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RoseWind [281]
3 years ago
14

Describe the relationship between force and mass when the acceleration of two objects remain constant

Physics
1 answer:
LekaFEV [45]3 years ago
6 0

Answer:

If they both remain constant they ain't moving, It states that the rate of change of velocity of an object is directly proportional to the force applied and takes place in the direction of the force. It is summarized by the equation

Explanation:

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During summer months, your ceiling fan blades should be set to spin counterclockwise. When your ceiling fan spins quickly in this direction, it pushes air down and creates a cool breeze.
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3 years ago
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What is the force on a 1670kg elevator accelerating at 6m/s square?
marshall27 [118]

As per Newton's 2nd law

we know that

F = ma

it is product of mass and acceleration

here we know that

m = 1670 kg

also we know that

a = 6 m/s^2

so from above equation we have

F = 1670 * 6

F = 10020 N

so the force here will be 10020 N

7 0
3 years ago
A crate is lifted vertically 1.5 m and then held at rest. The crate has weight 100 N (i.e., it is reese (enr647) – HS OnRamps 04
icang [17]

Answer:

W = 0 J

Explanation:

It is given that,

Weight of the crate, W = 100 N

Distance moved by the crate, d = 1.5 m

Let W is the work done to hold the crate 1.5 m above the ground in this way. It is given by the product of force and the displacement. Its formula is given by :

W=F\times d\times cos\theta

Here, \theta=90^{\circ} as it is lifted vertically

W=100\ N\times 1.5\ m\ cos(90)      

W = 0

So, the work done to hold the crate is 0. Hence, this is the required solution.

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3 years ago
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Please select the word from the list that best fits the definition
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Answer:

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3 years ago
An object falls a distance h from rest. If it travels 0.560h in the last 1.00 s, find (a) the time and (b) the height of its fal
VashaNatasha [74]

Answer:

(a) t = 2.97s

(b) h = 43.3 m

Explanation:

Let t be the time it takes to fall a distance h, then t - 1 (s) is the time it takes to fall a distance of h - 0.56h = 0.44 h

For the ball to fall from rest a distance of h after time t

h = gt^2/2

Also for the ball to fall from rest a distance of 0.44h after time (t-1)

0.44h = g(t-1)^2/2

We can substitute the 1st equation into the 2nd one

0.44gt^2/2 = g(t-1)^2/2

and divide both sides by g/2

0.44t^2 = (t-1)^2

0.44t^2 = t^2 - 2t + 1

0.56t^2 - 2t + 1 = 0

t= \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

t= \frac{2\pm \sqrt{(-2)^2 - 4*(0.56)*(1)}}{2*(0.56)}

t= \frac{2\pm1.33}{1.12}

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Since t can only be > 1 s we will pick t = 2.97 s

(b) h = gt^2/2 = 43.3 m

6 0
3 years ago
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