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2 years ago

7

A women who normally weighs 400 N stands on top of a ladder so high she is one full additional earth radius above the surface of

the earth. How much does she weigh now

1 answer:

AveGali [126]2 years ago

3 0

The woman weighs 100N now.

Explanation:

Given:

F is given as 400N

one full additional earth radius formed

To Find:

Weight of the woman=?

Solution:

We know that gravitational force is inversely proportional to the square of radius.

$f_g=\frac{1}{r^2}$

where

$f_g$ is the gravitational force

$r^2$ is the radius square

So when radius is doubled, the force becomes one fourth of the original.

Therefore $\frac{1}{4} \times400 = 100N$

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You are given a vector in the xy plane that has a magnitude of 84.0 units and a y-component of -67.0 units.

melomori [17]

Answer:

Explanation:

a)Magnitude = $\sqrt{(x1-y2)^{2} + (x1-x2)^{2} }$

84= $\sqrt{(0- (-67))^{2} + (x-0)^{2} }$

x= +50.67 or -50.67 units

b) We are given that the resultant is entirely in the -ve x direction which means that the y-component of the resultant is 0; It means that the y-component of the next vector = -ve of the y component of the initial vector i.e 67.

To make the magnitude 80 units in the negative x direction where the y component is 0, the x component must be -130.67(-50.67 - 80) as the x component is + 50.67units.

Magnitude = $\sqrt{(0- (67))^{2} + (-130.67)^{2} }$ = 146.85 units

c) The direction vector = 67/146.85 i - 130.67/146.85 j where i corresponds to the vector in y direction and j corresponds to the vector in x direction. Or this vector is at an angle of 180 - $Tan^{-1}(67/130.67)degrees$ i.e 152.85 degrees from the +ve x-axis.

5 0

2 years ago

To start a lawn mower, you must pull on a rope wound around theperimeter of a flywheel. After you pull the rope for 0.95 s, thef

Molodets [167]

Answer:

29.76245 rad/s², -117.80972 rad/s²

28.2743 rad/s

3.95833

Explanation:

$\omega_f$ = Final angular velocity

$\omega_i$ = Initial angular velocity

$\alpha$ = Angular acceleration

$\theta$ = Angle of rotation

t = Time taken

Equation of rotational motion

$\omega_f=\omega_i+\alpha t\\\Rightarrow \alpha=\frac{\omega_f-\omega_i}{t}\\\Rightarrow \alpha=\frac{4.5\times 2\pi-0}{0.95}\\\Rightarrow \alpha=29.76245\ rad/s^2$

Angular acceleration during speed up is 29.76245 rad/s²

$\omega_f=\omega_i+\alpha t\\\Rightarrow \alpha=\frac{\omega_f-\omega_i}{t}\\\Rightarrow \alpha=\frac{0-4.5\times 2\pi}{0.24}\\\Rightarrow \alpha=-117.80972\ rad/s^2$

Angular acceleration during spin down is -117.80972 rad/s²

Angular speed is given by

$\omega=2\pi 4.5=28.2743\ rad/s$

Maximum angular speed reached by the flywheel is 28.2743 rad/s

$\theta=\omega_it+\frac{1}{2}\alpha t^2\\\Rightarrow \theta=0\times t+\frac{1}{2}\times 29.76245\times 0.95^2\\\Rightarrow \theta=13.4303\ rad$

$\theta=\omega_it+\frac{1}{2}\alpha t^2\\\Rightarrow \theta=2\pi 4.5\times 0.24+\frac{1}{2}\times -117.80972\times 0.24^2\\\Rightarrow \theta=3.39292\ rad$

The ratio would be $\dfrac{13.4303}{3.39292}=3.95833$

3 0

2 years ago

A large truck and a small car have a head-on collision. Which scenario is true?

solmaris [256]

B. is the answer dude

7 0

2 years ago

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Compare and Contrast How is a tug - of - war similar to unequal sharing of electron

laiz [17]

Imagine that the two atoms are playing tug of war with the electron, but one is pulling with a stronger force than the other. This inequality results in the majority of the rope being pulled toward the stronger atom. In the bond, the electron is pulled closer to the stronger atom too. This results in each side having a partial charge, one being more negative than the other.

7 0

2 years ago

There is a naturally occurring vertical electric field near the Earth’s surface that points toward the ground. In fair weather c

Neko [114]

Answer:

0.00110616446408

Explanation:

r = Radius of sphere = 11 μm

V = Volume = $\dfrac{4}{3}\pi r^3$

$\rho$ = Density of pollen = 1000 kg/m³

q = Charge = − 0.550 fC

E = Electric field = 110 N/C

g = Acceleration due to gravity = 9.81 m/s²

Mass is given by

$m=\rho V\\\Rightarrow m=1000\times \dfrac{4}{3}\pi (11\times 10^{-6})^3\ kg$

Force due to gravity is given by

$F_g=mg\\\Rightarrow F_g=1000\times \dfrac{4}{3}\pi (11\times 10^{-6})^3\times 9.81\\\Rightarrow F_g=5.4693494471\times 10^{-11}\ N$

Electric force is given by

$F_e=qE\\\Rightarrow F_e=-0.55\times 10^{-15}\times 110\\\Rightarrow F_e=-6.05\times 10^{-14}\ N$

The ratio is

$\dfrac{F_e}{F_g}=\dfrac{6.05\times 10^{-14}}{5.4693494471\times 10^{-11}}\\\Rightarrow \dfrac{F_e}{F_g}=0.00110616446408$

The ratio is 0.00110616446408

7 0

2 years ago