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3 years ago

If an object is not moving are the forces acting on it balanced? Yes or no?why?

Physics

1 answer:

xxMikexx [17]3 years ago

6 0

This is another time to look at Newton's 2nd law of motion:

Net Force = (mass) x (acceleration)

If the object is not moving, then its acceleration is certainly zero, and Newton's law looks like this:

Net Force = (mass) x (zero)

or Net Force = (zero) .

"Net Force = zero" means that if there ARE any forces acting on the object, then they add up to zero, and we call them "balanced" forces.

So the answer is '<em>yes</em>', and that's why.

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3 years ago

A car comes to a bridge during a storm and finds the bridge washed out. The driver must get to the other side, so he decides to

aksik [14]

Answer:

The answer is below

Explanation:

a) The vertical displacement = Δy = 21.5 m - 1.5 m = 20 m

The horizontal displacement = Δx = 69 m wide

Using the formula:

$\Delta y = u_yt+ \frac{1}{2}a_yt^2\\ \\u_y=initial\ velocity\ of \ car\ in\ y\ direction = 0,a_y=g=acceleration\ due\ to\ gravity\\=10m/s^2\\\\\Delta y = \frac{1}{2}a_yt^2\\\\\Delta y=\frac{1}{2}a_yt^2\\\\t=\sqrt{\frac{2\Delta y}{a_y} }=\sqrt{\frac{2*20}{10} } =2\ m/s$

Also:

$\Delta x = u_xt+ \frac{1}{2}a_xt^2\\ \\u_x=initial\ velocity\ of \ car\ in\ x\ direction = 0,a_x=acceleration=0\\\\\Delta x = u_xt\\\\u_x=\frac{\Delta x}{t}=\frac{69}{2} =34.5\ m/s$

b)The car is moving at a constant speed in the horizontal direction, hence the initial velocity = final velocity

$v_x=u_x=34.5\ m/s\\\\v_y=u_y+a_yt\\\\v_y=0+gt\\\\v_y=10(2)=20\ m/s\\\\v=\sqrt{v_x^2+v_y^2}=\sqrt{34.5^2+20^2}=39.9\ m/s\\ v=39.9\ m/s$

4 0

4 years ago

A satellite orbiting the earth is directly over a point on the equator at 12:00 midnight every four days. It is not over that po

Mrrafil [7]

Answer:

106417026.88435 m

Explanation:

T = Time period of the satellite = 4 days

m = Mass of the Earth = 5.972 × 10²⁴ kg

G = Gravitational constant = 6.67 × 10⁻¹¹ m³/kgs²

Time period is given by

$T=2\pi\sqrt{\dfrac{r^3}{GM}}\\\Rightarrow r=\left(\dfrac{T^2GM}{4\pi ^2}\right)^{\dfrac{1}{3}}\\\Rightarrow r=\left(\dfrac{(4\times 24\times 60\times 60)^2\times 6.67\times 10^{-11}\times 5.972\times 10^{24}}{4\pi ^2}\right)^{\dfrac{1}{3}}\\\Rightarrow r=106417026.88435\ m$

The radius of the satellite's orbit is 106417026.88435 m

5 0

3 years ago