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3 years ago

14

Find the ratio of the radii of a baseball to the Earth, knowing that the radius of a baseball is .09 m, and that the Earth's rad

ius is 6.37 x 10^6 m.

1 answer:

Xelga [282]3 years ago

6 0

0.09 / 6.37 x 10⁶ = 1.4129 x 10⁻⁸

The radius of the baseball is 1.4129 x 10⁻⁸ the radius of the Earth.

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while driving his sports car at 20.0 m/s down a four lane highway, eddie comes up behind a slow moving dump truck and decides to

Arte-miy333 [17]

Explanation:

Given:

u = 20 m/s

a = 5 m/s^2

v = 30 m/s

t = ?

Use the first kinematic equation of motion:

v = u + at

t = (v - u)/a = 10/5 = 2 seconds

3 0

3 years ago

Vector A with arrow, which is directed along an x axis, is to be added to vector B with arrow, which has a magnitude of 5.5 m. T

ohaa [14]

Answer:

Magnitude of vector A = 0.904

Explanation:

Vector A , which is directed along an x axis, that is

$\vec{A}=x_A\hat{i}$

Vector B , which has a magnitude of 5.5 m

$\vec{B}=x_B\hat{i}+y_B\hat{j}$

$\sqrt{x_{B}^{2}+y_{B}^{2}}=5.5\\\\x_{B}^{2}+y_{B}^{2}=30.25$

The sum is a third vector that is directed along the y axis, with a magnitude that is 6.0 times that of vector A $\vec{A}+\vec{B}=6x_A\hat{j}\\\\x_A\hat{i}+x_B\hat{i}+y_B\hat{j}=6x_A\hat{j}$

Comparing we will get

$x_A=-x_B\\\\y_B=6x_A$

Substituting in $x_{B}^{2}+y_{B}^{2}=30.25$

$\left (-x_{A} \right )^{2}+\left (6x_{A} \right )^{2}=30.25\\\\37x_{A}^2=30.25\\\\x_{A}=0.904$

So we have

$\vec{A}=0.904\hat{i}$

Magnitude of vector A = 0.904

8 0

4 years ago

Which clouds are often associated with thunder and lightning?

ehidna [41]

Your answer is cumulonimbus clouds

3 0

3 years ago

Gelneren [198K]

Answer:

response

Explanation:

Acceleration is your changing Velocity. An object that is ACCELERATING is experiencing a change in velocity. usually positive. if an object such as a car reduces velocity, it is called deceleration

3 0

3 years ago

A 9-μC positive point charge is located at the origin and a 6 μC positive point charge is located at x = 0.00 m, y = 1.0 m. Find

sukhopar [10]

Answer:

The coordinates of the point is (0,0.55).

Explanation:

Given that,

First charge $q_{1}=9\times10^{-6}\ C$ at origin

Second charge $q_{2}=6\times10^{-6}\ C$

Second charge at point P = (0,1)

We assume that,

The net electric field between the charges is zero at mid point.

Using formula of electric field

$E=\dfrac{kq}{r^2}$

$0=\dfrac{k\times9\times10^{-6}}{d^2}+\dfrac{k\times6\times10^{-6}}{(1-d)^2}$

$\dfrac{(1-d)}{d}=\sqrt{\dfrac{6}{9}}$

$\dfrac{1}{d}=\dfrac{\sqrt{6}}{3}+1$

$\dfrac{1}{d}=1.82$

$d=\dfrac{1}{1.82}$

$d=0.55\ m$

Hence, The coordinates of the point is (0,0.55).

3 0

3 years ago