^{2}in^{2} } \\2991.47[cm^{2}]

3 0
3 years ago
He wrote a book of aerobics
vampirchik [111]

Answer:

George Washington

Explanation:

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4 0
2 years ago
Read 2 more answers
A horizontal 810-N merry-go-round of radius 1.60 m is started from rest by a constant horizontal force of 55 N applied tangentia
Sloan [31]

Answer:

576 joules

Explanation:

From the question we are given the following:

weight = 810 N

radius (r) = 1.6 m

horizontal force (F) = 55 N

time (t) = 4 s

acceleration due to gravity (g) = 9.8 m/s^{2}

K.E = 0.5 x MI x ω^{2}

where MI is the moment of inertia and ω is the angular velocity

MI = 0.5 x m x r^2

mass = weight ÷ g = 810 ÷ 9.8 = 82.65 kg

MI = 0.5 x 82.65 x 1.6^{2}

MI = 105.8 kg.m^{2}

angular velocity (ω) = a x t

angular acceleration (a) = torque ÷ MI

where torque = F x r = 55 x 1.6 = 88 N.m

a= 88 ÷ 105.8 = 0.83 rad /s^{2}

therefore

angular velocity (ω) = a x t = 0.83 x 4 = 3.33 rad/s

K.E = 0.5 x MI x ω^{2}

K.E = 0.5 x 105.8 x 3.33^{2} = 576 joules

6 0
3 years ago
When the weight of the object increase block what is the force of friction applied? Explanation?
erik [133]

Answer:

There is absolutely No relationship between the weight of an object (which is constant) and the frictional force. If a block is sliding on a surface, that surface will be exerting a force on the block. That force can be resolved into a component parallel to the surface (which we call the frictional component), and a component perpendicular to the surface (called the normal component). For many situations, we find experimentally that the frictional component is approximately proportional to the normal component. The frictional component divided by the normal component is defined to be a quantity called the coefficient of kinetic or sliding friction. The coefficient of kinetic friction obviously depends on the nature of the surfaces involved. The normal component on an object can be decreased if you pull in the direction of the normal component (the weight does not change). However pulling this way on the object not only decreases the normal component, but it also decreases the frictional component since they are proportional. This is why it is easier to slide something if you pull up on it while you push it. If you push down, the normal and frictional components increase so it is harder to slide the object. The weight of an object is the downward force exerted by Earth’s gravity on that object, and it does not change no matter how you push or pull on the object.

8 0
2 years ago
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