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Ierofanga [76]
3 years ago
12

Suppose the radius of the Earth is given to be 6378.01 km. Express the circumference of the Earth in m with 5 significant figure

s.
Round to 5 sig figs with trailing zeros --> 40074000
Physics
1 answer:
bixtya [17]3 years ago
7 0

The mathematical description that fits to find the circumference of a circle (Approximation we will make to the earth considering it Uniform) is,

\phi = 2\pi r

Here,

r = Radius

The radius of the earth is 6378.01 km or 6378010m

Replacing we have that the circumference of the Earth is

\phi = 2\pi (6378010m)

\phi = 40074000 m

\phi = 40074*10^3 m

Therefore the circumference of the Earth in m with 5 significant figures is 40074*10^3 m and using only trailing zeros the answer will be 40074000m

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Say you want to make a sling by swinging a mass M of 2.3 kg in a horizontal circle of radius 0.034 m, using a string of length 0
ycow [4]

Answer:

T = 764.41 N

Explanation:

In this case the tension of the string is determined by the centripetal force. The formula to calculate the centripetal force is given by:

F_c=m\frac{v^2}{r}  (1)

m: mass object = 2.3 kg

r: radius of the circular orbit = 0.034 m

v: tangential speed of the object

However, it is necessary to calculate the velocity v first. To find v you use the formula for the kinetic energy:

K=\frac{1}{2}mv^2

You have the value of the kinetic energy (13.0 J), then, you replace the values of K and m, and solve for v^2:

v^2=\frac{2K}{m}=\frac{2(13.0J)}{2.3kg}=11.3\frac{m^2}{s^2}

you replace this value of v in the equation (1). Also, you replace the values of r and m:

F_c=(2.3kg)(\frac{11.3m^2/s^2}{0.034})=764.41N

hence, the tension in the string must be T =  Fc = 764.41 N

5 0
3 years ago
Please help ASAP!!
inessss [21]

Answer:

at t=46/22, x=24 699/1210 ≈ 24.56m

Explanation:

The general equation for location is:

x(t) = x₀ + v₀·t + 1/2 a·t²

Where:

x(t) is the location at time t. Let's say this is the height above the base of the cliff.

x₀ is the starting position. At the base of the cliff we'll take x₀=0 and at the top x₀=46.0

v₀ is the initial velocity. For the ball it is 0, for the stone it is 22.0.

a is the standard gravity. In this example it is pointed downwards at -9.8 m/s².

Now that we have this formula, we have to write it two times, once for the ball and once for the stone, and then figure out for which t they are equal, which is the point of collision.

Ball: x(t) = 46.0 + 0 - 1/2*9.8 t²

Stone: x(t) = 0 + 22·t - 1/2*9.8 t²

Since both objects are subject to the same gravity, the 1/2 a·t² term cancels out on both side, and what we're left with is actually quite a simple equation:

46 = 22·t

so t = 46/22 ≈ 2.09

Put this t back into either original (i.e., with the quadratic term) equation and get:

x(46/22) = 46 - 1/2 * 9.806 * (46/22)² ≈ 24.56 m

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