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Sav [38]
1 year ago
10

What is the mass of an object that requires 100N (kg-m/s2) of force in order to accelerate it at 10m/s2 (Please use G-R-E-S-A)

Physics
1 answer:
anastassius [24]1 year ago
3 0

Answer:

10kg

Explanation:

(I'm not super familiar with the GRESA method so apologies for any inaccuracies)

Given: We are given values for Force: 100N, and Acceleration: 10m/s2.

Required: We are trying to find Mass (m)

Equation: The best equation to use to solve this problem is F=ma,

Force = Mass x Acceleration. We can rearrange this for mass: m = F/a.

Solution: By substituting in the values we have: m = \frac{100}{10}

Answer: Mass = 10kg

Hope this helped!

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Initially its moving with tail wind so here the speed of wind will support the motion of the plane

so we can say

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now when its moving with head wind we can say that wind is opposite to the motion of the plane

V_{plane} - v_{wind} = \frac{distance}{time}

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Explains this: a light source can emit more than one type of light
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6 0
2 years ago
A 15x10^-6c charge is placed at the origin and a 9x10^-6C charge is placed on the x-axis at x=1.00m. where, on the x-axis is the
Harlamova29_29 [7]

The Electric field is zero at a distance 2.492 cm from the origin.

Let A be point where the charge 15\times10^-6 C is placed which is the origin.

Let B be the point where the charge 9\times 10^-6 C is placed. Given that B is at a distance 1 cm from the origin.

Both the charges are positive. But charge at origin is greater than that of B. So we can conclude that the point on the x-axis where the electric field = 0 is after B on x - axis.

i.e., at distance 'x' from B.

Using Coulomb's law, \frac{kQ_A^2}{d_A^2} = \frac{kQ_B^2}{d_B^2},

Q_A = 15\times 10^-6 C

Q_B=9\times10^-6C

d_A = 1+x cm

d_B=x cm

k is the Coulomb's law constant.

On substituting the values into the above equation, we get,

\frac{(15\times10^-6)^2}{(1+x)^2} =\frac{(9\times10^-6)^2}{x^2}

Taking square roots on both sides and simplifying and solving for x, we get,

1.67x = 1+x

Therefore, x = 1.492 cm

Hence the electric field is zero at a distance 1+1.492 = 2.492 cm from the origin.

Learn more about Electric fields and Coulomb's Law at brainly.com/question/506926

#SPJ4

3 0
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