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katrin2010 [14]
2 years ago
11

How is force related to math

Physics
2 answers:
dybincka [34]2 years ago
8 0

Answer:

Newton's second law of motion describes the relationship between force and acceleration. They are directly proportional. If you increase the force applied to an object, the acceleration of that object increases by the same factor. In short, force equals mass times acceleration.

Explanation:

OLga [1]2 years ago
7 0

Answer:

Force is push or pull. Unbalanced forces make an object accelerate. The unit of force is the Newton (abbreviation is N). A Newton is the force it takes to make 1 kg change its velocity by 1 m/s (meter per second) every second.

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A point charge of -3.0 x 10-5C is placed at the origin of coordinates. Find the electric field at the point 3. r= 50 m on the x-
Snezhnost [94]

Answer: -5×10-3

Explanation:

E=kq/r

4 0
2 years ago
Suppose a car travels 106 km at a speed of 28 m/s and uses 1.9 gals of gasoline in the process. Only 30% of the gasoline goes in
USPshnik [31]

Answer:

a) The magnitude of the force is 968 N

b) For a constant speed of 30 m/s, the magnitude of the force is 1,037 N

Explanation:

<em>NOTE: The question b) will be changed in other to give a meaningful answer, because it is the same speed as the original (the gallons would be 1.9, as in the original).</em>

Information given:

d = 106 km = 106,000 m

v1 = 28 m/s

G = 1.9 gal

η = 0.3

Eff = 1.2 x 10^8 J/gal

a) We can express the energy used as the work done. This work has the following expression:

W=F\cdot d

Then, we can derive the magnitude of the force as:

F=\frac{W}{d}=\frac{\eta\cdot (G\cdot Eff)}{d}=\frac{0.3*1.9*(1.8*10^8)}{106*10^3} =968\,N

b) We will calculate the force for a speed of 30 m/s.

If the force is proportional to the speed, we have:

F_2=F_1(\frac{v_2}{v_1} )=968(\frac{30}{28} )=968*1.0714=1,037\,N

6 0
3 years ago
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kotykmax [81]
A is the correct answer I think
6 0
2 years ago
Field of psychology that studies the behavior of shoppers
Kay [80]

Answer:

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Explanation:

5 0
3 years ago
You are asked to design a cylindrical steel rod 50.0 cm long, with a circular cross section, that will conduct 170.0 J/s from a
zepelin [54]

Answer:

You are asked to design a cylindrical steel rod 50.0 cm long, with a circular cross section, that will conduct 170.0 J/s from a furnace at 350.0 ∘C to a container of boiling water under 1 atmosphere.

Explanation:

Given Values:

L = 50 cm = 0.5 m

H = 170 j/s

To find the diameter of the rod, we have to find the area of the rod using the following formula.

Here Tc = 100.0° C

        k  = 50.2

       H = k × A × \frac{[T_{H -}T_{C} ] }{L}

Solving for A

       A  =  \frac{H * L }{k * [ T_{H}- T_{C} ] }

       A  = \frac{170 * 0.5}{50.2 * [ 350 - 100 ]}

       A  = \frac{85}{12550} = 6.77 ×10^{-3} m²

Now Area of cylinder is :

     A =  \frac{\pi }{4} d²

solving for d:

    d =  \sqrt{\frac{4 * 0.00677 }{\pi } }

    d  = 9.28 cm

5 0
3 years ago
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